NEET Physics Multiple Choice Question and Answers

General Physics

Heat and Thermodynamics ‘

Sound Waves

Electrostatics

Electricity and Magnetism

Optics

Modern Physics

Semiconductor Devices and Communication Systems

Miscellaneous Objective Questions

  • General Physics
  • Heat and Thermodynamics
  • Wave Motion
  • Electrostatics
  • Electricity and Magnetism
  • Optics
  • Modern Physics
  • Semiconductor Devices and Communication Systems

 

 

General Physics Multiple Choice Question And Answers

General Physics

Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.

Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.

  1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
  2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
  3. Statement 1 is true and Statement 2 is false.
  4. Statement 1 is false and Statement 2 is true.
  5. Both Statement 1 and Statement 2 are false.

Some questions are preceded, by a supporting paragraph in addition to the statements.

Units Dimensions And Errors

Question 1. Statement 1 The dimensional formula for the product of resistance and conductance is the same as that for the dielectric constant.

Statement 2 Both have the dimension of the time constant.

Answer: 3. Statement 1 is true and Statement 2 is false.

Resistance and conductance are reciprocal to each other, hence their product has no units and no dimensions. The same is true for the dielectric constant (K = ε/ε0), which is a ratio of two permittivities. Hence, Statement 1 is true but Statement 2 is false.

Question 2. Statement 1 SI units are logical and coherent.

Statement 2 This system of units is a rationalized system.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A coherent system of units is based on a system of base units from which all derived units are obtained by multiplication or division without introducing numerical factors. Hence, SI units are logical, coherent, and rationalized.

Question 3. Statement 1 The power of an engine depends on mass, angular speed, torque, and angular momentum, so the formula for power is not derived with the help of the dimensional method.

Statement 2 In mechanics, if a particular physical quantity depends on more than three quantities, we cannot derive the formula of that physical quantity using the dimensional method.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In mechanics, the physical quantities have dimensions in terms of M, L, and T only. With dependence on more than three factors, the required relation cannot be derived.

Question 4. Statement 1 The order of accuracy in a measurement depends on the least count of the measuring instrument.

Statement 2 The smaller the least count, the greater the number of significant figures in the measured value.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Both the statements are independently correct.

Question 5. Statement 1 The mass of an object is 24.6 kg. This value contains three significant digits.

Statement 2 The value of the same mass when expressed as 24600 g has five significant digits.

Answer: 3. Statement 1 is true and Statement 2 is false.

The degree of accuracy and hence the number of significant digits of a measurement cannot be increased by changing the unit of a physical quantity.

Question 6. Statement 1 The dimensional method cannot be used to derive the relationship among the work done (W) by a force (F), the displacement (s), and the angle (0) between the directions of \(\vec{F}\) and \(\vec{s}\).

Statement 2 All trigonometric functions are dimensionless.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Work done = W = Fs cos θ. Since cos θ is dimensionless, the dependence of W on cos θ cannot be determined.

Question 7. Statement 1 The dimensional formula for RC is the same as that for L/R, where L, C, and R represent, capacitance and resistance respectively.

Statement 2 Both RC and L/R have the dimensions of frequency.

Answer: 3. Statement 1 is true and Statement 2 is false.

The dimensions of both L/R and CR are that of time and not of frequencies, as given in Statement 2.

Motion In A Straight Line

Question 1. Statement 1 The speedometer of an automobile measures the average speed of the automobile.

Statement 2 The average velocity is equal to the total distance divided by the total time taken.

Answer: 5. Both Statement 1 and Statement 2 are false.

The speedometer measures the instantaneous speed of an automobile.

The average velocity is a vector quantity and is measured by

∴ \(\vec{v}_{\mathrm{av}}=\frac{\text { net displacement }}{\text { time taken }}\)

Question 2. Statement 1 The relative velocity between any two objects moving in opposite directions is equal to the sum of the velocities of the two objects.

Statement 2 Sometimes the relative velocity between two objects is equal to the difference in velocities of the two objects.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

When two objects move in opposite directions, the magnitude of relative velocity is equal to the sum of their velocities. When their motion is in the same direction, the relative velocity is equal to the difference of the two velocities.

Question 3. Statement 1 The displacement of a particle may be zero, whereas the distance traveled by it is not zero.

Statement 2 The displacement is the longest distance between the initial and final positions.

Answer: 3. Statement 1 is true and Statement 2 is false.

When a particle completes a closed loop, its displacement is zero but the distance covered is nonzero! The displacement is measured by the areal distance, or the shortest distance between the two points, not the longest distance.

Question 4. Statement 1 A body can have an acceleration even if its velocity is zero at that instant of time.

Statement 2 A body comes to a momentary rest when it reverses its direction of motion.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Consider the motion of an object projected vertically upwards. At the highest point, the object comes to a momentary rest (v = 0) but gravity still acts (so acceleration ≠ 0). This is given as Statement 2.

Question 5. Statement 1 When a body is thrown horizontally or dropped from the same height, it reaches the ground at the same time.

Statement 2 The horizontal velocity has no effect on the vertical direction.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The component of the horizontal velocity along the vertical is zero. So, both the bodies reach the ground at the same time.

Question 6. Statement 1 A particle moving in a straight line at a constant speed must have zero acceleration.

Statement 2 A particle moving along a curve at a constant speed may have zero acceleration.

Answer: 3. Statement 1 is true and Statement 2 is false.

A body moving along a curved path changes its direction continuously and represents an accelerated motion.

Question 7. Statement 1 If the displacement-time graph of the motion of a body is a straight line parallel to the time axis, it follows that the body is at rest.

Statement 2 The time rate of change of displacement measures the instantaneous velocity.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The instantaneous velocity is the slope of the displacement-time graph \((\vec{v}=\overrightarrow{d x} / d t)\)

For the zero slope, \(\vec{v}=\overrightarrow{0}\)

Question 8. Statement 1 If the velocity-time graph of a particle in motion is a curve then the particle is either uniformly accelerated or uniformly retarded.

Statement 2 The slope of the velocity-time graph gives the instantaneous acceleration.

Answer: 4. Statement 1 is false and Statement 2 is true.

The slope of the v-t graph gives the instantaneous acceleration. The slope changes continuously if the v-t graph is a curve.

Question 9. Statement 1 In a uniform circular motion, the kinetic energy of a body remains constant.

Statement 2 The momentum of the body does not change with time.

Answer: 3. Statement 1 is true and Statement 2 is false.

The kinetic energy is given by

∴ \(E_{\mathrm{k}}=\frac{1}{2}\left(\frac{\vec{p} \cdot \vec{p}}{m}\right)=\frac{p^2}{2 m}\), which remains constant with time.

But the momentum \((\vec{p})\) is a vector which changes with a change in
direction.

Motion In A Plane

Question 1. Statement 1 Two balls of different masses are projected vertically upwards with the same speed. They will pass through their point of projection simultaneously in the downward direction.

Statement 2 The maximum height and the downward velocity attained at the point of projection are independent of the masses of the balls.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A motion under gravity has a constant acceleration \(\vec{g}\) The maximum height and the downward velocity at the point of projection are independent of the mass of the body.

Question 2. Statement 1 Electric current and time both have directions as well as magnitudes. But they are not considered to be vectors.

Statement 2 They do not follow the laws of vector addition.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Electric current and time add up scalarly and not like vectors. Hence, they are both scalars and not vectors.

Question 3. Statement 1 When a ball is projected at an angle of 45°, its range is maximum.

Statement 2 For the range to be maximum, the value of sin 20 should be equal to unity.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In a projectile motion, the horizontal range is given by R = \(\left(u^2 \sin 2 \theta\right) / g\), and R is maximum when sin 2θ is maximum (= 1), i.e., when θ = 45°.

Question 4. Statement 1 A stone of mass m is tied to a stone of radius R and is whirled in a vertical circle. The minimum speed at the topmost point is \(\sqrt{g R}\).

Statement 2 At the topmost point, the centripetal force is provided partly by the weight (mg) of the stone and partly by the tension in the string.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

At the topmost point in the vertical circle, the weight and tension taken together provide the required centripetal force. For the minimum speed, tension T = 0. So, \(v_{\min }=\sqrt{g R}\)

General Physics Objective Questions And Answers Motion In A Plane Vertical Circle Q4

Question 5. Statement 1 In javelin throwing, the athlete throws the projectile at an angle slightly more than 45°.

Statement 2 The maximum horizontal range depends on the angle of projection.

Answer: 4. Statement 1 is false and Statement 2 is true.

The maximum range R = \(\left(u^2 \sin 2 \theta\right) / g\) does depend on the angle of projection θ (= 45°).

Question 6. Statement 1 When the velocity of projection of a body is made n times, its time of flight becomes n times.

Statement 2 The range of a projectile does not depend on the initial velocity of the projectile.

Answer: 3. Statement 1 is true and Statement 2 is false.

Time of flight = \(T=\frac{2 u \sin \theta}{g}\)

If \(u^{\prime}=n u, T^{\prime}=\frac{2(n u) \sin \theta}{g}=n T\)

Thus, Statement 1 is true but Statement 2 is false, as the range (R) depends on u and θ both.

Question 7. Statement 1 Generally, the path of a projectile from the earth is parabolic. But it is elliptical for a projectile going up to a very great height.

Statement 2 Up to an ordinary height, a projectile moves under a uniform gravitational force. But for a great height, a projectile moves under a variable force.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The path followed by a projectile is parabolic for the heights h«RE, where RE is the radius of the earth. The gravity g remains constant. But for the heights h» RE,g decreases significantly, leading to an elliptical path.

Question 8. Statement 1 A cyclist leans inwards while taking a turn, whereas a person sitting in a car leans outwards on a curve.

Statement 2 A centripetal acceleration is acting toward the center of the curve.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In taking a turn, the cyclist needs a centripetal force towards the center. This force is provided by the frictional force acting between the tires and the ground.

The person sitting in a car is in an accelerated frame and leans outwards due to the pseudo force generated.

Question 9. Statement 1 The position of a particle in the xy-plane is expressed by x = a sin ωt and y = a(1- cos ωt), where a and ω are constants. Then, the particle will have a parabolic motion.

Statement 2 A particle under the influence of two mutually perpendicular velocities does not take a parabolic path.

Answer: 4. Statement 1 is false and Statement 2 is true.

The path followed by the particle will be a circle of radius centered on (0, a), as given below.

⇒ \(\sin \omega t=\frac{x}{a} \text { and } \cos \omega t=1-\frac{y}{a}\)

∴ \(\left(\frac{x}{a}\right)^2+\left(1-\frac{y}{a}\right)^2=1\)

⇒ x2 + (y-a)2 = a2.

So, Statement 1 is false. Under two mutually perpendicular velocities, the path followed will not be a parabola.

Question 10. Statement 1 In a uniform circular motion, the centripetal force is always perpendicular to the velocity vector.

Statement 2 The velocity vector at a point is always along the tangent to the trajectory at the point.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In a uniform circular motion, the speed is constant, so KE = constant.

Hence, work done = \(\vec{F} \cdot \vec{s}=0\)

So, \(\vec{F}\) is perpendicular to the velocity vector.

For any trajectory, the velocity vector has its direction tangential to the trajectory.

Laws Of Motion

Question 1. Statement 1 Frictional forces are nonconservative forces.

Statement 2 A potential energy can be associated with a frictional force.

Answer: 3. Statement 1 is true and Statement 2 is false.

Frictional forces dissipate energy and are thus nonconservative. Potential energies can be stored in conservative forces. So, no potential energy can be assigned to a frictional force.

Question 2. Statement 1 The apparent weight of a body in an elevator moving with some downward acceleration is less than the true weight of the body.

Statement 2 A part of the body’s weight is spent in providing the downward acceleration when the body is in the elevator.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

We feel our weight by the normal reaction \((\mathcal{N})\) offered by the floor.

For a lift accelerating down, mg – \((\mathcal{N})\) = ma.

∴ apparent weight = \((\mathcal{N})\) = mg- ma, which is less than the true weight mg.

Since mg = \((\mathcal{N})\) + ma, part of mg is spent as ma.

Question 3. Statement 1 Two bodies of masses M and m (where M > m) are allowed to fall from the same height. If the air resistance for each is the same then both the bodies will reach the ground simultaneously.

Statement 2 For the same air resistance, the accelerations of both bodies will not be equal.

Answer: 4. Statement 1 is false and Statement 2 is true.

For the mass M,

∴ \(M g-f=M a_1 \Rightarrow a_1=g-\frac{f}{M}\)

For m, \(a_2=g-\frac{f}{m}\)

Since M>m, a1>a2. Hence, M will reach first and then m. So, Statement 1 is false and Statement 2 is true, since a1 a2.

Question 4. Statement 1 The use of ball bearings between two moving parts of a machine is a common practice.

Statement 2 The ball bearings reduce rolling friction and provide good stability.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Ball bearings reduce rolling friction, and hence using them in machines is a common practice.

Question 5. Statement 1 On a rainy day, it is difficult to drive a car or a bus at a high speed.

Statement 2 The value of the coefficient of friction is lowered on wetting the surface.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A wet road reduces friction, so the high-speed movement over such a road may lead to skidding.

Question 6. Statement 1 A horse has to pull a cart harder during the first few steps of its motion.

Statement 2 The first few steps are always difficult.

Answer: 3. Statement 1 is true and Statement 2 is false.

The first few steps are harder, as the horse has to overcome the static friction, which is greater than the kinetic friction.

Question 7. Statement 1 A centripetal force is always required for a motion along a curved path.

Statement 2 On a banked curved track, the vertical component of the normal reaction provides the required centripetal force.

Answer: 3. Statement 1 is true and Statement 2 is false.

A centripetal force is always required for a motion along a curved path. On a banked road, the horizontal component of the normal reaction provides the required centripetal force.

General Physics Objective Questions And Answers Law Of Motion Centripetal Force Q7

Question 8. Statement 1 A pseudo force is an imaginary force that is recognized only by a noninertial observer to explain the physical situation according to Newton’s laws of motion.

Statement 2 A pseudo force has no physical origin, i.e., it is not caused by the basic interactions in nature. It does not exist in an action-reaction force pair.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

In an accelerated (or a noninertial) frame, the observer experiences a pseudo force (= mass of the object x acceleration of the frame with a negative sign = \(-m \vec{a}_{\mathrm{fr}}\)).

However, it is not a field force like gravitational and elastic forces

Question 9. Statement 1 A body subjected to three coplanar concurrent forces may be in equilibrium.

Statement 2 For equilibrium, the vector sum of all the concurrent forces acting at a point should be

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A body will be in equilibrium if the vector sum of the three concurrent forces acting on it is zero.

Question 10. Statement 1 An animate object can accelerate in the absence of external forces.

Statement 2 Newton’s second law of motion is not applicable to animate objects.

Answer: 4. Statement 1 is false and Statement 2 is true.

When we apply the second law of motion, there is no distinction between whether the object is animate or inanimate. An animate object always requires an external force in order to produce an acceleration.

Question 11. Statement 1 It is difficult to move a cycle along the road with its brakes on.

Statement 2 The degree of sliding friction is always greater than that of rolling friction.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

With the brakes on, there is no rolling but sliding. Sliding friction is greater than rolling friction, which makes the movement difficult.

Question 12. Statement 1 An impulsive force is large and acts for a short period of time.

Statement 2 A finite change in momentum should be produced by the force.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A large force acting for a short duration is called an impulsive force. According to the impulse-momentum theorem, a change in momentum is produced by the impulse.

Question 13. Statement 1 The driver in a vehicle moving at a constant speed on a straight road is in a noninertial frame of reference.

Statement 2 A reference frame to which Newton’s laws of motion are applicable is noninertial.

Answer: 5. Both Statement 1 and Statement 2 are false.

A motion along a straight line with a uniform velocity has zero acceleration, and such a reference frame is called an inertial frame. A noninertial frame is an accelerated frame. Newton’s second and third laws are valid only when we consider a pseudo-force.

Question 14. Statement 1 It is easier to pull a heavy object than to push it on a level ground.

Statement 2 The magnitude of frictional force depends on the nature of the two surfaces in contact.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

While pulling, the component F cos 0 of the pulling force acts against the weight mg, so that the normal reaction is reduced to \(\mathcal{N}\) = mg- F sin 0.

This reduces the kinetic frictional force \(f_{\mathrm{k}}=\mu \mathcal{N}=\mu(m g-F \sin \theta)\)

While pushing, the normal reaction is increased to \(-\mathcal{N}=m g+F \sin \theta\)

Hence, the frictional force becomes \(f_{\mathrm{k}}=\mu(m g+F \sin \theta)\)

Now, more effort is required to overcome the friction.

General Physics Objective Questions And Answers Law Of Motion Kinetic Friction Force Q14

General Physics Objective Questions And Answers Law Of Motion Normal Reaction Q7

Work Energy Power And Collision

Question 1. Statement 1 In an elastic collision, kinetic energy is conserved.

Statement 2 Energy is always conserved.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

In an elastic collision, kinetic energy remains the same before and after the collision. The law of energy conservation is the basic conservation law in nature. Energy may change the form but remains unchanged in total.

Question 2. Statement 1 The work done in a uniform circular motion is zero.

Statement 2 Force is always directed along displacement.

Answer: 3. Statement 1 is true and Statement 2 is false.

In a uniform circular motion, the centripetal force is directed radially inwards, whereas the velocity and displacement vectors are tangential. Hence, work done = W = \(\vec{F} \cdot \vec{s}\) = 0.

General Physics Objective Questions And Answers Work Energy Power And Collision Uniform Circular Motion Q2

Question 3. Statement 1 If the momentum of a body increases by 50%, its kinetic energy will increase by 125%.

Statement 2 Kinetic energy is directly proportional to the square of velocity.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

⇒ \(\mathrm{KE}=\frac{1}{2} m v^2\)

If m is constant, KE oc v2.

Again, \(\mathrm{KE}=\frac{m^2 v^2}{2 m}=\frac{p^2}{2 m}=E_1\)

With a 50% increase in momentum,

∴ \(\mathrm{KE}=\frac{1}{2 m}\left(p+\frac{p}{2}\right)^2=\frac{9 p^2}{2 m \times 4}=E_2\)

% increase in \(\mathrm{KE}=\frac{\Delta \mathrm{KE}}{\mathrm{KE}} \times 100 \%=\left(\frac{9}{4}-1\right) 100 \%=125 \%\)

Question 4. Statement 1 The total energy is negative for a bound system.

Statement 2 The potential energy of a bound system is negative and more than the kinetic energy.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The total energy (KE + PE) of a bound system (like the electron-nucleus system in an atom, the earth-sun system, or an earth-satellite system) is always negative with a negative PE (greater than KE in magnitude). The energy required to detach the component is called the binding energy.

Question 5. Statement 1 A centripetal force does no work.

Statement 2 Force and displacement are mutually perpendicular to each other.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Centripetal force and displacement are mutually perpendicular, Hence, the work done is zero.

Question 6. Statement 1 A rigid body can be elastic.

Statement 2 If a force is applied to a rigid body, its dimensions may change.

Answer: 5. Both Statement 1 and Statement 2 are false.

A rigid body is a solid body in which the deformation is zero or so small that it can be neglected. The separation between any two given points remains constant regardless of the external forces exerted on it.

Question 7. Statement 1 The work done in moving a body over a closed loop is zero for every force in nature.

Statement 2 Work does not depend on the nature of force.

Answer: 4. Statement 1 is false and Statement 2 is true.

The work done over a closed loop is zero only for conservative forces (like gravitational and electrostatic forces). But it is nonzero for nonconservative forces (like friction).

Question 8. Statement 1 The kinetic energy is conserved at every instant during an elastic collision.

Statement 2 No deformation of matter occurs during an elastic collision.

Answer: 4. Statement 1 is false and Statement 2 is true.

A collision is always associated with a deformation for a very short duration. In the process, the KE is converted into PE (during compression) and thereafter back to the same amount of KE during an elastic collision.

Question 9. Statement 1 A quick collision between two bodies is more violent than a slow collision even when the initial and final velocities are identical.

Statement 2 The rate of change of momentum is greater in the first case.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

According to the impulse-momentum theorem, FΔt = Δp, where F = impact force, Δt = time of impact, and Δp = change in momentum. For the same value of Δp, FΔt = F’Δt’. For a quick collision, Δt is small, so F is large and more violent. For a large Af, F’ is small.

Since F > F’, the rate of change of momentum is

∴ \(\frac{\Delta p}{\Delta t}>\frac{\Delta p}{\Delta t^{\prime}}\)

Question 10. Statement 1 In an elastic collision between two billiard balls, the total kinetic energy is conserved during the short duration of the collision of balls (i.e., as long as they are in contact).

Statement 2 The energy spent against friction still follows the law of conservation of energy.

Answer: 4. Statement 1 is false and Statement 2 is true.

In an elastic collision, there is an exchange between the KE and the elastic PE during the short duration of the impact. So, Statement 1 is false. The energy spent by the frictional force is converted to heat energy. So, energy conservation still follows.

Question 11. Statement 1 A particle collides head-on with another stationary particle such that the first particle comes to rest after the collision. The collision should necessarily be elastic.

Statement 2 In an elastic collision, there is a loss in the momentum of the system of particles.

Answer: 3. Statement 1 is true and Statement 2 is false.

When two particles of equal masses collide elastically, their velocities are mutually interchanged. The linear momentum is always conserved and is valid for all types of collisions.

Question 12. Statement 1 During an elastic collision between two bodies, the relative speed of the approach before the collision is equal to the relative speed of separation after the collision.

Statement 2 In an elastic collision, the linear momentum of the system remains conserved.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The coefficient of restitution is given by

∴ \(e=\frac{\text { relative speed of separation }}{\text { relative speed of approach }}\)

For an elastic collision, e = l. The law of conservation of linear momentum is always true for all types of collisions.

Question 13. Statement 1 In a one-dimensional elastic head-on collision, the total linear momentum remains unchanged before, during, and after the collision.

Statement 2 In a one-dimensional elastic head-on collision, the total kinetic energy remains the same before, during, and after the collision.

Answer: 3. Statement 1 is true and Statement 2 is false.

At the moment the balls A and B collide, A has a larger velocity than that of B, i.e., v1 > v2. But the contact surfaces move through equal distances in equal intervals of time as long as the contact is maintained. During the contact, A and B push each other and the velocities change such that the total momentum still remains unchanged. When some deformation occurs, the total kinetic energy of the system decreases as some energy is stored as elastic compressional potential energy. When the deformation is maximum, the kinetic energy is minimum and the two balls attain equal velocities.

General Physics Objective Questions And Answers Work Energy Power And Collision Moment The Balls Q13

Rotation

Question 1. Statement 1 If the polar ice melts, days will be longer.

Statement 2 Moment of inertia increases and thus angular velocity decreases.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

When the polar ice melts, there is a shift of mass from the axis of rotation to away from the axis. This increases the moment of inertia. Since the angular momentum remains constant, ω will decrease and the duration of a day will increase.

Question 2. Statement 1 The earth is slowing down and consequently the moon is coming nearer to it.

Statement 2 The angular momentum of the earth-moon system is always conserved.

Answer: 4. Statement 1 is false and Statement 2 is true.

The earth is not slowing down and the angular momentum of the earth-moon system is always conserved.

Question 3. Statement 1 For a system of particles under a central force field, the angular momentum is conserved.

Statement 2 The torque acting on such a system is zero.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Under the central force, \(\vec{F}\) and \(\vec{r}\) are directed along the same line. So, torque = \(\vec{\tau}=\vec{r} \times \vec{F}=\overrightarrow{0}\) . Hence, the angular momentum is conserved.

Question 4. Statement 1 A solid sphere is rolling on a rough horizontal surface. The acceleration of the contact point is zero.

Statement 2 A solid sphere can not roll on a smooth surface.

Answer: 4. Statement 1 is false and Statement 2 is true.

For a pure rolling motion on a horizontal plane, the relative velocity of the point of contact and the tangential acceleration are both zero, but still, there is a radial acceleration directed towards the center. On a smooth surface, the sphere will undergo slipping due to the absence of friction.

Question 5. Statement 1 Moment of inertia is always constant.

Statement 2 Angular momentum is conserved which is why the moment of inertia is conserved.

Answer: 5. Both Statement 1 and Statement 2 are false.

The moment of inertia about a given axis is not constant but depends on the distribution of mass relative to the axis of rotation.

Further, angular momentum is conserved only in the absence of an external torque.

Question 6. Statement 1 The value of the radius of the gyration of a body depends on the axis of rotation.

Statement 2 The radius of gyration of a body is the root-mean-square distance of the particles of the body from the axis of rotation.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The radius of gyration (k) of a body depends on the position of the axis of rotation and the distribution of mass around it. Considering the particles of the same mass, we have

∴ \(k=\sqrt{\frac{r_1^2+r_2^2+\ldots+r_n^2}{n}}\) = rms distance.

Question 7. Statement 1 The size and shape of the rigid body remain unaffected under the effect of external forces.

Statement 2 The distance between two particles remains constant in a rigid body.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A rigid body is an ideal concept in which the separation between any of its two particles and the shape and size of the body remain unaffected by external forces and torques.

Question 8. Statement 1 The center of mass of a system does not move under the action of internal forces.

Statement 2 Internal forces are not conservative forces.

Answer: 3. Statement 1 is true and Statement 2 is false.

The acceleration of the center of mass of a system of particles is due only to the external forces acting on the system, and the interval forces have no contributions in accelerating the center of mass. Internal forces always act in pairs (like an action-reaction pair) and they add up to zero for the whole system. Hence, the motion of the center remains unaffected.

Inertial forces are pseudo forces and exist only in noninertial frames to ensure the validity of Newton’s second law.

Question 9. Statement 1 A judo fighter in order to throw his opponent onto the mat tries to bend his opponent initially and then rotate him around his hip.

Statement 2 As the mass of the opponent is brought closer to the fighter’s hip, the force required to throw the opponent is reduced.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The rotation of the opponent makes it difficult for the person rotated to resist. By bringing him closer to the hip, the moment of inertia of the opponent is reduced.

Question 10. Statement 1 A wheel moving down a perfectly smooth inclined plane will undergo a slipping (not rolling) motion.

Statement 2 For a perfect rolling motion, the work done against friction is zero.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

In the absence of friction on a smooth incline, no frictional torque required for rolling is available. So, the wheel slips.

For a perfectly rolling motion, friction is static and the work done against friction is zero.

Question 11. Statement 1 If the ice cap of the poles melts, the duration of a day will shorten.

Statement 2 Ice will flow towards the equator and increase the moment of inertia of the earth.

Answer: 4. Statement 1 is false and Statement 2 is true.

When the polar ice melts, it spreads over the surface and the moment of inertia increases. In order to conserve the angular momentum, the angular velocity decreases, which increases the duration of a day.

Question 12. Statement 1 The velocity of a body at the bottom of an inclined plane of a given height is more when it slides down the plane compared to when it is rolling down the same plane.

Statement 2 In rolling down, a body acquires kinetic energies of both translation and rotation.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

During sliding, the total potential energy is converted into kinetic energy. Thus, at the bottom, \(v=\sqrt{2 g h}\). When the body rolls down the plane, the PE is partly converted into KE of translation \(\left(\frac{1}{2} m v^2\right)\) and partly into KE of rotation \(\left(\frac{1}{2} I \omega^2\right)\)

∴ \(v^{\prime}=\sqrt{\frac{2 g h}{1+\frac{k^2}{R^2}}}=\frac{v}{\sqrt{1+\frac{k^2}{R^2}}} \Rightarrow v^{\prime}<v\)

Question 13. Statement 1 There are very small sporadic changes in the period of rotation of the earth.

Statement 2 Shifting of large air masses in the earth’s atmosphere produces a change in the moment of inertia of the earth, causing its period of rotation to change.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Shifting of large air masses produces a change in the earth’s moment of inertia, and to keep the angular momentum constant, the period of rotation is changed.

Gravitation

Question 1. Statement 1 An astronaut experiences weightlessness in a space satellite.

Statement 2 When a body falls freely, it does not experience gravity.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A satellite orbiting the Earth is falling (accelerating towards the Earth), but its tangential speed keeps it away from hitting the surface of the Earth.

Objects (or astronauts) in an orbit are said to experience weightlessness. They do have a gravitational force acting on them, but the satellite and all its contents are in free fall, and the absence of any normal reaction leads to the experience of weightlessness.

Question 2. Statement 1 Comets do not obey Kepler’s laws of planetary motion.

Statement 2 Comets have highly elliptical orbits.

Answer: 4. Statement 1 is false and Statement 2 is true.

Comets go around the sun in a highly elliptical orbit. Like all orbiting bodies, comets also follow Kepler’s second law: the closer they are to the sun, the faster they move.

Question 3. Statement 1 A planet is a heavenly body revolving around the sun.

Statement 2 A star is a luminous body made of gaseous material.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The planets in the solar system orbit the sun under its gravitational force.

Stars are luminous celestial objects containing hydrogen and helium held together by their own gravity.

Question 4. Statement 1 The difference in the values of the acceleration due to gravity at the poles and that at the equator is proportional to the square of the angular speed of the earth.

Statement 2 The value of acceleration due to gravity is minimum at the equator and maximum at the poles.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

At the poles, \(g_{\mathrm{P}}=g=\frac{G M_{\mathrm{E}}}{R_{\mathrm{E}}^2}\)

At the equator, \(g_{\mathrm{eq}}=g-R_{\mathrm{E}} \omega^2\)

∴ \(g_{\mathrm{p}}-g_{\mathrm{eq}}=R_{\mathrm{E}} \omega^2 \propto \omega^2\)

Hence, geq is the minimum value of the acceleration due to gravity.

Question 5. Statement 1 The earth without its atmosphere would be inhospitably cold.

Statement 2 All heat would escape the earth in the absence of the atmosphere.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The atmosphere protects life on the earth by absorbing the solar UV radiations warming the surface through heat retention and reducing the temperature extremes between day and night The moon has no atmosphere and its temperature in sunlight is around 122 °C. The moon’s surface freezes at -152.7 °C at night; and a similar would be the case with the earth, whose surface would be inhospitably cold without its atmosphere.

Question 6. Statement 1 Water kept in an open vessel will quickly evaporate on the surface of the moon.

Statement 2 The temperature at the surface of the moon is much higher than the boiling point of water.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Due to the negligibly low pressure on the moon, water evaporates quickly. During the daytime, the temperature is very high (above the water’s boiling point) and water will boil.

Question 7. Statement 1 The time period of revolution of a satellite around the planet is directly proportional to the radius of the orbit of the satellite.

Statement 2 Artificial satellites follow Kepler’s laws of planetary motion.

Answer: 4. Statement 1 is false and Statement 2 is true.

Kepler’s laws of planetary motion are followed by the orbital motion of artificial satellites, like all planetary motions. For such a motion, the square of the time of revolution is directly proportional to the cube of the radius of the orbit. Hence, T ∝ R is a false relation.

Question 8. Statement 1 The square of the period of revolution of a planet is proportional to the cube of the semimajor axis of its elliptical orbit.

Statement 2 The sun’s gravitational field is inversely proportional to the square of its distance from the planet.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Kepler’s law of periods \(\left(T^2 \propto a^3\right)\) is true. This law is based on Newton’s law of gravitation \(\left(F \propto \frac{1}{r^2}\right)\).

Question 9. Statement 1 If a pendulum falls freely under gravity then its time period becomes infinite.

Statement 2 A freely falling body has an acceleration equal to g.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In free fall, there is no normal force or tension, so the bob is in a state of weightlessness, and geff = 0.

Hence, time period = \(T=2 \pi \sqrt{\frac{l}{g_{\text {eff }}}}=\infty\)

A body falling freely under gravity has an acceleration of g.

Question 10. Statement 1 The length of the day is slowly increasing.

Statement 2 The dominant effect causing a slowdown in the rotation of the Earth is the combined gravitational pull of other planets in the solar system.

Answer: 3. Statement 1 is true and Statement 2 is false.

The earth’s rotation is slowing down slightly with time. This is due to the tidal effects the moon has on the Earth’s rotation. This causes a slow increase in the length (duration) of the day. Gravitational pulls are central forces that do not produce any external torques, so angular momenta will remain conserved.

Question 11. Statement 1 A person in a closed cabin falling freely under gravity does not experience gravity.

Statement 2 The inertial mass and the gravitational mass have an equivalence.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

We feel our weight by. the normal force exerted by the surface in contact. From the adjoining figure,

⇒ \(F_{\text {net }}=m g-\mathcal{N} \text { but } F_{\text {net }}=m g\)

∴ \(\mathcal{N}=0\)

Hence, it is a state of effective weightlessness.

Mass is always equal to the ratio F/a. So, gravitational and inertial masses are equal.

General Physics Objective Questions And Answers Gravitation Adjoining Q11

Question 12. Statement 1 Kepler’s second law can be understood by the principle of conservation of angular momentum.

Statement 2 Kepler’s second law is related to areal velocity, which can further be proved to be based on the principle of conservation of angular momentum as \(\frac{d A}{d t}=\frac{1}{2} r^2 \omega\).

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Kepler’s second law (areal velocity = constant) leads to the principle of conservation of angular momentum.

Question 13. Statement 1 The angular speed of a planet orbiting around the sun increases when it is closer to the sun.

Statement 2 The total angular momentum of the system remains constant.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

From Kepler’s second law,

∴ \(\frac{1}{2} r^2-\frac{d \theta}{d t}=\cos x\),

Where \(\frac{d \theta}{d t}=\omega\) is the angular velocity.

When the planet is closer to the sun, r decreases, so ω increases.

This is linked with the principle of conservation of angular momentum.

Question 14. Statement 1 The total energy is conserved in moving a satellite to a higher orbit.

Statement 2 The sum of the changes in KE and PE is the same in magnitudes but opposite in nature.

Answer: 3. Statement 1 is true and Statement 2 is false.

The total mechanical energy (= KE + PE) of an artificial satellite orbiting the earth is \(E=-\frac{G M_E m}{2\left(R_E+h\right)}\), where ME is the mass of the earth and RE is the radius of the orbit. When raised to a higher orbit, h increases and E also increases (i.e., becomes less negative). The difference in energy is given by the system, so the energy is conserved.

Question 15. Statement 1 At the poles, the values of the acceleration due to gravity (g) are greater than that at the equator.

Statement 2 The earth rotates on its axis in addition to its revolution around the sun.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

At the equator, \(g_{\mathrm{eq}}=g-\omega^2 R_{\mathrm{E}}\) nd at the poles, gp = g. Thus, gp > geq. This change is due to the rotation (spin) of the earth about its own axis as well as to the unequal distances of the poles and the equator from the earth’s center.

Question 16. Statement 1 A geostationary satellite must be located in the equatorial plane at some point vertically above the equator.

Statement 2 The only external force acting on an artificial satellite is directed toward the centre of the earth.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A geostationary satellite is located in a circular orbit at a height of 35786 km above the Earth’s equator in the equatorial plane so that it appears motionless at a fixed position in the sky to an observer on the Earth.

For the circular motion of the satellite, the gravitational attraction by the earth provides the required centripetal force.

Question 17. Statement 1 The gravitational potential energy of a solid sphere is negative.

Statement 2 Two bodies attract each other

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In an attractive field, the potential energy is always negative. Gravitation is a conservative attractive force. So, a sphere requires some energy to pull apart all its fragments, so that the PE increases from the negative value to zero.

Simple Harmonic Motion

Question 1. Statement 1 The periodic time of a hard spring is less as compared to that of a soft spring.

Statement 2 The periodic time depends on the spring constant, which is larger for hard springs.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The time period of a spring-block system is given by

∴ \(T=2 \pi \sqrt{\frac{m}{k}}\)

where khard>ksoft.

Question 2. Statement 1 The time period of a pendulum on a satellite orbiting the earth is infinity.

Statement 2 The time period of a pendulum is inversely proportional to √g.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In a satellite orbiting the Earth, any object is in a state of weightlessness.

So, \(g_{\mathrm{eff}}=0 \text { and } T=2 \pi \sqrt{\frac{l}{g_{\mathrm{eff}}}}=\infty\)

Question 3. Statement 1 Soldiers are asked to break steps while crossing the bridge.

Statement 2 The frequency of marching may be equal to the natural frequency of the bridge and may lead to resonance which can break the bridge.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Statement 1 is true because the marching soldiers are asked not to be in steps so as to avoid any resonance. At resonance, the amplitude of oscillations becomes very large. This may damage the. bridge.

Question 4. Statement 1 In an SHM, the kinetic and potential energies become equal when the displacement is (1/√2) times the amplitude.

Statement 2 In an SHM, the kinetic energy is zero when the potential energy is maximum.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

In an SHM, \(\mathrm{KE}=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \text { and } \mathrm{PE}=\frac{1}{2} m \omega^2 x\)

Both the energies are equal at \(x= \pm \frac{A}{\sqrt{2}}\). When the KE is zero, the PE is maximum, so that their sum remains conserved.

Question 5. Statement 1 In an SHM, the motion is to and fro as well as periodic.

Statement 2 The velocity of the particle is \(v=\omega \sqrt{A^2-x^2}\), where x is the displacement and A is the amplitude.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

An SHM is a periodic motion about the mean position with the velocity \(v=A \omega \cos \omega t=\omega \sqrt{A^2-x^2}\).

Question 6. Statement 1 The resonance is a special case of forced vibrations in which the natural frequency of vibrations of the body is the same as the impressed frequency and the amplitude of the forced vibrations is maximum.

Statement 2 The amplitude of the forced vibrations of a body increases with an increase in the frequency of an externally impressed periodic force.

Answer: 3. Statement 1 is true and Statement 2 is false.

In a forced SHM, resonance is a special case when the natural frequency of the oscillator is the same as that of the external periodic force. Before resonance, the amplitude increases with an increase in frequency; and beyond resonance, the amplitude decreases with an increase in frequency.

Question 7. Statement 1 The graph of the potential energy and kinetic energy of a particle executing an SHM with respect to its position (x) is a parabola.

Statement 2 The potential energy and kinetic energy do not vary linearly with the position.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The graphs showing the variations of the KE and PE versus the position (x) is a parabola. The KE and PE vary as x2. The curve is nonlinear.

General Physics Objective Questions And Answers Simple Harmonic Motion Parabola Q7

Question 8. Statement 1 Let the bob of a simple pendulum be a ball full of water. If a fine hole is made in the bottom of the ball, the time period first increases and then decreases.

Statement 2 As water flows out of the ball, its weight decreases.

Answer: 3. Statement 1 is true and Statement 2 is false.

When water gets drained out of the shell, the center of mass is lowered, thus increasing the length of the pendulum. Consequently, the time period increases. When the shell is more than half-empty, the center of mass again rises up with a decrease in the length and a decrease in the time period of the pendulum.

The time period \(T=2 \pi \sqrt{\frac{l}{g}}\) is independent of the weight of the bob.

Question 9. Statement 1 The amplitude of an oscillating pendulum decreases gradually with time.

Statement 2 The frequency of oscillations of the pendulum decreases with time.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In the presence of damping (air friction), the displacement amplitude decreases exponentially with time. The frequency of oscillations also decreases with time, so that the oscillations become slow.

General Physics Objective Questions And Answers Simple Harmonic Motion Damping Q9

Elasticity Fluids

Question 1. Statement 1 Stress is the internal force per unit area of a body.

Statement 2 Rubber is more elastic than steel.

Answer: 3. Statement 1 is true and Statement 2 is false.

Stress is developed in a body due to the internal forces per unit area. Steel is more elastic than rubber.

Question 2. Statement 1 A hollow shaft is found to be stronger than a solid shaft made of the same material.

Statement 2 The torque required to produce a given twist in a hollow cylinder is greater than that required to twist a solid cylinder of the same material and of the same dimensions.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

If two cylinders of the same length, the same mass, and the same material are made such that one is solid while the other is hollow then it can be shown that the twisting couple per unit twist for a hollow cylinder is greater than that for a solid cylinder. Hence, hollow shafts are stronger.

Question 3. Statement 1 Lead is more elastic than rubber.

Statement 2 If the same load is attached to lead and rubber wires of the same cross-sectional area, the strain in the lead wire is very much less than that in the rubber wire.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The material for which the stress at the elastic limit is more is said to be more elastic. This corresponds to Statement 2.

Question 4. Statement 1 A hydrogen-filled balloon stops rising after it has attained a certain height in the sky.

Statement 2 The atmospheric pressure decreases with height and becomes zero when the maximum height is attained.

Answer: 3. Statement 1 is true and Statement 2 is false.

Hydrogen is lighter than air, so it rises up due to a buoyant force. It stops rising further after some height. In the upper atmosphere, the pressure is low, so the balloon expands, and its buoyancy increases but not the pull of gravity. It stops in equilibrium. At the maximum height, the atmospheric pressure is not zero.

Question 5. Statement 1 The specific gravity of a liquid is a dimensionless quantity.

Statement 2 It is the ratio of the density of the fluid to the density of water.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The specific gravity of a substance is its density relative to water. So, it is dimensionless.

Question 6. Statement 1 For a Reynolds number Re > 2000, the nature of the flow of a fluid is turbulent.

Statement 2 Inertial forces are dominant compared to the viscous forces at such a high Reynolds number.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The Reynolds number is defined as the ratio

∴ \(R e=\frac{\text { inertial forces }}{\text { viscous forces }}\)

When the ratio exceeds 2000, the nature of flow becomes turbulent due to the dominance of inertial forces over viscous forces.

Question 7. Statement 1 Machine parts get jammed in winter.

Statement 2 The viscosity of a lubricant used in machine parts increases at low temperatures.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In winter, when the temperature considerably drops, the viscosity of lubricating oil increases, so the oil does not reach the machine parts easily, leading to a jam inside.

Question 8. Statement 1 A thin stainless-steel needle can lay floating on a still water surface.

Statement 2 Any object floats when the buoyant force balances the weight of the body.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A steel needle presses the water surface and experiences forces due to the surface tension to keep floating. For equilibrium, the weight and the buoyant force must balance each other.

General Physics Objective Questions And Answers Elasticity Fluids Steel Needle Q8

Question 9. Statement 1 The shape of an automobile is so designed that its front resembles the streamlined pattern of the fluid through which it moves.

Statement 2 The resistance offered by the fluid on streamlined objects is maximum.

Answer: 3. Statement 1 is true and Statement 2 is false.

The automobiles are designed to have a streamlined shape in the front part. This reduces the aerodynamic drag so as to minimize the resistance offered by the fluid.

Question 10. Statement 1 Turbulence is always dissipative.

Statement 2 A high Reynolds number promotes turbulence.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

In a turbulent flow, an energy dissipation occurs. During the turbulent flow, the Reynolds number is high.

Question 11. Statement 1 A needle placed carefully on the surface of still water may float, whereas the ball of the same material will always sink.

Statement 2 The buoyancy of an object depends both on the material and shape of the object.

Answer: 3. Statement 1 is true and Statement 2 is false.

A needle floats on a water surface due to the force of surface tension balancing the weight. A solid sinks because the buoyant force is not sufficient to balance the weight. Buoyancy depends only on the shape.

Question 12. Statement 1 The smaller drops of a liquid resist deforming forces better than the larger drops.

Statement 2 The excess pressure inside a drop is directly proportional to its surface area.

Answer: 3. Statement 1 is true and Statement 2 is false.

The excess pressure \(\left(\Delta p=\frac{2 T}{R}\right)\) inside a smaller drop is larger, hence it offers more resistive force for deformation than a larger drop. The excess pressure varies as 1 /R and it does not depend on the surface area.

Question 13. Statement 1 The surface energy of an oil drop is the same whether placed on a glass or a water surface.

Statement 2 The surface energy is dependent only on the properties of the oil.

Answer: 5. Both Statement 1 and Statement 2 are false.

The surface tension is a molecular phenomenon and depends on cohesive and adhesive forces. So, the surface energy for an oil-glass pair is different than that for an oil-water pair. It depends on both the surfaces in contact.

Question 14. Statement 1 When the height of a capillary tube is less than the liquid rise in the capillary tube, the liquid does not overflow.

Statement 2 The product of the radius of a concave meniscus and the height of liquid in a capillary tube always remains constant.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

When the length of the capillary tube is insufficient, the liquid rises to the top, and the radius of curvature of the concave meniscus changes to ensure that hR remains constant.

Electrostatics Multiple Choice Question And Answers

Electrostatics

Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.

Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.

  1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
  2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
  3. Statement 1 is true and Statement 2 is false.
  4. Statement 1 is false and Statement 2 is true.
  5. Both Statement 1 and Statement 2 are false.

Some questions are preceded, by a supporting paragraph in addition to the statements.

Question 1. Statement 1 A charged particle free to move in an electric field always moves along the electric field lines.

Statement 2 The electric field lines diverge from a positive charge and converge to a negative charge.

Answer: 4. Statement 1 is false and Statement 2 is true.

A charged particle when released from rest in an electrostatic field always travels along the field. But when its initial velocity makes some angle 0 with the field direction, it follows a parabolic path. Electric field lines diverge from a positive charge.

Question 2. Statement 1 If a conductor is given charge then no excess inner charge appears;

Statement 2 The electric field inside a conductor is always zero.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The charge given to a conductor totally remains on the outer surface and no part of it appears on the inner surface. The electric field inside a conductor is zero as a consequence of Gauss’s theorem and the absence of an inner charge.

Question 3. Statement 1 An electrostatic field line never forms a closed loop.

Statement 2 An electrostatic field is a conservative force field.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Electrostatic field lines do not form a closed loop. In a dipole, they start at the positive charge and terminate at the negative charge. This corresponds to the conservative nature of an electrostatic field for which the work done for a closed path is zero.

Question 4. Statement 1 Gauss’s law cannot be used to calculate the electric field near an electric dipole.

Statement 2 An electric dipole does not have symmetrical charge distribution.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Gauss’s law is applicable when the charge distribution is cylindrical, spherical, or planar. It cannot be applied for an asymmetric charge distribution like that of an electric dipole.

Question 5. Statement 1 The Nuclear force is a dominating force in nature.

Statement 2 The Coulomb force is weaker than the gravitational force.

Answer: 3. Statement 1 is true and Statement 2 is false.

The nuclear forces are the strongest in nature, whereas the Coulomb force is much stronger than the gravitational force.

Question 6. Statement 1 On going away from a point charge or a small electric dipole, the electric field decreases at the same rate in both cases.

Statement 2 An electric field is inversely proportional to the square of the distance from the charge and the cube of the distance from the electric dipole.

Answer: 4. Statement 1 is false and Statement 2 is true.

The electric field due to a dipole \(\propto \frac{1}{r^3}\), and that due to a monopole \(\propto \frac{1}{r^2}\).

∴ \(\left(\frac{d E}{d r}\right)_{\text {dipole }} \propto \frac{1}{r^4} \text { and }\left(\frac{d E}{d r}\right)_{\text {monopole }} \propto \frac{1}{r^3}\)

Question 7. Statement 1 A Gaussian surface is considered carefully.

Statement 2 The point where the electric field is to be calculated should be within the surface.

Answer: 5. Both Statement 1 and Statement 2 are false.

A Gaussian surface is symmetrical about the given charge distribution, and the point where the field is to be found is outside the Gaussian surface.

Question 8. Statement 1 In the cavity within a conductor, the electric field is zero.

Statement 2 Charges with a conductor always reside only at its outer surface.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The electric field in the cavity of a charged conductor is always zero because no charge exists on the inner surface. Both of these facts are consequences of Gauss’s theorem.

Question 9. Statement 1 Gauss’s law shows a diversion when the inverse square law is not obeyed.

Statement 2 Gauss’s law is a consequence of the conservation of charges.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Gauss’s law is a consequence of the inverse-square law. Charge cannot be created nor be destroyed, it always remains conserved.

Question 10. Statement 1 The acceleration of a charged particle in a uniform electric field does not depend on the velocity of the charged particle.

Statement 2 Charge is an invariant quantity, which means that the amount of charge with a particle does not depend on the frame of reference.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Force = F = qE; acceleration = \(a=\frac{F}{m}=\frac{q E}{m}\)

Since q, E, and m are constants, the acceleration a is also a constant and independent of the velocity of the charged particle. Charge is always invariant.

Question 11. Statement 1 If an electron and a proton possessing equal KEs enter an electric field in a particular direction, the path of the electron will be more curved than that of the proton.

Statement 2 An electron describes a larger curve due to its small mass.

Answer: 5. Both Statement 1 and Statement 2 are false.

When a charged particle is projected at some angle θ (≠ 0° or π), it follows a parabolic path, for which the curvature varies from point to point.

At a given point, \(F=q E=\frac{m v^2}{r}=\frac{2 \mathrm{KE}}{r}\)

An electron and a proton have the same KE and the same charge, so the radius of the curvature r will be the same and the path followed will be the same.

Question 12. Statement 1 Electrons move away from a region of lower potential to a region of higher potential.

Statement 2 An electron has a negative charge.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Electrons are negatively charged particles that move against the direction of the electric field. An electric field is directed from a higher potential to a lower potential.

Question 13. Statement 1 The electrical potential of the earth is taken as zero.

Statement 2 No electric field exists on the earth’s surface.

Answer: 3. Statement 1 is true and Statement 2 is false.

The earth is a very large conducting body where the addition or subtraction of charges is insignificant. Hence, its potential is taken to be zero. Near the surface of the earth, the magnitude of the electric field is around 100 V m-1.

Question 14. Statement 1 Electric field lines are perpendicular to a conducting surface.

Statement 2 Electric field lines are perpendicular to an equipotential surface.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A conducting surface is equipotential, on which the electrical field is directed perpendicular to its surface.

Question 15. Statement 1 Two adjacent spherical conductors carrying the same amount of positive charge have a potential difference between them.

Statement 2 The potential to which a conductor is raised depends on the charge.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The potential of a spherical conductor of radius R is \(V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\).

If the charge is the same and the radii of the spheres are unequal then their potentials will be different. The potential V depends on both the charge Q and the radius R.

Question 16. Statement 1 An electric field is preferred to a magnetic field for deflecting the electron beam in a television picture tube.

Statement 2 An electric field requires a low voltage.

Answer: 4. Statement 1 is false and Statement 2 is true.

In television picture tubes, a magnetic field is preferred to an electric field for deflecting the electron beam.

The electric field \(E=\frac{V}{d}\) requires a high voltage to produce the required field.

Question 17. Statement 1 If a dielectric is placed in an external field then the field inside the dielectric will be less than the field outside.

Statement 2 The electric field will induce a dipole moment opposite to the field direction.

Answer: 3. Statement 1 is true and Statement 2 is false.

Dielectric polarization produces a field in a direction opposite to the externally applied field.

Polarization produces dipoles whose moments (-ve to +ve) are directed along the externally applied field.

Question 18. Statement 1 The electrostatic potential energy of a spherical shell with a uniformly distributed negative charge is positive.

Statement 2 Two similar charges repel each other.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Potential energy = work done

∴ \(\int d W=\int V d q=\int_0^{-Q_1} \frac{1}{4 \pi \varepsilon_0 R}(-q)(-d q)=\frac{Q^2}{8 \pi \varepsilon_0 R}=\frac{( \pm Q)^2}{8 \pi \varepsilon_0 R}\)

This is positive for both positive and negative charges. Potential energy is positive for a repulsive field (repulsive force between two like charges).

Question 19. Statement 1 When an uncharged parallel-plate capacitor is charged by connecting it to a cell, the heat produced in the circuit is equal to the energy stored in the capacitor.

Statement 2 The charge on a parallel-plate capacitor means equal and opposite charges on its inner faces.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Total energy delivered by the cell = Qε .

Energy stored in the capacitor = \(\frac{Q^2}{2 C}=\frac{Q \varepsilon}{2}\).

Heat developed = \(\mathrm{Q} \varepsilon-\frac{\mathrm{Q} \varepsilon}{2}=\frac{\mathrm{Q} \varepsilon}{2}\).

A charged capacitor has equal and opposite charges, which produce an electrostatic field between the plates.

Question 20. Statement 1 It is possible for a charged particle to move in a circular path around a uniformly charged, long, straight conductor.

Statement 2 The electrostatic force on a moving charged particle is directed toward the conductor.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The electrostatic field due to a long, straight-charged conductor is radial, which can provide a centripetal force to the charged particle moving along the circle/with the axis as the conductor.

Modern Physics Multiple Choice Questions And Answers

Modern Physics

Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.

Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.

  1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
  2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
  3. Statement 1 is true and Statement 2 is false.
  4. Statement 1 is false and Statement 2 is true.
  5. Both Statement 1 and Statement 2 are false.

Some questions are preceded, by a supporting paragraph in addition to the statements.

Photoelectric Effect Matter Waves

Question 1. Statement 1 A photon has no rest mass, yet it carries a definite momentum.

Statement 2 The momentum of a photon is due to its energy and henceit has an equivalent mass.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A particle that travels at the speed of light (a photon) has a zero rest mass but has an energy of E =hf and momentum of \(p=\frac{h}{\lambda}=\frac{E}{c}\)

Question 2. Statement 1 The energy (E) and momentum (p) of a photon are related by \(p=\frac{E}{c}\)

Statement 2 A photon behaves like a particle.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The energy of a photon of frequency f is E = hf and its momentum is \(p=\frac{h}{\lambda}=\frac{E}{c}\).

A photon has a particle nature to describe the photoelectric effect.

Question 3. Statement 1 When light of a certain wavelength falls on a metal surface, it ejects electrons.

Statement 2 Light has a wave nature.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

When a light of a wavelength less than the threshold wavelength strikes a photosensitive surface, photoelectrons are ejected. This is explained by the photon (particle) theory. But light has a dual (both particle and wave) nature.

Question 4. Statement 1 On increasing the frequency of light, a larger number of photoelectrons are emitted.

Statement 2 The maximum KE of electrons emitted is directly proportional to the intensity of the incident light.

Answer: 5. Both Statement 1 and Statement 2 are false.

The number of photoelectrons ejected is directly proportional to the intensity of the incident light and not to the frequency of the incident light. The maximum KE of the emitted photoelectrons depends on the frequency and not on the intensity of the incident radiation.

Question 5. Statement 1 There is a physical significance of matter waves.

Statement 2 Both interference and diffraction are displayed by matter waves.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Particles exhibit a wave nature as well. They can be diffracted (as in the Davisson-Germer experiment) and the phenomenon of matter-wave interference lies at the heart of quantum physics.

Question 6. Statement 1 The relative velocity of two photons travelling in opposite directions is the velocity of light.

Statement 2 The rest mass of a photon is zero.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

According to the theory of relativity, speed cannot exceed that of light in a vacuum (c = 3 x102 m s-1). Photons travel at the speed of light, and their relative velocity when moving apart will be c. The rest mass of a particle traveling at the speed of light is zero. This follows from the equation

∴ \(m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

Question 7. Statement 1 An electron microscope is based on the de Broglie hypothesis.

Statement 2 A beam of electrons behaves as a wave that is converged by electric and magnetic lenses.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In an electron microscope, a beam of accelerated electrons is used. Its de Broglie wavelength,

∴ \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m e V}}\),

can be around 105 times shorter than that of the visible light (photons). Electric and magnetic fields are used to converge or diverge the electron beam. These fields are analogous to the glass lenses of an optical microscope.

Question 8. Statement 1 The de Broglie equation has significance for microscopic or submicroscopic particles.

Statement 2 The de Broglie wavelength is inversely proportional to the mass of the objectifies velocity is constant.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The de Broglie wavelength \(\left(\lambda=\frac{h}{m v}\right)\) has an appreciable value for a particle with a small mass like that of a subatomic particle.

Question 9. Statement 1 The second postulate of the special theory of relativity refers to the velocity of light in a vacuum.

Statement 2 A Cerenkov radiation results when particles move through a transparent medium at a speed greater than that of light in that medium. It does not violate the second postulate of the special theory.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Cerenkov radiation occurs when some particles move through a transparent medium at a speed more than the speed of light in the medium.

Still, the particle’s speed is less than the speed of light in a vacuum. So, the second postulate of special relativity is not violated.

Question 10. Statement 1 Stopping potential depends upon the frequency of the incident light but is independent of the intensity of light.

Statement 2 The maximum kinetic energy of photoelectrons is proportional to the stopping potential.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

According to Einstein’s equation of photoelectric effect, \(h v=\phi_0+\mathrm{KE}_{\max } \text {, where } \mathrm{KE}_{\max }=e V_{\mathrm{st}}\).

Thus, the stopping potential Vst depends upon v, the frequency of the incident light. Further, KEmax is proportional to Vst.

Question 11. Statement 1 The photoelectric effect demonstrates the wave nature of light.

Statement 2 The number of photoelectrons is proportional to the frequency of light.

Answer: 5. Both Statement 1 and Statement 2 are false.

The photoelectric effect corresponds to the particle nature (photon) of light, not its wave nature.

The photoelectric current is proportional to the intensity of light and not to the frequency of light.

Question 12. Statement 1 The maximum kinetic energy of the emitted photoelectrons changes only, with a change in the frequency of the incident radiation.

Statement 2 The kinetic energy of the emitted photoelectrons depends on the intensity of the incident light.

Answer: 3. Statement 1 is true and Statement 2 is false.

From the equation, hv = Φ0 + KEmax, the work function (Φ0) is constant for a metal.

KEmax changes only with a change in the frequency (v) of light.

The intensity of the incident light can change only the photoelectric current and not the maximum kinetic energy of the emitted electrons.

Question 13. Statement 1 The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread of their kinetic energies.

Statement 2 The work function of the metal varies as a function of the depth from the surface.

Answer: 3. Statement 1 is true and Statement 2 is false.

The spread of kinetic energy of the emitted photoelectrons is due to the collisions of free electrons before emissions. The work function (Φ0) is constant for a metal.

Bohr Model Hydrogen Spectrum

Question 1. Statement 1 The total energy of an electron in a hydrogen atom is negative.

Statement 2 It is bound to the nucleus.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The electron in the H atom is bound to the nucleus by an electrostatic force and hence has a negative energy.

Question 2. Statement 1 In a metal, all the free electrons have the same energy.

Statement 2 Electrons do not obey Pauli’s exclusion principle.

Answer: 3. Statement 1 is true and Statement 2 is false.

Free electrons in a metal occupy a single energy band and so have the same energy. Electrons in a metal always follow the Pauli exclusion principle.

Question 3. Statement 1 Lasers are used to measure the distances of distant objects such as the moon.

Statement 2 They are highly coherent sources of light.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Lasers are focused, intense, narrow, and highly energetic coherent sources of light, which can travel through large distances before divergence. Hence, these are used to measure large distances.

Question 4. Statement 1 Bohr postulated that the orbital electrons in the stationary orbits around the nucleus do not radiate.

Statement 2 According to classical physics, all moving electrons radiate.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The Bohr postulate refers to ‘stationary orbits’ for the electrons which are non-radiating.

According to the classical theory, an accelerated charge always radiates. But this is not true in the Bohr theory, as circular motion is an accelerated motion.

Question 5. Statement 1 Crystalline solids can cause X-rays to diffract.

Statement 2 Interatomic spacings in crystalline solids are of the order of 0.1 nm.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The lattice spacing crystals (≈ 0.1 nm) are of the order of wavelengths of X-rays and hence act as a three-dimensional grating to produce diffraction.

Question 6. Statement 1 A laser beam of 0.2 W can drill holes through a metal sheet, whereas a 1000-W torchlight cannot.

Statement 2 The frequency of a laser beam is much higher than that of a torchlight.

Answer: 3. Statement 1 is true and Statement 2 is false.

A laser beam is a highly intense, coherent, monochromatic, and parallel beam of light, which can be converted to a very small spot by a convex lens to drill a hole. Light from a 1000-W source cannot drill a hole, because such light has less intensity and the beam is not a parallel one.

Question 7. Statement 1 It is essential that all the lines available in the emission spectrum will also be available in the absorption spectrum.

Statement 2 The spectrum of a hydrogen atom is only an absorption spectrum.

Answer: 5. Both Statement 1 and Statement 2 are false.

In the emission spectrum, a transition occurs from a higher energy state to a lower energy state—which corresponds to n(n-1)/2 transitions. However, this is not true for the absorption spectrum.

The spectrum from an H atom may be an emission spectrum or an absorption spectrum.

Question 8. Statement 1 The Bohr model cannot be used to explain multiple electron spectra.

Statement 2 It does not take interelectronic interactions into account.

Answer: 5. Both Statement 1 and Statement 2 are false.

Multiple electronic spectra can be explained for various transitions from a higher energy state to a lower energy state.

Question 9. Statement 1 Electrons in the atom are held due to Coulomb forces.

Statement 2 The atom is stable only because the centripetal force due to the Coulomb force is balanced by the centrifugal force.

Answer: 3. Statement 1 is true and Statement 2 is false.

An electrostatic attraction on the orbital electron provides the required centripetal force and keeps the atom stable, and the orbits are non-radiating (stationary).

Question 10. Statement 1 X-ray astronomy is possible only from the satellites orbiting the Earth.

Statement 2 The efficiency of an X-ray telescope is larger than compared to any other telescope.

Answer: 3. Statement 1 is true and Statement 2 is false.

X-rays are absorbed in the Earth’s atmosphere, so X-ray astronomy will not be suitable from the Earth but from its satellites, where there are no atmospheres.

Question 11. Statement 1 A fluorescent tube emits white light.

Statement 2 An emission of light in a fluorescent tube takes place at a very high temperature.

Answer: 5. Both Statement 1 and Statement 2 are false.

The light emitted by a fluorescent tube is of multiple colors depending on the nature of the gas enclosed. Thus, the emission occurs at the normal temperature.

The Nucleus Radioactivity

Question 1. Statement 1 Nuclei having a mass number around 60 are the most stable.

Statement 2 When two or more light nuclei are combined into a heavier nucleus, the binding energy per nucleon will increase.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

In the binding-energy (BE) curve, the peak is at around A = 60, for which BE/4 is the maximum. This corresponds to the most stable nuclei. In a fusion reaction, the resultant nucleus is more stable due to the increased binding energy.

Question 2. Statement 1 Mass defects in nuclear reactions are less than 1%.

Statement 2 In nuclear reactions, the change in binding energy per nucleon (BE/A) is generally less than 1%.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The increase in binding energy is due to the mass defect (BE = ΔMc2).

Question 3. Statement 1 Cobalt-60 is useful in cancer therapy.

Statement 2 Cobalt-60 is a source of γ – radiation, capable of killing cancerous cells.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Cobalt-60 \(\left({ }^{60} \mathrm{Co}\right)\) is a synthetic radioisotope of cobalt which is produced artificially in nuclear reactors. It is a source of γ-rays, which are used to kill cancerous cells.

Question 4. Statement 1 Neutrons penetrate matter more readily as compared to protons.

Statement 2 Neutrons are slightly more massive than protons.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A neutron has a charge and a mass slightly more than that of a proton. Being electrically neutral, it can penetrate matter more than an electron can.

Question 5. Statement 1 Radioactive nuclei emit β-particles.

Statement 2 Electrons exist inside the nucleus.

Answer: 3. Statement 1 is true and Statement 2 is false.

A nucleus is composed of protons and neutrons only. Emissions of β- particles are due to the conversion of a neutron into a proton:

\({ }_0^1 \mathrm{n} \rightarrow{ }_1^1 \mathrm{p}+{ }_{-1}^0 \mathrm{\beta}+\overline{\mathrm{v}}\)

Question 6. Statement 1 The densities of all nuclei are the same.

Statement 2 The radius of a nucleus is directly proportional to the cube root of the mass number.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The density of a nucleus is constant (2.3 x l017 kg m-3) and is independent of the size of the nucleus. This follows from the condition

R = R0A1/3

Question 7. Statement 1 γ-radiation emissions occur after α- and β-decays.

Statement 2 Energy levels occur in a nucleus.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

After a β-emission, the daughter nucleus is left in an excited state. It de-excites itself with emissions of gamma rays. The same is true for an α-decay.

Thus, nuclei have discrete energy levels, and gamma-ray emissions occur due to transitions between these levels.

Question 8. Statement 1 Isotopes of an element can be separated by using a mass spectrometer.

Statement 2 Separation of isotopes is possible because of the difference in the electron numbers in the isotopes.

Answer: 3. Statement 1 is true and Statement 2 is false.

Isotopes of an element can be separated by mass spectrometers, in which electric and magnetic fields are used. Isotopes have the same number of protons but different numbers of neutrons.

Question 9. Statement 1 Heavy water is a better moderator than normal water.

Statement 2 Heavy water absorbs neutrons more efficiently than normal water.

Answer: 3. Statement 1 is true and Statement 2 is false.

Heavy water (D20) is used in nuclear reactors to slow down fast-moving neutrons by collisions so that they can produce fission reactions with U-235.

Question 10. Statement 1 For a nuclear reactor, it is desirable to have a multiplication factor of k = 1.

Statement 2 Sustained chain reactions occur in this critical condition.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

For controlled chain reactions, the multiplication factor is k= 1. For k >1, the chain reaction is uncontrolled, and an enormous amount of energy is released.

Question 11. Statement 1 The binding energy per nucleon for nuclei with atomic mass numbers A > 100 decreases with A.

Statement 2 The nuclear forces are weak for heavier nuclei.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

From the binding-energy curve, BE/A decreases with an increase in A for nuclei having A > 100. A decrease in BE indicates a decrease in stability and an increase in Coulomb repulsion between protons.

Question 12. Statement 1 A beam of charged particles is employed in the treatment of cancer.

Statement 2 Charged particles on passing through a material medium lose their energy by causing ionization of atoms along their paths.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A beam of charged particles (protons) is used to treat cancer. During their passage to the target, they ionize the material medium.

Question 13. Statement 1 In the process of nuclear fission, the fragments emit two or three neutrons as soon as they are formed, and they subsequently emit particles.

Statement 2 As the fragments contain an excess of neutrons over protons, emissions of neutrons and particles bring their neutron-proton ratios to stable values.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In nuclear fission, a heavy nucleus breaks up into two lighter nuclei with emissions of 2-3 fast-moving neutrons. This emission is due to the comparatively larger value of the neutron-proton ratio in the products.

Question 14. Statement 1 Heavy water is preferred to ordinary water as a moderator in reactors.

Statement 2 Heavy water, used for slowing down the fast-moving neutrons, has a lesser absorption probability of neutrons than ordinary water.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Heavy water is a better moderator because, in addition to slowing down fast-moving neutrons, its absorption probability is comparatively less.

Question 15. Statement 1 The Rydberg constant varies as the mass number of a given element.

Statement 2 The reduced mass of the electron is dependent on the mass of the nucleus only.

Answer: 4. Statement 1 is false and Statement 2 is true.

The Rydberg constant (\(\left.R_{\infty}=m e^4 / 8 \varepsilon_0^2 h^3 c\right)\)) is independent of the mass number of a given element. The reduced mass for a heavy nucleus is

∴ \(\mu=\frac{m_{\mathrm{e}} M}{m_{\mathrm{e}}+M} \approx m e\)

Question 16. Statement 1 X-rays can penetrate through the flesh but not through the bones.

Statement 2 The penetrating power of X-rays depends on the accelerating voltage.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Bones are much harder than flesh. So, the X-rays can pass through the flesh but not through the bones. Higher accelerating voltages emit X-ray photons of higher energy, called hard X-rays.

Question 17. Statement 1 More energy is released in a fusion than in a fission.

Statement 2 More nucleons take part a fission.

Answer: 3. Statement 1 is true and Statement 2 is false.

A fusion reaction releases 3-4 times more energy than a fission reaction. The number of nuclei participating in a fusion is more than that in a fission.

Question 18. Statement 1 Generally, heavy nuclei are unstable.

Statement 2 It has more neutrons than protons.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Heavy nuclei are unstable and they decay, because they have less binding energy per nucleon. The neutron-proton ratio increases with an increase in mass number.

Question 19. Statement 1 Heavy water is used as a moderator in nuclear reactors.

Statement 2 Water cools down the fast neutrons.

Answer: 3. Statement 1 is true and Statement 2 is false.

Moderation is the process of the reduction of the initial high kinetic energy of fast neutrons during fission.

Since neutrons are light compared to most nuclei, the moderator must be such that it has a nearly identical mass.

This is achieved by heavy water (D20), which has a moderating efficiency of 80 times that of ordinary water.

Question 20. Statement 1 Thermonuclear fusion reactions may become sources of unlimited power for mankind.

Statement 2 A single fusion event involving isotopes of hydrogen produces more energy than the energy from the nuclear fission of \({ }_{92}^{235} \mathrm{U}\).

Answer: 3. Statement 1 is true and Statement 2 is false.

In a fusion reaction, light nuclei (like H2) fuse together with a release of clean energy suitable for mankind.

One fusion event produces less energy compared to that produced in a single fission.

Question 21. Statement 1 If a heavy nucleus is split up into two medium-sized parts, each of the nuclei will have more binding energy per nucleon than the original nucleus.

Statement 2 Joining two lighter nuclei together to form a single nucleus of medium size means more binding energy per nucleon than the two light nuclei.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Heavy nuclei are less stable and have a lower value of BE/A. They break up so that the lighter nuclei are more stable with greater values of BE/A.

From the binding-energy curve, lighter nuclei have less binding energy and hence are unstable. When they fuse to form medium-sized nuclei, they become more stable and have a higher binding energy.

Question 22. Statement 1 It is desirable to slow down fast-moving neutrons to sustain controlled chain reactions.

Statement 2 Slow-moving neutrons efficiently collide with \({ }^{235} \mathrm{U}\) nuclei.

Answer: 3. Statement 1 is true and Statement 2 is false.

Fast neutrons are produced during nuclear fission. A fission of \({ }^{235} \mathrm{U}\) is possible with thermal neutrons. So, for a sustained reaction, neutrons are slowed down by moderators.

Question 23. Statement 1 \({ }^{90} \mathrm{Sr}\) from the radioactive fallout from a nuclear explosion ends up in the bones of human beings through the milk consumed by them. It leads to an impaired production of red blood cells.

Statement 2 The energetic β-particles emitted in the decay of \({ }^{90} \mathrm{Sr}\) damage the bone marrow.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Strontium-90 is a radioactive isotope produced by nuclear fission. It undergoes a β-decay into yttrium-90 with a decay energy of 0.546 MeV.

Strontium-90 gets deposited in our bones and bone marrow. Its presence in the bones can cause bone cancer.

Question 24. Statement 1 It is very easy to detect neutrinos in nature.

Statement 2 It has a high affinity to interact with matter.

Answer: 5. Both Statement 1 and Statement 2 are false.

A neutrino is a subatomic elementary particle with no electric charge and very little mass (≈ 0.2 x 10-5 x me). They are most penetrating and capable of passing through an enormous number of atoms without causing any reaction.

Hence, neutrinos are very difficult to defect and have an extremely low affinity to interact with matter.

Question 25. Statement 1 \({ }_{27}^{60} \mathrm{Co}\) is a source of gamma radiation.

Statement 2 Gamma emissions are due to nuclear decay.

Answer: 3. Statement 1 is true and Statement 2 is false.

Cobalt-60 is a radioactive isotope and a source of gamma radiation, used to destroy tumors.

A nucleus has discrete energy levels and transitions from higher to lower energy levels to emit γ-rays.

Question 26. Statement 1 Magnetic resonance imaging (MRI) is a useful diagnostic tool for producing images of various parts of the human body.

Statement 2 Protons in various tissues of the human body play a role in MRI.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

MRI is a medical imaging technique used in radiology to produce images of various parts of the human body. The human body is mostly composed of water, which contains hydrogen nuclei (protons). These nuclei play a significant role in the MRI process.

Question 27. Statement 1 A radium nucleus at rest decayed into of nucleus and an α-particle. They both move back to each other.

Statement 2 law of conservation of linear momentum.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Radium-226 decays into radon-222 through an alpha emission. It follows from the principle of conservation of linear momentum that both the products must move in opposite directions with equal momenta.

Question 28. Statement 1 The ratio of the time taken for light emissions from an atom to that for the release of nuclear energy in a fission is 1: 100.

Statement 2 The time taken for the light emission from an atom is of the order of 10-8 s.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The time taken for a light emission from an atom is around 10-8 seconds. The time taken for the release of energy during a fission reaction is around 10-6 s.

∴ required ratio = \(\frac{10^{-8} s}{10^{-6} s}=\frac{1}{100}\)

Question 29. Statement 1 The number of active nuclei present in a radioactive sample decreases exponentially with time.

Statement 2 The activity of a radioactive sample decreases linearly with time.

Answer: 3. Statement 1 is true and Statement 2 is false.

According to the law of radioactive disintegration, \(-\frac{d N}{d t}=\lambda N\), which leads to \(N=N_0 \mathrm{e}^{-\lambda t}\). Thus, the number of active nuclei decreases exponentially with time.

Activity = \(|A|=\left|\frac{d N}{d t}\right|=\lambda N=\lambda N_0 \mathrm{e}^{-\lambda t}=A_0 \mathrm{e}^{-\lambda t}\),

which decreases exponentially with time.

Question 30. Statement 1 When a- and β-particles pass through external electric fields, β-particles are deflected much more than a-particles.

Statement 2 β-particles have much larger velocities than a-particles.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Force = F = qE and acceleration = \(a=\frac{q E}{m}\)

Let the particles enter the electric field perpendicularly so that the displacement along the field be

⇒ \(y=\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{q E}{m}\right) t^2\)

∴\(\frac{y_{\mathrm{e}}}{y_\alpha}=\frac{m_\alpha e}{2 e m_{\mathrm{e}}}=\frac{m_\alpha}{2 m_{\mathrm{e}}} \gg 1\)

∴ Hence, \(y_{\mathrm{e}} \gg y_\alpha\)

Question 31. Statement 1 In an X-ray tube, the wavelength of the characteristic X-rays depends on the metal used as the target.

Statement 2 Metals of large atomic numbers are best suited for the production of X-rays.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Characteristic X-rays are emitted due to transitions of electrons from the outer shell to a vacancy in the inner shell of an atom. These X-rays are characteristics of the target element and vary with the metal used. Tungsten (an element of a high atomic number) is most commonly used in an X-ray tube as the target because its melting point is high.

Question 32. Statement 1 Characteristic X-rays are produced by transitions of orbital electrons in the target atom.

Statement 2 In characteristic X-rays, Kα X-rays are of a smaller wavelength than that of Kβ X-rays for the same element.

Answer: 3. Statement 1 is true and Statement 2 is false.

Characteristic X-rays correspond to transitions of electrons from higher-energy shells to lower-energy shells.

The wavelength of characteristic X-rays depends on the difference in energy in the two shells.

For \(\mathrm{K}_\alpha, \lambda=\frac{h c}{E_{\mathrm{K}}-E_{\mathrm{L}}}\)

For \(\mathrm{K}_\beta, \lambda=\frac{h c}{E_{\mathrm{K}}-E_{\mathrm{M}}}\)

∵ \(\Delta E_{\mathrm{KL}}<\Delta E_{\mathrm{KM}}\)

∴ \(\lambda_{\mathrm{K} \alpha}>\lambda_{\mathrm{K} \beta}\)

So, the reason is false.

Modern Physics The Nucleus The Nucleus Radioactivity Wave length Q32

Question 33. Statement 1 In radioactivity, the nature of a sample is identified by its half-life or by its average life but not by its total life.

Statement 2 The total life of any radioactive sample is infinity.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The half-life and mean life are characteristics of the radioactive sample but the total life for any such sample is infinity. Since the decay is exponential, the number of active nuclei will reduce to zero after an infinite time.

Question 34. Statement 1 If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change.

Statement 2 When an electron beam strikes the target in an X-ray tube, part of its kinetic energy is converted into X-ray energy.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Emissions of characteristic X-rays are due to electronic transitions from higher-energy shells to lower-energy shells, in which the energy values are fixed, discrete, and thus independent of accelerating potential.

X-rays are produced due to sudden decelerations of fast-moving electrons when they collide and interact with the target (anode). Only 1% of the KE of electrons is converted into X-rays and the rest into heat.

Semiconductor Devices and Communication Systems Multiple Questions And Answers

Semiconductor Devices and Communication Systems

Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.

Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.

  1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
  2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
  3. Statement 1 is true and Statement 2 is false.
  4. Statement 1 is false and Statement 2 is true.
  5. Both Statement 1 and Statement 2 are false.

Some questions are preceded, by a supporting paragraph in addition to the statements.

Semiconductors Transistors Logic Gates

Question 1. Statement 1 The resistance of a semiconductor decreases with a rise in temperature.

Statement 2 The energy gap between the conduction band and the valence band is very small.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The energy gap between the conduction band and the valence band in a semiconductor is very small. When the temperature is increased, electrons get sufficient thermal energy to jump over to the conduction band and participate in electrical conduction. Thus, an increase in temperature increases the current, thereby reducing the resistance.

Question 2. Statement 1 The resistivity of a semiconductor increases with temperature.

Statement 2 The atoms of a semiconductor vibrate with a large amplitude at higher temperatures, thereby increasing resistivity.

Answer: 5. Both Statement 1 and Statement 2 are false.

The resistivity of a semiconductor decreases with a rise in temperature due to the transfer of electrons from the valence band to the conduction band. This increases the number of electron-hole pairs.

Question 3. Statement 1 An n-type semiconductor has a large number of electrons, but still, it is electrically neutral.

Statement 2 An n-type semiconductor is obtained by doping an intrinsic semiconductor with a pentavalent impurity.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The doping of germanium (tetravalent) with a pentavalent impurity (phosphorus, arsenic, or antimony) makes an n-type semiconductor. Since the atom on the whole is electrically neutral, the extrinsic (n-type or p-type) semiconductor is also neutral.

Question 4. Statement 1 The energy gap between the valence band and the conduction band is greater in silicon than in germanium.

Statement 2 Thermal energy produces fewer minority carriers in silicon than in germanium.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The energy-band gap in silicon is 1.14 eV and that in germanium is 0.67 eV. In silicon atoms, electrons are more tightly bound to the nucleus, so there is more energy gap.

Question 5. Statement 1 The number of electrons in a p-type silicon semiconductor is less than that in a pure silicon semiconductor at room temperature.

Statement 2 It is due to the law of mass action.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

According to the law of mass action,

∴ \(n_{\mathrm{i}}^2=n_{\mathrm{e}} n_{\mathrm{h}}\)

In a p-type semiconductor, nh>ne.

Question 6. Statement 1 The temperature coefficient of resistance is positive for metals and negative for p-type semiconductors.

Statement 2 The effective charge carriers in metals are negatively charged, whereas they are positively charged in p-type semiconductors.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The resistance of a metallic conductor increases with temperature due to an increase in the amplitude of vibrations of atoms. This obstructs the electron flow. Thus, metals have a positive temperature coefficient of resistance.

In metals, the charge carriers are electrons, whereas, in p-type semiconductors, the majority of carriers are holes.

Question 7. Statement 1 Insulators do not allow the flow of current through themselves.

Statement 2 They have no free-charge carriers.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In an insulator, the conduction band is completely empty and the valence band is completely filled. Further, the energy gap (Eg)is greater than kT. So, the thermal energy is insufficient for the electrons to jump over to the conduction band.

Question 8. Statement 1 A p-n junction with a reverse bias can be used as a photodiode to measure the intensity of light.

Statement 2 In a reverse-bias condition, the current is small but more sensitive to a change in the intensity of the incident light.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A photodiode converts light into electrical energy. It is a p-n junction operated in the reverse bias. When the device is exposed to illumination, the reverse current increases linearly with the intensity of the incident light.

Question 9. Statement 1 During reverse-biasing, a diode does not conduct current.

Statement 2 It decreases the thickness of the depletion layer.

Answer: 3. Statement 1 is true and Statement 2 is false.

A p-n junction in the reverse bias offers a large resistance to the current. The thickness of the depletion layer widens.

Question 10. Statement 1 A pure semiconductor has a negative temperature coefficient of resistance.

Statement 2 On raising the temperature, more charge carriers are released. This increases the conductance and decreases the resistance of a semiconductor.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

When the temperature of an intrinsic (pure) semiconductor is raised, more electrons acquire sufficient thermal energy to jump over to the conduction band, so the conductivity increases and the resistivity decreases. This corresponds to a negative temperature coefficient of resistance.

Question 11. Statement 1 At a fixed temperature, silicon will have a minimum conductivity when it has a smaller acceptor doping.

Statement 2 The conductivity of an intrinsic semiconductor is slightly higher than that of a lightly doped p-type semiconductor.

Answer: 3. Statement 1 is true and Statement 2 is false.

The conductivity of an intrinsic semiconductor (pure silicon) is less than that of a semiconductor lightly doped with a p- or n-type impurity.

Question 12. Statement 1 A photocell is called an electric eye.

Statement 2 When light is incident on some semiconductor, its electrical resistance is reduced.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A photoresistor, or light-dependent resistor (LDR), is a variable resistor whose resistance decreases with increasing intensity of the incident light. Thus, it exhibits photoconductivity. These are used as sensors to detect light and are hence called electric eyes.

Question 13. Statement 1 Diode lasers are used as optical sources in optical communications.

Statement 2 Diode lasers consume less energy.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A diode laser uses semiconductor technology that produces coherent light between the visible and infrared ranges. These are used in optical communications for high-speed data transmissions. They are efficient in converting power into optical power with a minimum loss of energy.

Question 14. Statement 1 A photodiode and a photovoltaic cell are both based on the same principle.

Statement 2 Both use the same method of operation to work.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Both a photodiode and a photovoltaic (solar) cell generate emfs and convert light into electrical energy when solar radiations fall on its p-n junction.

Question 15. Statement 1 In a common-emitter (CE) amplifier, the load resistance of the output circuit is 1000 times the load resistance of the input circuit. If α = 0.98 then the voltage gain is 49 x 103.

Statement 2 \(\alpha=\frac{\beta}{1-\beta}\), where the symbols have their usual meanings.

Answer: 3. Statement 1 is true and Statement 2 is false.

⇒ \(I_{\mathrm{e}}=I_{\mathrm{c}}+I_{\mathrm{b}} \Rightarrow \frac{I_{\mathrm{e}}}{I_{\mathrm{c}}}=1+\frac{I_{\mathrm{b}}}{I_{\mathrm{c}}}\)

⇒ \(\frac{1}{\alpha}=1+\frac{1}{\beta} \Rightarrow \alpha=\frac{\beta}{1+\beta}\)

Voltage gain = \(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}=\frac{R_{\mathrm{o}} I_{\mathrm{c}}}{R_{\mathrm{i}} I_{\mathrm{b}}}=\beta\left(\frac{R_{\mathrm{o}}}{R_{\mathrm{i}}}\right)\)

∴ \(\frac{\alpha}{1-\alpha}\left(\frac{R_{\mathrm{o}}}{R_{\mathrm{i}}}\right)=\frac{0.98}{0.02} \times 1000=49 \times 10^3\)

Question 16. Statement 1 In a transistor, the base is made thin.

Statement 2 A thin base makes a transistor stable.

Answer: 3. Statement 1 is true and Statement 2 is false.

In a transistor, the base is very thin and lightly doped, so that very few holes combine with electrons in the base region and most of the holes diffuse through the base to the collector region.

Question 17. Statement 1 Most amplifiers use the common-emitter (CE) circuit configuration.

Statement 2 The input resistance of the CE amplifier is comparatively higher.

Answer: 3. Statement 1 is true and Statement 2 is false.

Transistor amplifiers in the common-emitter (CE) mode are mostly preferred due to the following features:

  1. High current gain,
  2. High voltage gain,
  3. Low input resistance,
  4. High output resistance.

Question 18. Statement 1 In a common-emitter transistor amplifier, the input current is much less than the output current.

Statement 2 A common-emitter transistor amplifier has a very high input impedance.

Answer: 3. Statement 1 is true and Statement 2 is false.

A transistor amplifier in the CE mode has a high current gain. Hence, the input current is much less than the output current and has a low input impedance.

Question 19. Statement 1 A transistor amplifier in the CE mode has a low input impedance.

Statement 2 The base-emitter region is forward-biased.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A transistor in the CE configuration has its input junction (emitter-base junction) forward-biased. So, it has a low input impedance.

Question 20. Statement 1 In the common-base (CB) configuration, the current gain of the transistor is less than unity.

Statement 2 The collector terminal is reverse-biased for amplification.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The current gain in the common-base mode of a transistor is very small because the output region (collector-base side) is reverse-biased for amplification.

Question 21. Statement 1 NAND is a universal gate.

Statement 2 It can be used to describe all other logic gates.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

NAND and NOR gates are called universal gates, as all other gates can be obtained by using them.

Question 22. Statement 1 A NOT gate can be built using a diode.

Statement 2 The output voltage and the input voltage of the diode have a 180° phase difference.

Answer: 5. Both Statement 1 and Statement 2 are false.

A NOT logic gate can be built not by a p-n junction diode but by an n-p-n transistor in which the output is in the opposite phase with the input.

Communication Systems

Question 1. Statement 1 In a communication system based on amplitude modulation, the modulation index is kept less than 1.

Statement 2 It ensures the minimum distortion of signals.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The modulation index is defined as the ratio of the amplitude of the modulating signal (Am) to that of the carrier wave (Ac). If the modulation index (μ = Am/Ac) is greater than 1, this will result in a distortion due to overmodulation. Thus, for the minimum distortion, μ<1.

Question 2. Statement 1 In the transmission of long-distance radio signals, the short-wave band is used.

Statement 2 For shorter wavelengths, attenuation is very low.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In long-distance transmissions, the short-waveband is used, so that the dissipation (attenuation) of energy is minimum.

Question 3. Statement 1 Television signals are received through sky-wave propagations.

Statement 2 The ionosphere reflects electromagnetic waves of frequencies greater than a certain critical frequency.

Answer: 5. Both Statement 1 and Statement 2 are false.

Signals of frequencies more than 40 MHz are not reflected back to the earth. So, they cannot be used for long-distance transmissions. In TV transmissions, the frequency range used is much higher (≈ 100-200 MHz), and they do not get reflected.

Question 4. Statement 1 Short-wave communications over long distances are not possible via ground waves.

Statement 2 Ground waves can bend around the corners of the objects on the earth.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Ground-wave propagations are useful only at low frequencies (not short waves) because the absorption of the waves increases considerably with frequency.

Question 5. Statement 1 Communications in the UHF and VHF regions can be established by space waves or tropospheric waves.

Statement 2 Communications in the UHF and VHF regions are limited to the line-of-sight distances.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The UHF/VHF frequency range is not reflected by the ionosphere. So, they are limited to the direct, or LoS (line-of-sight), path.

Question 6. Statement 1 Sky-wave signals are used for long-distance radio communications. These signals are in general less stable than ground-wave signals.

Statement 2 The state of the ionosphere varies from hour to hour, from day to day, and from season to season.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Sky-wave transmissions are applicable for the frequency range (3-30 MHz). These radio waves can propagate through the atmosphere and are reflected by the ionosphere.

Due to the variation of properties of the ionospheric layers from time to time, sky waves are less stable compared to ground waves.

Question 7. Statement 1 Electromagnetic waves with frequencies smaller than the critical frequencies of the ionosphere cannot be used for communications using sky-wave propagations.

Statement 2 The refractive index of the ionosphere becomes very high for frequencies higher than the critical frequency.

Answer: 5. Both Statement 1 and Statement 2 are false.

The critical frequency is the highest frequency above which electromagnetic waves get transmitted through the ionosphere and below which the waves are reflected from the ionosphere. The critical frequency is given by \(f_{\mathrm{c}}=9 \sqrt{N_{\max }}\), where Nmax is the electron density in the ionosphere. With frequencies greater than fc, there are transmissions with a small refractive index.

Heat And Thermodynamics Objective Questions And Answers

Heat And Thermodynamics

Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.

Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.

  1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
  2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
  3. Statement 1 is true and Statement 2 is false.
  4. Statement 1 is false and Statement 2 is true.
  5. Both Statement 1 and Statement 2 are false.

Some questions are preceded, by a supporting paragraph in addition to the statements.

Kinetic Theory

Question 1. Statement 1 Woolen clothes keep the body warm in winter.

Statement 2 Air is a bad conductor of heat.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Woolen clothes and air are both bad conductors of heat.

Question 2. Statement 1 The temperature near the seaside is moderate.

Statement 2 Water has a high thermal capacity.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Water has a high specific heat capacity (= 4200 J kg-1 K-1), so it cools slowly and also heats up slowly.

Consequently, the amount of heat exchanged in each case is large. Hence, the coastal area has a moderate temperature: neither too hot nor too cold.

Question 3. Statement 1 At room temperature, water does not sublimate from ice to steam.

Statement 2 The critical point of water is much above the room temperature.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

At the triple point, all three phases (solid, liquid, and vapor) coexist. The liquid-vapor boundary terminates at the critical temperature Tc and the critical pressure pc. In water, the critical temperature is around 647 K (374 °C), much higher than room temperature, and at a pressure of around 218 atm.

Question 4. Statement 1 A blue star is at a higher temperature than a red star.

Statement 2 Wien’s displacement law states that \(T \propto \frac{1}{\lambda_{\mathrm{m}}}\)

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

According to Wien’s displacement law, the wavelength λm corresponding to maximum intensity in the radiation graph shifts towards the lower wavelength side with the increase in temperature. Blue light has a lower wavelength compared to the red one, i.e., λblue < λred, hence the temperature of a blue star is more than that of a red star.

Question 5. Statement 1 In a temperature-pressure (p-T) phase diagram of water, the slope of the melting curve is found to be negative.

Statement 2 Ice contracts on melting into water.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The slope of (p-T) phase diagrams for water is negative because during the melting process (ice → water) ice contracts.

Question 6. Statement 1 A piece of ice, with a stone frozen inside it, floats on water in a beaker. When the ice melts, the level of water in the breaker decreases.

Statement 2 The density of stone is more than that of water.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Ice frozen with a stone inside floats on water because the buoyant force is sufficient to balance the weight of (ice + stone). When ice melts, the stone sinks due to its higher density.

Question 7. Statement 1 For a higher temperature, the peak emission wavelength of a black body shifts to the lower-wavelength side.

Statement 2 Peak emission wavelengths of a black body are proportional to the fourth power of temperature.

Answer: 3. Statement 1 is true and Statement 2 is false.

According to Wien’s law, the wavelength, corresponding to the peak emission (λm) shifts towards the lower-wavelength side with an increase in temperature\(\left(\lambda_{\mathrm{m}} \propto \frac{1}{T}\right)\). According to the Stefan-Boltzmann law, the total radiant heat energy emitted is proportional to the fourth power of its absolute temperature.

Question 8. Statement 1 A hollow metallic closed container with a small opening maintained at a high temperature can act as a source of black-body radiation.

Statement 2 All metals act as black-body radiators.

Answer: 3. Statement 1 is true and Statement 2 is false.

Metals have different emissive and absorptive powers, hence they do not act as black bodies. A closed metallic conductor with a cavity and hole (Ferry’s black body) is a near approximation of a black body.

9. Statement 1 Water kept in an open vessel will quickly evaporate on the tire surface of the moon.

Statement 2 The temperature at the surface of the moon is much higher than the boiling point of water.

Answer: 3. Statement 1 is true and Statement 2 is false.

The moon has no atmosphere due to its low escape velocity, so quick evaporation of water is also due to the same reason and not due to the high temperature at the moon’s surface alone.

Question 10. Statement 1 When hot water is poured into a beaker of thick glass, the beaker cracks.

Statement 2 The outer surface of the beaker expands suddenly.

Answer: 3. Statement 1 is true and Statement 2 is false.

Glass has a low thermal conductivity. So, its inner surface absorbs heat and expands, whereas the outer surface does not. This leads to cracking.

Question 11. Statement 1 The radiation from the sun’s surface varies as the fourth power of its absolute temperature.

Statement 2 The sun is not a black body.

Answer: 3. Statement 1 is true and Statement 2 is false.

The sun with an effective temperature of approximately 5800 K is an approximate black body and hence follows Stefan’s law (H ∝ T4) and Wien’s law (λmT = constant)

Question 12. Statement 1 The maximum airflow due to convection does not occur at the north pole but occurs at 30° N.

Statement 2 There is a maximum temperature difference between the equator and 30° N.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The earth’s atmosphere is fixed to the earth by gravity and rotates with it. The heating of the earth’s surface by the sun creates circulation. Due to the earth’s rotation, there is a buildup of air at about 30° N leading to the maximum convection.

Question 13. Statement 1 Perspiration from the human body helps in cooling the body.

Statement 2 A thin layer of water on the skin enhances heat emissivity.

Answer: 3. Statement 1 is true and Statement 2 is false.

Perspiration gives the sensation of cooling because the latent heat required for the evaporation of sweat is derived from the body.

Question 14. Statement 1 The air at some distance above a fire is hotter than the same distance below it.

Statement 2 The air surrounding the fire carries heat upwards.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The air surrounding the fire becomes hot, expands, and becomes lighter. A convection current is built up and cold air comes down.

Question 15. Statement 1 For an ideal gas, at a constant temperature, the product of pressure and volume is constant.

Statement 2 The mean-square velocity of the molecules of a gas is inversely proportional to its mass.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

According to Boyle’s law, pV = constant at a constant temperature. Hence, Statement 1 is true. According to the kinetic theory, the pressure (p) exerted by a gas is

∴ \(p=\frac{1}{3} \frac{M}{V} c_{\mathrm{rms}}^2\)

⇒ mean-square velocity \(\propto \frac{1}{M}\).

Question16. Statement 1 The root-mean-square velocity and the most probable velocity of the molecules in a gas are the same.

Statement 2 The Maxwell distribution for the speed of molecules in a gas is asymmetric.

Answer: 4. Statement 1 is false and Statement 2 is true.

For a gas, RMS velocity = \(\sqrt{\frac{3 R T}{M}}\)

and most-probable velocity = \(\sqrt{\frac{2 R T}{M}}\)

So, Statement 1 is false.

The graph showing the velocity distribution among the molecules is asymmetric, so the reason is true.

Question 17. Statement 1 The size of a hydrogen balloon increases as it rises in air.

Statement 2 The material of the balloon can be easily stretched.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The balloon expands due to a decrease in atmospheric pressure with height.

Question 18. Statement 1 The ratio cp/cv for a diatomic gas is more than that for a monatomic gas.

Statement 2 The molecules of a monatomic gas have fewer degrees of freedom than those of a diatomic gas.

Answer: 4. Statement 1 is false and Statement 2 is true.

The ratio cp/cv (=γ) is given by \(\gamma=1+\frac{2}{f}\) where f is the degree of freedom.

For monatomic gases, \(\gamma_{\text {mono }}=1+\frac{2}{3}=\frac{5}{3}\) and for diatomic gases, \(\gamma_{\mathrm{di}}=1+\frac{2}{5}=\frac{7}{5}\)

∴γmono > γdi

Question 19. Statement 1 The vibrational energy of a diatomic molecule corresponding to each degree of freedom is kBT.

Statement 2 For every molecule, the vibrational degree of freedom is 2.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Vibrational motion in a diatomic molecule has two types of energy: kinetic and potential, which contributes to 2 degrees of freedom. According to the equipartition law, the energy of vibrational motion is

∴ \(E_{\mathrm{vib}}=2 \cdot \frac{1}{2} k_B T=k_B T\)

Thermodynamics

Question 1. Statement 1 Thermodynamic processes in nature are irreversible.

Statement 2 Dissipative effects cannot be eliminated.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Most of the thermodynamic processes in nature are irreversible due to the presence of dissipative forces like friction, viscosity, etc.

Question 2. Statement 1 In an isolated system, the entropy increases.

Statement 2 The processes in an isolated system are adiabatic.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

According to the second law of thermodynamics, the entropy of an isolated system never decreases because thermodynamic systems always lead toward the thermodynamic equilibrium, a state with the maximum entropy.

In an adiabatic process, no energy or mass transfer takes place between the system and the surroundings.

Question 3. Statement 1 It is not possible for a system unaided by an external agency to transfer heat from one body at a lower temperature to another at a higher temperature.

Statement 2 It is not possible to violate the second law of thermodynamics.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The statement given in Statement 1 is itself the second law of thermodynamics, which cannot be violated.

Question 4. Statement 1 For an isothermal process in an ideal gas, the heat absorbed by the gas is entirely used in the work done by the gas.

Statement 2 During a process taking place in a system, if the temperature remains constant then the process is isothermal.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

According to the first law of thermodynamics,

dQ = dU + dW = ncv dT + dW.

In an isothermal process, the temperature T is constant, so dT = 0; and the total heat given (dQ) is spent in the work done (dW).

Question 5. Statement 1 cp is always greater than cv in gases.

Statement 2 The work done at a constant pressure is more than that at a constant volume.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

cp > cv because more heat is required to increase the temperature by dT, keeping the pressure constant rather than keeping the volume constant. In the former case, a part of the heat energy is spent in doing the work; but in the latter, the entire heat is spent in increasing the temperature, and no work is done.

Question 6. Statement 1 Reversible systems are difficult to find in the real world.

Statement 2 Most processes are dissipative in nature.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Reversible systems are associated with nondissipative processes, which are ideal and impracticable. Dissipative processes correspond to the irreversible nature of forces.

Question 7. Statement 1 The melting point of ice decreases with an increase in pressure.

Statement 2 Ice contracts on melting.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

An increase in pressure decreases the melting point of a solid, which contracts on melting (like ice).

Question 8. Statement 1 Air quickly leaking out of a balloon becomes cooler.

Statement 2 The leaking air undergoes adiabatic expansion.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The rapid leaking of air is an adiabatic expansion, which is associated with cooling.

Question 9. Statement 1 When a glass of hot milk is placed in a room and allowed to cool, its entropy increases.

Statement 2 Allowing hot objects to cool down does not violate the second law of thermodynamics.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

According to the second law of thermodynamics, the total entropy of an isolated system never decreases: it always increases for irreversible processes like the cooling of milk in a room.

Question 10. Statement 1 When a bottle of cold carbonated drink is opened a slight fog forms around the opening.

Statement 2 An adiabatic expansion of the gas causes a lowering of temperature and condensation of water vapor.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A sudden and fast expansion (an adiabatic process) of dissolved gases in a cold drink is associated with cooling, which causes condensation of water vapor and the formation of fog around the opening.

Question 11. Statement 1 The Carnot cycle is useful in understanding the performance of heat engines.

Statement 2 The Carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of a given temperature.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The Carnot cycle is a theoretical thermodynamic cyclic process that gives the maximum possible efficiency during the conversion of heat into work.

Question 12. Statement 1 The specific heat of a gas in an adiabatic process is zero and in an isothermal process is infinite.

Statement 2 The specific heat of gas is directly proportional to the change of heat in a system and inversely proportional to the change in temperature.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

When the heat ΔQ is either absorbed or expelled, ΔQ = nCΔT, where C is the molar heat capacity, which is process-dependent. For an adiabatic process, ΔQ = 0, so C = 0. For an isothermal process, ΔT = 0, so \(C=\frac{\Delta Q}{n \Delta T}=\infty\).

In both cases, C ∝ ΔQ and \(C \propto \frac{1}{\Delta T}\).

Question 13. Statement 1 During the rapid pumping of air in tires, the air inside the tire becomes hotter than the atmospheric air.

Statement 2 An adiabatic process occurs at a highly rapid rate.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Rapid pumping of air into a tire is an adiabatic compression, which is associated with a rise in temperature.

Question 14. Statement 1 In an adiabatic process, the change in internal energy of a gas is equal to the work done on or by the gas in the process.

Statement 2 The temperature of a gas remains constant in an adiabatic process.

Answer: 3. Statement 1 is true and Statement 2 is false.

In an adiabatic process, the heat exchanged is ΔQ = 0, so ΔU + ΔW = 0, ΔU = -ΔW. This corresponds to the statements of assertion. In an adiabatic process, there is either cooling (during expansion) or heating (during compression).

Question 15. Statement 1 In a pressure cooker, some water is brought to a boil. The cooker is then removed from the stove. Now, after removing the lid of the pressure cooker, the water starts to boil again.

Statement 2 The impurities in water bring down its boiling point.

Answer: 3. Statement 1 is true and Statement 2 is false.

In a pressure cooker, the steam inside builds up a high pressure. This high-pressure steam raises the boiling point of water above the normal 100 °C. When the cooker is removed from the stove and the lid is opened, the pressure again decreases, due to which the boiling point decreases and water starts to boil again.

Question 16. Statement 1 In a gas, any rapid change must be adiabatic whereas a slow change may be adiabatic.

Statement 2 In a p-V diagram, the magnitude of the slope is greater for an adiabatic process than for an isothermal process.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The process is adiabatic only when there is no exchange of heat between the system and the surroundings (ΔQ = 0). The change may be rapid or slow:

∴ \(\left|\frac{d p}{d V}\right|_{\text {adiabatic }}=\gamma p\).

and \(\left|\frac{d p}{d V}\right|_{\text {isothermal }}=p \text {, with } \gamma=\frac{C_p}{C_V}>1\).

Wave Motion Objective Questions And Answers

Wave Motion

Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.

Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.

  1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
  2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
  3. Statement 1 is true and Statement 2 is false.
  4. Statement 1 is false and Statement 2 is true.
  5. Both Statement 1 and Statement 2 are false.

Some questions are preceded, by a supporting paragraph in addition to the statements.

Sound Waves

Question 1. Statement 1 The flash of lightning is seen before the sound of thunder is heard.

Statement 2 The speed of sound is greater than the speed of light.

Answer: 3. Statement 1 is true and Statement 2 is false.

The speed of light (= 3 x 108 m s-1) is much greater than the speed of sound (= 330 m s-1). So, the flash is observed much earlier.

Question 2. Statement 1 Sound waves cannot propagate through a vacuum but light waves can.

Statement 2 Sound waves cannot be polarized but light waves can be.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Light waves, which are electromagnetic in nature, can travel through a vacuum. But sound waves are mechanical elastic waves, which need an elastic medium that can undergo periodic compression and expansion. Polarization occurs only in transverse waves (like light) and not in longitudinal waves (like sound waves).

Question 3. Statement 1 To hear distinct beats, the difference in frequencies of two sources should be less than ten.

Statement 2 The more the number of beats per second, the more difficult it is to hear them.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The sensation of hearing of any sound persists in our brain for 0.1 s, so for distinct hearing of beats the difference in frequencies (= number of beats per second) must be less than 10.

Question 4. Statement 1 Transverse sound waves do not occur in gases.

Statement 2 Gases cannot sustain shearing strain.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Sound is a longitudinal wave motion. Gases can undergo only volume strain and not shear strain.

Question 5. Statement 1 The fundamental frequency of an organpipe increases as the temperature is increased.

Statement 2 This is because as the temperature increases, the velocity of sound increases more rapidly than the length of the pipe.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The fundamental frequency (η) of an open organ pipe, \(\eta=\frac{v}{2 l}\) increases with temperature as the speed of sound v increases faster than length l.

Question 6. Statement 1 When two vibrating tuning forks having frequencies 256 Hz and 512 Hz are held near each other, beats cannot be heard.

Statement 2 The principle of superposition is valid only if the frequencies of the oscillators are nearly equal.

Answer: 3. Statement 1 is true and Statement 2 is false.

Beats can be heard only if the beat frequency is less than 10 s-1. In this case, Δn = 256 s-1. The principle of superposition holds for any difference in frequency.

Question 7. Statement 1 In a stationary wave system, the displacement nodes are pressure antinodes and displacement antinodes are pressure nodes.

Statement 2 When a closed organ pipe is set into vibration, the pressure of the gas at the closed end remains constant.

Answer: 3. Statement 1 is true and Statement 2 is false.

When stationary waves are set up in a closed organ pipe, displacement nodes are produced at the closed end where pressure antinodes are formed. At pressure antinodes, pressure/density undergoes the maximum variation.

Question 8. Statement 1 When two identical strings stretched to slightly different tensions vibrate together, the loudness of sound heard changes periodically.

Statement 2 Interference can occur in all wave motions under suitable conditions.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Interference is a wave phenomenon exhibited by both longitudinal (sound) and transverse (light) waves.

The vibration of strings with slightly different frequencies produces waves that produce beats due to interference in time. This causes the periodic maximum and minimum of loudness.

Question 9. Statement 1 When two sounds of slightly different frequencies are heard together, periodic variations in intensity (called beats) are observed. A similar phenomenon is not observed when two lights of slightly different wavelengths reach a point and superpose.

Statement 2 Sound waves are longitudinal in nature, while light waves are transverse.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The superposition of sound waves from two independent sources having a small frequency difference produces beats, which is a rhythmic variation in intensity with time. A similar phenomenon with light is not normally observed because the light we have around us comes from incoherent sources. For such a production in light, we need a laser source. The laser beam can be split into two components. The frequency of one beam can be changed appropriately. Their superposition will lead to the desired result.

Question 10. Statement 1 An open organ pipe can be used as a musical instrument but not a closed organ pipe.

Statement 2 The fundamental frequency of an open organ pipe is twice the fundamental frequency of a closed organ pipe of the same length.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

In an open organ pipe, all the harmonics of the fundamental are formed. This richness in overtone is the cause of the melodious note. In a closed organ pipe, only odd harmonics are present. The absence of (even) harmonics makes the note nonmelodious. Fundamental frequencies in closed and open pipes are \(f_{\mathrm{c}}=\frac{v}{4 l} \text { and } f_{\mathrm{o}}=\frac{v}{2 l}\), where v =speed of sound.

Question 11. Statement 1 The Doppler effect occurs in all wave motions.

Statement 2 The Doppler effect can be explained by the principle of superposition of waves.

Answer: 3. Statement 1 is true and Statement 2 is false.

The Doppler effect corresponds to an apparent change in the frequency of both sound (longitudinal) and light (transverse) wave motions. This effect has nothing to do with the superposition but it explains how fast the crest or trough reaches the observer due to the relative motion between the source and the observer.

Electricity and Magnetism Objective Questions And Answers

Electricity and Magnetism

Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.

Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.

  1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
  2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
  3. Statement 1 is true and Statement 2 is false.
  4. Statement 1 is false and Statement 2 is true.
  5. Both Statement 1 and Statement 2 are false.

Some questions are preceded, by a supporting paragraph in addition to the statements.

Current Electricity

Question 1. Statement 1 The total current entering and appointing the circuit is equal to the total current leaving the circuit, according to Kirchhoff’s law.

Statement 2 It is based on the law of conservation of energy.

Answer: 3. Statement 1 is true and Statement 2 is false.

Since the charge does not accumulate at any point, the total incoming current is equal to the total outgoing current. This corresponds to the conservation of charge and not of energy.

Question 2. Statement 1 A current continues to flow through a superconducting coil even after its supply is switched off.

Statement 2 Superconducting coils show the Meissner effect.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The Meissner effect is the expulsion of the magnetic field from a superconductor during its transition to the superconducting when cooled below the transition temperature Tc (usually close to absolute zero). Due to its zero resistance to the current flow, current continues to flow when switched off.

Electricity And Magnetism Objective Questions And Answers Current Electricity Meissner Effect Q2

Question 3. Statement 1 An electric bulb glows instantly as it is switched on.

Statement 2 The drift speed of electrons in a metallic wire is very large.

Answer: 3. Statement 1 is true and Statement 2 is false.

A conductor is full of free electrons. Any disturbance anywhere is instantly transmitted to the bulb which glows. The drift speed is small due to the collisions between the electrons and the ions.

Question 4. Statement 1 Ohm’s law is applicable to all conducting elements.

Statement 2 Ohm’s law is a fundamental law.

Answer: 5. Both Statement 1 and Statement 2 are false.

Ohm’s law is valid for circuits with a constant resistance, for which the V-I graph is a straight line with a positive slope. It is not valid for semiconductors, diodes, and superconductors. Ohm’s law is not a universal (fundamental) law but an empirical law.

Question 5. Statement 1 It is advantageous to transmit electric power at a high voltage.

Statement 2 A high voltage implies a high current.

Answer: 3. Statement 1 is true and Statement 2 is false.

In an AC circuit, the voltage is stepped up by a transformer. At a high voltage, the current is reduced, so the power loss is reduced.

Question 6. Statement 1 Current is passed through a metallic wire so that it becomes red hot. When cold water is poured on one half of its portion, the other half becomes hotter.

Statement 2 Resistance decreases due to a decrease in temperature.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

When cold water is poured on one half of the red-hot conductor, its temperature is reduced, resistance is reduced and the current through it is increased so that the other half becomes hotter.

Question 7. Statement 1 In the given circuit, the emf is 2 V and the internal resistance of the cell is 1 Ω. When R = 1Ω, the reading of the voltmeter is 1 V.

Statement 2 V= ε -IR, where ε = 2 V.

Hence, \(I=\frac{2 \mathrm{~V}}{2 \Omega}=1 \mathrm{~A}\)

Electricity And Magnetism Objective Questions And Answers Current Electricity Circuit Q7

Answer: 3. Statement 1 is true and Statement 2 is false.

The voltmeter reading gives the terminal voltage across the cell, so V= ε -Ir (and not ε-IR).

Main current = \(I=\frac{\varepsilon}{R+r}=\frac{2 \mathrm{~V}}{2 \Omega}=1 \mathrm{~A}\)

V = 2 V-(1A)(1Ω)=1V

Question 8. Statement 1 In a simple battery circuit, the point of the lowest potential is the positive terminal of the battery.

Statement 2 Current flows towards the point of the higher potential, as it does such a circuit from the negative to the positive terminal.

Answer: 5. Both Statement 1 and Statement 2 are false.

The point of lowest potential of the circuit is the negative terminal of the cell. Current flows from the higher potential (positive terminal) to the negative terminal (lower potential) through the outer circuit.

Question 9. Statement 1 The conductivity of an electrolyte is very low as compared to a metal at room temperature.

Statement 2 The number density of free ions in an electrolyte is much smaller as compared to the number density of free electrons in metals. Further, ions, being heavier, drift much more slowly.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Conduction in an electrolyte is due to the movement of free ions which are comparatively much heavier than electrons and have a smaller drift speed. Hence, metals have higher conductivity than electrolytes.

Question 10. Statement 1 A domestic electrical appliance working on a threepin plug will continue working even if the top pin is removed.

Statement 2 The top pin (earth) is only a safety measure.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The top pin is used for earthling as a safety measure. The remaining two terminals are active (live) and neutral which supply power to a device.

Question 11. Statement 1 Electronic appliances with metallic bodies, for example, heaters and electric irons, have three-pin connections, whereas an electric bulb has a two-pin connection.

Statement 2 Three-pin connections reduce the heating of connecting cables.

Answer: 3. Statement 1 is true and Statement 2 is false.

Devices with metallic bodies are connected to the ground (top pin) as a safety measure. No grounding is required for bulbs.

Question 12. Statement 1 Good conductors of heat are also good conductors of electricity, and vice versa.

Statement 2 Mainlyelectrons are responsible for these connections.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Conduction of heat and current electricity are both linked to the motion of free electrons. According to Wiedemann-Franz law, the ratio of the electronic contribution to thermal conductivity (K) to the electrical conductivity (σ) of a metal is proportional to the temperature (T). This explains that good thermal conductors are also good conductors of electricity.

Question 13. Statement 1 A larger dry cell has a higher emf.

Statement 2 The emf of a dry cell is proportional to its size.

Answer: 5. Both Statement 1 and Statement 2 are false.

The large plate area of cells is responsible for its capacity (in ampere-hour) and not the emf.

Question 14. Statement 1 A bird perches on a high-power line and nothing happens to the bird.

Statement 2 The level of the bird is very high from the ground. So, it is not grounded through any conducting device.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Electricity flows through the path of least resistance. Birds do not get an electrical shock while sitting on live electrical wires because they are not good conductors of electricity. Further, the two points of contact are at the same potential and do not form a closed circuit for current to flow through the bird sitting on a power line.

Magnetic Effect of Current

Question 1. Statement 1 A voltmeter is connected in parallel with a circuit.

Statement 2 The resistance of a voltmeter is very high.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A voltmeter is connected in parallel to measure the potential difference between two points. It has high resistance, so it draws a small current and does not change the main current appreciably.

Question 2. Statement 1 A charge, whether stationary or in motion, produces a magnetic field around it.

Statement 2 Moving charges produce only an electric field in the surrounding space.

Answer: 5. Both Statement 1 and Statement 2 are false.

A moving charge \(\left(d q v=\frac{d q}{d t} d l=I d l\right)\) is equivalent to a current element that produces a magnetic field. A charge at rest can produce an electric field.

Question 3. Statement 1 A planar circular loop of area A and carrying current I is equivalent to a magnetic dipole of dipole moment m = IA.

Statement 2 At large distances, the magnetic field due to a current-carrying circular loop and that due to a magnetic dipole are the same.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A ‘current-carrying loop is equivalent to a magnetic dipole whose dipole moment \(|\vec{m}|\) is IA ampere metre2.

The magnetic field at a large distance from a circular current loop is

⇒ \(B=\frac{\mu_0 I R^2}{2\left(R^2+r^2\right)^{3 / 2}}=\frac{\mu_0 I R^2}{2 r^3} \quad(\text { for } r \gg R)\)

∴ \(\frac{\mu_0 I \pi R^2}{2 \pi r^3}=\frac{\mu_0}{2 \pi} \frac{I A}{r^3}=\frac{\mu_0 m}{2 \pi r^3}\)

Where m = IA = magnetic dipole moment.

Question 4. Statement 1 When the radius of a current-carrying loop is doubled, its magnetic moment becomes four times.

Statement 2 The magnetic moment depends on the area of the loop.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The magnetic dipole moment is m =IπR2.

When R is doubled, m’ = Iπ(2R)2 = 4 m.

This is because m ∝ area of the loop.

Question 5. Statement 1 A charged particle can be accelerated in a cyclotron by the alternate variation of the electric field.

Statement 2 The energy of a charged particle is increased by the electric field applied.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

In a cyclotron,both \(\vec{E}\) and \(\vec{B}\) fields act together. The \(\vec{B}\) field provides the circular path while the \(\vec{E}\) field gives an impulse in between the gaps.

Question 6. Statement 1 A cyclotron does not accelerate electrons.

Statement 2 The mass of an electron is very small.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The frequency of the alternating electric field is kept constant, given by \(f=\frac{q B}{2 \pi m}\).

The electron mass is small and it changes considerably with speed given by \(m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\), so f is not synchronized for acceleration.

Question 7. Statement 1 A spark occurs between the poles of a switch when the switch is opened.

Statement 2 Current flowing in a conductor produces a magnetic field.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

When the switch is opened, the magnetic energy stored in the coil or any inductive component gets discharged. This causes a huge current through the two terminals of the switch. This voltage across the terminal gap of the switch exceeds the dielectric strength of air and produces a spark.

Question 8. Statement 1 If a proton and an α-particle enter a uniform magnetic field perpendicularly with the same speed, the time period of revolution of the α-particle is double that of the proton.

Statement 2 In a magnetic field, the period of revolution of a charged particle is directly proportional to the mass of the particle and is inversely proportional to the charge of the particle.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The frequency of revolution of a charged particle in a magnetic field is

⇒ \(f=\frac{1}{T}=\frac{q B}{2 \pi m} \Rightarrow T=\frac{2 \pi m}{q B}\)

∴ \(T_{\text {proton }}=\frac{2 \pi m_{\text {proton }}}{q B} \text { and } T_\alpha=\frac{2 \pi \cdot 4 m_{\text {proton }}}{2 q B}=2 T_{\text {proton }}\)

Time of revolution \(\propto \frac{m}{q}\)

Question 9. Statement 1 The energy of a charged particle moving in a uniform magnetic field does not change.

Statement 2 The work done by the magnetic field on the charge is zero.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Force on a charged particle moving in a magnetic field = \(\vec{F}=q \vec{v} \times \vec{B}\) This force only provides centripetal force for circular motion and does no work. Hence, work done is zero and KE remains constant.

Question 10. Statement 1 When a magnetic dipole is placed in a nonuniform magnetic field, only a torque acts on the dipole.

Statement 2 A force would act on a dipole if the magnetic field is uniform.

Answer: 5. Both Statement 1 and Statement 2 are false.

In a nonuniform magnetic field, the two poles experience unequal forces as well as a torque. Hence, the torque and the net force act together producing translational and rotational motions.

Question 11. Statement 1 Magnetic field lines form a closed loop in nature.

Statement 2 Monomagnetic poles do not exist in nature.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Magnetic field lines always form a closed loop in nature because no magnetic monopole exists in nature. Inside the magnet they travel from the south pole to the north pole and outside, from the north pole to the south pole, thereby forming a closed loop.

Electromagnetic Induction

Question 1. Statement 1 The possibility of an electric bulb fusing is higher at the time of switching on and off.

Statement 2 Inductive effects produce a surge in switching off and switching on.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

During switching on or off a sudden change in magnetic flux induces high voltage \(\left(|e|=\frac{d \phi}{d t}\right)\) which may cause an electric bulb to fuse.

Question 2. Statement 1 Only a change in magnetic flux will maintain the induced current in the coil.

Statement 2 The presence of a large magnetic flux through a coil maintains a current in the coil if the circuit is continuous.

Answer: 3. Statement 1 is true and Statement 2 is false.

When the magnetic flux linked with a coil changes with time, the induced current is maintained. The presence of a steady magnetic flux will not induce current in a closed coil.

Question 3. Statement 1 An emf ε is induced in a closed loop where magnetic flux is varied. The induced electric field \(\vec{E}\) is not a conservative field.

Statement 2 The line integral \(\int \vec{E} \cdot \overrightarrow{d l}\) around a closed loop is nonzero.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Induced electric fields form closed loops, so that a charge moving along the loop under the induced field will experience a force parallel to its displacement, and nonzero work is done, unlike conservative forces, where the work done in a closed loop is zero. The non – conservative nature of the induced electric field is also expressed mathematically as \(\oint \vec{E} \cdot \overrightarrow{d l} \neq 0\)

Question 4. Statement 1 If the current is flowing through a machine of iron, eddy currents are induced.

Statement 2 The change in magnetic flux through an area causes eddy currents.

Answer: 5. Both Statement 1 and Statement 2 are false.

Eddy currents are induced by the time-varying magnetic flux linked with a conductor and not by the current flowing through the conductor itself.

Question 5. Statement 1 In a moving train, a small potential difference arises across the axles of the wheels due to the earth’s magnetic field. This potential difference vanishes at the equator.

Statement 2 At the equator, the earth’s magnetic field is horizontal.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A motional emf ε = Blv is produced when the moving conductor cuts the magnetic field lines. At the equator, the axles do not cut the field lines.

Question 6. Statement 1 When a charged particle enters a magnetic field from outside, it cannot complete one rotation inside the field.

Statement 2 The entry and exit of a charged particle into and out of a uniform magnetic field are symmetrical.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The charged particle that enters the magnetic field at P will experience a force \(\vec{F}=q(\vec{v} \times \vec{B})\). This force, perpendicular to \(\vec{v}\) and \(\vec{B}\), provides a centripetal force. The path followed is a semicircle. The entry and exit at P and Q are symmetrical about the center O.

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Electromagnetic Induction

Question 1. Statement 1 The possibility of an electric bulb fusing is higher at the time of switching on and off.

Statement 2 Inductive effects produce a surge in switching off and switching on.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

During switching on or off a sudden change in magnetic flux induces high voltage \(\left(|e|=\frac{d \phi}{d t}\right)\) which may cause an electric bulb to fuse.

Question 2. Statement 1 Only a change in magnetic flux will maintain the induced current in the coil.

Statement 2 The presence of a large magnetic flux through a coil maintains a current in the coil if the circuit is continuous.

Answer: 3. Statement 1 is true and Statement 2 is false.

When the magnetic flux linked with a coil changes with time, the induced current is maintained. The presence of a steady magnetic flux will not induce current in a closed coil.

Question 3. Statement 1 An emf 8 is induced in a closed loop where magnetic flux is varied. The induced electric field \(\vec{E}\) is not a conservative field.

Statement 2 The line integral \(\int \vec{E} \cdot \overrightarrow{d l}\) around a closed loop is nonzero.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Induced electric fields form closed loops, so that a charge moving along the loop under the induced field will experience a force parallel to its displacement, and nonzero work is done, unlike conservative forces, where the work done in a closed loop is zero. The non – conservative nature of the induced electric field is also expressed mathematically as \(\oint \vec{E} \cdot \overrightarrow{d l} \neq 0\)

Question 4. Statement 1 If the current is flowing through a machine of iron, eddy currents are induced.

Statement 2 The change in magnetic flux through an area causes eddy currents.

Answer: 5. Both Statement 1 and Statement 2 are false.

Eddy currents are induced by the time-varying magnetic flux linked with a conductor and not by the current flowing through the conductor itself.

Question 5. Statement 1 In a moving train, a small potential difference arises across the axles of the wheels due to the earth’s magnetic field. This potential difference vanishes at the equator.

Statement 2 At the equator, the magnetic field is horizontal.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A motional emf ε = Blv is produced when the moving conductor cuts the magnetic field lines. At the equator, the axles do not cut the field lines.

Question 6. Statement 1 When a charged particle enters a magnetic field from outside, it cannot complete one rotation inside the field.

Statement 2 The entry and exit of a charged particle into and out of a uniform magnetic field are symmetrical.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The charged particle that enters the magnetic field at P will experience a force \(\vec{F}=q(\vec{v} \times \vec{B})\). This force, perpendicular to \(\vec{v}\) and \(\vec{B}\), provides a centripetal force. The path followed is a semicircle. The entry and exit at P and Q are symmetrical about the center O.

Electricity And Magnetism Objective Questions And Answers Electromagnetic Induction Charged Particles Q6

Alternating Current

Question 1. Statement 1 Faraday’s laws are consequences of the conservation of energy.

Statement 2 In a purely resistive AC circuit the current lags behind the voltage.

Answer: 3. Statement 1 is true and Statement 2 is false.

Faraday’s laws of EMI follow the principle of conservation of energy. In pure resistance in an AC circuit, current and voltage are always in the same phase.

Question 2. Statement 1 While flowing through an inductor, an alternating current lags behind the voltage by a phase angle of \(\frac{\pi}{2}\), when AC flows through an inductor.

Statement 2 The inductive reactance increases as the frequency of an AC source decreases.

Answer: 3. Statement 1 is true and Statement 2 is false.

In an inductor, the current lags behind the voltage by \(\frac{\pi}{2}\).

Inductive reactance = XL = ωL = 2πfL increases with an increase in frequency.

Question 3. Statement 1 The quantity L/R possesses a dimension of time.

Statement 2 To reduce the rate, of increase of current through a solenoid, we should increase the time constant (L/R).

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The induced emf in an inductor is \(|\varepsilon|=L \frac{d I}{d t}\) which has the dimensions of [I.R].

Thus, \(\left[\frac{L}{R}\right] \frac{d I}{d t}=[I]\)

So, \(\frac{L}{R}\) has dimention Of Time.

Further, \(\frac{d I}{d t}=\frac{I}{\left(\frac{L}{R}\right)}\)

To reduce \(\frac{d I}{d t}\) the value of \(\frac{L}{R}\) should be large.

Question 4. Statement 1 In a series LCR circuit, the resonance occurs at one frequency only.

Statement 2 At resonance, the inductive reactance is equal and opposite to the capacitive reactance.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

At resonance, the circuit is purely resistive and has a single resonance frequency = \(f=\frac{1}{2 \pi \sqrt{L C}}\).

This corresponds to the equal and opposite values of the inductive reactance and capacitive reactance.

Question 5. Statement 1 No power loss is associated with pure capacitors in an AC circuit.

Statement 2 No current flows in this circuit.

Answer: 3. Statement 1 is true and Statement 2 is false.

In an AC circuit with a pure capacitor, the phase difference between the current and the voltage is \(\frac{\pi}{2}\), so power dissipated \(\left(\frac{1}{2} I_0 V_0 \cos \frac{\pi}{2}\right)\) is zero.

Question 6. Statement 1 At resonance, a series LCR circuit has a current.

Statement 2 At resonance, in a series LCR circuit, the current and the emf are not in phase with each other.

Answer: 5. Both Statement 1 and Statement 2 are false.

At resonance, in an LCR circuit, the current is maximum since the impedance is minimum (=R). At resonance, the circuit is purely resistive, so the current. and the voltage is in the same phase.

Question 7. Statement 1 Induction coils are usually made of a thick copper wire.

Statement 2 Induced current is more in resistance.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Induction coils have high inductance and low resistance. For low resistance, the copper wire has to be thick. For more induced current, resistance has to be low according to Ohm’s law.

Question 8. Statement 1 Transformers are used only in alternating current, not in direct current.

Statement 2 Only alternating current can be stepped up or down by means of transformers.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

With a DC source, the periodic change in magnetic flux does not occur to induce voltage in the secondary coil. Hence, only an AC source can be used to raise or lower the voltage by adjusting the turn ratio Ns/Np.

Question 9. Statement 1 We use a thick wire in the secondary coil of a stepdown transformer to reduce heat production.

Statement 2 When the plane of the armature is parallel to the magnetic field lines, the magnitude of induced emf is maximum.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Thick wires have low resistance, hence there is less loss of energy by heat production.

Instantaneous magnetic flux linked with the coil of armature in a B-field is \(\phi=N \vec{A} \cdot \vec{B}=N A B \cos \theta\).

Induced emf = \(\varepsilon=-\frac{d \phi}{d t}=N A B \omega \sin \theta=\varepsilon_{\max } \sin \theta\)

Given, \(\theta=\frac{\pi}{2}\) so emf = maximum = NABω.

Magnetism And Matter

Question 1. Statement 1 We cannot think of a magnetic field configuration with three poles.

Statement 2 A bar magnet does exert a torque on itself due to its own field.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Monopolesin magnetism do not exist. Magnets always exist as dipoles. A system like a bar magnet can produce a magnetic field but does not experience a force or a torque in its own field.

Question 2. Statement 1 Magnetic field lines are continuous and closed.

Statement 2 Magnetic monopoles do not exist.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Magnetic field lines form a closed loop. This is a consequence of the nonexistence of magnetic monopole.

Question 3. Statement 1 The magnetic force between two short magnets follows the inverse square law of distance when they are coaxial.

Statement 2 The magnetic force between two poles does not follow the inverse square law of distance.

Answer: 5. Both Statement 1 and Statement 2 are false.

The magnetic force between two coaxial short magnets is given by \(F \propto \frac{1}{r^4}\). The reason is also false as magnetic force between two poles does follow the inverse square law of distance.

Question 4. Statement 1 Magnetic susceptibility is a pure number.

Statement 2 The value of magnetic susceptibility for vacuum is one.

Answer: 3. Statement 1 is true and Statement 2 is false.

Magnetic susceptibility % is the ratio of magnetization (= magnetic moment per unit volume) to the applied magnetizing field H. Both M and H have the same unit, so it is a pure number (dimensionless). Susceptibility for vacuum is zero since magnetization M in a vacuum is zero.

Question 5. Statement 1 A superconductor is a perfect diamagnetic substance.

Statement 2 A superconductor is a perfect conductor.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A superconductor is a perfect diamagnetic substance since in its interior the magnetic field \(\vec{B}\) = 0 and is completely screened. A superconductor is a perfect conductor as its resistance is zero.

Question 6. Statement 1 The magnetic poles of Earth do not coincide with the geographical poles.

Statement 2 The discrepancy between the orientation of a compass and true north-south direction is known as magnetic declination.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The axis about which the earth rotates passes through the geographical north and south poles whereas the magnetic axis of the earth (assumed to be a magnetized sphere) is somewhat inclined to the rotational axis. Magnetic declination is the angle between the magnetic meridian and the geographical meridian.

Question 7. Statement 1 Diamagnetic materials can exhibit magnetism.

Statement 2 Diamagnetic materials have permanent dipole moments.

Answer: 5. Both Statement 1 and Statement 2 are false.

Diamagnetic materials neither exhibit magnetism nor do they have permanent dipole moments.

Question 8. Statement 1 The true geographic north directions were found by using a magnetic needle.

Statement 2 The magnetic meridian of the earth is along the axis of rotation of the earth.

Answer: 5. Both Statement 1 and Statement 2 are false.

A magnetic needle is used to find the magnetic north but not the geographic north. A magnetic meridian is not aligned along the axis of rotation of the earth.

Question 9. Statement 1 A ferromagnetic substance becomes paramagnetic above the Curie temperature.

Statement 2 Domains are destroyed at high temperatures.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The Curie temperature is the temperature above which a ferromagnetic material becomes paramagnetic due to the breaking up of magnetic domains.

Question 10. Statement 1 A disc-shaped magnet is levitated above a superconducting material that has been cooled by liquid nitrogen.

Statement 2 Superconductors repel magnets.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Superconductors exist only below a certain critical temperature but above that temperature, they behave like normal materials. When a magnet is placed above a superconductor (which is cooled using liquid nitrogen), a magnetic field is induced within it which is exactly equal and opposite to the applied external magnetic field due to the magnet. The polarities are such that they repel each other and the repulsive force is enough to float the magnet.

Question 11. Statement 1 In water, the value of the magnetic field decreases.

Statement 2 Water is a diamagnetic substance.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Water has relative permeability \(\left(\mu_{\mathrm{r}}=\frac{\mu}{\mu_0}\right)\) less than 1 and thus a negative magnetic susceptibility. Water is thus a diamagnetic substance in which magnetic field strength decreases.

Optics Objective Questions And Answers

Optics

Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.

Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.

  1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
  2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
  3. Statement 1 is true and Statement 2 is false.
  4. Statement 1 is false and Statement 2 is true.
  5. Both Statement 1 and Statement 2 are false.

Some questions are preceded, by a supporting paragraph in addition to the statements.

Ray Optics

Question 1. Statement 1 A thick lens shows more chromatic aberration.

Statement 2 A thick lens behaves asifitis composed of many thin lenses.

Answer: 3. Statement 1 is true and Statement 2 is false.

Chromatic aberration is a defect in lenses in not focusing all the colors at the same point. It is caused by dispersion and is prominent with thick lenses. A thick lens is like a combination of truncated prisms and not composed of thin lenses.

Optics Objective Questions

Question 2. Statement 1 A total reflecting prism is used to erect the inverted image without any deviation.

Statement 2 The rays of light incident perpendicular to the base of the prism emerge as parallel rays.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

An erecting prism is a right-angled isosceles prism used to erect an inverted image without any deviation, as shown in the ray diagram.

Optics Objective Questions And Answers Erecting Prism Q2

Question 3. Statement 1 The color of a green flower seen through a red glass appears to be dark.

Statement 2 A red glass transmits only red light.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The green flower emits green light which is absorbed by the red glass and appears dark. This is true as a red glass emits only red light.

Question 4. Statement 1 If a convex lens is kept in water, its converging power decreases.

Statement 2 The focal length of a convex lens increases in water.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Optics Objective Questions

The focal length of a lens depends on the refractive index of the surrounding medium.

When in air, \(\frac{1}{f_{\mathrm{a}}}=\left({ }^{\mathrm{a}} \mu_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=k\left({ }^{\mathrm{a}} \mu_{\mathrm{g}}-1\right)\)

When in water, \(\frac{1}{f_{\mathrm{w}}}=\left({ }^{\mathrm{w}} \mu_{\mathrm{g}}-1\right) k=k\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)\)

As \(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}<{ }^{w_{\mathrm{g}}} \mu_{\mathrm{g}}, \frac{1}{f_{\mathrm{a}}}>\frac{1}{f_{\mathrm{w}}}\)

Hence, the converging power is reduced when in water. This is the same as /w>/a.

Question 5. Statement 1 Diamonds glitter brilliantly.

Statement 2 Diamonds absorb sunlight.

Answer: 3. Statement 1 is true and Statement 2 is false.

The sparkling of a diamond is due to its high refractive power (refractive index = 2.4) and its cut shape. Due to the high refractive index, the critical angle is small and the light inside the diamond gets internally reflected several times before emergence, thus making ‘ the diamond glitter brilliantly. The glitter of a diamond is not due to the absorption of light but due to total internal reflection.

Question 6. Statement 1 Red light is used as the danger signal.

Optics Objective Questions

Statement 2 The velocity of red light is maximum and thus it offers more visibility in the dark.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Red light has the longest wavelength in the visible region and undergoes the least scattering according to Rayleigh’s law,\(I \propto \frac{1}{\lambda^4}\). Hence, a red signal is visible from a long distance. Scattering spreads the light in a wider area but reduces the intensity. Red light is scattered the least and has the maximum intensity as well.

Question 7. Statement 1 If the optical density of a substance is more than that of water then the mass density of the substance can be less than water.

Statement 2 Optical density and mass density are not related.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A greater optical density does not always mean a greater mass density (mass per unit volume). For example, kerosene is lighter than water, but it has a higher refractive index, so it is optically denser than water. In most cases though, a greater optical density is associated with a greater mass density.

Question 8. Statement 1 A secondary rainbow has an inverted color sequence, while a primary rainbow does not.

Statement 2 A secondary rainbow is formed by a single total internal reflection.

Answer: 3. Statement 1 is true and Statement 2 is false.

In the formation of a secondary rainbow, light undergoes two internal reflections inside water droplets. The color sequence in a secondary rainbow is thus inverted as compared to that in a primary one.

Question 9. Statement 1 When an object is placed between two plane mirrors kept parallelly, all the images formed are of the same intensity.

Statement 2 In the case of parallel plane mirrors, only two images are possible.

Answer: 5. Both Statement 1 and Statement 2 are false.

When an object is placed between two parallel plane mirrors, an infinite number of images are formed with gradually decreasing intensity due to absorption and multiple reflections.

Optics Objective Questions

Question 10. Statement 1 A single lens produces a coloured image of an object illuminated by white light.

Statement 2 The refractive index of the material of a lens is different for different wavelengths of light.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A single lens produces a colored image of an object in white light (called chromatic aberration). Different components of white light have different values of focal length and do not get focused at a single position.

Question 11. Statement 1 Endoscopy involves the use of optical fibers to study internal organs.

Statement 2 The working of optical fibers is based on the phenomenon of total internal reflection.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

An endoscope is used to examine internal organs. In the process, optical fibers are used which work on the principle of total internal reflection.

Question 12. Statement 1 A double convex lens (μ = 1.5) has a focal length of 10 cm. When the lens is immersed in water(μ = 4/3), its focal length becomes 39 cm.

Statement 2 \(\frac{1}{f}=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\left(\frac{\mu_1-\mu_w}{\mu_w}\right)\)

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

When air, the focal length fa is given by

⇒ \(\frac{1}{f_{\mathrm{a}}}=\frac{1}{10 \mathrm{~cm}}=\left(\mu_1-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

⇒ \((1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) (1)

When immersed in water, the focal length /w is given

∴ \(\frac{1}{f_{\mathrm{w}}}=\left(\frac{\mu_1}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{1.5}{1.33}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) (2)

Dividing (1) by (2),

∴ \(\frac{f_{\mathrm{w}}}{10 \mathrm{~cm}}=\frac{0.5}{\frac{1.5}{1.33}-1} \Rightarrow f_{\mathrm{w}}=39.1 \mathrm{~cm}\)

Optics Objective Questions

Question 13. Statement 1 A red-colored object appears dark in yellow light.

Statement 2 Red light is scattered less.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A red-colored object reflects only red light and absorbs yellow light. The scattering of red light is the least because its wavelength is maximum.

Question 14. Statement 1 By roughening the surface of a glass sheet, its transparency can be reduced.

Statement 2 A rough glass surface absorbs more heat.

Answer: 3. Statement 1 is true and Statement 2 is false.

Roughening a glass surface reduces its transparency because of irregular reflection and not due to absorption.

Question 15. Statement 1 A normal human eye can clearly see all the objects beyond a certain minimum distance.

Statement 2 The human eye has the capacity to suitably adjust the focal length of the eye lens to a certain extent.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The human eye can adjust its focal length to see objects clearly beyond 25 cm.

Question 16. Statement 1 A microscope forms the magnified image of an object.

Statement 2 The angle subtended by the image is more than that subtended by the object.

Optics Objective Questions

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A microscope (simple or compound) increases the visual angle of an image so that the final image formed at 25 cm from the eye appears magnified as compared to the object.

Question 17. Statement 1 If the objective and eye lenses of a microscope are interchanged then it can work as a telescope.

Statement 2 The objective lens of a telescope has a small focal length.

Answer: 5. Both Statement 1 and Statement 2 are false.

The objective and eye lenses of a microscope cannot be interchanged to convert it into a telescope. This is because a microscope’s lens has a small focal length (around a few cm), whereas a telescope’s objective has a large focal length.

Question 18. Statement 1 Magnification by a convex mirror is always positive, but that by a concave mirror may be either positive or negative.

Statement 2 It depends on the sign convention chosen.

Answer: 3. Statement 1 is true and Statement 2 is false.

In a convex mirror, the image formed is virtual, diminished, and erect (positive magnification).

In a concave mirror, the magnification is positive for an object lying between the focus and the pole (u < f) and negative for u >f when the image formed is real and inverted.

Question 19. Statement 1 Owls can move freely during the night.

Statement 2 They have a large number of rods on their retina.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Owls have appreciably better night visions because they have comparatively more rods and not as many cones. Thus, they lose color vision but see a lot better at night when it is almost dark. But in total darkness, they can see nothing. They always need some light to see.

Question 20. Statement 1 The formula connecting u, v, and / for a spherical mirror is valid only for those mirrors whose sizes are very small compared to their radii of curvature.

Statement 2 The laws of reflection are strictly valid for plane surfaces but not for large spherical surfaces.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The relation connecting u, v, and f in ray optics is valid for paraxial rays, and the laws of reflection are true for plane surfaces.

Question 21. Statement 1 The edge of the image of a white object formed by a concave mirror on the screen appears white.

Statement 2 A concave mirror does not suffer from chromatic aberration.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The image formed due to a concave mirror does not suffer from chromatic aberrations, so it appears white.

Question 22. Statement 1 A ray of light is incident from outside on a glass sphere surrounded by air. This ray may suffer a total internal reflection at the second interface.

Statement 2 If a ray of light goes from a denser to a rarer medium, it bends away from the normal.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

A ray of light incident on a glass sphere gets refracted and strikes the surface again. It may be totally reflected if the surface angle of incidence is greater than the critical angle.

Question 23. Statement 1 A real image is formed when two rays starting from a point and moving along different paths meet again at some other point.

Statement 2 A real image can never be formed by reflection from a convex mirror or refraction through a concave lens.

Answer: 3. Statement 1 is true and Statement 2 is false.

Real images are formed by the actual intersection of rays after reflection or refraction.

Real images can be formed by a convex mirror or a concave lens, as shown in the following figure.

Optics Objective Questions And Answers Real Image Q30

Question 24. Statement 1 In high-quality optical devices, such as cameras, binoculars, and periscopes, prisms instead of plane mirrors are used to reflect light.

Statement 2 In-plane mirrors, reflections occur both at the front and rear (silvered) surfaces.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A reflection from a prism is actually a total internal reflection, where there is 100% reflection. This is not true for a plane mirror.

Question 25. Statement 1 The sun rises before the actual sunrise.

Statement 2 During sunrise, refraction takes place through different atmospheric layers.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The time delay (advanced sunrise and delayed sunset) is due to atmospheric refraction. The sun is visible to us 2 minutes before the actual sunrise and 2 minutes after the actual sunset.

Question 26. Statement 1 Sometimes the rainbow is seen in the sky when it is raining. When one sees a rainbow, one’s back is towards the sun.

Statement 2 Internal reflection from water droplets causes dispersion. The final ray is in the backward direction.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A rainbow is caused by refraction, dispersion, and total internal reflection of sunrays through water droplets suspended in the atmosphere. Rainbows always appear in the section of the sky directly opposite the sun.

Question 27. Statement 1 The frequencies of the incident reflected and refracted beams of light incident from one medium to another are the same.

Statement 2 The incident, reflected, and refracted rays are coplanar.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

During reflection and refraction, the frequency of photon (E =hf) does not change because E and B are continuous at the boundary of the two media. The planes of incidence and refraction are actually the same plane

Question 28. Statement 1 The corpuscular theory fails to explain the velocities of light in air and water.

Statement 2 According to the corpuscular theory, light should travel faster in a denser medium than in a rarer medium.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

According to Newton’s corpuscular theory, the speed of light in a denser medium is greater than the speed in a rarer medium, which is • not true.

Question 29. Statement 1 Optical fibers are used for telecommunications.

Statement 2 The optical fibers work on the principle of total internal reflection.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Optical fibers work on the principle of total internal reflection. They are used in telecommunications because of their higher bandwidths, lower attenuations, and no interference from stray signals.

Question 30. Statement 1 When a beam of white light is passed through a lens, violet light gets more refracted than red light.

Statement 2 The focal length of the red light is greater than that of the violet light.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

White light passing through a lens undergoes dispersion. Violet light is the most refracted and red light is the least refracted because \(\mu_v>\mu_r\). This is why \(f_{\mathrm{v}}<f_{\mathrm{r}}\).

Optics Objective Questions And Answers Ray Optics Q30

Question 31. Statement 1 Goggles (sunglasses) have zero power.

Statement 2 The radius of curvature of both sides of the lens is the same.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Power depends on the radii of curvature of the curved surfaces of a lens and this is given by

∴ \(P=\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

P = 0 when R1 =R2.

Wave Optics And Electromagnetic Waves

Question 1. Statement 1 In Young’s double-slit experiment (YDSE), bright and dark fringes are equally spaced.

Statement 2 It only depends upon phase differences.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The separation between two consecutive dark or two consecutive bright fringes is equally spaced. The spacing, known as the fringe width, is given by \(\beta=\frac{D \lambda}{d}\). The position of the minima or maxima depends on the phase difference.

Question 2. Statement 1 A thin film such as a soap bubble or a thin layer of oil on water shows beautiful colors when illuminated by white light.

Statement 2 It happens due to the interference of light reflected from the upper Surface of the thin film.

Answer: 3. Statement 1 is true and Statement 2 is false.

Colors in a thin film arise due to interference of light waves reflected from the top and bottom surfaces of the thin film. The colors that appear depend upon the thickness of the film and the angle of incidence of the incident white light.

Question 3. Statement 1 The sun looks bigger in size at sunrise and sunset than during the day.

Statement 2 The phenomenon of diffraction bends light rays.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The appearance of the bigger size of the sun during sunrise or sunset is due to atmospheric refraction. It is not due to diffraction, which is a wave phenomenon of bending of light around the comer.

Question 4. Statement 1 A colored spectrum is seen when we look through a muslin cloth.

Statement 2 It is due to the diffraction of white light passing through fine slits.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A piece of muslin cloth effectively acts as a series combination of fine slits (like a grating). White rays undergo diffraction and produce a colored spectrum.

Question 5. Statement 1 Clouds in the sky generally appear to be whitish.

Statement 2 Diffractions due to clouds are efficient in equal measures at all wavelengths.

Answer: 3. Statement 1 is true and Statement 2 is false.

Clouds consist of tiny water droplets and dust particles whose sizes are very large compared to the wavelength of light from the sun So, there is the least scattering and we receive unscattered sunlight, which is white.

Question 6. Statement 1 An electron microscope has more resolving power than an optical microscope.

Statement 2 We can control the energy of electrons.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The resolving power of a microscope is inversely proportional to wavelength. The de Broglie wavelength \(\left(\lambda=\frac{h}{m v}\right)\) of the electron beam used in an electron microscope is much shorter than the wavelength of the visible light used in an optical microscope. Hence, an electron microscope has a greater resolving power. The energy of an electron can be controlled by accelerating the voltage.

Question 7. Statement 1 The resolving power of a telescope is more if the diameter of the objective lens is more.

Statement 2 An objective lens of a larger diameter collects more light.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The angular position of the first minimum in the diffraction pattern of a circular aperture of diameter D is \(\theta_1=1.22 \frac{\lambda}{D}\)

The reciprocal of θ1 is the measure of resolving power, i.e., \(\frac{D}{1.22 \lambda}\)

With an increase in diameter, the resolving power will increase.

Question 8. Statement 1 The focal length of the objective lens in a telescope is longer than that of the eyepiece.

Statement 2 A telescope has a high resolving power due to the large focal length of its objective.

Answer: 3. Statement 1 is true and Statement 2 is false.

The large focal length of an objective lens produces a high magnification but for a high resolving power, the aperture (diameter) has to be large.

Question 9. Statement 1 The resolving power of a telescope depends only on the wavelength of the diffracted light.

Statement 2 The resolving power is proportional to the square of the wavelength of the diffracted light.

Answer: 5. Both Statement 1 and Statement 2 are false.

Resolving power is inversely proportional to the wavelength of light used and directly proportional to the aperture (diameter) of the objective lens.

Question 10. Statement 1 The standard optical diffraction gratings cannot be used to discriminate between different X-ray wavelengths.

Statement 2 The grating spacing is not of the order of X-ray wavelengths.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Diffraction effects are appreciable when the wavelength is of the order of the size of the diffracting element.

The wavelengths of X-rays are much smaller than the spacing of the grating elements, so diffraction is insignificant in discriminating X-ray wavelengths.

Question 11. Statement 1 Two coherent point sources of light S1 and S2 are placed on a line as shown. P and Q are two points on that line. If the maximum intensity is observed at point P then, the maximum intensity should also be observed at Q.

Statement 2 In the figure given with Statement 1, the distance |S1P- S2P| is equal to |S2Q- S1Q|

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Waves originating from the coherent sources S1 and S2 superpose at P to produce the maximum intensity. This is true when the path difference (S1P- S2P) is nk, where n = 0, 1, 2,…

The same path difference (S2Q- S1Q) is also given for Q, so Q will also be the position of maximum intensity.

Optics Objective Questions And Answers Wave Optics And Electromagnetic Waves Two Coherent Point Q11

Question 12. Statement 1 When a tiny circular obstacle is placed in the path of a beam of light some distance away, a bright spot is seen at the center of the shadow of the obstacle.

Statement 2 A destructive interference occurs at the center of the shadow.

Answer: 3. Statement 1 is true and Statement 2 is false.

A plane wavefront from a distant source incident on a circular obstacle gets diffracted and interferes constructively at the center of the shadow region and forms a bright spot.

Question 13. Statement 1 Radio waves can be polarized.

Statement 2 Sound waves in air are longitudinal in nature.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

Radio waves are electromagnetic (transverse) waves and can be polarized.

Question 14. Statement 1 When a ray of light is incident at a polarizing angle on a glass, the refracted ray is partially polarized.

Statement 2 The intensity of light decreases during polarization.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

When unpolarized light is incident on a transparent medium (glass or water) at a polarizing angle, the reflected light is completely polarized, whereas the refracted ray is partially polarized.

When unpolarized light is polarized at reflection, the energy is partly associated with the reflected and partly with the refracted waves. During transmission through a polaroid, the intensity is reduced according to Malus’s law.

Question 15. Statement 1 Skiers use polarized glasses.

Statement 2 Light reflected by snow is partially polarized.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Skiers use polarized sunglasses to eliminate the glare from the plane surface office. The reflected light from plane surfaces is polarized and is blocked by the polarized glasses, thus reducing the glare.

Question 16. Statement 1 A famous painting was made not by using brush strokes in the usual manner but rather by putting a myriad of small color dots. In this painting, the color you see at any given area on the painting changes as you move away.

Statement 2 The angular separation of adjacent dots changes with the distance from the painting.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

The angular separation between two closely spaced dots changes with distance. Due to the finite resolving power of the eye, the two dots appear as one and lead to an apparent change in color.

Question 17. Statement 1 At first glance, the top surface of a morpho butterfly’s wings appears blue-green. If the wings move, the color changes.

Statement 2 Different pigments in the wings reflect light at different angles.

Answer: 3. Statement 1 is true and Statement 2 is false.

The wings of a morpho butterfly create colors by causing light waves to diffract and interfere with its wings, which are composed of nano and micro-scale structures.

Question 18. Statement 1 Dipole oscillations produce electromagnetic waves.

Statement 2 Accelerated charge produces electromagnetic waves.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A dipole oscillation consists of accelerated charge which, according to Maxwell’s theory, radiates electromagnetic waves.

Question 19. Statement 1 The electromagnetic waves are transverse in nature.

Statement 2 Waves of wavelength 10 mm are radio waves and microwaves.

Answer: 3. Statement 1 is true and Statement 2 is false.

Electromagnetic waves constitute oscillating electric and magnetic fields, which are mutually perpendicular as well as perpendicular to the direction of propagation.

Electromagnetic waves of wavelength in the range 1-30 mm lie in the far-infrared region.

Question 20. Statement 1 Electromagnetic radiation exerts pressure.

Statement 2 Electromagnetic waves carry both momentum and energy.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

The radiant energy falling on a surface exerts pressure \(p=\frac{I}{c}\) for 100% absorption and \(p=\frac{2 I}{c}\) for total reflection, where I = intensity of the incident wave and c = speed of light.

Both energy (E = hv) and momentum \(\left(p=\frac{E}{c}\right)\) are associated with electromagnetic waves.

Question 21. Statement 1 When a charged particle moves in a circular path, it produces electromagnetic waves.

Statement 2 A charged particle has an acceleration.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A charged particle undergoing circular motion is accelerated, which produces electromagnetic waves.

Question 22. Statement 1 Electromagnetic waves are transverse in nature.

Statement 2 The electric and magnetic fields of an electromagnetic wave are mutually perpendicular to each other and also perpendicular to the direction of propagation.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Electromagnetic waves are transverse in nature because their components — \(\vec{E}\) vector and \(\vec{B}\) vector—are mutually perpendicular and also perpendicular to the direction of propagation.

Question 23. Statement 1 In electromagnetic waves, electric and magnetic field lines are perpendicular to each other.

Statement 2 Electric and magnetic fields are self-sustaining.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

Electromagnetic waves are transverse waves with \(\vec{E}\)– and \(\vec{B}\)– components undergoing sinusoidal variations along mutually perpendicular directions. These oscillations are self-sustaining.

Question 24. Statement 1 UV radiations cause photodissociation of ozone (O3) into O2 and O, thus causing damage to the stratospheric ozone layer.

Statement 2 An ozone hole results from global warming and climate damage.

Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.

When high-energy ultraviolet rays are absorbed by ozone, the ozone molecules are split into ordinary oxygen molecules and free oxygen atoms:

\(\mathrm{O}_3 \stackrel{\mathrm{uv}}{\longrightarrow} \mathrm{O}_2+\mathrm{O}\)

The decline of ozone in the stratosphere (called the ozone hole) results in reduced absorption of ultraviolet radiation from the sun. This is the cause of global warming and climate damage.

Question 25. Statement 1 Microwave communications are preferred to optical communications.

Statement 2 Microwaves provide a large number of channels and bandwidths compared to optical signals.

Answer: 5. Both Statement 1 and Statement 2 are false.

Optical communication systems are preferred to microwave communication systems for the following reasons. An optical communications system has

  1. A wider bandwidth,
  2. A larger capacity,
  3. A lower power consumption and
  4. An immunity from interference.

Question 26. Statement 1 A quasar emits radio waves more than a radio galaxy.

Statement 2 A quasar is a very small size.

Answer: 3. Statement 1 is true and Statement 2 is false.

A quasar (quasi-stellar radio source) is a massive and extremely remote celestial object that emits large amounts of radiant energy, much larger than a radio galaxy. The average size of quasars is around 0.1 x 106 AU, which is about 1.5 light-years.

Question 27. Statement 1 A pulsar is a source of radio waves that vary in intensity at regular intervals.

Statement 2 A pulsar is a rotating neutron star.

Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.

A pulsar (pulsating star) is a rotating neutron star that emits radio waves that sweep through the earth and vary in intensity with time.

Question 28. Statement 1 The environmental damage has increased the amount of ozone in the atmosphere.

Statement 2 An increase in ozone increases the amount of ultraviolet radiation on the earth.

Answer: 5. Both Statement 1 and Statement 2 are false.

The ozone layer absorbs harmful ultraviolet radiation from the sun and saves life on Earth.

Ray Optics Multiple Choice Questions And Answers

Ray Optics

Question 1. An air bubble in a glass slab with refractive index1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 deep when viewed from the opposite face. The thickness of the slab is

  1. 8 cm
  2. 10 cm
  3. 12 cm
  4. 16 cm

Answer: 3. 12 cm

Refractive index = \(\mu=\frac{\text { real depth }}{\text { apparent depth }}\)

In the first case, \(\mu=\frac{x}{5 \mathrm{~cm}}\),

and in the second case, \(\mu=\frac{D-x}{3 \mathrm{~cm}}\)

∴ x+(D-x) = μ(5 cm + 3 cm)

⇒ D = μ(8 cm) =1.5(8 cm) =12 cm

Ray Optics Multiple Choice Questions And Answers Refractive Index Q1

Question 2. A ray of light traveling in a transparent medium of refractive index μ. falls on a surface separating the medium from the air at an angle of incidence of 45°. For which of the following values of p can the ray undergo total internal reflection?

  1. μ =1.33
  2. μ =1.50
  3. μ =1.25
  4. μ =1.40

Answer: 2. μ = 1.50

For total internal reflection, the angle of incidence (i) must be greater than the critical angle (θC).

But, \(\mu=\frac{1}{\sin \theta_c} \Rightarrow \sin \theta_c=\frac{1}{\mu}\)

⇒ \(\sin i>\sin \theta_c \Rightarrow \sin 45^{\circ}>\frac{1}{\mu}\)

∴ \(\frac{1}{\sqrt{2}}>\frac{1}{\mu} \Rightarrow \mu>1.4\)

Hence, μ =1.50.

Question 3. A small coin is resting on the bottom of a beaker filled with a transparent liquid. A ray of light from the coin travels up to the 4 cm surface of the liquid and moves along the surface. How fast is the light traveling in the liquid?

Ray Optics Multiple Choice Questions And Answers Small Coin Q3

  1. 1.8 x 108 m s-1
  2. 2.4 x 108 m s-1
  3. 3.0 x 108 m s-1
  4. 1.2 x 108 m s-1

Answer: 1. 1.8 x 108 m s-1

In the given situation, Medical the angle Entrance of incidence Physics is the critical angle θC, for which \(\theta_c=\frac{3 \mathrm{~cm}}{5 \mathrm{~cm}}=\frac{3}{5}\)

The refractive index is

⇒ \(\mu=\frac{1}{\sin \theta_c}=\frac{5}{3}\)

In terms of the speed of light,

⇒ \(\mu=\frac{c_0}{c_1}\)

The speed of light in the liquid is

∴ \(c_1=\frac{c_0}{\mu}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{5 / 3}=\frac{9}{5} \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}=1.8 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\).

Ray Optics Multiple Choice Questions And Answers Critical Angle Q3

Question 4. For the given incident ray as shown in the figure, the condition for the total internal reflection of the ray will be satisfied if the refractive index of the block is

Ray Optics Multiple Choice Questions And Answers Incident Ray Q4

  1. \(\frac{\sqrt{3}+1}{2}\)
  2. \(\frac{\sqrt{2}+1}{2}\)
  3. \(\sqrt{\frac{3}{2}}\)
  4. \(\sqrt{\frac{7}{6}}\)

Answer: 3. \(\sqrt{\frac{3}{2}}\)

The refractive index of the block is

⇒ \(\mu=\frac{\sin 45^{\circ}}{\sin r}=\frac{1 / \sqrt{2}}{\sin \left(90^{\circ}-\theta\right)}=\frac{1}{\sqrt{2} \cos \theta}\)

⇒ \(\mu^2=\frac{1}{2 \cos ^2 \theta}=\frac{1}{2\left(1-\sin ^2 \theta\right)}\)

⇒\(\sin ^2 \theta=1-\frac{1}{2 \mu^2}\) (1)

For total internal reflection, θ > θC

⇒ \(\sin ^2 \theta>\sin ^2 \theta_c\left(=\frac{1}{\mu^2}\right)\) (2)

Comparing (1) and (2),

⇒ \(1-\frac{1}{2 \mu^2}>\frac{1}{\mu^2} \Rightarrow \frac{3}{2 \mu^2}<1\)

∴ \(\frac{2 \mu^2}{3}>1 \text {, hence } \mu \geq \sqrt{\frac{3}{2}}\).

Question 5. Light enters at an angle of incidence in a transparent rod of refractive index μ. For what value of the refractive index of the material of the rod will the light once entered into it not leave it through its lateral face whatever be the value of the angle of incidence?

  1. μ = l
  2. μ = 1.4
  3. μ > √2
  4. μ = 1.3

Answer: 3. μ > √2

For refraction at P, from Snell’s law,

⇒ \(\mu=\frac{\sin i}{\sin r}=\frac{\sin i}{\sin \left(90^{\circ}-\theta\right)}=\frac{\sin i}{\cos \theta}\) (1)

For total internal reflection at Q, the angle of incidence (i) must be greater than the critical angle \(\theta_c\left[=\sin ^{-1}\left(\frac{1}{\mu}\right)\right]\)

From (1),

⇒ \(\cos \theta=\frac{\sin i}{\mu}\)

⇒ \(1-\sin ^2 \theta=\frac{\sin ^2 i}{\mu^2}\)

⇒ \(\sin ^2 \theta=1-\frac{\sin ^2 i}{\mu^2}\) → (2)

But \(\sin \theta>\sin \theta_c \Rightarrow \sin ^2 \theta>\frac{1}{\mu^2}\) → (3)

From (2) and (3),

⇒ \(\frac{1}{\mu^2}<1-\frac{\sin ^2 i}{\mu^2}\)

⇒ \(1<\mu^2-\sin ^2 i\)

⇒ \(1+\sin ^2 i<\mu^2\)

∴ \(\mu^2>1+\sin ^2 i\)

The maximum value of sin i is 1, so

μ2 > 2 or μ > √2.

Ray Optics Multiple Choice Questions And Answers Snells Law Q5

Question 6. One face of a 6-cm-thick rectangular glass plate is silvered. An object held at 8 cm in front of the first face forms an image 12 cm behind the silvered face. The refractive index of the glass is

  1. 1.2
  2. 1.6
  3. 1.8
  4. 1.4

Answer: 1. 1.2

Let AB be the true position of the plane mirror and A’B’ be its displaced apparent position due to refraction.

For a plane mirror, the object and image are at an equal distance from the mirror, so

8 cm + x = (6 cm – x) +12 cm

x = 5 cm

Now, the refractive index of glass is

∴ \(\mu=\frac{\text { real depth }}{\text { apparent depth }}=\frac{6 \mathrm{~cm}}{5 \mathrm{~cm}}=1.2\).

Ray Optics Multiple Choice Questions And Answers Plane Mirror Q6

Question 7. The apparent depth of water in a cylindrical water tank of diameter 2R cm is reduced at the rate of x cm min-1 when water is being drained out at a constant rate. The amount of water drained in cm3 min-1 is (n1 = refractive index of air, n2 = refractive index of water)

  1. \(\frac{x \pi R^2 n_1}{n_2}\)
  2. \(\frac{x \pi R^2 n_2}{n_1}\)
  3. \(\frac{2 \pi R x n_1}{n_2}\)
  4. \(\frac{2 \pi R x n_2}{n_1}\)

Answer: 2. \(\frac{x \pi R^2 n_2}{n_1}\)

The refractive index of water is

⇒ \(\frac{n_2}{n_1}=\frac{\text { real depth }}{\text { apparent depth }}\)

⇒ \(\frac{n_2}{n_1}=\frac{H}{h} \Rightarrow \frac{d h}{d t}=\frac{n_1}{n_2} \frac{d H}{d t}\)

Given, \(\frac{d h}{d t}=x \mathrm{~cm} \mathrm{~min}^{-1}, \text { so } \frac{d H}{d t}=\frac{n_2}{n_1} x \mathrm{~cm} \mathrm{~min}^{-1}\)

∴ Volume drained out per min = \(A\left(\frac{d H}{d t}\right)=\pi R^2 \frac{n_2}{n_1} x\).

Question 8. Light waves enter from medium 1 to medium 2. Its velocity in medium 2 is double that in medium 1. For total internal reflection, the angle of incidence must be greater than

  1. 90°
  2. 40°
  3. 30°
  4. 60°

Answer: 3. 30°

The refractive index of medium 1 is greater than that of medium 2 since the speed in medium 2 is greater than that in 1.

Now, \({ }^2 n_1=\frac{n_1}{n_3}=\frac{c_0 / c_1}{c_0 / c_3}=\frac{c_2}{c_1}=\frac{2 c_1}{c_1}=2\)

If θC is the critical angle,

⇒ \(\theta_c=\frac{1}{2 n_1}=\frac{1}{2}=\sin 30^{\circ}\)

θC = 30°.

Question 9. Which of the following is not due to total internal reflection?

  1. Working with optical fiber
  2. Mirage on hot summer days
  3. Difference between the real and apparent depth of a pond
  4. Brilliance of diamond

Answer: 3. Difference between the real and apparent depth of a pond

Workings of optical fiber, formation of mirages, and brilliance of diamond can be explained on the basis of total internal reflection.

Question 10. In the given figure, the apparent depth of the liquid column is

Ray Optics Multiple Choice Questions And Answers The Apparent Depth Of The Liquid Q10

  1. \(\frac{d_1}{\mu_2}+\frac{d_2}{\mu_1}\)
  2. \(\sqrt{\frac{d_1 d_2}{\mu_1 \mu_2}}\)
  3. \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\)
  4. \(\frac{d_1+d_2}{\mu_1+\mu_2}\)

Answer: 3. \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\)

Apparent shift of the bottom in the lower medium is x2 = \(d_2\left(1-\frac{1}{\mu_2}\right)\)

and in the upper medium is x1 = \(d_1\left(1-\frac{1}{\mu_1}\right)\)

Total shift is x1 + x2 = \(\left(d_1+d_2\right)-\left(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\right)\)

Apparent depth = total depth- shift

∴ \(\left(d_1+d_2\right)-\left(x_1+x_2\right)=\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\).

Question 11. Considering the normal incidence of the ray, the equivalent refractive index of the combination of two slabs shown in the figure is

Ray Optics Multiple Choice Questions And Answers Combination Of Two Slabs Q11

  1. 1.8
  2. 1.43
  3. 2
  4. None of these

Answer: 2. 1.43

The apparent depth for the given pair of media is

⇒ \(d=\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\)

For equivalent refractive index,

⇒ \(\mu=\frac{\text { real depth }}{\text { apparent depth }}=\frac{d_1+d_2}{\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}}\)

Given, \(\mu_1=\frac{4}{3}, d_1=10 \mathrm{~cm}, \mu_2=\frac{3}{2}, d_2=15 \mathrm{~cm}\)

∴ \(\mu=\frac{10 \mathrm{~cm}+15 \mathrm{~cm}}{(10 \mathrm{~cm}) \frac{3}{4}+(15 \mathrm{~cm}) \frac{2}{3}}=\frac{25}{17.5}=1.43\)

Question 12. A microscope is focused on a mark on a piece of paper and a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again?

  1. 2 cm upward
  2. 1 cm upward
  3. 4.5 cm downward
  4. 1 cm downward

Answer: 2. 1 cm upward

In the given figure, O is the position of the mark and I is its apparent position.

Refractive index, \(\mu=\frac{\mathrm{AO}}{\mathrm{AI}}\)

∴ \(1.5=\frac{3 \mathrm{~cm}}{\mathrm{AI}} \Rightarrow \mathrm{AI}=2 \mathrm{~cm}\)

Shift of the image

= OI = 3 cm- 2 cm =1 cm.

The microscope must be raised upward from P to Q by 1 cm.

Ray Optics Multiple Choice Questions And Answers Shift Of The Image Q12

Question 13. For the situation shown in the figure, what should be the refractive index of glass so that the ray will suffer total internal reflection at the vertical face?

Ray Optics Multiple Choice Questions And Answers Refractive Index Of Glass Q13

  1. 1.732
  2. 1.5
  3. 1.32
  4. 1.6

Answer: 3. 1.32

Refractive index,

⇒ \(\mu=\frac{\sin 60^{\circ}}{\sin r}=\frac{\sin 60^{\circ}}{\sin \left(90^{\circ}-\theta\right)}=\frac{\sqrt{3}}{2 \cos \theta}\)

For total internal reflection,

θ ≥ θC,

⇒ \(\cos \theta_c=\frac{\sqrt{3}}{2 \mu}=\frac{\sqrt{3}}{2\left(1 / \sin \theta_c\right)}\)

⇒ \(\tan \theta_c=\frac{2}{\sqrt{3}}\)

Hence, μ = \({cosec} \theta_c=\frac{\sqrt{7}}{2}=1.32\)

Ray Optics Multiple Choice Questions And Answers Total Internal Reflection Q13(1)

Ray Optics Multiple Choice Questions And Answers Total Internal Reflection Q13(2)

Question 14. In a concave mirror, an object is placed at a distance d1 from the focus and the image is formed at a distance d2 from the focus. The focal length of the mirror is

  1. \(\sqrt{d_1 d_2}\)
  2. \(d_1 d_2\)
  3. \(\frac{d_1+d_2}{2}\)
  4. \(\sqrt{\frac{d_1}{d_2}}\)

Answer: 1. \(\sqrt{d_1 d_2}\)

From the mirror formula, \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

Given, u = d1 + f, v = d2 + f

⇒ \(\frac{1}{f}=\frac{1}{d_1+f}+\frac{1}{d_2+f}=\frac{\left(d_1+d_2\right)+2 f}{\left(d_1+f\right)\left(d_2+f\right)}\)

⇒ d1d2 + f(d1+f2)+f2 = (d1+d2)f +2f2

∴ \(f^2=d_1 d_2 \Rightarrow f=\sqrt{d_1 d_2}\).

Question 15. A short linear object of length L lies along the axis of a concave mirror of focal length f at a distance d from the pole of the mirror. The size of the image is then (nearly)

  1. \(\frac{L f}{d-f}\)
  2. \(\frac{d-f}{L f}\)
  3. \(\frac{L f^2}{(d-f)^2}\)
  4. \(\frac{L(d-f)^2}{f^2}\)

Answer: 3. \(\frac{L f^2}{(d-f)^2}\)

Since \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

Differentiating, \(\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}=0\)

Size of the object placed along the axis is L = AM, SO size of the image = \(|\Delta v|=\frac{v^2}{u^2} \cdot L\)

Now, \(1+\frac{u}{v}=\frac{u}{f} \Rightarrow \frac{u}{v}=\frac{u}{f}-1=\frac{u-f}{f}=\frac{d-f}{f}\)

∴ \(|\Delta v|=L\left(\frac{v}{u}\right)^2=\frac{L f^2}{(d-f)^2}\)

Question 16. A 2.0 cm-tall object is placed 15 cm in front of a concave mirror of a focal length of 10 cm. What is the size and nature of the image?

  1. 4 cm, real
  2. 4 cm, virtual
  3. 1 cm, real
  4. None of these

Answer: 1. 4 cm, real

Given,f =10 cm and M =15 cm. Since u >f, the image will be real and magnified.

∵ \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \Rightarrow \frac{1}{15 \mathrm{~cm}}+\frac{1}{v}=\frac{1}{10 \mathrm{~cm}} \Rightarrow \frac{1}{v}=\frac{5}{150 \mathrm{~cm}}\)

⇒ v = 30 cm.

∴ magnification is m = \(\frac{v}{u}=\frac{30 \mathrm{~cm}}{15 \mathrm{~cm}}=2\)

∴ Size of the image = 2 x 2 cm = 4 cm.

Question 17. A concave mirror of focal length f1 is placed at a distance d from a convex lens of focal length f2. A beam of light coming from infinity and incident on this convex lens-concave mirror combination returns to infinity. The distance d must be equal to

  1. f1+f2
  2. -f1+f2
  3. 2f1+f2
  4. -2f1+f2

Answer: 3. 2f1+f2

Parallel rays from infinity incident on the convex lens L get focused at its focus F, where OF = f2. Since the rays retrace their paths after reflection from the concave mirror, the rays must be incident normally (coming from its center) and therefore F must be the center of curvature of the concave mirror. So, PF = R = 2f1

The distance between the lens and the mirror is

d = OF + FP = f2 + 2f1.

Ray Optics Multiple Choice Questions And Answers Parallel Rays Q17

Question 18. The angle of incidence for a ray of light at a refracting surface of a prism is 45°. The angle of the prism is 60°. If the ray suffers a minimum deviation through the prism, the angle of minimum deviation and the refractive index of the material of the prism respectively are

  1. \(30^{\circ}, \sqrt{2}\)
  2. \(45^{\circ}, \sqrt{2}\)
  3. \(45^{\circ}, \frac{1}{\sqrt{2}}\)
  4. \(30^{\circ}, \frac{1}{\sqrt{2}}\)

Answer: 1. [late]30^{\circ}, \sqrt{2}[/latex]

Given, A = 60° and i = 45°.

Since the ray suffers minimum deviation, i = i’ and r = r’ with r + r’ = A.

Hence, r + r’ = r+r = A = 60° ⇒ r = 30°.

Again, i+i’ = A+8 ⇒ i+i=A+δm.

the angle of minimum deviation,

δm = 2i-A = 2(45°)- 60° = 30°.

Refractive index,

∴ \(\mu=\frac{\sin i}{\sin r}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\sqrt{2}\).

Question 19. The refracting angle of a prism is A and the refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is

  1. 180°- 3A
  2. 180°- 2A
  3. 90°- A
  4. 180° + 2A

Answer: 2. 180°- 2A

For a prism,

⇒ \(\mu=\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}=\cot \frac{A}{2} \text { (given) }\)

⇒ \(\cos \frac{A}{2}=\sin \left(90^{\circ}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right)\)

Hence, \(\frac{A+\delta_m}{2}=\frac{\pi}{2}-\frac{A}{2}\).

So, the angle of minimum deviation, δm = π – 2A = 180°- 2A

Question 20. A ray of light is incident at an angle of incidence i on one face of a prism of angle A (assumed to be small) and emerges normally from the opposite face. If the refractive index of the material of the prism is μ, the angle of incidence is nearly equal to

  1. μA
  2. \(\frac{\mu A}{2}\)
  3. \(\frac{A}{\mu}\)
  4. \(\frac{A}{2 \mu}\)

Answer: 1. μA

Since the ray emerges normally out of the prism, angle r = A.

⇒ \(\mu=\frac{\sin i}{\sin r}=\frac{i}{r}=\frac{i}{A}\) (for small angles).

Hence, the angle of incidence is i = μA

Ray Optics Multiple Choice Questions And Answers Prism Q20

Question 21. The refractive index of the material of a prism is √2 and its refracting angle is 30°. One of the refracting surfaces of the prism is silvered outward. A beam of monochromatic light entering the prism from the other face retraces its path after reflection from the silvered surface. The angle of incidence on the prism is

  1. 45°
  2. 60°
  3. 30°

Answer: 1. 45°

Given, μ ⇒ √2 and A = 30°.

Since the ray retraces its path after reflection from the second face, it must be normally incident. So, r = A = 30°.

Now, \(\mu=\frac{\sin i}{\sin r} \Rightarrow \sqrt{2}=\frac{\sin i}{\sin 30^{\circ}}\)

⇒ \(\sin i=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}=\sin 45^{\circ}\)

∴ i = 45°.

Ray Optics Multiple Choice Questions And Answers Path After Reflection From The Secnd Face Q21

Question 22. If the refractive index of the material of an equilateral prism is √3 then the angle of minimum deviation from the prism is

  1. 30°
  2. 45°
  3. 60°
  4. 75°

Answer: 3. 60°

Given, A = 60° and μ = √3.

⇒ \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)

⇒ \(\sqrt{3}=2 \sin \left(\frac{60^{\circ}+\delta_m}{2}\right)\)

⇒ \(\sin \left(\frac{60^{\circ}+\delta_m}{2}\right)=\frac{\sqrt{3}}{2}=\sin 60^{\circ}\).

∴ The angle of minimum deviation = δm = 60°.

Question 23. The angle of deviation δ through a dun prism (A small) whose material has its refractive index μ can be expressed by

  1. \(\delta=(\mu-1) A\)
  2. \(\delta=(\mu+1) A\)
  3. \(\delta=\frac{\sin \frac{A+\delta}{2}}{\sin \frac{A}{2}}\)
  4. \(\delta=\left(\frac{\mu-1}{\mu+1}\right) A\)

Answer: 1. \(\delta=(\mu-1) A\)

For a thin-angled prism, A = small.

So i, i, i’, r, r’ must be small for a ray to emerge.

⇒ \(\mu=\frac{\sin i}{\sin r} \approx \frac{i}{r} \Rightarrow i=\mu r\)

Similarly, i’ = μr’.

Adding, i+i’ = μ{r + r’)

⇒  A + δ = μA ⇒ δ = (μ.-1)A.

Question 24. One face of a prism of refracting angle A is silvered. Light rays incident at an angle of incidence 2A on the first surface retrace their path after suffering reflection at the silvered surface. The refractive index of the die material of the prism is

  1. 2 sinA
  2. 2 cosA
  3. \(\frac{1}{2} \cos A\)
  4. tan A

Answer: 2. 2 cosA

Since the ray is reflected from the second face, it must be incident normally. Hence, the angle of refraction r-A. Refractive index is

⇒ \(\mu=\frac{\sin i}{\sin r}=\frac{\sin 2 A}{\sin A}=\frac{2 \sin A \cos A}{\sin A}\)

μ = 2cos A

Ray Optics Multiple Choice Questions And Answers Refractive Index Q24

Question 25. The angle of minimum deviation for a glass prism is equal to its refracting angle. The refractive index of glass is 1.5. Then, the angle of the prism is

  1. \(2 \cos ^{-1}\left(\frac{3}{4}\right)\)
  2. \(\sin ^{-1}\left(\frac{3}{4}\right)\)
  3. \(2 \sin ^{-1}\left(\frac{3}{2}\right)\)
  4. \(\cos ^{-1}\left(\frac{3}{2}\right)\)

Answer: 1. \(2 \cos ^{-1}\left(\frac{3}{4}\right)\)

Given, the angle of prism = A = δm, refractive index = μ =1.5.

From the prism formula,

⇒ \(\mu=\frac{\sin \left(\frac{A+\delta_{\mathrm{m}}}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin A}{\sin \frac{A}{2}}\)

⇒ \(\mu=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \cos \frac{A}{2}=\frac{\mu}{2}=\frac{3}{4}\)

∴ \(\frac{A}{2}=\cos ^{-1}\left(\frac{3}{4}\right) \Rightarrow A=2 \cos ^{-1}\left(\frac{3}{4}\right)\).

Question 26. For the angle of minimum deviation through a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index

  1. Lies between √2 and 1
  2. Lies between 2 and √2
  3. Is less than 1
  4. Is greater than 2

Answer: 2. Lies between 2 and √2

Given, δm = A

Hence, \(\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin A}{\sin \frac{A}{2}}=2 \cos \frac{A}{2}\)

Since i + i’ = A+8,

hence at minimum deviation,

i + i = 2A ⇒ i = A

Let us consider the limiting values of i:

⇒ For \(i_{\min }=0, A_{\min }=0 \Rightarrow \mu_{\max }=\left(2 \cos \frac{A}{2}\right)=2 \cos 0=2\)

⇒ Again, \(i_{\max }=90^{\circ}, A_{\max }=90^{\circ}, \text { so } \mu_{\min }=2 \times \cos \frac{90^{\circ}}{2}=\sqrt{2}\)

∴ μ lies between 2 and √2

Question 27. A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first face of the prism is

  1. Zero
  2. 30°
  3. 45°
  4. 60°

Answer: 2. 30°

For a prism, r+r’ = A.

For minimum deviation, r = r’.

r + r = A = 60° ⇒ r = 30°.

Question 28. A thin prism having a refracting angle of 10° is made of crown glass with a refractive index of 1.42. This prism is combined with another thin prism of flint glass with a refractive index 1.7. This combination produces dispersion without deviation. The refracting angle of the second prism should be

  1. 10°

Answer: 1. 6°

For a crown prism, AC =10° and μC =1.42.

For the flint prism, μF =1.7.

For null deviation, δC + δF = 0

⇒ \(\left(\mu_C-1\right) A_C+\left(\mu_F-1\right) A_F=0\)

∴ \(\left|A_F\right|=\frac{\left(\mu_C-1\right) A_C}{\left(\mu_F-1\right)}=\frac{(1.42-1) 10^{\circ}}{(1.7-1)}=\frac{0.42}{0.70} \times 10^{\circ}=6^{\circ}\).

Question 29. A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If fine lenses are made of different materials of refractive indices μ1 and μ2, and R is the radius of curvature of the curved surface of the lenses then the focal length of the combination is

  1. \(\frac{R}{2\left(\mu_1+\mu_2\right)}\)
  2. \(\frac{R}{2\left(\mu_1-\mu_2\right)}\)
  3. \(\frac{R}{\left(\mu_1-\mu_2\right)}\)
  4. \(\frac{2 R}{\left(\mu_2-\mu_1\right)}\)

Answer: 3. \(\frac{R}{\left(\mu_1-\mu_2\right)}\)

For a convex lens,

⇒ \(\frac{1}{f_1}=\left(\mu_1-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

⇒ \(\left(\mu_1-1\right)\left(\frac{1}{\infty}-\frac{1}{-R}\right)=\frac{\left(\mu_1-1\right)}{R}\)

For a concave lens,

⇒ \(\frac{1}{f_2}=\left(\mu_2-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\mu_2-1\right)\left(-\frac{1}{R}-\frac{1}{\infty}\right)=-\frac{\left(\mu_2-1\right)}{R}\)

For the combination with equivalent focal length F

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{\left(\mu_1-1\right)-\left(\mu_2-1\right)}{R}=\frac{\mu_1-\mu_2}{R}\)

∴ \(F=\frac{R}{\left(\mu_1-\mu_2\right)}\)

Ray Optics Multiple Choice Questions And Answers Combination With Equal Focal Length Q29

Question 30. A biconvex lens of glass (μ =15) has a radius of curvature of 20 cm. Which one of the following options describes best the image formed of a dying object of height 2 cm placed 30 cm from the lens?

  1. Virtual, erect, height = 0.5 cm
  2. Real, inverted, height = 4 cm
  3. Real, inverted, height =1 cm
  4. Virtual, erect, height =1 cm

Answer: 2. Real, inverted, height = 4 cm

Given, a convex lens,

μ =1.5, R1 = R2 = 20 cm, h0, = 2 cm, u = 30 cm.

From the lens formula,

⇒ \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{20 \mathrm{~cm}}-\frac{1}{-20 \mathrm{~cm}}\right)\)

⇒ \(0.5\left(\frac{2}{20 \mathrm{~cm}}\right)=\frac{1}{20 \mathrm{~cm}}\)

Again \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

⇒ \(\frac{1}{20 \mathrm{~cm}}=\frac{1}{v}-\frac{1}{(-30 \mathrm{~cm})} \Rightarrow \frac{1}{v}=\frac{1}{60 \mathrm{~cm}}\)

⇒ v = 60 cm.

Magnification = \(m=\frac{v}{u}=\frac{60 \mathrm{~cm}}{-30 \mathrm{~cm}}=-2\)

Negative magnification means the image is real, inverted and whose height is

⇒ \(h_{\mathrm{I}}=|m| h_0=2 \times 2 \mathrm{~cm}=4 \mathrm{~cm}\)

Question 31. A lens having focal length f and an aperture of diameter D forms an image of intensity I. An aperture of diameter D/2 in the central region of the lens is covered by black paper. The focal length of the lens and intensity of the image now will be respectively

  1. \(f \text { and } \frac{I}{4}\)
  2. \(\frac{3 f}{4} \text { and } \frac{I}{2}\)
  3. \(f \text { and } \frac{3 I}{4}\)
  4. \(\frac{f}{2} \text { and } \frac{I}{2}\)

Answer: 3. \(f \text { and } \frac{3 I}{4}\)

The focal length of a lens does not depend on the area exposed to the incoming incident rays, so the focal length remains unchanged at f.

The intensity of the image formed is proportional to the area exposed.

Thus, \(\frac{I_2}{I_1}=\frac{A_2}{A_1}=\frac{\pi\left(\frac{D}{2}\right)^2-\pi\left(\frac{D}{4}\right)^2}{\pi\left(\frac{D}{2}\right)^2}=\frac{3}{4}\)

∴ \(I_2=\frac{3}{4} I_1=\frac{3}{4} I\)

Question 32. A biconvex lens is cut into two halves along

  • XOX’ and
  • YOY’ as shown in the figure. Let f,f’,f” be the focal lengths of the complete lens, of each half in case (1) and of each half in case (2) respectively.

Choose the correct statement from the following.

Ray Optics Multiple Choice Questions And Answers Equiconvex Lens Q32

  1. f’ = f, f” = f
  2. f’ = 2f, f” = 2f
  3. f’ = f, f” = 2f
  4. f’ = 2f, f” = f

Answer: 3. f’ = f, f” = 2f

For the complete lens,

⇒ \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2(\mu-1)}{R}\)

The focal length of each half-cut lens along X’OX will be

⇒ \(\frac{1}{f^{\prime}}=(\mu-1) \frac{2}{R}, \text { hence } f^{\prime}=f\)

The focal length of each half-cut lens along YOY’ is

⇒ \(\frac{1}{f^{\prime \prime}}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)=\frac{\mu-1}{R}=\frac{1}{2}\left[\frac{2(\mu-1)}{R}\right]=\frac{1}{2 f} \Rightarrow f^{\prime \prime}=2 f\)

∴ \(f^{\prime \prime}=2 f \text { and } f^{\prime}=f\)

Question 33. An object is located on a wall. An image of the same size is to be obtained on the parallel wall with the help of a convex lens. The lens is placed at a distance d ahead of the second wall. The required focal length of the lens should be

  1. \(\text { only } \frac{d}{4}\)
  2. \(\text { only } \frac{d}{2}\)
  3. \(\text { more than } \frac{d}{4} \text { but less than } \frac{d}{2}\)
  4. \(\text { less than } \frac{d}{4}\)

Answer: 2. \(\text { only } \frac{d}{2}\)

Since the object and its image are of the same size, magnification is

⇒ \(m=\frac{|v|}{|u|}=1\)

Since v = d,

u = v = d.

From the lens formula,

⇒ \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{d}-\frac{1}{(-d)}=\frac{2}{d}\)

⇒ \(f=\frac{d}{2}\)

Ray Optics Multiple Choice Questions And Answers Magnification Q33

Question 34. A plano-convex lens is made of a material of refractive index μ, =1.5. The radius of curvature of the curved surface of the lens is 20 cm. If its plane surface is silvered, the focal length of the silvered lens will be

  1. 10 cm
  2. 20 cm
  3. 40 cm
  4. 60cm

Answer: 2. 20 cm

When the plane surface is not silvered, its focal length is f1 where

⇒ \(\frac{1}{f_1}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)=\frac{1}{2 R}\)

f1 = 2R = 2(20 cm) = 40 cm.

When the plane surface is silvered, rays from focus F will be parallel after refraction, fall normally on the silvered face, and retrace its path, forming the real image at F.

Thus, the silvered lens acts as a concave mirror with a center at F whose radius of curvature = f1 = 4O cm.

⇒ \(f \text { of mirror }=\frac{40 \mathrm{~cm}}{2}=20 \mathrm{~cm}\).

Ray Optics Multiple Choice Questions And Answers Silvered Lens Q34

Alternative method

In the silvered lens, there are two refractions from the curved surface and one reflection from the plane surface.

Hence, the power of the silvered lens

= power of the lens + power of the mirror

⇒ \(\frac{1}{F}=\frac{2}{f_l}+\frac{1}{f_{\mathrm{m}}}=\frac{2}{40 \mathrm{~cm}}+\frac{1}{\infty}\)

F = 20 cm.

Question 35. A plano-convex lens is made of a material with a refractive index of 1.6. The radius of curvature of the curved surface is 60 cm. The focal length of the lens is

  1. 50 cm
  2. 100 cm
  3. 200 cm
  4. 400 cm

Answer: 2. 100 cm

Given, μ =1.6, R1 = 60 cm, and R2 = ∞ (for plane surface).

From the lens maker’s formula,

⇒ \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

∴ \((1.6-1)\left(\frac{1}{60 \mathrm{~cm}}-\frac{1}{\infty}\right)=\frac{0.6}{60 \mathrm{~cm}}=\frac{1}{100 \mathrm{~cm}}\)

∴ f = 100 cm.

Question 36. A luminous object is placed at a distance of 30 cm from a convex lens of focal length 20 cm. On the other side of the lens, at what distance from the lens should a convex mirror of radius of curvature 10 cm be placed in order to have an upright image of the object coincident with it?

  1. 20 cm
  2. 30 cm
  3. 50 cm
  4. 60 cm

Answer: 3. 50 cm

For the lens,

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

⇒ \(\frac{1}{v}-\frac{1}{-30 \mathrm{~cm}}=\frac{1}{20 \mathrm{~cm}}\)

⇒ \(\frac{1}{v}=\frac{1}{20 \mathrm{~cm}}-\frac{1}{30 \mathrm{~cm}}=\frac{1}{60 \mathrm{~cm}}\)

v = 60 cm = OC

Rays emerging from the lens must be directed towards the center of curvature C of the convex mirror so as to retrace its path such that the object and the image are coincident.

The distance between the lens and the mirror is OP = OC- PC = v-R = 60 cm-10 cm = 50 cm.

Ray Optics Multiple Choice Questions And Answers Distance Between The Lens And The Mirror Q36

Question 37. A lens is made of flint glass (refractive index = 1.5). When the lens is immersed in a liquid with a refractive index of 1.25, its focal length

  1. Increases by a factor of 1.25
  2. Increases by a factor of 2.5
  3. Increases by a factor of 1.2
  4. Decreases by a factor of 1.2

Answer: 2. Increases by a factor of 2.5

⇒ \(\frac{1}{f_{\mathrm{a}}}=\left({ }^{\mathrm{a}} \mu_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) → (1)

When the lens is immersed in the liquid (μ1 =1.25), the focal length is

⇒ \(\frac{1}{f_1}=\left({ }^1 \mu_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) → (2)

Taking the ratio of (1) and (2),

⇒ \(\frac{f_1}{f_{\mathrm{a}}}=\frac{{ }^{\mathrm{a}} \mu_{\mathrm{g}}-1}{\left(\frac{\mu_{\mathrm{g}}}{\mu_1}-1\right)}=\frac{(1.5-1)}{\left(\frac{1.5}{1.25}-1\right)}=\frac{5}{2}\)

⇒ \(f_1=\frac{5}{2} f_{\mathrm{a}}\)

Hence, f increases by a factor of 2.5.

Question 38. A camera lens has an aperture of f and the exposure time is (1/60) s. What will be the exposure time when the aperture becomes 1.4?

  1. \(\frac{1}{42} \mathrm{~s}\)
  2. \(\frac{1}{31} \mathrm{~s}\)
  3. \(\frac{1}{72} \mathrm{~s}\)
  4. \(\frac{1}{56} \mathrm{~s}\)

Answer: 2. \(\frac{1}{31} \mathrm{~s}\)

The f-number of an optical system. (camera lens) is the ratio of the system’s focal length to the diameter of the aperture (hole) through which light enters. It is the reciprocal of the relative aperture.

Time of exposure (t) ∝ (f-number)2

⇒ \(\frac{t}{(1 / 60) \mathrm{s}}=\frac{(1.4)^2}{1}\)

∴ \(t=(1.4)^2\left(\frac{1}{60}\right) \mathrm{s}=\frac{14 \times 14}{6000} \approx \frac{1}{31} \mathrm{~s}\)

Question 39. A convex lens of refractive index 3/2 has a power of 2.5 D in it is immersed in a liquid of refractive index 2 then the new poof of the lens is

  1. 1.25 D
  2. -1.25 D
  3. -1.5D
  4. 1.5 D

Answer: 2. -1.25 D

The power of a lens is

⇒ \(P=\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

When in the air,

⇒ \(2.5 \mathrm{D}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

When immersed in the liquid,

⇒ \(P=\frac{1}{f^{\prime}}=\left({ }^{\prime} \mu_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{\dot{R}_2}\right)\)

⇒ \(\frac{P}{2.5 \mathrm{D}}=\frac{\left(\frac{\mu_{\mathrm{g}}}{\mu_1}-1\right)}{\left(\mu_{\mathrm{g}}-1\right)}=\frac{\left(\frac{1.5}{2}-1\right)}{(1.5-1)}\)

∴ \(P=-\frac{2.5}{2} \mathrm{D}=-1.25 \mathrm{D}\).

Question 40. In a convex lens of focal length f, the minimum distance between the object and its real image must be

  1. 3f
  2. 4f
  3. 1.5f
  4. 2f

Answer: 2. 4f

Let D be the separation between the object and its real image formed by a convex lens.

⇒ \(D=|u|+|v|=(\sqrt{u}-\sqrt{v})^2+2 \sqrt{u v}\)

For D to be minimum,

(√u-√v)2 = 0 = u = v.

From the lens formula (for real image),

⇒ \(\frac{1}{v}-\frac{1}{(-u)}=\frac{1}{f}\)

⇒ \(\frac{1}{v}+\frac{1}{v}=\frac{1}{f} \Rightarrow v=u=2 f\)

∴ \((D)_{\min }=2 \sqrt{u v}=2 \sqrt{(2 f)(2 f)}=4 f\)

Question 41. Two identical glass (μg = 1.5) biconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (μw = 4/3). The focal length of the combination is

  1. \(\frac{f}{3}\)
  2. f
  3. \(\frac{4 f}{3}\)
  4. \(\frac{3 f}{4}\)

Answer: 4. \(\frac{3 f}{4}\)

For a glass convex lens,

⇒ \(\frac{1}{f}=\frac{1}{f_1}=\frac{1}{f_2}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

⇒ \(\frac{1}{f}=(1.5-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{1}{R}\)

For water-biconcave lens,

⇒ \(\frac{1}{f_3}=\left(\mu_w-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

⇒ \(\left(\frac{4}{3}-1\right)\left(\frac{1}{-R}-\frac{1}{R}\right)\)

⇒ \(\left(\frac{1}{3}\right)\left(-\frac{2}{R}\right)=-\frac{2}{3 R}\)

For an equivalent focal length for the combination,

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{1}{R}+\frac{1}{R}-\frac{2}{3 R}=\frac{4}{3 R}=\frac{4}{3}\left(\frac{1}{f}\right)\)

∴ \(F=\frac{3 f}{4}\)

Ray Optics Multiple Choice Questions And Answers Equivalent Focal Length Q41

Question 42. Two identical thin plano-convex glass lenses (refractive index = 1.5) each having radius of curvature 20 cm are placed with their convex surfaces in contact at the center. The intervening space between them is filled with oil with a refractive index of 1.7. The focal length of the combination is

  1. -50 cm
  2. -25 cm
  3. 50 cm
  4. -20 cm

Answer: 1. -50 cm

For the two identical plano-convex lenses of glass,

⇒ \(\frac{1}{f}=\frac{1}{f_1}=\frac{1}{f_2}=\left(\mu_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

⇒ \((1.5-1)\left(\frac{1}{\infty}-\frac{1}{-R}\right)=\frac{1}{2 R}\)

For the biconcave liquid lens,

⇒ \(\frac{1}{f_3}=\left(\mu_l-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.7-1)\left(-\frac{1}{R}-\frac{1}{R}\right)=-\frac{1.4}{R}\)

For the equivalent focal length,

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{2}{f_1}+\frac{1}{f_3}=2\left(\frac{1}{2 R}\right)+\left(-\frac{1.4}{R}\right)=-\frac{0.4}{R}\)

∴ \(F=-\frac{R}{0.4}=-\frac{20 \mathrm{~cm}}{0.4}=-50 \mathrm{~cm}\)

Ray Optics Multiple Choice Questions And Answers Plano Convex Lenses Of Glass Q42

Question 43. A converging beam of rays is incident on a diverging lens. After passing through the lens, the rays converge at a point 15 cm from the lens on the opposite side. If the lens is removed, the point where the rays converge moves 5 cm closer to the lens. The focal length of the lens is

  1. 5 cm
  2. -10 cm
  3. -30 cm
  4. 20 cm

Answer: 3. -30 cm

In the given figure, rays converge at O without the lens, hence acting as an object. After passing through the concave lens, the rays get slightly diverged and meet at I, which is the position of the image. Thus,

PI = v =15 cm and OP = u =15 cm-5 cm =10 cm.

From the lens formula,

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

⇒ \(\frac{1}{f}=\frac{1}{15 \mathrm{~cm}}-\frac{1}{10 \mathrm{~cm}}=-\frac{1}{30 \mathrm{~cm}}\)

Hence, f = -30 cm.

Ray Optics Multiple Choice Questions And Answers Q43

Question 44. Two thin lenses of focal lengths f1 and f2 are in contact and coaxial. The power of the combination is

  1. \(\frac{f_1+f_2}{2}\)
  2. \(\frac{f_1+f_2}{f_1 f_2}\)
  3. \(\sqrt{\frac{f_1}{f_2}}\)
  4. \(\sqrt{\frac{f_2}{f_1}}\)

Answer: 2. \(\frac{f_1+f_2}{f_1 f_2}\)

For two thin lenses in contact,

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{f_1+f_2}{f_1 f_2}\)

Equivalent power of the combination,

⇒ \(P=\frac{1}{F}=\frac{f_1+f_2}{f_1 f_2}\)

Question 45. A person can see objects clearly only when they lie between 50 cm and 400 cm away from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens the person has to use will be

  1. Convex, +2.25 dioptre
  2. Concave, -0.25 dioptre
  3. Concave, -0.2 dioptre
  4. Convex, +0.15 dioptre

Answer: 2. Concave, -0.25 dioptre

The person can see distinctly up to 400 cm = 4 m. For distinct vision up to ∞, he needs a concave lens that can form a virtual image at 400 cm when the object is at ∞ as shown in the figure.

From the lens formula,

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

⇒ \(\frac{1}{-400 \mathrm{~cm}}-\frac{1}{\infty}=\frac{1}{f}\)

power of the lens is

∴ \(P=\frac{100}{f_{\mathrm{cm}}} \mathrm{D}=\frac{100}{-400} \mathrm{D}=-\frac{1}{4} \mathrm{D}=-0.25 \mathrm{D}\)

The negative sign is for the concave lens.

Ray Optics Multiple Choice Questions And Answers Object Is At Infinity Q45

Question 46. For a normal eye, the cornea of the eye provides a converging power of 40 D, and the least converging power of the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea for the eye lens can be estimated to be

  1. 5 cm
  2. 2.5 cm
  3. 1.67 cm
  4. 1.5 cm

Answer: 3. 1.67 cm

Given, the converging power of the cornea, PC = +40 D; the least converging power of the eye lens, Pe = +20 D.

equivalent power of eye lens,

P = Pe + PC = 20 D + 40 D = 60 D

But, \(P=\frac{1}{\text { focal length of eye lens }(f)}\)

⇒ \(f=\frac{1}{P}=\frac{1}{60 \mathrm{D}}=\frac{1}{60} \mathrm{~m}=\frac{100}{60} \mathrm{~cm}=\frac{5}{3} \mathrm{~cm}\)

Distance between the retina and cornea for the eye lens

= focal length of the eye lens

∴ \(\frac{5}{3} \mathrm{~cm}=1.67 \mathrm{~cm}\).

Question 47. For a person, the near point of vision is 100 cm. The power the correcting lens requires to provide normal vision is

  1. +1 D
  2. -lD
  3. +3D
  4. -3D

Answer: 3. +3D

The near point for normal vision is 25 cm.

Hence, the correction lens must be a convex lens as shown in the figure.

The object O placed at 25 cm from the lens forms its virtual image I at 100 cm, which the defective eye can distinctly see.

Thus, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{(-100 \mathrm{~cm})}-\frac{1}{(-25 \mathrm{~cm})}=\frac{3}{100 \mathrm{~cm}}\)

∴ Power = \(P=\frac{1}{f}=\frac{3}{1 \mathrm{~m}}=+3 \mathrm{D}\)

Ray Optics Multiple Choice Questions And Answers Normal Vision Q47

Question 48. A farsighted person has his near point at 50 cm. The power of the lens he should use to see at 25 cm clearly is

  1. +1D
  2. +2D
  3. -2D
  4. -1D

Answer: 2. +2D

Refer to the ray diagram of the preceding. Here, OP = u = 25 cm and IP = v = 50 cm.

⇒ \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-50 \mathrm{~cm}}-\frac{1}{-25 \mathrm{~cm}}=\frac{1}{50 \mathrm{~cm}}\)

∴ f = + 50 cm.

So, its power = \(P=\frac{100}{f_{\mathrm{cm}}}=\frac{100}{50} \mathrm{D}=+2 \mathrm{D}\)

Question 49. An astronomical telescope has its objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance of

  1. 46.0 cm
  2. 50.0 cm
  3. 54.0 cm
  4. 37.3 cm

Answer: 3. 54.0 cm

Given, f0 = 40 cm, fe = 4 cm.

Distance of the object AB from objective = u0 = -200 cm.

The image formed by the object is PQ which is formed at the focus of the eyepiece so that the final image is at infinity (for normal adjustment).

Now, \(\frac{1}{v_{\mathrm{o}}}-\frac{1}{u_{\mathrm{o}}}=\frac{1}{f_{\mathrm{o}}}\)

⇒ \(\frac{1}{v_{\mathrm{o}}}=\frac{1}{u_{\mathrm{o}}}+\frac{1}{f_{\mathrm{o}}}=\frac{1}{(-200 \mathrm{~cm})}+\frac{1}{40 \mathrm{~cm}}=\frac{1}{50 \mathrm{~cm}}\)

⇒ v0 = 50 cm.

Hence, the separation between the two lenses,

OQ + QE = 50 cm + 4 cm = 54 cm.

Ray Optics Multiple Choice Questions And Answers Q49

Question 50. In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the inner part of the objective lens. The eyepiece forms a real image of this line. The length of this line is l. The magnification of the telescope is

  1. \(\frac{L}{l}+1\)
  2. \(\frac{L}{l}-1\)
  3. \(\frac{L+1}{L-1}\)
  4. \(\frac{L}{l}\)

Answer: 4. \(\frac{L}{l}\)

Magnification produced by the eyepiece is

⇒ \(m_{\mathrm{e}}=\frac{\text { length of image }}{\text { length of object }}=\frac{l}{L}\)

Applying the lens formula,

⇒ \(\frac{1}{v_{\mathrm{e}}}-\frac{1}{\left(-u_{\mathrm{e}}\right)}=\frac{1}{f_{\mathrm{e}}} \Rightarrow \frac{u_{\mathrm{e}}}{v_{\mathrm{e}}}+1=\frac{u_{\mathrm{e}}}{f_{\mathrm{e}}}\)

the telescope is set for a normal adjustment,

\(u_{\mathrm{e}}=f_{\mathrm{o}}+f_{\mathrm{e}}\)

⇒ \(\frac{1}{m_{\mathrm{e}}}=\frac{f_{\mathrm{o}}+f_{\mathrm{e}}}{f_{\mathrm{e}}}-1=\frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}\)

⇒ \(m_{\mathrm{e}}=\frac{l}{L}=\frac{f_{\mathrm{e}}}{f_{\mathrm{o}}} \Rightarrow \frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}=\frac{L}{l}\) (1)

But magnification of telescope (in normal adjustment) is \(\frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}=\frac{L}{l}\) [from (1)].

Ray Optics Multiple Choice Questions And Answers Telescope Is Set For Normal Adjustment Q50

Question 51. If the focal length of the objective lens is increased then the magnifying power(s) of

  1. A microscope will increase but that of a telescope will decrease
  2. Both a microscope and a telescope will increase
  3. Both a microscope and a telescope will decrease
  4. A microscope will decrease but that of a telescope will increase

Answer: 4. A microscope will decrease but that of a telescope will increase

The magnifying power of a microscope is m = \(\frac{L}{f_{\mathrm{o}}} \frac{D}{f_{\mathrm{e}}}\), thus \(m \propto \frac{1}{f_0}\).

Hence, an increase in f0 will decrease the magnifying power of the microscope.

For a telescope, \(m=\frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}\), hence \(m \propto f_0\).

An increase in f0 will increase the magnifying power of the telescope.

Question 52. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is 20 cm. The focal lengths of the lenses are

  1. 10 cm and 10 cm
  2. 15 cm and 5 cm
  3. 18 cm and 2 cm
  4. 11 cm and 9 cm

Answer: 3. 18 cm and 2 cm

For a telescope in normal adjustment, tube length = L = f0 + fe = 20 cm

and magnifying power = \(\frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}\) = 9, so f0 = 9fe.

fe + 9fe= 20 cm ⇒ fe = 2 cm

and f0 = 9fe = 9(2 cm) = 18 cm.

Question 53. An astronomical telescope of tenfold angular magnification has a length of 44 cm. The focal length of the objective is

  1. 440 cm
  2. 44 cm
  3. 40 cm
  4. 4 cm

Answer: 3. 40 cm

Given, angular magnification,

⇒ \(M=\frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}=10 \text {, so } f_{\mathrm{o}}=10 f_{\mathrm{e}}\)

and tube length,

L = f0 + fe = 44 cm

⇒ fe + 10fe = 44 cm ⇒ fe = 4 cm.

A focal length of objective = f0 = 10fe = 40 cm.

Question 54. The astronomical telescope consists of an objective and an eyepiece. The focal length of the objective is

  1. Shorter than that of the eyepiece
  2. Greater than that of the eyepiece
  3. Equal to that of the eyepiece
  4. Five times shorter than that of the eyepiece

Answer: 2. Greater than that of the eyepiece

In an astronomical telescope the focal length of the objective (f0) is greater than that of the eyepiece (fe), i.e.,f0 > fe.

This produces higher magnification = m = \(\frac{f_0}{f_e}\).

Question 55. A telescope has an objective lens of focal length 200 cm and an eyepiece of focal length 2 cm. If this telescope is used to see a 50-meter-high building at a distance of 2 km, what is the height of the image of the building formed by the objective lens?

  1. 5 cm
  2. 10 cm
  3. 1 cm
  4. 2 cm

Answer: 1. 5 cm

For the objective lens,

focal length is f0 = 200 cm = 2 m;

object distance is u0= 2 km = 2000 m;

height of the object is h0= 50 m.

From the lens formula,

⇒ \(\frac{1}{v_0}-\frac{1}{\left(-u_0\right)}=\frac{1}{f_0}\)

Multiplying by u0 throughout

⇒ \(\frac{u_0}{v_0}+\frac{u_{\mathrm{o}}}{u_{\mathrm{o}}}=\frac{u_{\mathrm{o}}}{f_0}\)

⇒ \(\frac{1}{m_0}=\frac{u_0}{f_0}-1=\frac{2000 \mathrm{~cm}}{2 \mathrm{~cm}}-1=999\)

⇒ \(m_{\mathrm{o}}=\frac{1}{999}=\frac{h_1}{h_{\mathrm{o}}}=\frac{h_{\mathrm{I}}}{50 \mathrm{~m}}\)

height of the image formed by the objective,

∴ \(h_{\mathrm{I}}=\frac{50 \mathrm{~m}}{999} \approx \frac{50 \mathrm{~m}}{1000}=5 \mathrm{~cm}\).

Question 56. The focal lengths of the objective and the eyepiece of a microscope are 1.6 cm and 2.5 cm respectively. The distance between the two lenses is 21.7 cm. If the final image is formed at infinity, what is the linear magnification?

  1. 11
  2. 110
  3. 1.1
  4. 44

Answer: 2. 110

Given, that f0 =1.6 cm, fe = 2.5 cm, the distance between the lenses is OE = v0 + fe = 21.7 cm.

Since the final image is formed at infinity, the position of image PQ formed by the objective must be situated at the focus of the eyepiece, so ue = QE = fe = 2.5 cm.

Hence, the image distance for the objective,

v0 = OQ = 21.7 cm- 2.5 cm =19.2 cm.

Now, \(\frac{1}{v_0}-\frac{1}{\left(-u_0\right)}=\frac{1}{f_0}\)

⇒ \(\frac{v_0}{v_0}+\frac{v_0}{u_0}=\frac{v_0}{f_0}\)

So, magnification due to objective,

⇒ \(m_{\mathrm{o}}=\frac{v_{\mathrm{o}}}{u_{\mathrm{o}}}=\frac{v_{\mathrm{o}}}{f_{\mathrm{o}}}-1=\frac{19.2 \mathrm{~cm}}{1.6 \mathrm{~cm}}-1=11\)

The magnification when the final image is formed at infinity is given by

⇒\(m=\frac{v_{\mathrm{o}}}{u_{\mathrm{o}}}\left(\frac{D}{f_{\mathrm{e}}}\right)=11\left(\frac{25 \mathrm{~cm}}{2.5 \mathrm{~cm}}\right)=110\).

Ray Optics Multiple Choice Questions And Answers Focus Of Eyepiece Q56

Question 57. The magnifying power of a compound microscope increases a

  1. The focal length of the objective lens is increased and that of the eye lens is decreased
  2. The focal length of the eyelets is increased and that of the objective is decreased
  3. The focal lengths of both the objective and the eyepiece are increased
  4. The focal lengths of both the objective and the eyepiece are decreased

Answer: 4. The focal lengths of both the objective and the eyepiece are decreased

The magnifying power of a compound microscope is given by

⇒ \(M=\frac{v_{\mathrm{o}}}{u_{\mathrm{o}}}\left(1+\frac{\dot{D}}{f_{\mathrm{e}}}\right)=\frac{L}{f_{\mathrm{o}}}\left(1+\frac{D}{f_{\mathrm{e}}}\right)\)

Thus, M increases when both f0 and fe are decreased.

Question 58. The focal lengths of the objective and the eyepiece of a telescope are 200 cm and 4 cm respectively. What is the length of the telescope for normal adjustment?

  1. 196 cm
  2. 204 cm
  3. 250 cm
  4. 225 cm

Answer: 2. 204 cm

The tube length or the length of a telescope is given by

L = f0+fe =200 cm + 4 cm = 204 cm.

Question 59. A beam of red, green, and blue lights is incident on a right-angled prism. The material of the prism for the above three lights is 1.39, 1.47, and 1.44 respectively. The prism will

Ray Optics Multiple Choice Questions And Answers Right Angle Prism Q59

  1. Separate the blue light from the red and green lights
  2. Separate all title three lights from one another
  3. Not separate title three lights at all
  4. Separate the red light from the green and blue lights

Answer: 4. Separate the red light from the green and blue lights

The refractive index for the given prism when total reflection just occurs is

⇒ \(\mu=\frac{1}{\sin \theta_c}=\frac{1}{\sin 45^{\circ}}=\sqrt{2}\)

The component of the light beam for which μ = 2 will undergo total internal reflection.

Given, μred = 1.39,

μgreen = 1.47 > √2

μblue = 1.44 > √2

Thus, the red light will emerge while the blue and green lights will be totally reflected and separated from the red ones.

Ray Optics Multiple Choice Questions And Answers Total Reflection Q59

Question 60. Two similar thin biconvex lenses of focal lengths f  each are kept coaxially in contact with each other such that the focal length of the combination is F1. When the space between the two lenses is filled with glycerine which has the same refractive index (μ =1.5) as that of glass, the equivalent focal length is F2.. The ratio F1/F2 will be

  1. 2:1
  2. 1:2
  3. 2:3
  4. 3:1

Answer: 2. 1:2

The focal length of each biconvex lens is given by

⇒ \(\frac{1}{f_1}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=(1.5-1) \frac{2}{R}=\frac{1}{R}\)

For the biconcave lens, the focal length is given by

⇒ \(\frac{1}{f_2}=(\mu-1)\left(\frac{1}{-R}-\frac{1}{+R}\right)=-\frac{1}{R}\)

For combination → (1),

⇒ \(\frac{1}{F_1}=\frac{1}{f_1}+\frac{1}{f_1}=\frac{2}{f_1}=\frac{2}{R}\)

For combination → (2),

⇒ \(\frac{1}{F_2}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_1}\)

⇒ \(\frac{2}{f_1}+\frac{1}{f_2}=\frac{2}{R}-\frac{1}{R}=\frac{1}{R}\)

F2 = R

⇒ \(\frac{F_1}{F_2}=\frac{R / 2}{R}=\frac{1}{2}\)

Ray Optics Multiple Choice Questions And Answers Equiconcave Lens Q60(1)

Ray Optics Multiple Choice Questions And Answers Equiconcave Lens Q60(2)

Question 61. Pick the incorrect option in the context of a rainbow.

  1. When the light rays undergo two internal reflections in a water droplet, a secondary rainbow is formed.
  2. The sequence of colors is reversed in the secondary rainbow.
  3. An observer can see a rainbow when he is facing the sun.
  4. A rainbow is a combined effect of refraction and reflection of sun rays.

Answer: 3. An observer can see a rainbow when he is facing the sun.

We cannot see the rainbow while facing the sun. Rainbow is due to refraction, reflection, and dispersion of light from water droplets reaching our eyes from the opposite side.

Question 62. In a total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be the angle of refraction?

  1. 180°
  2. 90°
  3. Equal to the angle of incidence

Answer: 3. 90°

The critical angle for a given pair of media is the angle of incidence for which the corresponding angle of refraction is 90°.

Ray Optics Multiple Choice Questions And Answers Angle Of Incident Q62

Question 63. A person wears normal glasses of -5 D keeping the distance of glasses from the eyes at 2 cm. If he uses contact lenses then the required power of the lens will be

  1. +4.5 D
  2. -5.2 D
  3. -4.54 D
  4. +5.2 D

Answer: 3. -4.54 D

The focal length f of the tire concave lens is

⇒ \(\frac{1}{f}=-5 \mathrm{D} \Rightarrow f=-20 \mathrm{~cm}\)

Distance OF = 20cm + 2cm = 22 cm.

With the contact lens, OF = 22 cm, the final image is formed at the retina.

focal length of the contact lens used is fC = OF = -22 cm.

required power,

∴ \(P=-\frac{100}{22} \mathrm{D}=-4.54 \mathrm{D}\)

Ray Optics Multiple Choice Questions And Answers Focal Length Of Concave Lens Q63

Ray Optics Multiple Choice Questions And Answers Focal Length Of Concave Lens Q63(1)

Question 64. How can we change a camera from f/4 to f/5.6?

  1. By increasing the aperture two times keeping the focal length constant
  2. By increasing the aperture √2 times keeping the focal length constant
  3. By reducing the aperture to half keeping the focal length constant
  4. By reducing the aperture to l/√2 times keeping the focal length constant

Answer: 4. By reducing the aperture to l/√2 times keeping the focal length constant

The f-number of a camera is defined as \(\frac{\text { focal length }}{\text { diameter of aperture }}\)

Thus for an \(\frac{f}{4}\) camera, the F-number is 4, so 4 = \(\frac{f}{D_1}\)

For an \(\frac{f}{5.6}\) camera, the f-number is 5.6, so 5.6 = \(\frac{f}{D_2}\)

Hence, \(\frac{D_2}{D_1}=\frac{4}{5.6}=\frac{1}{1.4}=\frac{1}{\sqrt{2}} \Rightarrow D_2=\frac{D_1}{\sqrt{2}}\).

Question 65. One plano-convex and one plano-concave lens of the same radius of curvature R but of different materials are joined side by side as shown in the figure. The refractive indices of the materials are (μ1 and μ2)and the focal length of the combination is

Ray Optics Multiple Choice Questions And Answers Plano Convex And Concave Lenses Q65

  1. \(\frac{R}{2-\left(\mu_1-\mu_2\right)}\)
  2. \(\frac{2 R}{\mu_1-\mu_2}\)
  3. \(\frac{R}{2\left(\mu_1-\mu_2\right)}\)
  4. \(\frac{R}{\mu_1-\mu_2}\)

Answer: 4. \(\frac{R}{\mu_1-\mu_2}\)

For a convex lens,

⇒ \(\frac{1}{f_1}=\left(\mu_1-1\right)\left(\frac{1}{\infty}-\frac{1}{-R}\right)=\frac{\left(\mu_1-1\right)}{R}\)

For a concave lens,

⇒ \(\frac{1}{f_2}=\left(\mu_2-1\right)\left(\frac{1}{-R}-\frac{1}{\infty}\right)=-\frac{\left(\mu_2-1\right)}{R}\)

If F is the equivalent focal length of the combination then

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{R}\left(\mu_1-\mu_2\right)\)

∴ \(F=\frac{R}{\mu_1-\mu_2}\).

Question 66. A ray of light AO in a vacuum is incident on a glass slab at 60° and gets refracted at 30° along OB as shown in the figure. The optical path length of light rays from A to B is

Ray Optics Multiple Choice Questions And Answers Vacuum Is Incident In Glass Slab Q66

  1. 2a + 2b
  2. \(\frac{2 \sqrt{3}}{a}+2 b\)
  3. \(2 a+\frac{2 b}{3}\)
  4. \(2 a+\frac{2 b}{\sqrt{3}}\)

Answer: 1. 2a + 2b

⇒ \(\mathrm{AO}+\mu \cdot \mathrm{OB}=\frac{a}{\cos 60^{\circ}}+\left(\frac{\sin 60^{\circ}}{\sin 30^{\circ}}\right) \frac{b}{\cos 30^{\circ}}\)

= 2a + 2b.

Ray Optics Multiple Choice Questions And Answers Optical Path Q66

Question 67. An upright object is placed at a distance of 40 cm in front of a converging lens of focal length 20 cm. A converging mirror Of a focal length of 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be

  1. 40 cm from the converging mirror, the same size as the object
  2. 20 cm from the converging mirror, the same size as the size of the object
  3. 20 cm from the converging mirror, twice the size of the object
  4. 40 cm from the converging lens, twice the size of the object

Answer: 2. 20 cm from the converging mirror, the same size as the size of the object

The situation is shown in the diagram shown below.

Object PQ is placed at 40 cm from the convex lens which is also equal to 2f(= 2 x 20 cm) of the lens. Thus, a real, inverted image P’Q’ is also formed at 2f(= 40 cm) on the other side of the lens. The distance between P’Q’ and the converging mirror is 20 cm (= 2 x 10 cm) which is the position of its centre of curvature. Hence, a real and inverted image of P’Q’ of the same size will be formed at P’Q’ itself.

Ray Optics Multiple Choice Questions And Answers Q67

Question 68. A convex lens is placed 10 cm away from a light source and a sharp image is obtained on a screen 10 cm from the lens. Now a glass slab (refractive index = 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then, d is

  1. Zero
  2. 1.1 cm away from the lens
  3. 0.55 cm away from the lens
  4. 0.55 cm towards the lens

Answer: 3. 0.55 cm away from the lens

For the given convex lens,

⇒ \(|u|=|v|=10 \mathrm{~cm}=2 f \Rightarrow f=5 \mathrm{~cm}\)

When a glass slab (thickness t =1.5 cm, p = 1.5) is placed near the light source, the image of the source undergoes a shift

⇒ \(x=\left(1-\frac{1}{\mu}\right) t=\left(1-\frac{2}{3}\right) \frac{3}{2} \mathrm{~cm}=\frac{1}{2} \mathrm{~cm} \text { towards the lens }\)

Thus, u’ =10 cm-0.5 cm = 9.5 cm.

The corresponding image distance v’ is given by

⇒ \(\frac{1}{v^{\prime}}-\frac{1}{u^{\prime}}=\frac{1}{f}\)

⇒ \(\frac{1}{v^{\prime}}-\frac{1}{(-9.5 \mathrm{~cm})}=\frac{1}{5 \mathrm{~cm}}\)

⇒ \(\frac{1}{v^{\prime}}=\left(\frac{1}{5 \mathrm{~cm}}-\frac{1}{9.5 \mathrm{~cm}}\right) \Rightarrow v^{\prime}=10.55 \mathrm{~cm}\)

Thus, the screen has to be shifted by

(v’-v)- (10.55 cm-10 cm) = 0.55 cm

away from the lens to get a sharp image again

Question 69. A plano-convex lens of refractive index μ1 and focal length f1 is kept in contact with another plano-convex lens of refractive index μ2 and focal length f2. If the radius of curvature of their spherical faces is R each and \(f_1=2\left|f_2\right|\) then μ1 and μ2 are related as

  1. 2 – 2μ1 = 1
  2. 2 – μ1 =1
  3. 1 – μ2 = 1
  4. μ1 + μ2 = 3

Answer: 3. 2μ1 – μ2 = 1

For a plano-convex lens,

⇒ \(\frac{1}{f_1}=\left(\mu_1-1\right) \frac{1}{R} \text { and } \frac{1}{f_2}=\left(\mu_2-1\right) \frac{1}{R}\)

Since \(f_1=2 f_2 \text {, hence } \frac{\left(\mu_1-1\right)}{R}=\frac{\left(\mu_2-1\right)}{2 R}\)

⇒ 2μ1 – 2 = μ2 -1 ⇒ 2μ1 – μ2 =1.

Question 70. Two plane mirrors M1 and M2 are inclined to each other such that a ray of light incident on mirror M1 and parallel to mirror M2 is reflected from mirror M2 so as to be parallel to mirror M1. The angle between the two mirrors is

  1. 45°
  2. 75°
  3. 30°
  4. 60°

Answer: 4. 60°

The situation given in the question is shown by the diagram. Since AB is parallel to OC, the three angles are each equal to θ. Similarly, the CD is parallel to OB, and the angles θ are equal. In ΔOBC, 3θ =180°.

The angle between the two mirrors is \(\theta=\frac{180^{\circ}}{3}=60^{\circ}\)

Ray Optics Multiple Choice Questions And Answers Angle Betwwen Two Mirrors Q70

Question 71. The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea (7.8 mm). This surface separates two media of refractive indices 1.0 and 1.34. Calculate the distance from the refracting surface at which a parallel of light will come to focus.

  1. 0.5 cm
  2. 2.0 cm
  3. 3.1 cm
  4. 1 cm

Answer: 3. 3.1 cm

Given, μ1 =1, μ2 = 1.34 and R = +7.8 mm

Parallel rays from infinity converge at I, where PI = v.

For refraction through a spherical surface

⇒ \(\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}\)

⇒ \(\frac{1.34}{v}-\frac{1}{\infty}=\frac{1.34-1}{7.8 \mathrm{~mm}}\)

∴ Hence, \(v=\frac{1.34 \times 7.8 \mathrm{~nm}}{0.34}=30.74 \mathrm{~mm} \approx 3.1 \mathrm{~cm}\).

Ray Optics Multiple Choice Questions And Answers Parallel Rays Q71

Question 72. A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is √3, the angle of incidence is

  1. 45°
  2. 30°
  3. 90°
  4. 60°

Answer: 4. 60°

For a prism,

⇒ \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\) → (1)

Given, A = 60°.

For deviation to be minimum, i = i’, r = r’.

But A = 60° = r + r’ = 2r => r = 30°.

And i +i’ = A + δm => 2i = A+ δm . → (2)

Substituting (2) in (1),

⇒ \(\sqrt{3}=\frac{\sin i}{\sin 30^{\circ}}=2 \sin i\)

⇒ \(\sin i=\frac{\sqrt{3}}{2}=\sin 60^{\circ} \Rightarrow i=60^{\circ}\).

Question 73. A point source of light S is placed at a distance L in front of the center of a plane mirror of width d which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L as shown in the figure. The distance over which the man can see the image of the light source in the mirror is

Ray Optics Multiple Choice Questions And Answers Source Of Light Q73

 

  1. \(\frac{d}{2}\)
  2. 2d
  3. d
  4. 3d

Answer: 4. 3d

In the given diagram, the distance of light source S from the mirror AB is OS = L = SP ⇒ IP = 3L,

where I is the image of S in the plane mirror.

The observer moving along CD can see the image till the rays reach his eyes at C and D after reflection from the extreme points of the mirror. Thus, from similar triangles IAB and ICD,

∴ \(\frac{\mathrm{AB}}{\mathrm{CD}}=\frac{\mathrm{IO}}{\mathrm{IP}} \Rightarrow \frac{d}{D}=\frac{L}{3 L} \Rightarrow D=3 d\)

Ray Optics Multiple Choice Questions And Answers Plane Mirror Q73

Question 74. The formation of a real image using a biconvex lens is shown in the ray diagram. If the whole set-up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen?

Ray Optics Multiple Choice Questions And Answers Bioconvex Lens Q74

  1. No change
  2. Erect and real image
  3. Magnified image
  4. Image disappears

Answer: 4. Image disappears

The focal length of a lens depends on the refractive index of the material of the lens as well as on the refractive index of the surrounding medium.

When in the air,

⇒ \(\frac{1}{f_{\mathrm{a}}}=\left({ }^{\mathrm{a}} \mu_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

and when immersed in water,

⇒ \(\frac{1}{f_{\mathrm{w}}}=\left({ }^{\mathrm{w}} \mu_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\).

A change in focal length will cause the disappearance of the image from the screen.

Question 75. A ray of light passes through four transparent media with refractive indices μ1, μ2, μ3, and μ4 as shown in the figure. The surfaces of all the media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

Ray Optics Multiple Choice Questions And Answers Four Transparent Media Q75

  1. μ1 = μ2
  2. μ2 = μ3
  3. μ3 = μ4
  4. μ4 = μ1

Answer: 4. μ4 = μ1

According to Snells’ law, the incident and emergent rays through a slab will be parallel provided the media on both sides are identical. This follows for μ4 = μ1

Question 76. A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is placed at two distances x1 and x2 (x1 > x2) from the lens. The ratio of x1 and x2 is

  1. 4:3
  2. 3:1
  3. 5:3
  4. 2:1

Answer: 2. 3:1

When a real image is formed by a convex lens, v is +ve while u is -ve.

For magnification \(m=2, \frac{|v|}{|u|}=2\)

From the lens formula,

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{2 x_1}-\frac{1}{\left(-x_1\right)}=\frac{1}{f}\)

⇒ \(x_1=\frac{3}{2} f\).

When a virtual image is formed, u and v both are negative. Thus,

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{\left(-2 x_2\right)}-\frac{1}{\left(-x_2\right)}=\frac{1}{f}\)

⇒ \(x_2=\frac{f}{2}\)

Hence, \(\frac{x_1}{x_2}=3=3: 1\).

Question 77. A thin convex lens L (refractive index A’ 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself as shown in the figure. When a liquid of refractive index μ1 is put in between the lens and the mirror, the pin has to be moved to A’ such that OA’ = 27 cm to get its real inverted image at A’ itself. The value of μ1 will be

Ray Optics Multiple Choice Questions And Answers Thin Convex Lens Q77

  1. 3
  2. 2
  3. \(\frac{4}{3}\)
  4. \(\frac{3}{2}\)

Answer: 3. \(\frac{4}{3}\)

When the pin at A coincides with its own image, OA = f, the focal length of the convex lens, for which

⇒ \(\frac{1}{f}=\left(\mu_{\mathrm{g}}-1\right)\left\{\frac{1}{R}-\frac{1}{(-R)}\right\}=(1.5-1)\left(\frac{2}{R}\right)=\frac{1}{R}\)

f = R = OA =18 cm.

When the space between the lens and the plane mirror is filled with a liquid (refractive index \it), the liquid lens will be of focal length flt for which

⇒ \(\frac{1}{f_l}=\left(\mu_l-1\right)\left\{\frac{1}{R}-\frac{1}{\infty}\right\}=\frac{\mu_l-1}{-18 \mathrm{~cm}}\)

For the two lenses in contact, the equivalent focal length, F = OA’ = 27 cm.

⇒ \(\frac{1}{27 \mathrm{~cm}}=\frac{1}{18 \mathrm{~cm}}+\frac{\mu_l-1}{-18 \mathrm{~cm}} \Rightarrow \frac{1}{27}=\frac{1}{18}-\frac{\mu_l-1}{18}\)

⇒ \(\frac{1}{3}=\frac{1}{2}-\frac{\mu_l-1}{2} \Rightarrow \mu_l=\frac{4}{3}\)

∴ \(\mu_l=1+\frac{9}{27}=\frac{4}{3}\).

Question 78. The image of an object placed beyond the center of curvature of a concave mirror will be

Ray Optics Multiple Choice Questions And Answers Beyond The Centre Of Curvature Of A Concave Mirror Q78

  1. Virtual, diminished, and inverted
  2. Virtual, magnified, and inverted
  3. Reed, magnified and inverted
  4. Real, diminished, and inverted

Answer: 4. Real, diminished, and inverted

The image of an object placed beyond C of a concave mirror is

  1. Diminished,
  2. Inverted,
  3. Real and
  4. Lies between C and F, as shown in this ray diagram.

Object: AB.

Image: A’B’.

Ray Optics Multiple Choice Questions And Answers Q78

Question 79. A ray is incident at an angle of 60° on the surface of a transparent solid sphere of refractive index √3 at A. At Air Point B, there is partial reflection and emergence as shown. The angle θ between the reflected and emergent rays is

Ray Optics Multiple Choice Questions And Answers Solid Sphere Of Refractive Index Q79

  1. 60°
  2. 90°
  3. 80°
  4. 70°

Answer: 2. 90°

For refraction at A,

⇒ \(\mu=\frac{\sin i}{\sin r} \Rightarrow \sqrt{3}=\frac{\sin 60^{\circ}}{\sin r}=\frac{\sqrt{3}}{2 \sin r} \Rightarrow r=30^{\circ}\)

So, the angle of emergence ∠NBC = 60°.

The angle of reflection at B is

∠OBD = 30°.

= 180°- (60° + 30°) = 90°.

Ray Optics Multiple Choice Questions And Answers Angle Of Refraction Q79

Question 80. A compound microscope forms the final image of an object at 25 cm from the eyepiece. What is the focal length of the eyepiece if the focal length of the objective is 1 cm, the magnification produced is 100 and the tube length is 20 cm?

  1. 525 cm
  2. 6.25 cm
  3. 2.25 cm
  4. 4.25 cm

Answer: 2. 6.25 cm

For a compound microscope, magnification is

⇒ \(M=\frac{v_{\mathrm{o}}}{u_{\mathrm{o}}}\left(1+\frac{D}{f_{\mathrm{e}}}\right) \approx \frac{L}{f_{\mathrm{o}}}\left(1+\frac{D}{f_{\mathrm{e}}}\right)\)

⇒ \(100=\frac{20 \mathrm{~cm}}{1 \mathrm{~cm}}\left(1+\frac{25 \mathrm{~cm}}{f_{\mathrm{e}}}\right) \Rightarrow 5=1+\frac{25 \mathrm{~cm}}{f_{\mathrm{e}}}\)

⇒ \(f_{\mathrm{e}}=\frac{25 \mathrm{~cm}}{4}=6.25 \mathrm{~cm}\).

Question 81. In the displacement method for measuring the focal length of a convex lens, the distance between the object and the screen is 100 cm. The separation between the two positions of the lens for image formation was 40 cm. Find N, if the power of the lens is 100/N dioptre.

  1. 15
  2. 40
  3. 21
  4. 25

Answer: 3. 21

Focal length = \(f=\frac{D^2-x^2}{4 D}=\frac{(100 \mathrm{~cm})^2-(40 \mathrm{~cm})^2}{(400 \mathrm{~cm})}=21 \mathrm{~cm}\)

∴ Power: \(P=\frac{100}{f_{\mathrm{cm}}} \mathrm{D}=\frac{100}{N} \mathrm{D} \Rightarrow N=f=21\)

Question 82. An equiconvex lens of radius R has power P. A plano-convex lens with a radius of curvature R’ and made of the same material as power 1.5P. The value of R’ is

  1. \(\frac{3}{2} R\)
  2. 2R
  3. \(\frac{R}{2}\)
  4. \(\frac{R}{3}\)

Answer: 4. \(\frac{R}{3}\)

For the equiconvex lens,

⇒ \(P=\frac{1}{f}=(\mu-1)\left\{\frac{1}{R}-\frac{1}{(-R)}\right\}=(\mu-1) \frac{2}{R}\)

For the plano-convex lens

⇒ \(P^{\prime}=\frac{3}{2} P=(\mu-1)\left(\frac{1}{R^{\prime}}-\frac{1}{\infty}\right)=\frac{(\mu-1)}{R^{\prime}}\)

∴ \(\frac{3}{2}(\mu-1) \frac{2}{R}=\frac{(\mu-1)}{R^{\prime}} \Rightarrow R^{\prime}=\frac{R}{3}\)