Model Paper Question And Answers
Physics Model Paper Question 1.
- A carbon resistor is Coloured with four different bands red, green, orange and silver respectively. Find the range of its probable resistance.
- The EMF of an electrical cell is 2 volts. A 10Ω resistance is joined at its two ends the potential difference is measured at 1.6 volts. Find out the internal resistance and lost volt,
Answer:
1. Red: 2, Green: 5, Orange: 3, zeroes, Silver: 10% error
∴ Resistance = 25000 ±10% Ω
10% of 25000 Ω = 250011Ω
∴ Range ofthe resistance
= (25000-2500) to (25000 + 2500)
= 22500 Ω to 27500 Ω
2. Lost volt = 2-1.6 = 0.4 V
Current through this circuit I = \(\frac{1.6 \mathrm{~V}}{10 \Omega}\)
= 0.16 A
Physics Model Paper
Model Paper Solutions
Internal resistance, r = \(\frac{I r}{I}=\frac{\text { lost volt }}{I}\)
= \(\frac{0.4 \mathrm{~V}}{0.16 \mathrm{~A}}\)
= 2.5 Ω
Question 2. A copper wire of length l metre is bent to form a circular loop. If I amp current flows through the loop, find out the magnitude of the magnetic moment ofthe loop.
Answer:
Circumference = 2πr = l
Or , r= \(\frac{l}{2 \pi} \mathrm{m}\)
So, the area of the loop, A = πr²
= \(\pi\left(\frac{l}{2 \pi}\right)^2=\frac{l^2}{4 \pi} \mathrm{m}^2\)
Magnitude of magnetic moment pm = \(i A=\frac{i l^2}{4 \pi} A\)
Question 3. How many α and β -particles are emitted when U238 changes to Pb206 due to radioactivity? Atomic numbers of U238 and Pb206 are 92 and 82 respectively.
Answer:
Let, x = number of α -particles emitted
y = number of β -particles emitted
Decrease in mass number = 4x
Decrease in atomic number = (2x-y)
Here, 4x = 238- 206 = 32 or, x = 8
And, 2x-y = 92-82 = 10
y = 2x- 10 = 2 × 8- 10
= 6
Model Paper Solutions
Question 4. A TV tower is 120 m high. How much more height is to be added to it if its coverage range is to double?
Answer:
Coverage range, d = \(\sqrt{2 h R}\) where h = height of the TV tower
As R = radius ofthe earth = constant,d ∝ √h, So d becomes double when h becomes 4 times.
New height ofthe tower =4 × 120 = 480m
Increase in height = (480-120) = 360m
Question 5. Define electrical dipole moment. An electrical dipole is placed within a uniform electric field (E) and is rotated to an angle ∠θ = 180°. Find out the work done
Answer:
Two equal and opposite charges ±q, separated by a distance vector l directed from -q to +q, form an electric dipole. Its dipole moment is \(\vec{p}=q \vec{l}\)
Torque on an electric dipole in an electric field \(\vec{E}\) is, \(\vec{\tau}=\vec{p} \times \vec{E}\)
Its magnitude is, r = pEsind , where 6 = angle between p and \(\)
Work done to rotate the dipole from 0° to 180° is
W = \(\int d W=\int_0^{180} \tau d \theta\)
= \(=p E \int_0^{180} \sin \theta d \theta=p E[-\cos \theta]_0^{180}\)
= 2pE
Physics Model Paper
Model Paper Solutions
Question 6.
- On what factors does the capacitance of a capacitor depend
- Two capacitors of capacitances 20μF and 60μF are connected in series. If the potential difference between the two ends of the combination is 40 volts, calculate the terminal potential difference of each capacitor.
Answer:
1. The area of each plate, and the distance between the capacitor plates and the intermediate medium are the factors on which the capacitance depends
2. In series combination, the charge q on both the capacitors is the same. As V = q/c we have V ∝ 1/C. Here, the ratio between the capacitances = \(\frac{20}{60}\) = \(\frac{1}{3}\), so the ratio between the terminal potential differences \(\frac{3}{I}\)
So, the total voltage of 40V will be distributed as 30V and 10V
Question 7. What do you mean by the angle of dip at a place? At what place on the earth’s surface will the horizontal and vertical components of the earth’s magnetic field be equal?
Answer:
At a place where the angle ofdip is 45° , horizontal and vertical components are H = I cos45° = \(\frac{I}{\sqrt{2}}\)and V = I sin45° = \(\frac{I}{\sqrt{2}}\) . So , V = H
Question 8. In Young’s double-slit experiment, what is the effect on the interference pattern if
- The distance between the two slits is halved.
- The distance between the screen and the plane of limits is doubled.
- One ofthe slits is covered with translucent paper
Answer:
Fringe width, y = \(\frac{D \lambda}{2 d}\)
1. Distance 2d between the slits is halved, then y is doubled.
2. Distance D between the screen and the plane of the slits is doubled, and then again y is doubled.
3. Due to an increase in the optical path for rays passing through the covered slit, the fringe pattern will be displaced sideways.
Model Paper Solutions
Question 9.
- An object of height 2.5 cm is placed perpendicularly on the principal axis of a concave mirror of focal length f at a distance of ¾ f What will be the nature of the image of the object and its height?
Or,
- A person uses spectacles of power +2D
Answer:
1. For concave mirrors, the focal length f is -f, using the proper sign
Also u = – ¾
So the relation \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
Gives, \(\frac{1}{v}+\left(-\frac{4}{3 f}\right)=\frac{1}{-f}\)
2. \(\frac{1}{v}=\frac{4}{3 f}-\frac{1}{f}=\frac{1}{3 f}\)
Or, v = 3f
The positive sign of v means that the image is formed on the opposite side, so it is a virtual image.
Magnification = \(-\frac{v}{u}=-\frac{3 f}{-(3 f / 4)}\)
Image height =4 × 2.5 = 10cm
The lens is convex. So the defect is long sight or hypermetropia
Model Paper Solutions
Question 10.
- Under what potential difference should an electron be accelerated to obtain a de Broglie wavelength of 0.6 Å (h = . 6.62 × 10-34 J . s, me = 9.1 × 1-31 kg)
- Give an example of the production of electrons by photons.
- Light rays of wavelength λ and λ/2 are incident photosensitive metal surfaces.’ If the maximum kinetic energy of the emitted photoelectrons from the metal surface in 2nd -case is 3 times the maximum kinetic energy of emitted photoelectrons in the 1st case, then determine the work function of the metal.
Answer:
1. de Broglie wavelength \(\lambda=\frac{h}{p}\)
⇒ \(\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m e V}}\)
2 meV = \(\frac{h^2}{\lambda^2}\)
Or, = \(=\frac{h^2}{2 m e \lambda^2}\)
= \(\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times\left(9.1 \times 10^{-31}\right) \times\left(1.6 \times 10^{-19}\right)}\times\left(0.6 \times 10^{-10}\right)^2\)
= 418 V
2. Example: Photoelectric Effect
3. Emax = \(h f-W_0=\frac{h c}{\lambda}-W_0\)
1st case,\(E_{\max }=\frac{h c}{\lambda}-W_0\)
2nd case, \(3 E_{\max }=\frac{h c}{\lambda / 2}-W_0=\frac{2 h c}{\lambda}-W_0\)
Or, \(E_{\max }=\frac{2}{3} \frac{h c}{\lambda}-\frac{1}{3} W_0\)
⇒ \(\frac{h c}{\lambda}-W_0=\frac{2}{3} \frac{h c}{\lambda}-\frac{1}{3} W_0\)
\(\frac{h c}{\lambda}\left(1-\frac{2}{3}\right)=W_0\left(1-\frac{1}{3}\right)\)Or, \(W_0=\frac{\frac{1}{3} \frac{h c}{\lambda}}{2 / 3}=\frac{1}{2} \frac{h c}{\lambda}\)
Model Paper Solutions
Question 11. How is the characteristic X-ray spectrum formed?
Answer:
When high-energy cathode rays fall on the target of an X-ray tube, deep-rooted electrons of the target atoms may come out. If, say, a K-shell electron comes out of an atom, the vacancy is filled up by an electron of the immediately outer shells (L, M, …………). This electron transition produces an X-ray photon, whose frequency characterises the target atom. This is a characteristic X-ray.
Question 12. What are the majority and minority carriers in a p-type semiconductor?
Answer:
p -type semiconductor: majority carriers are holes and minority carriers are electrons.
Question 13. Convert the binary number 10011 into decimal equivalent
Answer:
(10011)2 =(1 × 24) + (0 × 23) + (0 × 22) + (1 × 21)+ (1 × 20)
= (16 + 0 + 0 + 2 + 1)
= (19)10
Model Paper Solutions
Question 14.
- In a potentiometer experiment, why is it necessary to use a long wire? The length and resistance of a potentiometer wire are 4 m and 10Ω respectively. It is connected to a cell of emf 2 volt. Another cell when joined to this potentiometer and null point is measured at 250 cm. Find out the emf ofthe second cell.
- In a metre bridge when the resistance in the left gap is 2Ω and an unknown resistance in the right gap, the balance point is obtained at 40 cm from zero end. On shunting the unknown resistance with 2Ω, find the shift of the balance point on the bridge.
- 36 cells each of internal resistance 0.5fl and emf 1.5 V each are used to send current through an external circuit of 2fl resistance. Find the best mode of grouping them for maximum current and the current through the external circuit,
Answer:
1. In different measurements using a potentiometer, the ratio 1/L often comes in the working formula. Here, L = length of the potentiometer wire, and l = length at which the null point is obtained.
Physics Model Paper
This ratio of 1/L is naturally accurate if both l and L are sufficiently large. So, a long wire is used in a potentiometer experiment. l = 250cm = 2.5m ; L = 4m ; emf ofthe potentiome¬ter cell, EQ = 2 V
So, the emf of the second cell = 1/L E0 = \(\frac{2.5}{4} \times 2\)
= 1.25 V
2. The unknown resistance (S) is on the right gap. So at the null point,
S = \(R \frac{100-l}{l}=2 \times \frac{100-40}{40}\)
= 3
Model Paper Solutions
If the balance is obtained at a distance l’ , then
⇒ \(S^{\prime}=R \frac{100-l^{\prime}}{l^{\prime}}=R\left(\frac{100}{l^{\prime}}-1\right)\)
⇒ \(\frac{100}{l^{\prime}}=1+\frac{S^{\prime}}{R}=\frac{R+S^{\prime}}{R}\)
= \(\frac{100 R}{R+S^{\prime}}=\frac{100 \times 2}{2+1.2}\)
= 62. 5 cm
Shift in balance point = 62.5 – 40 = 22.5cm
3.
Let, 36 cells are grouped in m rows, each row containing n cells. Then, mn = 36
Internal resistance in each row = nr; here, r = 0.5 × 1
Internal resistance for m rows = nr/m
Total resistance in the circuit = \(\left(R+\frac{n r}{m}\right)\): here
R = 2
Model Paper Solutions
The emf of each row = nE = effective emf of m rows;
here, E = 1.5 V
Current in an external circuit
I = \(\frac{n E}{R+\frac{n r}{m}}=\frac{m n E}{m R+n r}\)
⇒ \(\frac{m n E}{(\sqrt{m R}-\sqrt{n r})^2+2 \sqrt{m n R r}}\)
As mnR constant I is maximum \((\sqrt{m R}-\sqrt{n r})^2\) = 0
mR = nr
Or, \(\frac{36}{n} \times 2=n \times 0.5\)
= n x 0.5
Or, \(n^2=\frac{36 \times 2}{0.5}\) = 44
Then , n= 12 and m = \(\frac{36}{12}\)
= 3
So, the best mode of grouping is in 3 rows, each row having 12 cells
Question 15. Determine the resonance frequency ωr of a series LCR circuit with L = 2.0H, C = 32μF and R = 10 Ω. What is the Q -value of this circuit?
Answer:
L = 2.0 H, C = 32 μ F = 32 × 106 F , R = 10 × 1 Resonance frequency,
⇒ \(\omega_r=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{2 \times 32 \times 10^{-6}}}\)
= \(\frac{1000}{8}\)
= 125 Hz
Q = \(=\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{10} \sqrt{\frac{2}{32 \times 10^{-6}}}\)
= \(\frac{1}{10} \times \frac{1000}{4}\)
= 25
Model Paper Solutions
Question 16.
- Two convex lenses of focal lengths f1 and f2 respectively are placed in contact with each other. Then what will be the power of their equivalent lens?
- The refractive index of glass is 1.55. What is its polaris¬ ing angle? Determine the angle of refraction for the polarising angle
Answer:
1. Powers of the two lenses are P = \(=\frac{1}{f_1}\) and P = \(=\frac{1}{f_2}\)they are placed in contact, the power of the equivalent lens is
P = \(P_1+P_2=\frac{1}{f_1}+\frac{1}{f_2}\)
2. tan ip = n or, ip = tan-1(1.55) = 57.2°
The angle of refraction,
r = 90°- ip = 90°- 57.2°
= 32.8
Physics Model Paper
Question 17.
- After that time will the direction of current in an electric supply line of frequency 50 Hz be reversed?
- Which physical quantity, has unit Wb. m-2 it scalar or vector?
Answer:
1. Time = \(\frac{T}{2}=\frac{1}{2}\left(\frac{1}{50}\right)\)
= 0.01 s
2. A magnetic field, is a vector quantity
Model Paper Solutions
Question 18.
- How is the direction of a magnetic field \(\vec{B}\) at a point related to the magnetic line of force passing through that point?
- A long straight wire of length l is moving within a uniform magnetic field B with a velocity v perpendicular to the field. How much emf will induce?
Answer:
1. Magnetic field B acts in the direction of the line of force.
2. Induced emf = Blv
Question 19. At which temperature is a semiconductor completely transformed into an insulator?
Answer: 0K
Question 20.
- On what condition does a convex lens produce real images of an object on a screen at two positions ofthe lens?
- Sunray, sodium light and headlight of an automobile-which of these lights are polarised
Answer:
1. The distance (D) between the object and the screen should be greater than four times the focal length of (f) the convex lens. That is, D > 4f
2. None of the three is polarised
Multiple Choice Questions
Question 1. The dimension of electric potential is
- \(M L^2 T^2 A^{-1}\)
- \(M L^2 T^{-3} A^{-1}\)
- \(M L T^{-2} A^{-1}\)
- \(M L^2 T^3 A^{-1}\)
Answer: 2. \(M L^2 T^{-3} A^{-1}\)
Power, P = VI; so V = P/I
Dimension of V \(\frac{M L^2 \mathrm{~T}^{-3}}{\mathrm{~A}}=\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-1}\)
Model Paper Solutions
Question 2. A wire is stretched by 1% but volume remains constant, then
- Resistivity increases by 1%
- Resistance increases by 2%
- Resistivity decreases by 1%
- Resistance decreases by 2%
Answer: 2. Resistance increases by 2%
Physics Model Paper
Volume, V = αl = constant, where a = cross-sectional area.
Resistivity does not change due to elongation.
Resistance, R = \(\rho \frac{l}{\alpha}=\rho \frac{l^2}{\alpha l}=\rho \frac{l^2}{V}\)
As p and V are constants,
⇒ \(\frac{\Delta R}{R}=2 \frac{\Delta l}{l}=2 \times 1 \%\)
= 2%
Question 3. The energy of an electron revolving around the nucleus of an H-atom is 1.51 eV. The angular momentum of this electron will be
- \(\frac{h}{2 \pi}\)
- \(\frac{2 h}{2 \pi}\)
- \(\frac{3 h}{2 \pi}\)
- \(\frac{4 h}{2 \pi}\)
Answer: 3. \(\frac{3 h}{2 \pi}\)
For ground state (n = 1) , E1 = -13.6eV
Given, En = 1.51eV
∴ \(\frac{E_1}{E_n}=\frac{-13.6}{-1.51}\)
= 9
As \(E_n \propto \frac{1}{n^2}\) we have n
= 3
Angular momentum of this electron = \(3 \frac{h}{2 \pi}\)
Model Paper Solutions
Question 4. The angle of dip is 90°N at
- Magnetic south pole
- Magnetic north pole
- Geographic south pole
- Geographic north pole
Answer: 2. Magnetic north pole
The angle of dip is 90°N at the magnetic north pole
Question 5. The equation of an ac is i = 3sinωt + 4cosωt. Then rms value of this current will be
- \(\frac{3}{\sqrt{2}}\)
- \(\frac{4}{\sqrt{2}}\)
- \(\frac{7}{\sqrt{2}}\)
- \(\frac{5}{\sqrt{2}}\)
Answer: 4. \(\frac{5}{\sqrt{2}}\)
Physics Model Paper
i = \(3 \sin \omega t+4 \cos \omega t=5\left[\frac{3}{5} \sin \omega t+\frac{4}{5} \cos \omega t\right]\)
Since \(\sqrt{3^2+4^2}=5\)
= 5 \((\sin \omega t \cos \theta+\cos \omega t \sin \theta)\)
= \(\text { where, } \theta=\cos ^{-1} \frac{3}{5}=\sin ^{-1} \frac{4}{5}\)
= 5 sin(ωt+θ)
∴ Peak value, i0 = 5 unit; rms value = \(\frac{i_0}{\sqrt{2}}=\frac{5}{\sqrt{2}}\) unit
Question 6. A biconvex lens behaves like a convergent in air but behaves like a divergent in water. Then refractive index (μL) of the lens will be
- μL= 1
- μL > 1.33
- μL = 1-33
- 1<μL< 1.33
Answer: 4. 1<μL< 1.33
A biconvex lens is divergent when its refracting index is less than that of the surrounding medium
Model Paper Solutions
Question 7. The process in which the amplitude of the carrier wave is made proportional to the instantaneous amplitude of the signal wave is called
- Amplitude modulation
- Demodulation
- Rectification
- Amplification
Answer: 1. Amplitude modulation
Physics Model Paper
In the process of amplitude modulation, the amplitude of the carrier wave is made proportional to the instantaneous amplitude of the signal wave
Question 8. Two point charges separated by a distance d apart from each other with a repulsion force of 9 N. If the separation between them becomes 3d, the force of repulsion will
- 1N
- 3N
- 6N
- 27 N
Answer: 1. 1N
Force = \(\propto \frac{1}{d^2}\)
As the distance increases 3 times, the force decreases 9 times
Question 9. In the circuit AB = 6Ω, BC = 3HΩ, CD = 6Ω , DA = 12Ω, G = 10Ω. Current through galvanometer is
Physics Model Paper
- 8.7 mA
- 7.8 mA
- 8.7 A
- 0
Answer: 1.8.7 mA
As, \(\frac{6}{3}=\frac{12}{6}\) , the Wheatstone bridge is balanced
Model Paper Solutions
Question 10. +q point charge is placed at the centre of a hemispherical surface. The amount of electrical flux crossing through the surface will be
- \(\frac{q}{\epsilon_0}\)
- \(\frac{q}{2 \epsilon_0}\)
- \(\frac{q}{3 \epsilon_0}\)
- \(\frac{2 q}{\epsilon_0}\)
Answer: 2. \(\frac{q}{2 \epsilon_0}\)
Considering all directions, the total flux around the point charge = \(\frac{q}{\epsilon_0}\)
The hemispherical surface surrounds half the entire space. So the flux through the surface = \(\frac{1}{2} \frac{q}{\epsilon_0}\)
Question 11. An α -particle and a proton having the same momentum enter into a region of uniform magnetic field and move in circular paths. The ratio of the radii of curvature of their circular paths \(\frac{r_g}{r_p}\) in the field Is
- 1
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)
- 4
Answer: 3.\(\frac{1}{4}\)
The centripetal force for the rotation is supplied by the magnetic force on a moving particle.
Physics Model Paper
⇒ \(\frac{m v^2}{r}=q v B\)
Or, \(=\frac{m v^2}{q v B}=\frac{m v}{q B}\)
Hence for α -particle and proton, momentum mu and magnetic field B are the same.
⇒ \(\frac{r_\alpha}{r_p}=\frac{q_p}{q_\alpha}=\frac{+e}{+2 e}=\frac{1}{2}\)
Model Paper Solutions
Question 12. A straight conductor of length 0.5 m Is placed in a magnetic field \(\vec{B}\) = (2i + 4J) T . It carries a current 1 A along +ve x-axis. The magnitude and direction of force acting on the conductor respectively are
- 2N along +ve z-axis
- √18 N along +ve z-axis
- 4 N along +ve y-axis
- √2 along +ve x-axis
Answer: 1. 2N along +ve z-axis
As this current I = 1 A is flowing in the positive x direction, the length vector of the conductor is,
⇒ \(\vec{l}=0.5 \hat{i} \mathrm{~m}\)
∴ Force, F = \(\vec{F}=\overrightarrow{I l} \times \vec{B}=1(0.5 \hat{i}) \times(2 \hat{i}+4 \hat{j})=2 \hat{k N}\)
Question 13. Two waves, whose intensities are 9: 16 are made to interfere. The ratio of maximum and minimum intensi¬ ties in the interference pattern is
- 49: 1
- 49: 16
- 7:1
- 4: 3
Answer: 1. 49: 1
Intensity ratio = 9: 16, so amplitude ratio =3:4.
∴ Ratio of maximum and minimum amplitudes after interference = (4 + 3): (4- 3) = 7:1
∴ The ratio of maximum and minimum intensities = 49: 1
Model Paper Solutions
Question 14. If V1 increases from 2 V to 6 V then a change of current will be
- Zero
- 20 mA
- \(\frac{80}{3}\) mA
- 40 mA
Answer: 2. 20 mA
For V1, = 2 V (reverse bias), current = 0
Physics Model Paper
For V2= 6 V (forward bias),
Current = \(\frac{(6-3) V}{150 \Omega}=\frac{1}{50}\)
= 20 mA
Change of current = (20-0) = 20mA