WBCHSE Class 11 Physics Vector Question and Answers

Vector Long Answer Type Questions

Question 1. Under which condition will the magnitude of the scalar sum be equal to the magnitude of the vector sum?
Answer:

A scalar has only magnitude and therefore, the addition of scalars is the addition of magnitudes. When some vec¬tors have the same direction, then only their magnitudes are added in vector addition. In such a case, the magnitude of the vector sum is equal to the sum of the scalars.

Question 2. If \(|\vec{A}| \neq|\vec{B}|\), then is it possible that \(\vec{A}\) + \(\vec{B}\) =0? Explain.
Solution:

If \(\vec{A}\) + \(\vec{B}\) = 0 or, \(\vec{A}\) = –\(\vec{B}\); it means that \(\vec{A}\) and \(\vec{B}\) are equal in magnitude, but opposite in direction.

But it is given that \(|\vec{A}| \neq|\vec{B}|\), i.e., the magnitudes are not equal. So, \(\vec{A}\) + \(\vec{B}\) +0.

Understanding Vector Concepts Questions

Question 3. Can the sum of three vectors, i.e., their resultant, be equal to zero? Explain.
Answer:

Yes, if the resultant of any two of the vectors is equal and opposite to the third one, the resultant of the three vectors will be zero. Let a, b, c be three vectors related as \(\vec{a}, \vec{b}, \vec{c}\)

∴ \(\vec{a}+\vec{b}=-\vec{c} \text {. Then } \vec{a}+\vec{b}+\vec{c}=0\)

Question 4. Two stones are dropped at the same time from a height. The first is dropped from rest while the other is released with a small horizontal velocity. Which stone will reach the ground first?
Answer:

The initial velocity of the first stone is zero and that of the second stone is in the horizontal direction. Hence, the vertical component of the initial velocity of the second stone is also zero.

Consequently, the two stones will touch the ground at the same instant, but not at the same point. Due to an initial horizontal velocity, the second stone will have some horizontal displacement.

Question 5. A boy throws a ball vertically upward from a vehicle moving with a constant acceleration. Where would the ball land?
Answer:

The ball will fall behind the boy if he is facing the direction of acceleration.

• When the boy releases the ball, the horizontal velocities of both the ball and the vehicle are the same. The distance the ball covers after being released has nothing to do with its vertical velocity.
• However, the ball’s horizontal velocity remains the same, whereas that of the vehicle increases as it is moving with a constant acceleration. Therefore, the ball will travel a shorter distance than the vehicle will. Thus, it will land behind the boy.

WBBSE Class 11 Vector Q&A

Question 6. Can the value of a component of a vector be greater than the value of the vector Itself? Discuss the case of rectangular components in this context
Answer:

By the law of parallelogram of vectors, two components of a vector form two adjacent sides of a parallelogram with the vector as the diagonal. Since both the adjacent sides of a parallelogram can be longer than the diagonal, the value of the components can be greater than that of the vector.

But the rectangular components of a vector, being two adjacent sides of a rectangle, cannot be greater than the value of the vector, because the diagonal of a rectangle is always longer than any of the sides.

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Question 7. At zero wind speed, rainwater falls vertically with velocity V cm • s^-1 and is collected in a pot at a fixed rate. How will the rate of collection of rainwater change when the wind is blowing with a velocity of W cm s^-1 perpendicular to V?
Answer:

There will be no change in the rate of collection of rainwater due to wind velocity.

Rainwater falls vertically in the pot when there is no wind. If A cm² is the area of the cross-section of the pot, the rate of collection of rainwater is, x = AV cm³ · s^-1 …..(1)

Because of the wind, water will be filled in the pot obliquely. The component of the cross-sectional area of the pot perpendicular to the velocity of rain, is A₁ = A cosθ.

Here, the effective velocity of rain is \(V_1=\frac{V}{\cos \theta}\)

Hence, the rate of collection of water, \(x_1=A_1 \cdot V_1\)

= \(A \cos \theta \cdot \frac{V}{\cos \theta}=A V \mathrm{~cm}^3 \cdot \mathrm{s}^{-1}\)………(2)

Thus, from (1) and (2), it is inferred that the rate of collection of rainwater remains the same.

Question 8. Are the magnitudes of the two vectors (\(\vec{A}-\vec{B}\)) and (\(\vec{B}-\vec{A}\)) the same?
Answer:

The magnitudes of these two vectors are the same. But since \(\vec{B}-\vec{A}\) = –\(\vec{A}-\vec{B}\), their directions are opposite.

Question 9. Show that, if three forces acting on a particle can be taken sequentially to form the three sides of a trian¬gle, their resultant Is zero.
Answer:

Let \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) be the three forces acting at a point O. Now, taking these forces sequentially and keeping their magnitudes and directions the same, we obtain the three sides of a triangle, \(\overrightarrow{P Q}, \overrightarrow{Q R} \text { and } \overrightarrow{R P}\)

We have to show that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)

According to the triangle law of vector addition, \(\vec{a}+\vec{b}=\overrightarrow{P R}=-\overrightarrow{R P} \text { or, } \vec{a}+\vec{b}=-\vec{c}\)

or, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)

∴ The resultant of the three forces \(\vec{a}, \vec{b}, and \vec{c}\) is zero.

Applications of Vectors in Physics Q&A

Question 10. If the position coordinates of the points A and B are (x₁, y₁, z₁) and (x₂, y₂, z₂) respectively, determine the magnitude and direction of the vector \(\overrightarrow{A B}\).
Answer:

The position coordinates of A and B are (x₁, y₁, z₁) and (x₂, y₂, z₂) respectively.

∴ \(\overrightarrow{O A}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\) and \(\overrightarrow{O B}=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\)

From ΔOAB, we get, \(\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}\)

or \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}\)

= \(\left(x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\right)-\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right)=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)

The magnitude of the vector \(\overrightarrow{A B}\) is, AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

The direction of \(\overrightarrow{A B}\) is obtained from its direction cosines \(\frac{x_2-x_1}{A B}, \frac{y_2-y_1}{A B} \text { and } \frac{z_2-z_1}{A B} \text {. }\)

Question 11. A person leans out of a train moving with uniform velocity and drops a coin. How does the path of motion of the coin appear to a co-passenger and a person standing outside the train near the rail tracks?
Answer:

When the coin is dropped from the train, due to inertia of motion it has a horizontal component of velocity equal to that of the train. Simultaneously, it experiences a vertical downward acceleration due to the force of gravity. If the train’s velocity is v, the velocity of the co-passenger is also v. Therefore, with respect to this person, the horizontal component of the velocity of the coin is, v- v = 0.

So, the passenger will see the coin drop vertically downwards due to the action of gravity. With respect to any person standing outside the train near the rail tracks, the coin will have both horizontal and vertical movements. Thus, it is essentially a projectile motion, and the coin follows a parabolic path.

Vector Addition and Subtraction Q&A

Question 12. State whether any physical quantity having magni¬tude and direction is a vector.
Answer:

All physical quantities with magnitude and direction are not vectors. To be a vector, the physical quantity must obey the rules of vector addition. For example, the flow of current is not a vector quantity though it has both magni¬tude and direction.

Question 13. Can the magnitude of the resultant of two vectors be less than either of them? Explain.
Answer:

\(\vec{a}\) = \(\overrightarrow{A B}\) and \(\vec{b}\) = \(\overrightarrow{B C}\). If the resultant of these two vectors is \(\vec{c}\), then according to the triangle law of vectors, \(\vec{c}\) = \(\overrightarrow{A C}\).

Obviously, the side AC of the triangle ABC may be smaller than AB or BC, or both. Thus, the magnitude of the resultant vector \(\vec{c}\) may be smaller than that of \(\vec{a}\) or \(\vec{b}\) or both.

Question 14. By adding three unit vectors is it possible to get a unit vector?
Answer:

If any two unit vectors out of the three are equal in magnitude but opposite in direction, then adding all the three vectors we obtain a resultant equal to the third unit vector. For instance, on adding \(\vec{a}\), –\(\vec{a}\) and \(\vec{b}\), we get, \(\vec{a}\)+(\(\vec{a}\)) + \(\vec{b}\) = \(\vec{b}\).

Question 15. Resultant of two vectors \(\vec{F}_1\) and \(\vec{F}_2\) is \(\vec{P}\) When \(\vec{F}_2\) is reversed, the resultant is \(\vec{Q}\). Show that \(\left(P^2+Q^2\right)=2\left(F_1^2+F_2^2\right)\)
Answer:

When \(\vec{F}_2\) is reversed it becomes –\(\vec{F}_2\).

From given conditions, \(\vec{F}_1+\vec{F}_2=\vec{P}, \vec{F}_1+\left(-\vec{F}_2\right)=\vec{F}_1-\vec{F}_2=\vec{Q}\)

∴ \(P^2+Q^2=\left(\vec{F}_1+\vec{F}_2\right)^2+\left(\vec{F}_1-\vec{F}_2\right)^2\)

= \(F_1^2+F_2^2+2 \vec{F}_1 \cdot \vec{F}_2+F_1^2+F_2^2-2 \vec{F}_1 \cdot \vec{F}_2\)

= \(2\left(F_1^2+F_2^2\right)\)

Question 16. How does the change of acceleration due to gravity affect the path of a projectile?
Answer:

The locus of a projectile is denoted by y = \(x \tan \alpha-\frac{g}{2 u^2 \cos ^2 \alpha} x^2\)…..(1)

Let u and α be constants.

Now from equation (1) we can say that both the maximum height and the horizontal range of a projectile decrease with increasing g. In this case, the locus of the projectile is shown by line a.

On the other hand, both quantities increase with decreasing g. In this case, the locus of the projectile is shown by line b.

Real-Life Examples of Vector Applications

Question 17. Can four non-coplanar vectors produce equilibrium? Give reasons.
Answer:

Four non-coplanar vectors can produce equilib¬rium when they obey the following condition: The resultant of any two vectors must be equal in magnitude but opposite in direction to the resultant of the other two vectors.

Question 18. Show that a stretched wire cannot remain horizontal when weight is suspended from its mid-point.
Answer:

Let the two ends A and B of the wire be rigidly fixed and a weight W be suspended from the mid-point O. In this condition, the two parts of the string OA and OB make equal angles θ with the horizontal and the tension on each part is T.

For equilibrium, the vertical components of T on each wire together balance the weight W, and the horizontal components balance each other.

∴ 2T sinθ = W or, sinθ = \(\frac{W}{2T}\)

Since, T cannot be infinitely large and W ≠ 0,

sinθ ≠ 0 i.e., θ ≠ 0°

Hence, the wire cannot remain horizontal when a weight is suspended from its mid-point (or from any other point along its length).

Question 9. Two wooden blocks are falling from the same height. One is falling down an inclined plane and the other is In a free fall. Out of the two

1. Which one will reach the ground first and 
2. Which one will have a higher velocity when it touches the ground?

Answer:

1. Let the length, height, and angle of inclination of the inclined plane be l, h, and θ respectively.

The acceleration of the block falling along the inclined Plane = g sinθ. Let the time taken by the first and the second blocks respectively to touch the ground be t₁ and t₂

For the first block, \(l=\frac{1}{2} g \sin \theta \cdot t_1^2\)

or, \(t_1^2=\frac{2 l}{g \sin \theta}=\frac{2 h}{g \sin ^2 \theta}\) (because \(\sin \theta=\frac{h}{l}\))

or, \(t_1=\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)…..(1)

For the second block, h = \(\frac{1}{2} g t_2^2 \quad \text { or, } t_2=\sqrt{\frac{2 h}{g}}\)……(2)

As \(\sin \theta<1\)(because \(\theta<90^{\circ}\)), from equations (1) and (2) it is seen that \(t_1>t_2\), i.e., the block falling freely touches the ground first.

2. Let the velocities acquired by the first and the second blocks respectively be \(v_1\) and \(v_2\).

For the first block, \(v_1^2=2 g \sin \theta \cdot l=2 g \cdot \frac{h}{l} \cdot l=2 g h \text { or, } v_1=\sqrt{2 g h}\)

For the second block, \(v_2^2=2 g h \text { or, } v_2=\sqrt{2 g h}\)

∴ \(v_1=v_2\)

Hence, both the blocks touch the ground with the same velocity.

Step-by-Step Solutions to Vector Problems

Question 20. In a circus, a joker stands on a highly elevated plank with a ball in his hand. Another joker also stands with a rifle in his hand pointing it directly at the ball. If the rifle is fired precisely at the moment when the ball is released, will the bullet hit the bail? Air resis¬tance is negligible.
Answer:

The gravitational acceleration acting on the bullet and the ball are equal in magnitude and act vertically downwards. As the bullet is fired at the very moment the ball is released, the vertical displacements of both the bullet and the ball are the same. Therefore, the bullet will hit the ball.

Alternative method: The bullet from the rifle is directed straight to the ball. Let v₀ be the velocity with which the bullet leaves the rifle at an angle θ with the horizontal. Let us consider, the time taken by the bullet to pass across the vertical line MB at A is t.

Horizontal distance travelled by the bullet = OB = x

x = \(v_0 \cos \theta \times t \quad \text { or, } t=\frac{x}{v_0 \cos \theta}\)

∴ \(t^2=\frac{x^2}{v_0^2 \cos ^2 \theta}\)

For the vertical motion of the bullet, \(A B =v_0 \sin \theta \times t-\frac{1}{2} g t^2\)

= \(v_0 \sin \theta \times \frac{x}{v_0 \cos \theta}-\frac{1}{2} g \times \frac{x^2}{v_0^2 \cos ^2 \theta}\)

= \(x \tan \theta-\frac{1}{2} \frac{g x^2}{v_0^2 \cos ^2 \theta}\)

From the ΔOMB, \(\tan \theta=\frac{M B}{O B}=\frac{M B}{x}\) M B = \(x \tan \theta\)

Now, MA = \(M B-A B=x \tan \theta-\left[x \tan \theta-\frac{1}{2} \cdot \frac{g x^2}{v_0^2 \cos ^2 \theta}\right]\)

= \(\frac{1}{2} \frac{g x^2}{v_0^2 \cos ^2 \theta}=\frac{1}{2} g t^2\)

Thus in a time t, the bullet falls through a vertical distance \(\frac{1}{2} g t^2\) below M.

The vertical distance fallen by the ball is

h = \(u t+\frac{1}{2} g t^2=\frac{1}{2} g t^2\) [because u=0]

Thus the bullet and the ball will always reach point A at the same time. Hence the bullet will always hit the ball whatever be the velocity of the bullet.

Question 21. Under which condition will the magnitude of the resultant of two vectors be equal to that of any one of the constituent vectors?
Answer:

Suppose, a and b are the magnitudes of two vectors and the angle between them is θ. According to the problem let the magnitude of the resultant of \(\vec{a}\) and \(\vec{b}\) be equal to a.

Therefore, \(a^2=a^2+b^2+2 a b \cos \theta \text { or, } b(b+2 a \cos \theta)=0\)

∴ \(b+2 a \cos \theta\) =0(b≠0]

or, \(\theta=\cos ^{-1}\left(-\frac{b}{2 a}\right)\)

that is if the angle between \(\vec{a}\) and \(\vec{b}\) is = \(\cos ^{-1}\left(-\frac{b}{2 a}\right)\) or, \(\cos ^{-1}\left(-\frac{a}{2 b}\right)\), the magnitude of the resultant vector is equal to a or b respectively.

Vector Components and Resolution Questions

Question 22. If the angle between two vectors is slowly increased from 0 then what changes will be found in the resultant?
Answer:

Let \(\vec{a}\) and \(\vec{b}\) be the two vectors whose resultant is \(\vec{R}\). \(\vec{a}\) is directed along the positive x-axis and is constant, while \(\vec{b}\) moves from the positive x-axis in the anticlockwise direction.

Let at any moment, the angle between \(\vec{a}\) and \(\vec{b}\) be θ

∴ R = (a² + b² + 2abcosθ)½

From this equation we get,

when, θ = 0, R = a+ b

when, θ = 90°, R = (a²+b²)  when, 9 = 180°, R = |a-b|

when, θ = 270°, R = (a² + b²)½

when, θ = 360°, R = a+b

So, when the angle increases from 0° to 180° gradually, the magnitude of R decreases from (a+b) to |a-b|. Again, when θ increases from 180° to 360°, the magnitude of R increases from |a-b| to (a+b).

Question 23. If \(|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}\), them what us tghe angle between \(\vec{a}\) and \(\vec{b}\)?
Answer:

According to the question, \(|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}\)

∴ absinθ = abcosθ

[Let the angle between \(\vec{a}\) and \(\vec{b}\) be θ]

or, tan θ = 1 [a and b≠0]

∴ θ = 45°.

Question 24. What is the vector product of two equal vectors?
Answer:

The vector product of two equal vectors leads to a zero vector or null vector, i.e., the resultant vector has a magnitude equal to zero without any fixed direction. For any vector \(\vec{A},|\vec{A} \times \vec{A}|=(A)(A) \sin 0^{\circ}=0\).

∴\(\vec{A}\) x \(\vec{A}\)= \(\vec{0}\)

Question 25. Show that the projection or component of a vector \(\vec{R}\) on another vector \(\vec{A}\) is \(\vec{R}\) \(\hat{a}\), where a is a unit vector along \(\vec{A}\).
Answer:

In general, the component of a vector \(\vec{R}\) along a direction of another vector making an angle θ with it, is Rcosθ which is the projection of \(\vec{R}\) along the direction of that vector.

Consider two planes passing through the initial and terminal points of \(\vec{R}\), which are perpendicular to \(\vec{A}\) at F and H respectively.

Then, projection of \(\vec{R}\) on \(\vec{A}\) is FH = EG = Rcosθ = \(\vec{R}\) – \(\hat{a}\)

Question 26. If \(\vec{A}\) is a constant vector, then show that \(\frac{d \vec{A}}{d t}\) is perpendicular to \(\vec{A}\).
Answer:

⇒\(|\vec{A}|=\) constant

∴ \(\vec{A} \cdot \vec{A}=A^2=\) constant

∴ \(\frac{d}{d t}(\vec{A} \cdot \vec{A})=0\)

or, \(\vec{A} \cdot \frac{d \vec{A}}{d t}+\frac{d \vec{A}}{d t} \cdot \vec{A}=0 \text { or, } 2 \vec{A} \cdot \frac{d \vec{A}}{d t}=0 \text { or, } \vec{A} \cdot \frac{d \vec{A}}{d t}=0\)

So, \(\frac{d \vec{A}}{d t}\) is perpendicular to \(\vec{A}\).

Question 27. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, then what will be the total area around the fountain that gets wet?
Answer:

Let the angle with the horizontal be 6 and the hor¬izontal range of water coming out of the fountain be R.

∴ R = \(\frac{v^2 \sin 2 \theta}{g}\)

The value of R will be maximum when sin 2θ = 1

i.e., 2θ = 90° or, θ = 45°

∴ \(R_{\max }=\frac{v^2}{g}\)

So a circular area of radius R_max around the fountain gets wet.

∴ Total area, \(A=\pi R_{\max }^2=\pi\left(\frac{\nu^2}{g}\right)^2=\frac{\pi v^4}{g^2}\)