Physics Test

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Physics Test - 9

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Question 1
3 / -1

The driver of a train moving with a constant speed v₁ along a straight track sights another train at a distance d ahead of him on the same track moving in the same direction with a constant speed v₂. He at once applies the brakes and gives his train a constant retardation f. There will be a collision of the trains if:

A

B

C

D

Solution
Question 2
3 / -1

Two particles are projected under gravity with speed 4 m/s and 3 m/s simultaneously from same point and at angles 53° and 37° with the horizontal surface respectively as shown in figure. Then :

A
Their relative velocity is along vertical direction

B
Their relative acceleration is non-zero and it is along vertical direction

C
They will hit the surface simultaneously

D
None of these

Solution

Relative velocity in horizontal direction is zero.

Hence option A is correct answer.
Question 3
3 / -1

A projectile is projected with velocity v such that its range is twice the greatest height attained. Then its range is

A
1v²/5g

B
2v²/g

C
4v²/5g

D
2v/g

Solution

Here, we have to calculate the range of the projectile projected with velocity v such that its range is twice the greatest height attained.

Let,

v - initial velocity

θ - angle of projection

g - acceleration due to gravity

Then we have,

The horizontal range, R and the maximum height reached are,
Question 4
3 / -1

If error in measuring diameter of a circle is 4%, the error in circumference of the circle would be

A
2%

B
8%

C
4%

D
1%

Solution

Hence option C is the correct answer.
Question 5
3 / -1

Two stones are thrown vertically upwards simultaneously from the same point on the ground with initial speed u₁=30 m/sec and u₂=50 m/sec. Which of the curves represent correct variation (for the time interval in which both reach the ground) of relative position (x₂−x₁) and relative velocity (v₂−v₁) of second stone with respect to respect to first with time (t). Assume that stones do not rebound after hitting.

A

B

C

D
None of these

Solution

While both the stones are in flight, $a_{1}=g$ and $a_{2}=g$ So $a_{r e l}=0 \Rightarrow V_{r e l}=$ constant
$\Rightarrow X_{\text {rel}}=($ const $) t$
$\Rightarrow$ Curve of $x_{\text {rel }} v / s$ t will be straight line. After the first particle drops on ground, the seperation $\left(x_{\text {rel }}\right)$ ) will decrease parabolically (due to gravitational acceleration), and finally becomes zero. and $V_{\text {rel }}=$ slope of $x_{\text {rel }} v / s t$

Hence option A is the correct answer.
Question 6
3 / -1

Figure shows a mass M is attached to the middle of a string, the two ends of which are passing over two pulleys P₁ and P₂. The ends of strings are attached to masses M₁ and M₂. If the two parts of the string attached to M are inclined at an angle 2θ and the instantaneous downward velocity of the masses M₁ and M₂ be v, what is the upward velocity of M ?

A
2v cos θ

B
v/ cos θ

C
v

D
v/2

Solution

Here, we have to find out the upward velocity of the centre mass M

The given problem is described in the following figure.

From figure, it is clear that velocity of the masses $M_{1}$ and $M_{2}$ is $v=\frac{d y}{d t}$
And the velocity of the mass $M$ is $v^{\prime}=\frac{d x}{d t}$
Also $x^{2}+a^{2}=y^{2}$
Where, a is a constant Now differentiating with respect to $t$, we get $2 x \frac{d x}{d t}+0=2 y \frac{d y}{d t}$
ie, $x v^{1}=y v$
$v^{1}=\left(\frac{y}{x}\right) v$
$=\frac{v}{\frac{v}{v}}$
$=\frac{v}{\cos \theta}$
That is, the upward velocity of the mass $M$ is $v / \cos \theta$

Hence option B is the correct answer.
Question 7
3 / -1

Two bodies of different masses ma and mb are dropped from two different heights, a and b respectively. The ratio of times taken by the two bodies to drop through this distance is

A
a² : b²

B
a/m_b : b/m_a

C

D
a : b

Solution

Here, we have to calculate the time taken by two bodies to reach the ground which are dropped from different heights

Let g be the acceleration due to gravity, the the time taken and h be the height of a body

We have the time taken by a freely falling body to reach ground

$t = \sqrt{\frac{2 h}{g}}$

Then, for the first particle,

$t_{a} = \sqrt{\frac{2 a}{g}}$

And for the second particle,

$t_{b} = \sqrt{\frac{2 b}{g}}$

Now, if we take the ratio

$t_{a} : t_{b} = \sqrt{a} : \sqrt{b}$

That is, the ratio of the time taken by two bodies to reach the ground which are dropped from different heights is equal to the ratio of the square roots of the heights, a and b

ta : tb = Va: vb

Hence option C is the correct answer.
Question 8
3 / -1

Find the height above the surface of the earth where weight of a body becomes half.

A
R/2

B
(√2 -4)R

C
R/(√2 + 1)

D
R/√2

Solution

Given, weight of the body becomes half. Let that height be h

Hence option C is the correct answer.
Question 9
3 / -1

A constant force F is applied at the top of a ring as in figure. M is the mass and R is the radius of the ring. The change in angular momentum of particle about point of contact in a time t is.

A
Fr

B
a constant

C
FR/t

D
2FRt

Solution

Here, we have to find out the change in angular momentum of particle about point of contact in a time, t.
We have,
Angular momentum,
L = τt = r × F t
Where,
τ - torque
t - time
F - Force
r - perpendicular distance
Here,
r = 2R and F = F
L = 2RFt
That is, the change in angular momentum of particle about point of contact in a time, t is 2RFt.
Question 10
3 / -1

A particle of mass m is executing uniform circular motion on a path of radius r. If v is the speed and p the magnitude of its linear momentum, then the radial force acting on the particle is:

A
mv²/r

B
up/r

C
p²/mr

D
All of the above

Solution

Hence option D is the correct answer.
Question 11
3 / -1

In the figure shown all the surfaces are smooth. All the blocks A,B and C are movable X-axis is horizontal and y-axis vertical as shown. Just after the system is released from the position as shown:

A
Acceleration of 'A' relative to ground is in negative y-direction

B
Acceleration of 'A' relative to B is in positive xx-direction

C
The horizontal acceleration of 'B' relative to ground is in negative x-direction.

D
All of the above

Solution

Answers are A, B, C and D

There is no horizontal force on block A, therefore it does not move in x-direction. There is net downward force (mg -N) is acting on it, making its acceleration along negative y-direction. Hence option A is correct.

Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive x-direction. Hence option B is correct.

Due to the component of normal exerted by C on B, it moves in negative x-direction. Therefore the direction of acceleration is in negative x-axis. Hence option C is correct.

Hence Option D is correct answer.
Question 12
3 / -1

A body crosses the topmost point of a vertical circle with critical speed. What will be its centripetal acceleration when the string is horizontal:

A
6g

B
g

C
2g

D
3g

Solution

Hence option D is the correct answer.
Question 13
3 / -1

The potential energy of a particle is given by $U = \frac{a}{r^{2}} \frac{b}{r}$ where a and b are positive constants, r is the distance from the centre of the field. The stable equilibrium position of the particle corresponds to the distance given by

A
$r_{0} = \frac{2 a}{b}$

B
$r_{0} = \frac{a}{b}$

C
$r_{0} = \frac{a}{b}$

D
$r_{0} = \frac{a}{2 h}$

Solution
Question 14
3 / -1

If the radius of the earth were to shrink by one per cent, its mass remaining the same, the value of g on the earth's surface would.

A
increase by 0.5%

B
increase by 2%

C
decrease by 0.5%

D
decrease by 2%

Solution
Question 15
3 / -1

A block of mass M is at rest on a rough horizontal surface. The coefficient of friction between the block and the surface isμ. A force F=Mg acting at an angle θ with the vertical side of the block pulls it. In which of the following cases, the block can be pulled along the surface?

A
$\tan θ \geq μ$

B
$\cot θ \geq μ$

C
$\tan θ / 2 \geq μ$

D
$\cot θ / 2 \geq μ$

Solution

As shown in the figure the Normal reaction $N$ on the block and Vertical component of the force $F$ are opposite to weight.
We need only horizontal motion no vertical motion so vertical force should be balanced that is net force in vertical direction should be zero, i.e., $N+F \cos \theta-M g=0$ or $N=M g-F$ $\cos \theta,$ put $F=M g$ as given in the question. so by putting so we get $N=M g(1-\cos \theta)$
Now putting another condition to make $F$ able to pull the block horizontally the $F \sin \theta$ should be greater than or eual to Static friction and we know that maximum value of static friction is $\mu N$ so we have $F \sin \theta \geq \mu N$
putting value of $N$ we get $\sin \theta=\mu(1-\cos \theta)$
put $\cos \theta=1-2 \sin ^{2} \frac{\theta}{2}$ and $\sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$
we will get cot $\frac{\theta}{2} \geq \mu$ Option $D$ is correct