Physics Test

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Physics Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

A 60 kg boy lying on a surface of negligible friction throws a stone of mass 1 kg horizontally with a speed of 12 m/s away from him. With what kinetic energy does he move back due to result of this action?

A
2.4 J

B
7.2 J

C
1.2 J

D
36 J

Solution

The given surface is frictionless so no external force is acting on the system. Hence linear momentum of the system will remain conserved.
\(\Rightarrow\) Initial momentum \(=\) Final momentum
\(\Rightarrow 0=60 \times V+1 \times 12\)
\(\Rightarrow v=-\frac{1}{5} m / s(-v e \operatorname{sign} s h o w m\) an moves in opposite direction of stone)
Kinetic energy of the boy after separation \((K . E .)=\frac{1}{2} m v^{2}\) \(K . E .=\frac{1}{2} \times 60 \times \frac{1}{25}\)
\(K . E .=1.2 J\)
Hence, the correct option is (D)
Question 2
1 / -0

An elastic spring is compressed between two blocks of masses 1 kg and 2 kg resting on a smooth horizontal table as shown. If the spring has 12 J of energy and suddenly released, the velocity with which the larger block of 2 kg moves will be

A
2 m/s

B
4 m/s

C
1 m/s

D
8 m/s

Solution

Using momentum conservation
\(m_{A} \cdot v_{A}=m_{B} v_{B}\)
\(\Rightarrow v_{A}=2 v_{B}\)
Now total \(K . E=P . E\) of spring
\(\frac{1}{2} m_{A} v_{A}^{2}+\frac{1}{2} m_{B}\left(V_{B}\right)^{2}=12\)
\(\frac{1}{2} m_{A} \times 4 V_{B}^{2}+\frac{1}{2} m_{B}\left(V_{B}\right)^{2}=12\)
\(3\left(v_{B}\right)^{2}=12\)
\(v_{B}=2 m / s\)
Hence, the correct option is (A)
Question 3
1 / -0

Two small identical spheres each of mass m are projected slant-wise from the points A and B on the ground with equal velocities uu and making same angles θ and θ as shown in the figure. They collide with each other at the highest point C of the common path. If the collision is completely inelastic, how much time after the collision the particles come back to the ground?

A
4usinθ/g

B
3usinθ/g

C
2usinθ/g

D
usinθ/g

Solution

At the highest point, the velocity of the masses will only be horizontal and will be equal to \(u \cos \theta\)
After the masses stick together, the common velocity will be given as
\(v=\frac{m u \cos \theta-m u \cos \theta}{m+m}=0\)
So the body will be in free fall starting from rest.
For the vertical motion:
\(\frac{1}{2} g t^{2}=H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
Solving this, we get \(t=\frac{u \sin \theta}{g}\)
Hence, the correct option is (D)
Question 4
1 / -0

A ball of mass m at rest receives an impulse I₁ in the direction of north. After some time, it received another impulse I₂in the direction of south. The final kinetic energy of the ball is

A

B

C

D

Solution
Question 5
1 / -0

A force of 1 N acts upon a mass of 1 kg for 1 second and gives it a momentum P₁and kinetic energy E₁. The same force acts upon the same mass for a displacement of 1 m and gives it a momentum P₂ and kinetic energy E₂. Then, which of the following is correct?

A
P₁=P₂,E₁=E₂

B
P₁>P₂,E₁>E₂

C
P₁2,E₁2

D
P₁>P₂,E₁2

Solution

\(P_{1}=F t=1 \mathrm{N} \times 1 \mathrm{s}=1 \mathrm{Ns}\)
\(a=\frac{F}{m}\)
\(a=\frac{1}{1}=1 \mathrm{m} / \mathrm{s}^{2}\)
\(v=u+a t\)
\(v=0+(1)(1)=1 \mathrm{m} / \mathrm{s}^{2}\)
\(E_{1}=\frac{1}{2} m v^{2}\)
\(E_{1}=\frac{1}{2} \times 1 \times(a t)^{2}\)
\(E_{1}=\frac{1}{2} \times 1 \times 1=0.5 \mathrm{J}\)
\(s=u t+\frac{1}{2} a t^{2}\)
\(\Rightarrow 1=(0) t+\frac{1}{2}(1) t^{2}\)
\(\Rightarrow t=\sqrt{2} \mathrm{s}\)
\(v=u+a t\)
\(v=0+(1) \sqrt{2}=\sqrt{2} \mathrm{m} / \mathrm{s}\)
\(P_{2}=m v=(1)(\sqrt{2})=\sqrt{2} \mathrm{kgm} / \mathrm{s}\)
\(E_{2}=\frac{1}{2} m v^{2}\)
\(\Rightarrow E_{2}=\frac{1}{2} \times 1 \times(\sqrt{2})^{2}=1 \mathrm{J}\)
\(P_{1}Hence, the correct option is (C)
Question 6
1 / -0

A simple pendulum of length 'l' carries a bob of mass'm'. If the breaking strength of the string is 2mg. The maximum angular amplitude from the vertical can be:

A
0°

B
30°

C
60°

D
90°

Solution

\(T_{\max }=2 m g\)
Tension is max. at bottom \(T=m g+\frac{m v^{2}}{l}\)
\(2 m g=m g+\frac{m v^{2}}{l}\)
Then, \(v=\sqrt{l g}\)
Now by WET. \(m g l(1-\cos \theta)=\frac{m}{2}(\sqrt{l g})^{2}\)
\(g l-g l \cos \theta=\frac{l g}{2}\)
\(\frac{g l}{2}=g l \cos \theta\)
\(\cos \theta=\frac{1}{2}\)
\(\theta=60^{0}\)
Hence, the correct option is (C)
Question 7
1 / -0

The length of simple pendulum is 1m and mass of its bob is 50 gram. The bob is given sufficient velocity so that the bob describes vertical circle whose radius equal to length of pendulum. The maximum difference in the kinetic energy of bob during one revolution is:

A
0.98 J

B
1.96 J

C
4.9 J

D
9.8 J

Solution

\(L=1 m, m=0.05 k g\)
\(V_{b}=\sqrt{5 g R} m / s\)
\(V_{t}=\sqrt{g R}\)
the velocity is maximum at bottom the velocity is minimum at top \(\therefore \Delta K E=\frac{1}{2} m\left[(\sqrt{5 g R})^{2}-(\sqrt{g R})^{2}\right]\)
\(=\frac{m}{2}[4 g R]\)
\(=2 \times(0.05)(9.8)(1)\)
\(=0.98 J\)
Hence, the correct option is (A)
Question 8
1 / -0

A wheel having a radius of 10 cm is coupled by a belt to another wheel of radius 30 cm. 1st wheel increases its angular speed from rest at a uniform rate of 1.57 rad/s². The time for the 2nd wheel to reach a rotational speed of 100 rev/min is (assume that the belt does not slip):

A
20 s

B
10 s

C
1.5 s

D
15 s

Solution

\(\alpha_{1}=1.57 \mathrm{rad} / \mathrm{s}^{2}\)
\(\omega_{2}=100 \mathrm{rev} / \mathrm{m}\)
\(=100 \times \frac{2 \pi}{60}\)
\(=\frac{10 \pi}{3} \mathrm{rad} / \mathrm{s}\)
linear velocity of belt is equal \(\therefore \omega_{1} r_{1}=\omega_{2} r_{2}\) and \(\alpha_{1} r_{1}=\alpha_{2} r_{2}\)
\(\therefore \alpha_{2}=(1.57)\left(\frac{10}{30}\right)\)
\(=\frac{1.57}{3} r a d / s^{2}\)
\(\omega=\omega_{0}+\alpha t\)
\(\frac{10 \pi}{3}=0+\frac{1.57}{3} \times t\)
\(t=20 \mathrm{s}\)
Hence, the correct option is (A)
Question 9
1 / -0

The bob of a simple pendulum of mass ′m′ is performing oscillations such that the tension in the string is equal to twice the weight of the bob while it is crossing the mean position. The tension in the string when the bob reaches extreme position is :

A
mg/2

B
mg

C
3mg/2

D
zero

Solution

At bottom \(T=2 m g\) \(T-m g=\frac{m V^{2}}{R}\)
\(2 m g-m g=\frac{m V^{2}}{R}\)
\(\sqrt{R g}=V\)
Applying Work Energy Theorem between pt.
(1) and (2) \(m g R(1-\cos \theta)=\frac{1}{2} m V^{2}\)
\(m g R(1-\cos \theta)=\frac{m R g}{2}\)
here, \(\theta\) is the angle made by the particle with the axis at its extreme point,
\(1-\cos \theta=\frac{1}{2}\)
\(\theta=60^{\circ}\)
at, \(\theta=60^{\circ}\)
\(T=m g \cos \theta\)
\(\therefore T=m g \frac{1}{2}\)
Hence, the correct option is (A)
Question 10
1 / -0

A simple pendulum is oscillating with an angular amplitude 60^∘. The ratio of tensions in the string when the bob reaches the mean position and the extreme position respectively is :

A
2:1

B
3:1

C
1:3

D
4:1

Solution

At \(A\) \(T_{A}=m g \cos 60^{\circ}\)
\(T_{A}=\frac{m g}{2}\)
By Work Energy Theorem from A to B \(m g R\left(1-\cos 60^{0}\right)=\frac{1}{2} m V^{2}\)
\(m g \frac{R}{2}=\frac{1}{2} m V^{2}\)
\(V=\sqrt{R g}\)
At B \(T_{B}-m g=\frac{m V^{2}}{R}\)
\(T_{B}=m g+m g=2 m g\)
\(\therefore \frac{T_{B}}{T_{A}}=\frac{2 m g}{m g / 2}=4\)
Hence, the correct option is (D)