Physics Test

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Physics Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

A spring obeying the linear law F=−Kx is first compressed by 10cm and the work done is W₁. Next, it is compressed by another 10cm10cm, the work done now is W₂, then W₁:W₂is:

A
1:3

B
3:1

C
1:6

D
6:1

Solution

\(W_{1}=\Delta(P . E)\)
\(=\frac{1}{2} K\left(x_{2}^{2}-x_{1}^{2}\right)\)
\(=\frac{1}{2} K\left(10^{2}-0^{2}\right)\)
Similarly, \(W_{2}=\frac{1}{2} K\left(20^{2}-10^{2}\right)\)
So, \(W_{1}: W_{2}=100: 300=1: 3\)
Hence, the correct option is (A)
Question 2
1 / -0

A chain of mass m and length L is overhanging from the edge of a smooth horizontal table such that 3/4th of its length is lying on the table. The work done in pulling the chain completely on to the table is:

A
mgL/16

B
mgL/32

C
3mgL/32

D
mgL/8

Solution

\(\mathrm{W} \cdot \mathrm{D}=\Delta(P . E)\)
\(=\) mass of hanging part \(\times\) height of \(\mathrm{COM}\) of hanging part \(\times g\) \(=\frac{m}{4} \times \frac{L}{8} \times g=\frac{m g L}{32}\)
Hence, the correct option is (B)
Question 3
1 / -0

A uniform chain of length L hangs partly from a table which is kept in equilibrium by friction. The maximum length that can stand without slipping is l, then the coefficient of friction between the table and the chain is:

A
l²/L−l

B
l/L

C
l/L−l

D
L−l/l

Solution

Mass per unit length of the chain \(=\frac{M}{L}\) Mass of the hanging part \(=l\left(\frac{M}{L}\right)\) Mass of the part on the table \(=(L-l)\left(\frac{M}{L}\right)\) Force equation is:
\(\mu \frac{M}{L}(L-l) g=\frac{M}{L}(l) g\)
\(\mu=\frac{l}{L-l}\)
Hence, the correct option is (C)
Question 4
1 / -0

A body of mass m starting from rest is acted on by a force producing a velocity v=√(k×s) where k is a constant and s is displacement. The work done by the force in the first tt seconds is:

A

B

C

D

Solution

\(v=\frac{d s}{d t}=(k s)^{\frac{1}{2}} \ldots . .(1)\)
\(\therefore s^{-\frac{1}{2}} d s=k^{\frac{1}{2}} d t\)
Integrating both sides
\(s^{\frac{1}{2}}=\frac{1}{2} k^{\frac{1}{2}} t+\) constant
At \(t=0, s=0\)
\(\therefore\) constant \(=0\)
\(\therefore s^{\frac{1}{2}}=\frac{1}{2} k^{\frac{1}{2}} t\)
\(\Rightarrow s=\frac{1}{4} k t^{2}\)
Work done is equal to change in KE. \(\Delta(K E)=\frac{1}{2} m v^{2}=\frac{1}{2} m k s=\frac{1}{2} m k\left(\frac{1}{4} k t^{2}\right)=\frac{m k^{2} t^{2}}{8}\)
Hence, the correct option is (C)
Question 5
1 / -0

A ball of mass 0.2kg strikes an obstacle and moves at 60⁰to its original direction. If its speed also changes from 20m/s to 10m/s, the magnitude of the impulse received by the ball is

A
2√7Ns

B
2√3 Ns

C
2√5 Ns

D
3√2 Ns

Solution

Initial momentum, \(P_{i}=0.2 \times 20=4 \mathrm{m} / \mathrm{s} \hat{i}\)
Final momentum, \(P_{f}=0.2 \times 10\left(\cos 60^{\circ} \hat{i}+\sin 60^{\circ} \hat{j}\right)\)
Impulse \(=P_{f}-P_{i}=\hat{i}+\sqrt{3} \hat{j}-4 \hat{i}=-3 \hat{i}+\sqrt{3} \hat{j}\)
Magnitude of impulse \(=\sqrt{3^{2}+\sqrt{3}^{2}}=\sqrt{12}=2 \sqrt{3} N s\)
Hence, the correct option is (B)
Question 6
1 / -0

A ball reaches a wall at 60 m/s along +ve X direction, and leaves the wall in the opposite direction with the same speed. Assuming that the mass of the ball as 50gm and the contact time is 0.02 second, the force exerted by the wall on the ball is

A
300 N along + X direction

B
300 N along - X direction

C
3,00,000 N along + X direction

D
3,00,000 N along - X direction

Solution

Force exerted by the wall on the ball is the change in momentum of the ball per unit time. Initial momentum \(=m v=50 \times 10^{-3} \times 60=3 N s\)
So the momentum just after collision will be \(-3 N s\) \(\therefore\) change is momentum \(=3-(3)=6 N s\)
Force \(=\frac{6}{0.02}=300 N\)
The force exerted by the wall on the ball would be opposite to its initial velocity. Thus the direction of force would be negative X direction.
Hence, the correct option is (B)
Question 7
1 / -0

A 5000 kg rocket is set for vertical firing. The exhaust speed is 800ms⁻¹. To give an upward acceleration of 20ms⁻², the amount of gas ejected per second to supply the needed thrust is (g=10ms⁻²):

A
127.5 kg s⁻¹

B
137.5 kg s⁻¹

C
187.5kg s⁻¹

D
185.51kg s⁻¹

Solution

We know that \(F=\frac{\delta p}{\delta t}\) \(\delta p=\) rate of ejection \(\times \delta t \times v\)
\(F=\) rate of ejection \(\times v\) Also, \(F-m g=m a \Rightarrow F=m(g+a)\)
Therefore, rate of ejection \(=\frac{m(g+a)}{v}=\frac{5000 \times 30}{800}=187.5 \mathrm{kg} \mathrm{s}^{-1}\)
Hence, the correct option is (C)
Question 8
1 / -0

A body is acted on by a force given by F=(10+2t) N . The impulse received by the body during the first four seconds is

A
40 N s

B
56 N s

C
72 N s

D
32 N s

Solution

Given force, \(F=(10+2 t) N\)
Impulse, \(J=\int_{0}^{4} F \cdot d t\)
\(=\int_{0}^{4}(10+2 t) \cdot d t=\left[10 t+t^{2}\right]_{0}^{4}=\left[10(4)+4^{2}\right]=56 N s\)
Hence, the correct option is (B)
Question 9
1 / -0

An impulse I given to a body changes its velocity from v₁ to v₂. The increase in the kinetic energy of the body is given by:

A
I(v₁+v₂)

B
I(v₁+v₂)/2

C
I(v₁−v₂)

D
I(v₁−v₂)/2

Solution

Impulse, \(I=m\left(v_{2}-v_{1}\right)\)
Change in \(K E=\frac{1}{2} m\left(v_{2}^{2}-v_{1}^{2}\right)=\frac{1}{2} m\left(v_{2}-v_{1}\right)\left(v_{2}+v_{1}\right)=\frac{I\left(v_{2}+v_{1}\right)}{2}\)
Hence, the correct option is (B)
Question 10
1 / -0

A hammer of mass M strikes a nail of mass mm with velocity of u ms⁻¹ and drives it a meters into a fixed block of wood. The average resistance of the wood to the penetration of the nail is:

A

B

C

D

Solution

Using conservation of momentum we have \(v=\frac{M u}{M+m}\)
Thus we get \(F S=\frac{1}{2}(M+m) v^{2}\)
\(F a=\frac{1}{2}(M+m) v^{2}\)
\(F=\frac{1}{2 a}(M+m)\left(\frac{M u}{M+m}\right)^{2}=\left(\frac{M^{2}}{M+m}\right) \frac{u^{2}}{2 a}\)
Hence, the correct option is (D)