Physics Test-10

Result

Physics Test-10

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Power applied to a particle varies with time as \({P}=\left(3 {t}^{2}-2 {t}+1\right)\) watt, where \(t\) is in second. Find the change in its kinetic energy between time \(t=2 s\) and \(t=4 s\).

A

\(32 {~J}\)

B

\(46 {~J}\)

C

\(61 {~J}\)

D

\(100 {~J}\)

Solution

Energy supplied to the system \(=\) Change in energy of system.
Since, only energy of system is kinetic,
\(\int {Pdt}=\Delta {KE}\)
\(P=3 t^{2}-2 t+1=\frac{d E}{d t}\)
\(\therefore d E=\left(3 t^{2}-2 t+1\right) d t\)
\(E=\int_{t=2 s}^{t-4 s}\left(3 t^{2}-2 t+1\right) d t\)
\(=\left[\frac{3 t^{3}}{3}-\frac{2 t^{2}}{2}+t\right]_{t=25}^{t-4 {~s}}\)
\(=\left[t^{3}-t^{2}+t\right]_{2}^{4}\)
\(=\left[\left(4^{3}-2^{3}\right)-\left(4^{2}-2^{2}\right)+(4-2)\right]\)
\(E=56-12+2=46 {~J}\)
Question 2
4 / -1

A beam of light travelling along x-axis is described by the magnetic field \(B_{z}=5 \times 10^{-7}\) \({T} \sin \omega({t}-\frac{{x}}{{c}})\) If a beam of alpha particle moves in this filed along \({y}\) - axis with a speed of \(3 \times 10^{7} {~m} / {s}\), then the maximum electric and magnetic forces experienced by it will be:

A
\(4.80 \times 10^{-10} {~N}\) and \(4.8 \times 10^{-12} {~N}\) respectively.

B
\(4.80 \times 10^{-17} {~N}\) and \(4.8 \times 10^{-18} {~N}\) respectively.

C
\(2.80 \times 10^{-5} {~N}\) and \(2.1 \times 10^{-7} {~N}\) respectively.

D
\(1.5 \times 10^{-5} {~N}\) and \(4.1 \times 10^{-7} {~N}\) respectively.

Solution

Given, maximum magnetic field, \(B_{0}=5 \times 10^{-7} T\)
Charge on alpha particle,
\(q=+2 e=2 \times 1.6 \times 10^{-19}=3.2 \times 10^{-19} {C}\)
Speed of light,
\({c}=3 \times 10^{8} {~m} / {s}\)
Speed of alpha particle,
\(v=3 \times 10^{7} {~ms}^{-1}\)
Maximum electric field,
\(E_{0}=c B_{0}\)
\(=\left(3 \times 10^{8}\right) \times\left(5 \times 10^{-7}\right)=150 {Vm}^{-1}\)
Maximum force on alpha particle due to electric field \(=q \times E_{0}\)
\(=\left(3.2 \times 10^{-19}\right) \times 150\)
\(=4.80 \times 10^{-17} {~N}\)
Force on alpha particle due to magnetic field \(=q v B_{o}\)
\(=\left(3.2 \times 10^{-19}\right) \times\left(3 \times 10^{7}\right) \times\left(5 \times 10^{-7}\right) \\
=4.80 \times 10^{-18} {~N}\)
Question 3
4 / -1

A solid sphere of radius \({R}\) has a charge \({Q}\) distributed in its volume with a charge density \(\rho={kr}^{{a}}\), where \({k}\) and \({a}\) are constants and \({r}\) is the distance from its centre. If the electric field at \({r}=\frac{{R}}{2}\) is \(\frac {1}{8}\) times that at \(r=R,\) find the value of \(a\).

A

\(2\)

B

\(3.2\)

C

\(2.5\)

D

\(0.2\)

Solution

Using Gauss's law,
\(\oint \vec{E} . d \vec{A}=\frac{1}{\varepsilon_{0}} \int(\rho d v)=\frac{1}{\varepsilon_{0}} \int k r^{a} \times 4 \pi r^{2} d r\)
or \(E \times 4 \pi R^{2}=\left(\frac{4 \pi k}{\varepsilon_{0}}\right) \frac{R^{(a+3)}}{(a+3)}\)
\(\therefore E_{1}=\frac{k R^{(a+1)}}{\varepsilon_{0}(a+3)}\)
\(E\) For \(r=\frac{R}{2} \cdot E_{2}=\frac{k\left(\frac{R}{2}\right)^{a+1}}{\varepsilon_{0}(a+3)}\)
Given \(E_{2}=\frac{E_{1}}{8}\)
or \(\frac{k\left(\frac{R}{2}\right)^{a+1}}{\varepsilon_{0}(a+3)}=\frac{1}{8} \frac{k R^{(a+1)}}{\varepsilon_{0}(a+3)}\)
\(\therefore \quad \frac{1}{2^{a+3}}=\frac{1}{8}\)
or \(a=2\)
Question 4
4 / -1

In a rotational motion of a rigid body, all particles move with _____.

A

same linear velocity and same angular velocity

B

same linear velocity and different angular velocity

C

different linear velocities and same angular velocities

D

different linear velocities and different angular velocities

Solution

In a rotational motion of a rigid body, all particles move withdifferent linear velocities and same angular velocities.
But the linear velocity is also dependent on the distance of particles from axis of rotation. So, linear velocities will be different for all particles as the distances are different for all particles.
Question 5
4 / -1

A rupee coin starting from rest rolls down a distance of \(1 {~m}\) on an inclined plane at angle of \(30^{\circ}\) with the horizontal. Assuming that \({g}=\) \(9.81 {~ms}^{-2}\), time taken is

A

\(0.68\) sec

B

\(0.6\) sec

C

\(0.5 \) sec

D

\(0.78\) sec

Solution

Given,
Distance travelled by coin(s) \(= 1\) m
Plane inclined at an angle \((\theta)\) \(= 30^{\circ}\)
\(g = 9.81\) \(m/s^2\)
By doing force analysis of the system we find,
Torque of friction force \((f)\) about the centre of mass of coin
\(\Rightarrow f R=\frac{m R^{2}}{2} \alpha\)
\(\Rightarrow \quad f=\frac{m R \alpha}{2}\)
\(\Rightarrow \quad f=\frac{m a}{2} \quad(a=\alpha \cdot R)\)
On balancing the force about the centre of mass of coin
\(m g \sin \theta-f=ma\)
\(\Rightarrow m g \sin \theta-\frac{m g}{2}=m a\)
\(\Rightarrow \quad m g(1+ \frac{1}{2})=m g \sin \theta\)
\(\Rightarrow \quad a=\frac{g \sin \theta}{\left(1+\frac{1}{2}\right)}\)
\(\Rightarrow a = \frac{g \sin 30^{\circ}}{1+\frac{1}{2}}\)
\(=\frac{g \times \frac{1} 2}{1+\frac{1} 2}=\frac{2}{3} \times g \times\frac{1} 2\)
\(\Rightarrow a = \frac{g}3\)
According to motion's second law
Second law
\(S=u+ \frac{1} 2 a t^{2}\)
\(\Rightarrow S=0+\frac{1}{2} \times \frac{9}{3} \times t^{2}\)
\(\Rightarrow t^{2}=\frac{6}{g}\)
\(\Rightarrow t^{2}=\frac{6}{981}\)
\(\Rightarrow t^{2}=\frac{200}{327}\)
\(\Rightarrow t=\sqrt{0.61}\)
\(\Rightarrow t=0.78\) sec
Question 6
4 / -1

The rate of the steady volume flow of water through a capillary tube of length \(I\) and radius \({r},\) under a pressure difference of \({P}\) is \({V}\). This tube is connected with another tube of the same length but half the radius, in series. Then the rate of steady volume flow through them is (the pressure difference across the combination is \({P}\))-

A

\(\frac{V}{16}\)

B

\(\frac{V}{17}\)

C

\(\frac{16V}{17}\)

D

\(\frac{17V}{16}\)

Solution

The rate of flow of water inside a capillary,
\(V=\frac{\pi P r^{4}}{8 n l}\)
Pressure difference,
\(\quad P=\frac{V(8 n l)}{\pi r^{4}}\)
In series combination \(P=P_{1}+P_{2}\) where \(P_{1}\) and \(P_{2}\) are the pressure difference in the two tubes.
\(\therefore\ \frac{V(8 n l)}{\pi r^{4}}=\frac{V (8 n l)}{\pi r^{4}}+\frac{V^{\prime}(8 n l)}{\pi(r / 2)^{4}}\)
In series combination, rate of flow of water \(\left(V^{\prime}\right)\) will be same in both the tubes.
\(\frac{V}{r^{4}}=\frac{V^{\prime}}{r^{4}}+\frac{V^{\prime} \times 16}{r^{4}}\)
\(V=V^{\prime}+16 V^{\prime}\)
Therefore, \(V^{\prime}=\frac{V}{17}\)
Question 7
4 / -1

Two identical particles move towards each other with velocity \(2{v}\) and \(v\) respectively. The velocity of center of mass is:

A
\(\frac{{v}} { 2}\)

B
\(\frac{{v}} { 4}\)

C
\(\frac{{v}} { 8}\)

D
\(\frac{{v}} { 7}\)

Solution

Let the mass of each body be \(\mathrm{m}\). Their motion is represented as shown in the figure,
The direction of motion of first particle is taken as positive.
So the velocity of the center of mass is \(\vec{v}_{C M}=\frac{m_{1} \overrightarrow{v_{1}}+m_{2} \overrightarrow{v_{2}}}{m_{1}+m_{2}}\)
\(=\frac{m \times 2 v-m v}{m+m}\)
\(=\frac{v}{2}\)
Question 8
4 / -1

In a 10-bit PCM system, a message signal having maximum frequency of \(4 \mathrm{KHz}\) is to be transmitted. If the bit rate of this PCM system is 60 Kbits / sec, the appropriate sampling frequency is:

A
\(6 \mathrm{kHz}\)

B
\(7 \mathrm{kHz}\)

C
\(8 \mathrm{kHz}\)

D
\(9 \mathrm{kHz}\)

Solution

The maximum frequency for the signal will be \(4 \mathrm{kHz}\).
\(n=10\) bits, \(R_{b}=60 \mathrm{Kbits} / \mathrm{sec}\)
Bit rate \(=\mathrm{n} \mathrm{fs}\)
\mathrm{f}_{\mathrm{S}}=6 \mathrm{kHz}
And \(f_{m}\) is given as \(4 \mathrm{kHz}\) i.e
\(f_{s} \geq 2 f_{m} \)
\(f_{\mathrm{s}} \geq 8 \mathrm{kHz}\)
So, \(\mathrm{f}_{\mathrm{S}}=6 \mathrm{kHz}\) leads to undersampling.
\(\therefore\) The appropriate value of the sampling frequency \(\left(f_{s}\right)\) will be \(8 \mathrm{kHz}\)
Question 9
4 / -1

Which of the following figures represents the variation of particle momentum \(P\) and associated de-Broglie wavelength \(\lambda\)?

A

B

C

D

Solution

De-Broglie wavelength, \(\lambda=\frac{h}{p}\)
\(\Rightarrow \lambda \propto \frac{1}{p}\) i.e., graph will be a rectangular hyperbola.
Question 10
4 / -1

A small square loop of wire of side \(l\) is placed inside a large square loop of wire of side \(L(L>>l)\). The loops are coplanar and their centers coincide. The mutual inductance of the system is proportional to:

A
\(\frac{l}{L}\)

B

\(\frac {l^{2}}{L}\)

C

\(\frac{L^{}}{l}\)

D

\(\frac{L^{2}}{l}\)

Solution

Magnetic field produced by a current \(i\) in a large square loop at its centre \(B \propto \frac{i}{L}\) say \(B=K \frac{i}{L}\)
\(\therefore\) Magnetic flux linked with smaller loop,
\(\phi=B . S \phi=\left(K \frac{i}{L}\right)\left(l^{2}\right)\)
Therefore, the mutual inductance
\(M=\frac{\phi}{i}\)\(=K \frac{l^{2}}{L}\)
or \(M \propto \frac{l^{2}}{L}\)
Question 11
4 / -1

Two charges of \(+4 \mu \mathrm{C}\) and \(-16 \mu \mathrm{C}\) are separated from each other by a distance of \(0.6 \mathrm{~m}\). At what distance should a third charge of \(+6 \mu \mathrm{C}\) be placed from \(+4 \mu \mathrm{C}\) so that no force exerts on it will be zero?

A
\(0.4 \mathrm{~m}\)

B
\(0.6 \mathrm{~m}\)

C
\(1.2 \mathrm{~m}\)

D
\(0.3 \mathrm{~m}\)

Solution

Consider new charge \(+4 \mu \mathrm{C}\) is placed \(\mathrm{d} m\) apart from old \(+4 \mu \mathrm{C}\) charge and \((\mathrm{x}+0.6) \mathrm{m}\) apart from \(-16 \mu \mathrm{C}\) charge.
Let,
\(\mathrm{q}_{\mathrm{A}}=+4 \mu \mathrm{C}\) at point \(\mathrm{A}\)
\(\mathrm{q}_{\mathrm{B}}=-16 \mu \mathrm{C}\) at point \(\mathrm{B}\)
\(q_{C}=+6 \mu C\) at point \(C\)
Since, net force on charge \(q_{c}\) will zero.
\(\therefore\left|\mathrm{F}_{\mathrm{CA}}\right|=\left|\mathrm{F}_{\mathrm{CB}}\right|\)
From above concept,
\(\frac{K q_{A} q_{C}}{d^{2}}=\frac{K q_{B} q_{C}}{(0.6+d)^{2}} \)
\(\frac{24}{d^{2}}=\frac{96}{(0.6+d)^{2}} \)
\(4 d^{2}=(0.6+d)^{2} \)
\(4 d^{2}=0.36+d^{2}+1.2 d\)
\(3 d^{2}-1.2 d-0.36=0 \)
\(d_{1}=-0.2 \mathrm{~m} \)
\(d_{2}=+0.6 \mathrm{~m}\)
So, according to option \(+\mathbf{0 . 6} \mathrm{m}\) distance should a third charge of \(+6 \mu \mathrm{C}\) be placed from \(+4 \mu \mathrm{C}\) so that no force exerts on it will be zero.
Question 12
4 / -1

The height at which the weight of a body becomes \((\frac{1 }{ 16})\) th of its weight on the surface of the earth (radius \(R\) ) is:

A
\(4 R\)

B
\( 5 R\)

C
\(15 R\)

D
\(3 R\)

Solution

Given:
Weight of body at height \(\left(W^{\prime}\right)=m g\) ', the weight of the body on the surface of earth \((W)=m g\) and
Acceleration due to gravity at height \(h\) from the surface of the earth
\(\Rightarrow g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}\)
Multiply both sides by \(m\), then we get
\(\Rightarrow m g^{\prime}=\frac{m g}{\left(1+\frac{h}{R}\right)^{2}} \)
\(\Rightarrow W^{\prime}=\frac{W}{\left(1+\frac{h}{R}\right)^{2}}\)
According to question, \(W^{\prime}=W / 16\)
\(\Rightarrow \frac{W}{16}=\frac{W}{\left(1+\frac{h}{R}\right)^{2}}\)
\(\Rightarrow \frac{1}{16}=\frac{1}{\left(1+\frac{h}{R}\right)^{2}} \)
\(\Rightarrow\left(1+\frac{h}{R}\right)^{2}=16 \)
\(\Rightarrow\left(1+\frac{h}{R}\right)=4\)
\(\Rightarrow \frac{h}{R}=3 \)
\(\Rightarrow \mathrm{h}=3 \mathrm{R}\)
Question 13
4 / -1

The kinetic energy of a body is stated to increase by 300 percent. The corresponding increase in the momentum of the body will be:

A
\(50\)

B
\(100\)

C
\(200\)

D
\(300\)

Solution

Let initial Kinetic energy \(=\mathrm{KE}_{1}=\mathrm{E}\)
Given that:
Final kinetic energy \(\left(K . E_{2}\right)=K . E_{1}+300 \%\) of \(K E_{1}=E+3 E=4 E\)
The relation between the momentum and the kinetic energy is given by:
\(P=\sqrt{2 m(K E)}\)
Final momentum (P') will be:
\(P^{\prime}=\sqrt{2 m(K E)_{2}}\)
\(=\sqrt{2 m \times 4 E}=2 \sqrt{2 m E}=2 P\)
Increase in momentum \((\Delta \mathrm{P})=\mathrm{P}^{\prime}-\mathrm{P}\)
\(=2 \mathrm{P}-\mathrm{P}=\mathrm{P}\)
\(\% \text { Increase }=\frac{\Delta P}{P} \times 100\)
\(=\frac{2 P-P}{P} \times 100=100\)
Question 14
4 / -1

An object is placed at a distance of \(10 \mathrm{~cm}\) in front of a double convex lens made of glass of refractive index 1.5. Both the radii of curvature of the lens are \(20 \mathrm{~cm}\) in magnitude. What is the position of the image formed?

A
\(-35\) cm

B
\(10\) cm

C
\(-20\) cm

D
\(20\) cm

Solution

Given,
Distance of the object from the lens \(=u=-10 \mathrm{~cm}\)
Refractive index of the lens \(=\mu=1.5\)
Radii of curvature of the lens are \(20 \mathrm{~cm}\) in magnitude
\(R_{1}=20 \mathrm{~cm}\) and \(\mathrm{R}_{2}=-20 \mathrm{~cm} \quad\) (As per sign convention)
According to Len's Maker's formula
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right)\)
\(=0.5 \times \frac{2}{20}=\frac{1}{20}\) or,\(f=20 \mathrm{~cm}\)
From the Lens equation,
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \text { or },\)
\( v=\frac{f u}{u+f}\)
\(=\frac{20 \times(-10)}{-10+20}\)
\(=\frac{-200}{10}=-20 \mathrm{~cm}\)
Question 15
4 / -1

When we pull back the string of catapult (gulel) to throw a stone, it holds an energy. Which kind of energy is this?

A
Kinetic energy

B
Potential energy

C
Both (A) and (B)

D
None of the above

Solution

When we do work in stretching the rubber strings of a catapult (gulel), then the work done by us gets stored in the stretched rubber strings in the form of elastic potential energy.
The stretched strings of a catapult possess potential energy due to a change in their shape. This energy of the stretched strings of the catapult can be used to throw away a piece of stone with high speed.
Potential energy, stored energy that depends upon the relative position of various parts of a system. A spring has more potential energy when it is compressed or stretched. A steel ball has more potential energy raised above the ground than it has after falling to Earth.