Physics Test

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Physics Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

A steel ball of mass 60 g falls from a height 0.8 m on the horizontal surface of a massive slab. The coefficient of restitution between the ball and the slab is e = 0.4; the total momentum imparted to the slab by the ball after numerous bounces is

A
0.56 Ns

B
0.64 Ns

C
0.72 Ns

D
0.78 Ns

Solution

Let $v$ be the velocity of the ball on reaching the slab.
Then $v=\sqrt{2 g h}$
By Newton's law of collision, the velocity $v ^{\prime}$ after the first impact is $v ^{\prime}=$ ev (upward direction).
Similarly, $v^{\prime \prime}=e v^{\prime}=e^{2} v ; v^{\prime \prime}=e^{3} v$ and so on.
Momentum imparted to the slab after the first impact $= mv -\left(- mv ^{\prime}\right)= m \left( v + v ^{\prime}\right)$
$=\operatorname{mv}(1+e)$
Similarly, the momentum imparted in the second impact
$=m\left(v^{\prime}+v^{\prime \prime}\right)$
$=m\left(v^{\prime}+e v^{\prime}\right)$
$=m v^{\prime}(1+e)=m(1+e) e v$
Momentum imparted to the slab in the third impact
$= m \left( v ^{\prime \prime}+ v ^{\prime \prime \prime}\right)$
$=m\left(v^{\prime \prime}+e v^{\prime \prime}\right)$
$=m(1+e) v^{\prime \prime}$
$=m(1+e) e^{2} v$ and $s o$ on.
: total momentum imparted
$=m v(1+e)+m v e(1+e)+m v e^{2}(1+e)+\ldots$
$=m v(1+e)\left(1+e+e^{2}+e^{3}+\ldots\right)$
as $e<1$ $=m v(1+e)\left(\frac{1}{1-e}\right)$
$=m v\left(\frac{1+e}{1-e}\right)$
$=m \sqrt{2 g h}\left(\frac{1+e}{1-e}\right)$
$=0.06 \sqrt{2 \times 10 \times 0.8}\left(\frac{1+0.4}{1-0.4}\right)$
$=0.06 \sqrt{16} \times \frac{1.4}{0.6}$
$=4 \times 0.14=0.56 Ns$

Hence option A is correct.
Question 2
1 / -0

A sphere with some cavity has outer radius R. It rolls down an inclined plane without slipping and attains a speed v at the bottom. When this sphere slides down without rolling on the frictionless inclined plane of same height its speed at the bottom is . $\frac{5 v}{4}$ The radius of gyration of the sphere is

A
$\frac{3 R}{4}$

B
$\frac{3 R}{2}$

C
$\frac{R}{4}$

D
$\frac{R}{2}$

Solution

Let $k$ be the radius of gyration and $M$ the mass of the sphere. The kinetic energy should be the same in both the cases $=\frac{1}{2} M v^{2}+\frac{1}{2} I \omega^{2}=\frac{1}{2} M\left(\frac{5 v}{4}\right)^{2}$
i.e., $\frac{1}{2} M v^{2}+\frac{1}{2} M k^{2}\left(\frac{v^{2}}{R^{2}}\right)$
$=\frac{1}{2} M\left(\frac{5 v}{4}\right)^{2}$
$1+\frac{k^{2}}{R^{2}}=\frac{25}{16}$
$\Rightarrow \frac{k^{2}}{R^{2}}=\frac{9}{16}$
$\therefore k=\frac{3 R}{4}$

Hence option A is correct.
Question 3
1 / -0

A standard cell of emf E₁ = 6 V and negligible internal resistance is connected to the terminals of a potentiometer of resistance R = 20 Ω. A storage battery of emf E and negligible internal resistance is connected to a galvanometer as shown.

On moving the jockey along AB, when AN is 16 ohms, the galvanometer shows null deflection. The emf of the storage battery E is

A
4. 2 V

B
4.5 V

C
4.8 V

D
5.1 V

Solution

The e.m.f of the cell and the storage battery establish currents opposing each other along AN of the potentiometer wire.
When the circuit is in balance, no current is taken by the galvanometer. It may be assumed that the increase in the potentials at different points in the potentiometer is proportional to the increase in the resistance of the section AN. (i.e.) $E / 6= r / R$ where $r$ is the resistance of $AN$. $E / 6=16 / 20=4 / 5$
$\Rightarrow E=\frac{6 \times 4}{5}=\frac{24}{5}=4.8 V$

Hence option C is correct.
Question 4
1 / -0

An alternating voltage having frequency of 50 cycles/sec and maximum voltage 220 V is supplied to a circuit containing a pure inductance of 0.02 H and a pure resistance of 10 Ω in series. The value of maximum current in circuit is-

A
16.52 A

B
18.64 A

C
20.18 A

D
22.73 A

Solution

Impedance of L.R series circuit
$Z=\left(R^{2}+L^{2} \omega^{2}\right)^{\frac{1}{2}}$
$Z=\left[10^{2}+(0.02 \times 2 \times 3.14 \times 50)^{2}\right]^{\frac{1}{2}}$
$Z=(100+39.44)^{\frac{1}{2}}=11.8 \Omega$
value of current $=\frac{E_{\max }}{Z}=\frac{220}{11.8}=18.64 A$

Hence option B is correct.
Question 5
1 / -0

The antenna current of an AM transmitter is 8 A when only carrier wave is sent but the current increases to 8.88 A when the carrier wave is sinusoidally modulated. The percentage of modulation is

A
62%

B
65%

C
68%

D
71%

Solution

\begin{equation}\begin{array}{l}
\left(\frac{I_{T}}{I_{C}}\right)^{2}=1+\frac{m_{a}^{2}}{2} \\
I_{T}=8.88 A_{,} I_{C}=8 A ; m_{a}=? \\
\left(\frac{8.88}{8}\right)^{2}=1+\frac{m_{i}^{2}}{2} \\
1.232=1+\frac{m_{a}^{2}}{2} \\
\frac{m_{a}^{2}}{2}=1.232-1=0.232 \\
m_{a}=\sqrt{2 \times 0.232}=0.68 \text { or } 68 \%
\end{array}\end{equation}

Hence option C is correct.
Question 6
1 / -0

Figure shows the plot of a potential energy function of a conservative system U versus x. Which of the following statements is correct ?

A
The force has the greatest magnitude at point E₂.

B
The force is positive and greater than zero at points E₁, E₂, E₃ and E₄.

C
The system is in stable equilibrium at E₃.

D
The system is in neutral equilibrium at E₁.

Solution

$F=-\frac{\partial U}{\partial x}$
Force F versus $x$ is drawn in Figure. At points $E_{1}, E_{2}, E_{3}$ and $E_{4}$
$F=0$
F has the greatest magnitude nearer to point $E_{1.}$
If U is minimum l.e., $\frac{\partial U}{\partial x}=0$ and $\frac{\partial^{2} U}{\partial x^{2}}=$ positive, equilibrium is stable.
If U is maximum i.e., $\frac{\partial U}{\partial x}=0$ and $\frac{\partial^{2} U}{\partial x^{2}}=$ negative, equilibrium is unstable.
If U is constant i.e., $\frac{\partial U}{\partial x}=0$ and $\frac{\partial^{2} U}{\partial x^{2}}=0,$ equilibrium is neutral.
If U is constant i, e., $, \frac{\partial U}{\partial x}=0$ and $\frac{\partial^{2} U}{\partial x^{2}}=0,$ equilibrium is neutral.
Hence option D is correct.
Question 7
1 / -0

If a magnet is suspended at an angle of 30° to the magnetic meridian, and the dip needle shows 45° to the horizontal, the real dip is

A

\begin{equation}\tan ^{-1}(\sqrt{3} / 2)\end{equation}

B

\begin{equation}\tan ^{-1}(\sqrt{3})\end{equation}

C

\begin{equation}\tan ^{-1}(\sqrt{3} / \sqrt{2})\end{equation}

D

\begin{equation}\tan ^{-1}(2 / \sqrt{3})\end{equation}

Solution

Let $\delta$ and $\delta^{\prime}$ be the true dip and apparent dip respectively. Then we have the relation, $\tan \delta^{\prime}=\tan \delta \sec \theta$
where $\theta$ is the angle measured relative to the magnetic meridian. Given that $\theta=30 \circ, \delta^{\prime}=45^{\circ}$
$\tan \delta=\frac{\tan \delta^{\prime}}{\sec 30^{\circ}}$
$=\cos 30^{\circ} \tan 45^{\circ}$
$=\frac{\sqrt{3}}{2}$
$\therefore \delta=\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Hence option A is correct.
Question 8
1 / -0

The radius of a sphere is measured to be (2.1 ± 0.05) cm. Calculate the surface area (in cm²) with error limits.

A
54.44 ± 2.46

B
55.44 ± 4.64

C
52.03 ± 4.26

D
45.54 ± 2.68

Solution

$S=4 \pi r^{2}=4 \times \frac{22}{7} \times(2.1)^{2} $

$=55.44 cm ^{2} $

$\frac{\Delta S}{S}=2\left(\frac{\Delta r}{r}\right) $

$\Delta S=2\left(\frac{\Delta r}{r}\right) s $

$=\frac{2 \times 0.05 \times 55.44}{2.1}=2.64 $

$\therefore \Delta S=(55.44 \pm 2.64) cm ^{2}$

Hence option A is correct.
Question 9
1 / -0

When an UV light of 10¹⁵ Hz and intensity 2 W/m² is directed at a metal surface, photoelectrons emitted were found to have a maximum kinetic energy of 1.6 eV. If the work functions for different materials are as follows: Potassium 2.2 eV, Sodium 2.3 eV, Lithium 2.5 eV and Calcium 3.2 eV, identify the metal in the given problem.

A
Lithium

B
Calcium

C
Sodium

D
Potassium

Solution

$E=h v=6.63 \times 10^{-34} \times 10^{15}$
$=6.63 \times 10^{-19} J$
$=4.1 eV$
(dividing by $\left.e=1.6 \times 10^{-19}\right)$
$E_{\max }=h v-h v_{0}$
$\Rightarrow hv _{0}=4.1-1.6$
$=2.5 eV$
$\Rightarrow$ Lithium

Hence option A is correct.
Question 10
1 / -0

A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is

A
I/2

B
2I/11

C
I/10

D
I/5

Solution

Volume of the circular disc = surface area $*$ thickness
$=\pi r ^{2} \times \frac{ r }{6}=\frac{\pi r ^{3}}{6}$
The mass of the circular disc and solid sphere are the same $\frac{\pi r^{3}}{6} \times p=\frac{4}{3} \pi R^{3} \times p$
where R is the radius of the sphere, $\frac{r^{3}}{8}=R^{3}$
$\therefore R=\frac{r}{2}$
Moment of inertia of circular disc $I=\frac{1}{2} M r^{2}$
Moment of inertia of solid sphere $=\frac{2}{5} M R^{2}=\frac{2}{5} M \frac{r^{2}}{4}=\frac{1}{5} \times \frac{1}{2} m r^{2}=\frac{I}{5}$

Hence option D is correct.

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Physics Test - 5

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Physics Test - 5

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Question 1

0.56 Ns

0.64 Ns

0.72 Ns

0.78 Ns

Solution

Question 2

3R4

3R2

R4

R2

Solution

Question 3

4. 2 V

4.5 V

4.8 V

5.1 V

Solution

Question 4

16.52 A

18.64 A

20.18 A

22.73 A

Solution

Question 5

62%

65%

68%

71%

Solution

Question 6

The force has the greatest magnitude at point E2.

The force is positive and greater than zero at points E1, E2, E3 and E4.

The system is in stable equilibrium at E3.

The system is in neutral equilibrium at E1.

Solution

Question 7

\begin{equation}\tan ^{-1}(\sqrt{3} / 2)\end{equation}

\begin{equation}\tan ^{-1}(\sqrt{3})\end{equation}

\begin{equation}\tan ^{-1}(\sqrt{3} / \sqrt{2})\end{equation}

\begin{equation}\tan ^{-1}(2 / \sqrt{3})\end{equation}

Solution

Question 8

54.44 ± 2.46

55.44 ± 4.64

52.03 ± 4.26

45.54 ± 2.68

Solution

Question 9

Lithium

Calcium

Sodium

Potassium

Solution

Question 10

I/2

2I/11

I/10

I/5

Solution

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$\frac{3 R}{4}$

$\frac{3 R}{2}$

$\frac{R}{4}$

$\frac{R}{2}$

The force has the greatest magnitude at point E₂.

The force is positive and greater than zero at points E₁, E₂, E₃ and E₄.

The system is in stable equilibrium at E₃.

The system is in neutral equilibrium at E₁.