Physics Test

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Physics Test - 13

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Question 1
1 / -0

Two identical cars having like charges and placed at a distance R apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to 1/4^th their initial separation. The force of repulsion between them increases 20 times in comparison with the initial value. The ratio of their initial charges (Given Q₁>Q₂)

A
$\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

B
$\frac{\sqrt{5} + 2}{\sqrt{5} - 2}$

C
$\frac{\sqrt{5} + 1}{\sqrt{5} - 1}$

D
None of the above

Solution
Question 2
1 / -0

A wall made up of two layer X and Y. The thickness of the two layers is the same, but materials are different. The thermal conductivity of X is thrice than that of Y. In thermal equilibrium, the temperature difference between the two ends is 56^oC. Then the difference of temperature across wall X is

A
21°C

B
35°C

C
42°C

D
57°C

Solution

Suppose the thickness of each wall is 't. Then
$\left(\frac{Q}{T}\right)_{\text {combin }}$$=\frac{Q}{T}$
$\frac{K_{S} A\left(\theta_{1}-\theta_{2}\right)}{2 t}=\frac{2 K A\left(\theta_{1}-\theta\right)}{t}$
$K_{S}=\frac{2 \times 3 K \times K}{(3 K+K)}$
$K_{S}=\frac{3 K}{2}$
$\left(\theta_{1}-\theta_{2}\right)=56^{\circ} C$
$\Rightarrow \frac{3 K \times A \times 56}{2 \times 2 t}=\frac{2 K A\left(\theta_{1}-\theta\right)}{t}$
$\left(\theta_{1}-\theta\right)=21^{\circ} C$
Hence temperature difference across the walls $X$ is $\left(\theta_{1}-\theta\right)=21^{\circ} C$

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

A box is floating in the river of speed 5m/s. The position of the block is shown in the figure at t=0. A stone is thrown from point O at time t = 0 with a velocity $v = (v_{1} \hat{ı} + v_{2} \hat{ȷ} + 40 \hat{k}) m / s .$ Find the value of $v_{1}, v_{2}$ such that the stone hits the box.

A
$v_{1} = 3 m / s, v_{2} = 5 m / s$

B
$v_{1} = 2 m / s, v_{2} = 3 m / s$

C
$v_{1} = 5 m / s, v_{2} = 7.5 m / s$

D
$v_{1} = 4 m / s, v_{2} = 5 m / s$

Solution
Question 4
1 / -0

A concave mirror of radius of curvature h is placed at the bottom of a tank containing a liquid of refractive index µ up to a depth d. An object O is placed at height h above the bottom of the mirror. Outside the liquid, an observer M views the object and its image in the mirror. The apparent distance between these two will be

A
0 cm

B
3 cm

C
7 cm

D
9 cm

Solution
Question 5
1 / -0

A pendulum bob of mass M = 80 gm connected to the end of a string of length L = 5 m is released from rest from horizontal position as shown in the figure. At the lowest point, the bob makes an elastic collision with a stationary box of mass 8M, which is kept on a friction less surface. Then the tension in the string just after the impact is

A
6.1 N

B
3.4 N

C
1.7 N

D
8.6 N

Solution

According to law of motion $v^{2}=u^{2}+2 g h$
$v^{2}=2 g L$
$v=\sqrt{2 g L}$
So, the velocity of the bob just before the impact is $v=\sqrt{2 g L}$ along the horizontal direction. From conservation of momentum,
$M v=-M v_{1}+8 M v_{2}$.......(1)
From the coefficient of restitution equation,
$1=\frac{v_{1}+v_{2}}{v} \Rightarrow v_{1}+v_{2}=v$
From equation (1)
$M v_{1}+M v_{2}=-M v_{1}+8 M v_{2}$
$v_{1}=\frac{7}{9} v$ and $v_{2}=\frac{2}{9} v$
For tension in the string
\begin{equation}\begin{array}{l}
T-M g=\frac{M v_{1}^{2}}{L} \\
T=M g+\frac{49 M v^{2}}{81 L} \\
T=M g+\frac{49 M \times 2 g L}{81 L} \\
T=\frac{179}{81} M g \\
T=\frac{179}{81} \times 0.08 \times 10 \\
T=1.7 N
\end{array}\end{equation}

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

A stationary bomb explodes into two parts of masses 3 kg3 kg and 1 kg1 kg. The total KE of the two parts after explosion is 2400 J2400 J. The KE of the smaller parts is :

A
600 J

B
1800 J

C
1200 J

D
2160 J

Solution

KE = p22m

P22(1) + P22(3) = 2400 or

P2 = 2400 × 32

P22m = 2400 × 34 = 1800 J.

Hence option B is the correct answer.
Question 7
1 / -0

A point moves along an arc of a circle of radius $R$. Its velocity depends on the distance covered $s$ as $v=a \sqrt{s},$ where $a$ is a constant. The angle $\alpha$ between the
vector of the total acceleration and the vector of velocity as a function of $s$ is given as tan $\alpha=\frac{x s}{R}$. Find $x$

A
2s/R

B
R/s

C
s/R

D
1

Solution

From the equation $v=a \sqrt{s}$
$A_{t}=\frac{d v}{d t}=\frac{a}{2 \sqrt{s}} \frac{d s}{d t}=\frac{a}{2 \sqrt{s}} a \sqrt{s}=\frac{a^{2}}{2},$ and
$A_{n}=\frac{v^{2}}{R}=\frac{a^{2} s}{R}$
As $A_{t}$ is a positive constant, the speed of the particle increases with time, and the tangential
acceleration vector and velocity vector coincides in direction. Hence the angle between $\vec{v}$ and $\vec{A}$ is equal to that between $\vec{A}_{t}$ an $\vec{A}$, and $\alpha$ can be found by
means of the formula:
$\tan \alpha=\frac{\left|A_{n}\right|}{\left|A_{t}\right|}=\frac{\frac{a^{2} s}{R}}{\frac{a^{2}}{2}}=\frac{2 s}{R}$

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

Three identical cars A,B and C are moving at the same speed on three bridges. The car A goes on a plane bridge, B on a bridge convex upward and C goes on a bridge concave upward. Let FA, FB and FC be the normal force exerted by the cars on the bridges when they are at the middle of bridges.

A
FA is maximum of the three forces

B
FB is maximum of the three forces

C
FC is maximum of the three forces

D
FA=FB=FC

Solution

FA, FB and FC are the magnitude of the normal contact forces with which car and bridge exert forces on each other.
for car ${ }^{\prime} A^{\prime}$
$FA = mg \quad \ldots \ldots \ldots$ (I)
for car 'B'
$mv 2 r = mg - FB$
$\Rightarrow FB = mg - mv 2 r$..........(II)
for car 'C'
$mv 2 r = Fc - mg$
$\Rightarrow Fc = mg + mv 2 r$.......(III)
from equations (III), (II) and (I), we see that
$F _{ C }>FA >Fb$

Hence option C is the correct answer.
Question 9
1 / -0

The charged particle 5×1020 will flow per sec through a heater. What is the current passed through it ?

A
8 A

B
80 A

C
50 A

D
16 A

Solution

Magnitude of charge of electron is $1.6 \times 10-19 C$
So total charge, $q=5 \times 1020 \times 1.6 \times 10-19=80 C$
Now current $i = qt =80 A ,$ as time is $I$ second.

Hence option B is the correct answer.
Question 10
1 / -0

A wheel having a diameter of 3 m starts from rest and accelerates uniformly to an angular velocity of 210 r.p.m. in 5 seconds. Angular acceleration of the wheel is.

A
4.4πrads2

B
3.3πrads2

C
2.2πrads2

D
1.1πrads2

Solution

Initial angular velocity, $\omega 0=0$ Final angular velocity, $\omega \circ=21$ orpm $=210 \times 2 \Pi 6 orad / sec$
$=210 \times 2 \times 227 \times 60$
$=22 rad / sec$
Use formula, $\omega=\omega \circ+\alpha t$
$22=0+\alpha \times 5$
alpha $=225=4.4 rads 2$

Hence option A is the correct answer.