Physics Test

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Physics Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

In a hollow spherical shell, potential (V) changes with respect to distance (s) from centre as :

A
1

B
2

C
3

D
4

Solution

In shell, q charge is uniformly distributed over its surface, it behaves as a conductor.
$V=$ potential at surface $\frac{q}{4 \pi \epsilon_{o} R}$ and inside $V=\frac{q}{4 \pi \epsilon_{0} R}$
Because of this it behaves as an equipotential surface.
Question 2
1 / -0

A circular coil of 20 turns and radius 20 cm is mounted vertically with its plane in magnetic meridian. A small magnetic needle is placed at the centre of the coil. When a current is passed through the coil, the needle is deflected through 45^o . If the horizontal component of Earth's field is 0.34 x 10^-4 T, the current in the coil is:

A
0.6 A

B
6 A

C
6 x 10^-3 A

D
0.06 A

Solution

The arrangement is a tangent galvanometer
\[
\begin{array}{l}
I=K \tan \theta \\
=\frac{2 r B_{H}}{\mu_{0} n} \tan \theta \\
\therefore I=\frac{2 * 20 * 10^{-2} * 0.34 * 10^{-9}}{4 \pi * 10^{-7} * 20} * \tan 45^{\circ} \\
=\frac{2 * 3.4}{4 \pi}=\frac{17}{10 \pi} \text { ampere }
\end{array}
\]
=0.6 ampere
Question 3
1 / -0

The potential energy (in joules) of a mass of 2 kg of mass moving in the xy plane is given by U = 6x + 8y, where positions x and y are measured in meters. If the body is resting at a point (6 m, 4 m) at time t (0), it will cross the y-axis equal to time

A
1 s

B
2 s

C
3 s

D
4 s

Solution

Given that,
Mass $=2 \mathrm{kg}$
The potential energy is
$U=6 x+8 y$
We need to calculate the force
Using potential energy
$F=-\frac{d U}{d x}-\frac{d U}{d y}$
Put the value into the formula
$F=-6 i-8 j$
We need to calculate acceleration along $x$ axis
Using formula of acceleration
$a_{x}=\frac{F}{m}$
Put the value into the formula
$a_{x}=\frac{-6}{2}$
$a_{x}=-3 \mathrm{m} / \mathrm{s}^{2}$
We need to calculate acceleration along $x$ axis
Using formula of acceleration
$a_{y}=\frac{F}{m}$
Put the value into the formula
$a_{y}=\frac{-8}{2}$
$a_{y}=-4 \mathrm{m} / \mathrm{s}^{2}$
When the particle cross the y axis then $x=0$
So, The displacement of the object is $-6 \mathrm{m}$
$x_{\text {initial}}=6 \mathrm{m}$
We need to calculate the time
Using equation of motion $s=u t+\frac{1}{2} a t^{2}$
Put the value into the formula
$6=0+\frac{1}{2} \times 3 \times t^{2}$
$t=2 \sec$
Hence, The time will be 2 sec.
Question 4
1 / -0

When a material is subjected to a small magnetic field H, the intensity of magnetisation is proportional to

A
H^1/2

B
H

C
H²

D
H^-1/2

Solution

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

The speed of light will be minimum while passing through-

A
Glass

B
Vacuum

C
Water

D
Air

Solution

Any of the object or material which has the highest refractive index has the minimum speed of light. The glass has the highest refractive index among other like, vacuum, water and air.
Question 6
1 / -0

Under the action of a force F = Cx, the position of a body changes from 0 to x. The work done is

A
$\frac{1}{2} C x^{2}$

B
$C x^{2}$

C
$C x$

D
$\frac{1}{2} C x$

Solution

\begin{array}{l}
\text { Work done } W=\int_{0}^{x} F d x \\
W=\int_{0}^{x} C x d x \\
W=C \times\left.\frac{x^{2}}{2}\right|_{0} ^{x} \\
\therefore W=\frac{C}{2} \times\left(x^{2}-0\right) \quad \Longrightarrow W=\frac{1}{2} C x^{2}
\end{array}
Question 7
1 / -0

The unit of electrical resistance of a conductor is-

A
positive

B
volt

C
ampere

D
Ohm

Solution

Ohm is the SI unit of electrical resistance. 1 ohm is define as the resistance of a conductor when a potential difference of 1 volt is applied to its ends when a current of 1 ampere flows through it
Question 8
1 / -0

One astronomical unit is the average distance between

A
Earth and the sun

B
Earth and the moon

C
Jupiter and the sun

D
Pluto and the sun

Solution

One astronomical unit is the approximate mean distance between the Earth and sun. It’s about 93 million miles (150 million km), or 8 light-minutes.
Question 9
1 / -0

The escape velocity from Earth is 11.2 kms^-1. Another planet has mass 1000 times and radius 10 times that of Earth. The escape velocity at that planet will be

A
11.2 km/s

B
112 km/s

C
1.12 km/s

D
1120 km/s

Solution

$v_{e}=\sqrt{2 g R}=\sqrt{2 \frac{G M}{R^{2}}} \cdot R$ or $v \propto \sqrt{\frac{M}{R}}$
Mass is 1000 times and radius is 10 times $v_{e}^{\prime}=\sqrt{\frac{(1000 M) G}{100 R}}$
$\Rightarrow v_{e}^{\prime}=10 \sqrt{\frac{G M}{R}}$
$\Rightarrow v_{e}^{\prime}=10 V_{e} \Rightarrow v_{e}^{\prime}=10 \times 112 \Rightarrow v_{e}^{\prime}=112 \mathrm{kms}^{-1}$
Therefore, escape velocity will become 10 times.
Question 10
1 / -0

Current in a 60 W light bulb when it is connected to a 240 V power supply is___________.

A
0.26

B
0.28

C
0.25

D
1.25

Solution

According to ohm's law, Ohm's law equation (formula): V = I X R and the power law equation (formula): P = I X V. P = power
So, 60= I x 240
Therefore, I= 0.25 A

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Physics Test - 6

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Physics Test - 6

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Question 1

1

2

3

4

Solution

Question 2

0.6 A

6 A

6 x 10-3 A

0.06 A

Solution

Question 3

1 s

2 s

3 s

4 s

Solution

Question 4

H1/2

H

H2

H-1/2

Solution

Question 5

Glass

Vacuum

Water

Air

Solution

Question 6

12Cx2

Cx2

Cx

12Cx

Solution

Question 7

positive

volt

ampere

Ohm

Solution

Question 8

Earth and the sun

Earth and the moon

Jupiter and the sun

Pluto and the sun

Solution

Question 9

11.2 km/s

112 km/s

1.12 km/s

1120 km/s

Solution

Question 10

0.26

0.28

0.25

1.25

Solution

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6 x 10^-3 A

H^1/2

H²

H^-1/2

$\frac{1}{2} C x^{2}$

$C x^{2}$

$C x$

$\frac{1}{2} C x$