Physics Test

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Physics Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

Unit of solid angle is

A
Degree

B
Radian

C
Steradian

D
Radian-second

Solution

The steradian radian is the SI unit of solid angle. It is used in three-dimensional geometry, and is analogous to the radian, which quantifies planar angles.
Question 2
1 / -0

Lenz’s law is a consequence of the law of conservation of-

A
Charge

B
Momentum

C
Energy

D
Mass

Solution

According to Lenz law, the polarity of the induced emf is such that it opposes the change in magnetic flux responsible for its production.
Question 3
1 / -0

A flux of 0.2 Wb, linked with a coil having 50 turns, is reduced to zero in one second. What is the average emf induced in the coil?

A
10 V

B
20 V

C
25 V

D
30 V

Solution
Question 4
1 / -0

A wire of length $10 {~cm}$ has resistance $12 \Omega$. It is bent in the form of a circle. The effective resistance between the two points of its diameter is equal to:

A
$12 \Omega$

B
$6 \Omega$

C
$3 \Omega$

D
$24 \Omega$

Solution

Given the resistance $=12 \Omega$

If the total resistance of the wire is $R$, then after bending the total resistance will still be $R$.

Now, along any diameter, we can say that the whole circle is divided into two segments.

So, the resistance of each segment is half, i.e., $\frac{R}2$.

Now, these two segments are connected in parallel.

Resistance of the wire of a semicircle = $\frac{12}2=6 \Omega$

As the length of semicircle wire is half of the total length of wire and $R \propto l$.

There are two resistances $6 \Omega$ and $6 \Omega$ between two points on any diameter of the ring and they are in parallel.

Equivalent Resistance,

$\frac{1}{R_{e q}}=\frac{1}{6}+\frac{1}{6}$

$\Rightarrow R_{e q}=3 \Omega$
Question 5
1 / -0

Match List-I with List-II and select the correct answer with the help of codes given below:

A
2 1 3 4

B
1 2 4 3

C
1 2 3 4

D
1 2 3 4

Solution

The unit of temperature is kelvin, power is measured in watt, pressure in pascal and force in newton.
Question 6
1 / -0

A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

A
t^1/2

B
t^3/4

C
t^3/2

D
t²

Solution

Let's consider a body is moved along a straight line by a machine delivering a constant power $P$ The distance moved by the body is $S$.
Power, $P=F . v \ldots \ldots$ (1)
Force, $F=m a \ldots \ldots . .(2)$
where, $v=\frac{S}{t}$
acceleration, $a=\frac{S}{t^{2}}$
$m=m a s s$
from equation (1) and equation (2), we get
$P=\frac{m S}{t^{2}} \times \frac{S}{t}$
$S^{2}=\frac{P t^{3}}{m}$
From the above equation, we get
$S^{2} \propto t^{3}$
$S \propto t^{3 / 2}$
Question 7
1 / -0

Directions: The following question has four choices out of which only one is correct.
A bar magnet of length 3 cm has points A and B along its axis at distances of 24 cm and 48 cm, respectively, on the opposite sides.
Ratio of the magnitudes of the magnetic fields at these points will be

A
8 : 1

B
1 : 2 √2

C
3 : 1

D
4 : 1

Solution

For a short magnet at a point along its axial line, magnetic field is given by $B=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 M}{r^{3}}$
$\Rightarrow \quad \frac{B_{1}}{B_{2}}=\left(\frac{r_{2}}{r_{1}}\right)^{3}=\left(\frac{48}{24}\right)^{3}=\frac{8}{1}=8:1$
Question 8
1 / -0

Match List-I with List -II and select the correct answer using the codes given below the lists:

A
2 3 4 1

B
1 2 3 4

C
2 3 1 4

D
2 1 3 4

Solution

Light-year is a unit of length, joule is the unit of energy. Intensity of sound is measured in decibel. Frequency is measured in hertz (Hz).
Question 9
1 / -0

The mass of a planet is six times that of the Earth. The radius of the planet is twice that of the Earth. If the escape velocity from the Earth is $' v e'$ ,theescape velocity from the planet will be

A
$\sqrt{3} v_{e}$

B
$3 v_{e}$

C
$\sqrt{2} v_{e}$

D
$2 v_{e}$

Solution

Escape velocity is given by,
$v_{e}=\sqrt{\frac{2 G M_{e}}{R_{e}}}$
\[
\frac{v}{v_{e}}=\frac{\sqrt{\frac{2 G\left(6 M_{e}\right)}{2 R_{e}}}}{\sqrt{\frac{2 G M_{e}}{R_{e}}}}=\sqrt{3}
\]
$v=\sqrt{3} v_{e}$
Question 10
1 / -0

The potential energy in a spring, when stretched by 2 cm, is 'U'. Its potential energy, when stretched by 10 cm, is

A
$U 25$

B
$U 5$

C
25U

D
5U

Solution

$U=\frac{1}{2}\left(\frac{Y A}{L}\right) l^{2}$
$\because U \propto l^{2}$
$\frac{U_{2}}{U_{1}}=\left(\frac{l_{2}}{l_{1}}\right)^{2}=\left(\frac{10}{2}\right)^{2} \Rightarrow U_{2}=25 U_{1}$
i.e., potential energy of the spring will be $25 U$

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Physics Test - 8

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Physics Test - 8

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SHARING IS CARING

Question 1

Degree

Radian

Steradian

Radian-second

Solution

Question 2

Charge

Momentum

Energy

Mass

Solution

Question 3

10 V

20 V

25 V

30 V

Solution

Question 4

\(12 \Omega\)

\(6 \Omega\)

\(3 \Omega\)

\(24 \Omega\)

Solution

Question 5

2 1 3 4

1 2 4 3

1 2 3 4

1 2 3 4

Solution

Question 6

t1/2

t3/4

t3/2

t2

Solution

Question 7

8 : 1

1 : 2 √2

3 : 1

4 : 1

Solution

Question 8

2 3 4 1

1 2 3 4

2 3 1 4

2 1 3 4

Solution

Question 9

3ve

3ve

2ve

2ve

Solution

Question 10

U25

U5

25U

5U

Solution

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t^1/2

t^3/4

t^3/2

t²

$\sqrt{3} v_{e}$

$3 v_{e}$

$\sqrt{2} v_{e}$

$2 v_{e}$

$U 25$

$U 5$