Physics Test

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Physics Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The time period of a freely suspended magnet is 4 seconds. If it is broken in length into 2 similar parts and one part is suspended in the same way, then calculate its time period.

A
0.25 s

B
0.5 s

C
2 s

D
4 S

Solution

\[
\text { Time period, } T=2 \pi \sqrt{\frac{1}{M B_{H}}}=4
\]
When magnet is cut into two equal halves, then New magnetic moment \(M^{\prime}=\frac{M}{2}\)
New moment of inertia, \(l^{\prime}=\frac{(W / 2)(1 / 2)^{2}}{12}=\frac{1}{8} \cdot \frac{W^{2}}{12}\)
where "w' is the initial mass of the magnet.
\[
B u t, I=\frac{W I^{2}}{12}, \quad \therefore l^{\prime}=\frac{1}{8}
\]
Thus, New time period, \(T^{\prime}=2 \pi \sqrt{\frac{l^{\prime}}{M^{\prime} B_{H}}}\)
\[
\begin{array}{l}
\Rightarrow T^{\prime}=2 \pi \sqrt{\frac{1 / 8}{(M / 2) B_{H}}}=\frac{1}{2} 2 \pi \sqrt{\frac{1}{M_{H}}} \\
\Rightarrow T^{\prime}=\frac{1}{2} \times T=\frac{1}{2} \times 4=2 \mathrm{s}
\end{array}
\]
Question 2
1 / -0

Two charges +5μ and +10μ are kept 20 cm apart. Find the net electric field at the mid-point between the two charges.

A
13.5 x 10⁶ N/C directed towards +10μ

B
13.5 x 10⁶ N/C directed towards +5μ

C
4.5 x 10⁶ N/C directed towards +10μ

D
4.5 x 10⁶ N/C directed towards +5μ

Solution
Question 3
1 / -0

A current carrying rectangular coil is kept in a uniform magnetic field. In which orientation, the coil will not tend to rotate?

A
The magnetic field is perpendicular to the plane of the coil

B
The magnetic field is at 45^o with the plane of the coil

C
The magnetic field is parallel to the plane of the coil

D
Always in any orientation

Solution

The magnetic field is perpendicular to the plane of the coil
Question 4
1 / -0

As shown in following figure, four charges are kept on corners of a square having side of 5 cm. If Q is one microcoulomb, then what will be the electric field intensity at center?

A
2.04 x 10⁷ N/C downwards

B
2.04 x 10⁷ N/C upwards

C
1.02 x 10⁷ N/C downwards

D
1.02 x 10⁷ N/C upwards

Solution
Question 5
1 / -0

What is the acceleration of an electron in an electric field of magnitude 50 V/cm, if e/m value of the electron is 1.76 x 10¹¹ C/kg?

A
5.4 x 10¹² m/s²

B
6.4 x 10¹³ m/s²

C
8.8 x 10¹⁴ m/s²

D
Zero

Solution

Acceleration, \(a=\frac{e E}{m}\)
\(\Rightarrow a=1.76 \times 10^{11} \times 50 \times 10^{2}\)
\(\Rightarrow a=8.8 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}\)
Question 6
1 / -0

Masses of three wires of copper are in the ratio of 1:3:5 and their lengths are in the ratio of 5:3:1. What is the ratio of their electrical resistances?

A
125:15:1

B
1:15:125

C
5:3:1

D
1:3:5

Solution

\(\mathrm{R} \propto \frac{\left.\right|^{2}}{\mathrm{m}} \Rightarrow \mathrm{R}_{2}: \mathrm{R}_{2}: \mathrm{R}_{3}=\left(\frac{\mathrm{I}_{2}}{\mathrm{m}_{1}}\right)^{2}:\left(\frac{\mathrm{I}_{2}}{\mathrm{m}_{2}}\right)^{2}:\left(\frac{\mathrm{I}_{3}}{\mathrm{m}_{3}}\right)^{2}\)
\(=\frac{25}{1}: \frac{9}{3}: \frac{1}{5}=25: 3: \frac{1}{5} \Rightarrow 125: 15: 1\)
Question 7
1 / -0

The time period of a freely suspended magnet is 2 sec. If it is broken in length into two equal parts and one part is suspended in the same way, then calculate its time period.

A
1 s

B
√2 s

C
2 s

D
3 s

Solution

Here, \(T^{\prime}=\frac{T}{n} \Rightarrow T^{\prime}=\frac{2}{2}=1\)
Question 8
1 / -0

A charge (-q) and another charge (+Q) are placed at two points A and B respectively. Keeping the charge (+Q) fixed at B, the charge (- q) at A is moved to another point C such as ABC forms an equilateral triangle of side 'l'. What is the net work done in moving the charge (- q)?

A
zero

B

C

D

Solution

From the figure, potential at A and C are equal. Thus, work done in moving - q charge from A to C is zero.
Question 9
1 / -0

A
7 V/m

B
5√2 V/m

C
4√2 V/m

D
3√2 V/m

Solution
Question 10
1 / -0

A charged particle of mass 'm' and charge 'q' is released from rest in a uniform electric field E. What is the kinetic energy of the charged particle after 't' second, by neglecting the effect of gravity?

A
Eqm/t

B
E²q²t²/2m

C
2E²t²/mq

D
Eq²m/2t²

Solution

Then charge 'q' is released in an uniform electric field E then its acceleration a=qE/m (is constant

Thus, its motion will be uniformly accelerated motion and its velocity after time 't' is given by,