Physics Test

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Physics Test - 10

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Question 1
4 / -1

For the given uniform square lamina ABCD whose centre is O,

A
I_AC = I_EF

B
I_AC =√2I_EF

C
√2 I_AC = I_EF

D
I_AD= 3I_EF

Solution

Using perpendicular axis theorem:

2I_AC= I₀, where I₀ is the moment of inertia about an axis perpendicular to the plane of the lamina and passing through O.

Both diagonals are perpendicular to each other.

Further, 2I_EF = Io

Two medians (lines dividing the sides of the square equally) are also perpendicular to each other, also perpendicular to the axis passing through the centre and perpendicular to the plane of lamina.
Question 2
4 / -1

F₁ and F₂ are focal lengths of the objective and the eyepiece of a telescope, respectively. The angular magnification for this telescope is equal to

A
F₁/F₂

B
F₂/F₁

C
\(\frac{F_{1} F_{2}}{F_{1}+F_{2}}\)

D
\(\frac{F_{1}+F_{2}}{F_{1} F_{2}}\)

Solution

Angular magnification is the ratio of the focal length of the objective to the focal length of the eyepiece.Therefore, \(M=\frac{F_{1}}{F_{2}}\)
Question 3
4 / -1

The half-life of ¹³¹I is 8 days. If we have a sample of ¹³¹I at time t = 0, we can assert that

A
no nucleus will decay before t = 4 days

B
no nucleus will decay before t = 8 days

C
all nuclei will decay before t = 16 days

D
a given nucleus may decay any time after t = 0

Solution

A radioactive nucleus may decay any time after \(t=0 .\) The number of nuclei at any time t is given by \(\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-i t},\) where \(\lambda=\frac{0.693}{\mathrm{T}}\) where \(\mathrm{T}\) is the half-life.
Question 4
4 / -1

In the following figure, the height of the object is H₁ = + 2.5 cm. What is the height of the image of H₁?

A
- 5 cm

B
+ 5 cm

C
+ 7.5 cm

D
- 7.5 cm

Solution

Since the given mirror is a concave mirror, the image formed will be inverted and double the size of the object.

Height of the object = 2.5 cm

Height of the image = – 5 cm; -ve sign indicates that the image is inverted.
Question 5
4 / -1

A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of 75°. One of the fields has a magnitude of√2 x 10^-2 T. The dipole attains stable equilibrium at an angle of 30° with this field. What is the magnitude of the other field?

A
0.01 T

B
0.02 T

C
0.03 T

D
0.04 T

Solution

Refer to the figure given below: Let \(\theta_{1}\left(=30^{\circ}\right)\) be the angle between the magnetic
moment vector \(\mathrm{M}\) and the field vector \(\mathrm{B}_{1}\left(=1.5 \times 10^{-2} \mathrm{T}\right) .\) Then, as shown in the figure, the angle between \(\mathrm{M}\) and the other field \(\mathrm{B}_{2}\) will be \(\theta_{2}=75^{\circ}-30^{\circ}=45^{\circ}\)

The field exerts a torque \(\mathcal{I}_{1}=\mathrm{M} \times \mathrm{B}_{1}\) on the dipole and the field \(\mathrm{B}_{2}\) exerts a torque
\(\tau_{2}=\mathrm{M} \times \mathrm{B}_{2},\) where \(\mathrm{M}\) is the magnetic moment of the dipole. since the dipole is in
stable equilibrium, the net torque \(\tau\left(=\tau_{1}+\tau_{2}\right)\) must be zero, i.e. the two torques must be equal and opposite. In terms of magnitudes, we have \(\mathrm{mB}_{1} \sin \theta_{1}=\mathrm{mB}_{2} \mathrm{sin}\) \(\theta_{2}\)
or, \(\mathrm{B}_{2}=\frac{\mathrm{B}_{1} \sin \theta_{1}}{\sin \theta_{2}}\)
\(=\frac{\sqrt{2} \times 10^{-2} \times \sin 30^{\circ}}{\sin 45^{\circ}}=0.01 \mathrm{T}\)
Question 6
4 / -1

Two SHMs are respectively represented by y₁ = a sin (ωt - kx), and y₂ = b cos (ωt - kx). The phase difference between the two is:

A
π/2

B
π/4

C
π/6

D
3π/4

Solution

As

y1 = a sin (ωt - kx), and y2 = b cos (ωt - kx)

y2 = b sin (ωt - kx + (π/2))

Hence phase difference between them is π/2.
Question 7
4 / -1

A magnet of magnetic moment M and length 2l is bent at its mid-point such that the angle of bending is 60^o. Now, the magnetic moment is

A
M

B
2M

C
\(\frac{M}{2}\)

D
\(\frac{M}{\sqrt2}\)

Solution

In new situation we have

As the length of magnet is halved, Magnetic Moment \(M=m(\ell)=\frac{M}{2}\) Resultant Magnetic Moment
\(\begin{aligned} M_{R} &=\sqrt{\left(\frac{M}{2}\right)^{2}+\left(\frac{M}{2}\right)^{2}+2\left(\frac{M}{2}\right)\left(\frac{M}{2}\right) \cos 120^{\circ}} \\ &=\left(\frac{M}{2}\right) \sqrt{1+1-2 \times \cos 60^{\circ}} \\ &=\frac{M}{2} \sqrt{1+1-2 \times \frac{1}{2}} \\ & \Rightarrow M_{R}=\frac{M}{2} \end{aligned}\)
Question 8
4 / -1

A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m, then

A
each emitted electron carries half the initial energy

B
number of electrons emitted is a quarter of the initial number

C
each emitted electron carries one quarter of the initial energy

D
number of electrons emitted is half the initial number

Solution

Key Idea Number of photoelectrons einitted per second is directly proportional to intensity of light. Intensity of light source is
\[
I \propto \frac{1}{d^{2}}
\]
When distance is doubled, intensity becomes one-fourth. As number of photoelectrons \(\propto\) intensity, so number of photoelectrons is quarter of the initial number.
Question 9
4 / -1

A uniform metre scale of length 1 m is balanced on a fixed semi-circular cylinder of radius 30 cm as shown in the figure. One end of the scale is slightly depressed and released. The time period (in seconds) of the resulting simple harmonic motion is

(Take g = 10 ms^-2)

A
π

B
π/2

C
π/3

D
π/4

Solution

Refer to Fig. The magnitude of the restoring torque \(=\) force \(\times\) perpendicular distance

\[
\begin{array}{l}
=m g \times A B \\
=m g \times R \sin \theta
\end{array}
\]

since \(\theta\) is small, \(\sin \theta=\theta\). Here \(\theta\) is expressed in radian. The equation of motion of the scale is
\[
I \frac{d^{2} \theta}{d t^{2}}=-m g R \theta
\]
or \(\frac{d^{2} \theta}{d t^{2}}=\left(-\frac{m g R}{I}\right) \theta\)
\(\therefore\)
\(\omega=\sqrt{\frac{m g R}{I}}\) or \(\frac{2 \pi}{T}=\sqrt{\frac{m g R}{I}}\)
or \(T=2 \pi \sqrt{\frac{I}{m g R}} .\) Now \(I=\frac{m L^{2}}{12} .\) Hence
\(T=\frac{\pi L}{\sqrt{3 g R}}\)
Using the values \(L=1 \mathrm{m}, g=10 \mathrm{ms}^{-2}\) and \(R=0.3 \mathrm{m}\), we get \(T=\pi / 3\) second. Hence the correct choice is
(3).
Question 10
4 / -1

A magnet of length 10 cm and magnetic moment 1 Am² is placed along the side AB of an equilateral triangle ABC. If the length of side AB is 10 cm, then the magnetic field at point C is

A
10^-9 T

B
10^-7 T

C
10^-5 T

D
10^-4 T

Solution

Let m be the pole strength of each pole of the magnet figure. The magnetic field at C due to the N-pole is given by

\(B_{1}=\frac{\mu_{0}}{4 \pi} \frac{m}{(A C)^{2}}\)

direction along \(A C\) away from \(C .\) The magnetic induction at \(C\) due to the S-pole is given by
\[
B_{2}=\frac{\mu_{0}}{4 \pi} \cdot \frac{m}{(B C)^{2}}
\]
directed along \(C B\) towards \(B .\) since \(A C=B C, B_{1}=B_{2}\) The resultant magnetic induction at \(C\) is given by
\[
\begin{aligned}
B^{2} &=B_{1}^{2}+B_{2}^{2}+2 B_{1} B_{2} \cos 120^{\circ} \\
&=B_{1}^{2}+B_{2}^{2}-B_{1} B_{2} \\
&=2 B_{1}^{2}-B_{1}^{2}=B_{1}^{2} \quad\left(\because B_{2}=B_{1}\right)
\end{aligned}
\]
or
\[
\begin{aligned}
B=B_{1} &=\frac{\mu_{0} m}{4 \pi(A C)^{2}}=\frac{\mu_{0} m}{4 \pi a^{2}} \\
&=\frac{\mu_{0}}{4 \pi} \cdot \frac{(m a)}{a^{3}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{M}{a^{3}}
\end{aligned}
\]
Given: \(M=1 \mathrm{A} \mathrm{m}^{2}, a=10 \mathrm{cm}=0.1 \mathrm{m} .\) Also \(\mu_{0}=\)
\(4 \pi \times 10^{-7} \mathrm{T} \mathrm{A}^{-1} \mathrm{m} .\) Substituting these values in (1)
we get \(B=10^{-4} \mathrm{T},\) which is choice (4)