Physics Test

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Physics Test - 25

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Question 1
3 / -1

A particle in simple harmonic motion is described by displacement function $\mathrm{x}(\mathrm{t})=\mathrm{A} \cos (\omega \mathrm{t}+\phi)$ If the initial position of the particle is $1 \mathrm{~cm}$ and initial velocity is $\pi \mathrm{cm} / \mathrm{s}$, then what is its amplitude and initial phase? Let angular frequency of particle be $\pi$ per second.

A
1 cm

B
2 cm

C
$\sqrt 2$ cm

D
3 cm

Solution

Here, $\mathrm{x}(\mathrm{t})=\mathrm{A} \mathrm{A} \cos (\omega t+\phi)$
$x(0)=1 \mathrm{~cm}$ and $v(0)=\pi \mathrm{cm} / \mathrm{s}$
Differentiating $x(t)$ w.r.t. $t .$ $v(t)=-A \omega \sin (\omega t+\phi)$
At $t=0$ $x(0)=A \cos (\phi)$
and $v(0)=-A \omega \sin (\phi)$
Dividing (3) by (2) $\frac{v(0)}{x(0)}=-\omega \tan\phi$
or $\pi=-\omega tan\phi$ or $ \tan \phi=-1$
$\phi=\frac{3 \pi}{4}$ or $\frac{7 \pi}{4}$
Since at $t=0$ $x(0)=A \cos (\phi)=1$
$\Rightarrow \cos \phi$ is positive. Therefore $\phi=\frac{7 \pi}{4}$
$\therefore A \cos \left(\frac{7 \pi}{4}\right)=1$
or A. $A=\sqrt{2} \mathrm{~cm}$
Question 2
3 / -1

Three weights w, 2w, and 3w are connected to an identical spring suspended from a rigid horizontal rod. The assembly of the rod and the weights fall freely. The position of the weight from the rod is such that:

A
$3 w$ will be farthest

B
$w$ will be farthest

C
all will be at the same distance

D
$2 w$ will be farthest

Solution

Since all the weights fall freely under gravity along with the assembly of the rod, the acceleration of all objects is equal to $g$ . Hence, all the weights have the same relative distance between them at all points, and hence no stretching of spring occurs and thus the position (a distance of falling) of all the weights would be the same.
Question 3
3 / -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 $\sqrt{2} \times 10^{- 2} T$ . 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

The magnetic dipole attains stable equilibrium under the influence of these two fields making an angle $θ_{1} = 30 °$ with $B_{1}$ and $θ_{2} = 75 ° - 30 ° = 45 °$ with $B_{2}$ . For stable equilibrium, the net torque acting on the dipole must be zero, i.e., $τ_{1} + τ_{2} = 0$ or $τ_{1} = τ_{2}$
or $m B_{1} \sin θ_{1} = m B_{2} \sin θ_{2}$
$\Rightarrow B_{2} = B_{1} \frac{\sin θ_{1}}{\sin θ_{2}} = 15 m T \times \frac{\sin 30 °}{\sin 45 °}$
$= 15 m T \times \frac{1}{2} \times \sqrt{2} = 10.6 m T \approx 11 m T$
$= 0.01 T$
Question 4
3 / -1

In an oscillating LC circuit, the maximum charge on the capacitor is $Q$ . The charge on the capacitor, when energy is stored equally between the electric and magnetic fields, is:

A
$\frac{Q}{2}$

B
$\frac{Q}{\sqrt{2}}$

C
$\frac{Q}{\sqrt{3}}$

D
$\frac{Q}{3}$

Solution

If energy is stored equally in electric and magnetic fields, then the energy in an electric field will be half of the maximum value, which occurs when the charge on the capacitor is maximum.
Question 5
3 / -1

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

A
-5cm

B
+5cm

C
+7.5cm

D
-7.5cm

Solution
Question 6
3 / -1

For the given uniform square lamina $A B C D$ whose centre is $O$ , Identify the correct relation.

A
$I_{A C} = I_{E F}$

B
$I_{A C} = \sqrt{2} I_{E F}$

C
$\sqrt{2} I_{A C} = I_{E F}$

D
$I_{A D} = 3 I_{E F}$

Solution

Perpendicular Axis theorem:
$I_{z} = I_{x} + I_{y}$
(for a body in $X Y$ plane)
Where,
$I_{z} =$ moment of inertia about the $z$ -axis.
$I_{x}, I_{y} =$ moment of inertia about $x$ and $y$ -axis in the plane of the body respectively.
$I_{E F} = M \frac{(a^{2} + b^{2})}{12} [∴ a = b]$
$I_{E F} = M \frac{(a^{2} + a^{2})}{12} = \frac{M a^{2}}{6}$
$I_{Z} = \frac{M a^{2}}{6} + \frac{M a^{2}}{6} = \frac{M a^{2}}{6}$
By a theorem of the perpendicular axis,
$I_{A C} + I_{B D} = I_{Z} = I_{A C} = \frac{I_{2}}{2} = \frac{M a^{2}}{6}$
By a theorem of the perpendicular axis,
$I_{A C} + I_{B D} = I_{Z} = I_{A C} = \frac{I_{2}}{2} = \frac{M a^{2}}{6}$
Similarly, $I_{E F} = \frac{I_{Z}}{2} = \frac{M a^{2}}{6}$
$∴ I_{A C} = I_{E F}$
Question 7
3 / -1

A circular metal plate of radius 'R' is rotating with a uniform angular velocity 'ω' with its plane perpendicular to a uniform magnetic field B. Then, emf developed between the centre and the rim of the plate is

A
πωBR²

B
ωBR²

C
πωBR²/2

D
ωBR²/2

Solution

The change will accumulate on the rim due to the half effect until their fields are balanced by the magnetic field's force at a radius r.
Question 8
3 / -1

Directions: The following question has four choices out of which ONLY ONE is correct.

An open knife-edge of mass $M$ is dropped from a height $' h'$ on a wooden floor. If the blade penetrates $' s'$ into the wood, the average resistance offered by the wood to the blade is:

A

B

C

D

Solution
Question 9
3 / -1

A radioactive nuclide can decay simultaneously by two different processes, which have individual decay constants $λ_{1}$ and $λ_{2}$ . The effective decay constant of the nuclide $(λ)$ is given by:

A

B

C

D

Solution
Question 10
3 / -1

The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let $I_{a}, I_{b}$ , and $I_{c}$ be the intensities and $ν_{a}, ν_{b}$ and $ν_{c}$ be the frequencies for the curves $a, b$ , and $c$ respectively. Then,

A

B

C

D

Solution

From the graph, we note that the saturation current is the same for curves $b$ and $c$ but different for a curve $a$ . Therefore, intensities of $b$ and $c$ will be equal but different from that of $a$ i.e., $I_{a} \neq I_{b}$ but $I_{b} = I_{c}$ . As stopping potential is the same for curves $a$ and $b$ , hence $v_{a} = v_{b}$ .