1. Physics model

When the conducting strip moves on the U-shaped rails in a magnetic field, a motional emf is induced:

$\mathcal E = B\ell v$

where:

• $B = 0.1\,\text{T}$
• $\ell = 10\,\text{cm} = 0.1\,\text{m}$
• $v$ is the speed of the strip.

Since the total resistance is due to the strip only,

$R = 10\,\Omega$

so the induced current is

$i = \frac{\mathcal E}{R} = \frac{B\ell v}{R}$

This current-carrying strip in magnetic field experiences a retarding magnetic force:

$F = B\ell i = B\ell \left(\frac{B\ell v}{R}\right)=\frac{B^2\ell^2}{R}v$

Thus the damping force is proportional to velocity:

$F_d = -bv$

with damping constant

$b = \frac{B^2\ell^2}{R}$

2. Equation of motion

The strip attached to spring executes damped SHM:

$m\ddot x + b\dot x + kx=0$

Given:

• $m = 50\,\text{g} = 0.05\,\text{kg}$
• $k = 0.5\,\text{N m}^{-1}$

Compute $b$:

$b = \frac{(0.1)^2(0.1)^2}{10}$

$b = \frac{10^{-2}\cdot 10^{-2}}{10} = \frac{10^{-4}}{10}=10^{-5}\,\text{kg s}^{-1}$

3. Decay of amplitude

For damped SHM, amplitude decays as

$A(t)=A_0 e^{-\frac{b}{2m}t}$

We want amplitude to decrease by a factor of $e$:

$\frac{A}{A_0}=\frac{1}{e}$

So,

$e^{-\frac{b}{2m}t}=e^{-1} \implies \frac{b}{2m}t=1$

Hence,

$t=\frac{2m}{b}$

Substitute values:

$t=\frac{2(0.05)}{10^{-5}}=\frac{0.1}{10^{-5}}=10^4\,\text{s}$

4. Time period of oscillation

Since damping is very small, the period is approximately the natural period:

$T \approx 2\pi\sqrt{\frac{m}{k}}$

$T=2\pi\sqrt{\frac{0.05}{0.5}}=2\pi\sqrt{0.1}$

$\sqrt{0.1}\approx 0.316$

So,

$T \approx 2\pi(0.316) \approx 1.99\,\text{s} \approx 2\,\text{s}$

5. Number of oscillations

$N=\frac{t}{T} \approx \frac{10^4}{2}=5000$

6. Option check

The closest option is:

$\boxed{5000}$

So the correct option is C.