JEE Physics 2026 Question with Solution

Given: Two joined strings are under the same tension $T = 500 \, \text{N}$ with $\mu_A = 2 \times 10^{-4} \, \text{kg/m}$, $\mu_B = 4 \times 10^{-4} \, \text{kg/m}$, $L_A = 2.5 \, \text{m}$ and $L_B = 1.5 \, \text{m}$.

Find: The ratio $t_A/t_B$.

The speed of a transverse wave on a string is

$$v = \sqrt{\frac{T}{\mu}}$$

So the time taken by a pulse to travel length $L$ is

$$t = \frac{L}{v} = L\sqrt{\frac{\mu}{T}}$$

Hence,

$$t_A = L_A\sqrt{\frac{\mu_A}{T}}$$

and

$$t_B = L_B\sqrt{\frac{\mu_B}{T}}$$

Taking the ratio,

$$\frac{t_A}{t_B} = \frac{L_A\sqrt{\mu_A/T}}{L_B\sqrt{\mu_B/T}} = \frac{L_A}{L_B}\sqrt{\frac{\mu_A}{\mu_B}}$$

Substituting the given values,

$$\frac{t_A}{t_B} = \frac{2.5}{1.5}\sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}} = \frac{25}{15}\sqrt{\frac{2}{4}} = \frac{5}{3}\sqrt{\frac{1}{2}}$$

Thus,

$$\frac{t_A}{t_B} = \frac{5}{3\sqrt{2}}$$

Using $\sqrt{2} \approx 1.414$,

$$\frac{t_A}{t_B} \approx \frac{5}{3 \times 1.414} = \frac{5}{4.242} \approx 1.1786$$

Therefore, the ratio is approximately $1.18$, so the correct option is C.

Given: Both strings are under the same tension.

Find: $t_A/t_B$.

Since $t = L\sqrt{\mu/T}$, the common tension cancels immediately in the ratio:

$$\frac{t_A}{t_B} = \frac{L_A}{L_B}\sqrt{\frac{\mu_A}{\mu_B}}$$

Now use only lengths and linear densities:

$$\frac{t_A}{t_B} = \frac{2.5}{1.5}\sqrt{\frac{2}{4}} = \frac{5}{3\sqrt{2}} \approx 1.18$$

Therefore, the correct option is C. This shortcut works because wave speed depends on $\sqrt{T/\mu}$, and the same tension acts in both parts of the joined string.