Given data

• Length of pendulum: $L = 100\,\text{cm} = 1\,\text{m}$
• Mass of bob: $m = 250\,\text{g} = 0.25\,\text{kg}$
• Amplitude (linear): $A = 10\,\text{cm} = 0.1\,\text{m}$

Maximum tension in a simple pendulum

The tension is maximum at the mean position because the speed is maximum there.

At the lowest point, $T_{\max} = mg + \frac{mv^2}{L}$

Find maximum speed using SHM relation

For small oscillations, the pendulum performs SHM with angular frequency $\omega = \sqrt{\frac{g}{L}}$

Maximum speed in SHM is $v_{\max} = \omega A$

So, $v_{\max}^2 = \omega^2 A^2 = \frac{g}{L}A^2$

Substitute into tension formula

$T_{\max} = mg + \frac{m}{L}\left(\frac{g}{L}A^2\right)$ $T_{\max} = mg\left(1 + \frac{A^2}{L^2}\right)$

Put the values

$mg = 0.25 \times 9.8 = 2.45\,\text{N}$

Also, $\frac{A^2}{L^2} = \frac{(0.1)^2}{(1)^2} = 0.01$

Therefore, $T_{\max} = 2.45(1 + 0.01)$ $T_{\max} = 2.45 \times 1.01 = 2.4745\,\text{N}$

Given form of answer

It is given that $T_{\max} = \frac{x}{40}\,\text{N}$

Hence, $\frac{x}{40} = 2.4745$ $x = 2.4745 \times 40 = 98.98$

Since the answer is an integer, $x \approx 99$

Comparison with stored answer

My derived answer is $99$, whereas the stored correct answer is $101$.

The value $101$ would arise if one uses $g = 10\,\text{m/s}^2$: $T_{\max} = 0.25 \times 10 \times 1.01 = 2.525\,\text{N}$ $x = 40 \times 2.525 = 101$

So the stored answer assumes $g=10\,\text{m/s}^2$, which is standard in many JEE-style problems unless otherwise specified.