JEE Physics 2023 Question with Solution

Given: $m = \frac{1}{4} \, \text{kg}$, $L = 1 \, \text{m}$, $A = \frac{1}{10} \, \text{m}$, $g = 9.8 \, \text{m/s}^2$

Find: The value of $x$ if the maximum tension is $\dfrac{x}{40} \, \text{N}$.

For pendulum,

$$T_{\text{max}} = mg + \frac{mv^2}{L}$$

For SHM,

$$K_{\text{max}} = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2A^2$$

Using

$$\omega = \sqrt{\frac{g}{L}}$$

we get

$$mv^2 = m\left(\frac{g}{L}\right)A^2 = mg\frac{A^2}{L}$$

Substitute this into the tension formula:

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

Substituting the given values:

$$T_{\text{max}} = \frac{1}{4}\times 9.8 + \frac{1}{4}\times 9.8 \times \frac{1}{100}$$ $$T_{\text{max}} = \frac{98.98}{40}$$

Therefore, $x = 99$.

Given: The pendulum performs small oscillations with amplitude $A = \frac{1}{10} \, \text{m}$.

Find: Maximum tension in the string and then compare with $\dfrac{x}{40} \, \text{N}$.

At the mean position, speed is maximum. For SHM,

$$v_{\max} = \omega A$$

with

$$\omega^2 = \frac{g}{L}$$

So,

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

Now maximum tension occurs at the lowest point:

$$T_{\max} = mg + \frac{mv_{\max}^2}{L}$$

Substitute $v_{\max}^2 = \frac{g}{L}A^2$:

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

Using $m = \frac{1}{4} \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, $A = \frac{1}{10} \, \text{m}$, $L = 1 \, \text{m}$,

$$T_{\max} = \frac{1}{4}\times 9.8\left(1 + \frac{1}{100}\right)$$ $$T_{\max} = 2.4745 \, \text{N} = \frac{98.98}{40} \, \text{N}$$

Hence, the required value is $99$.