A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The maximum tension in the string is found to be x/40 N. The value of x is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 101. Tmax = mg[1 + (A/l)²] (tension at mean position). m = 0.25 kg, A/l = 10/100 = 0.1, take g = 10 m/s². Tmax = 0.25 × 10 × 1.01 = 2.525 N = 101/40 N. So x = 101.

A simple pendulum with length 100 cm and bob of mass 250 g is… | JEE Main 2023 (6 Apr, Shift 2)

Q50JEE Main 2023 · 6 APR · SHIFT 2Medium

Simple Pendulum

A simple pendulum with length $100\text{ cm}$ and bob of mass $250\text{ g}$ is executing S.H.M. of amplitude $10\text{ cm}$ . The maximum tension in the string is found to be $\frac{x}{40}\text{ N}$ . The value of $x$ is ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 101

$T_{max} = mg\left[1 + \left(\frac{A}{l}\right)^2\right]$ (tension at mean position).
$m = 0.25\text{ kg},\ A/l = 10/100 = 0.1$ , take $g = 10\text{ m/s}^2$ .
$T_{max} = 0.25 \times 10 \times 1.01 = 2.525\text{ N} = \frac{101}{40}\text{ N}$ .
So $x = 101$ .

Report a mistake

Simple Pendulum question from JEE Main 2023 (6 Apr, Shift 2)

More questions on Simple Pendulum