A small block of mass 100 g is tied to a spring of spring constant… | JEE Main 2023 (6 Apr, Shift 1)

Q23JEE Main 2023 · 6 APR · SHIFT 1Medium

Centripetal Acceleration & Force

A small block of mass $100\text{ g}$ is tied to a spring of spring constant $7.5\text{ N/m}$ and length $20\text{ cm}$ . The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity $5\text{ rad/s}$ about point $A$ , then tension in the spring is -

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Answer: (c)

Spring force gives the centripetal force: $kx = m\omega^2(l + x)$ .
$m\omega^2 = 0.1 \times 25 = 2.5$ , so $7.5x = 2.5(0.2 + x)$ .
$5x = 0.5 \Rightarrow x = 0.1\text{ m}$
Tension $= kx = 7.5 \times 0.1 = 0.75\text{ N}$ . So the answer is (c) 0.75 N.

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