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JEE Main 2013 (23 Apr, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from JEE Main 2013 (23 Apr, Shift 1)

Q49JEE Main 2013 · 23 APR · SHIFT 1Medium

Centripetal Acceleration & Force

A body of mass 'm' is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is 1 cm. If the angular velocity is doubled, the elongation in the spring is 5 cm. The original length of the spring is:

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Answer: (a)

Spring force gives the centripetal force: $kx = m\omega^2(l + x)$ .
Case 1: $k(1) = m\omega^2(l + 1)$ . Case 2: $k(5) = m(4\omega^2)(l + 5)$ .
Dividing: $5 = \frac{4(l + 5)}{l + 1} \Rightarrow 5l + 5 = 4l + 20$
$l = 15\text{ cm}$ . So the answer is (a) 15 cm.

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