JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Laws of Motion

One end of a massless spring of spring constant k and natural length

l_{0}

is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity

ω

about an axis passing through fixed end, then the elongation of the spring will be:

(A) $\frac{k - m ω ^{2} l _{0}}{m ω ^{2}}$
(B) $\frac{m ω l _{0}}{k + m ω ^{2}}$
(C) $\frac{m ω ^{2} l _{0}}{k - m ω ^{2}}$ ✓ Correct
(D) $\frac{k + m ω ^{2} l _{0}}{m ω ^{2}}$

Answer: (C)

Solution

The spring force provides the centripetal force. If elongation is

Δ x

, then

k Δ x = m (l_{0} + Δ x) ω^{2}

. Expanding:

k Δ x = m l_{0} ω^{2} + m ω^{2} Δ x

. Solving for

Δ x

Δ x (k - m ω^{2}) = m l_{0} ω^{2}

, so

Δ x = \frac{m ω ^{2} l _{0}}{k - m ω ^{2}}

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Laws of Motion

Solution

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