Verified Solution • 2025–26 Session moderate

QNEET-2016-Ph-1178: NEET 2016 Physics - Class 11 NEET-2016-Ph-1178 Solution: A light rod of length l has two masses \(m_1\) and \(m_2\)…

A light rod of length l has two masses m1m_1 and m2m_2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is m1m2m1+m2l2\frac{m_1m_2}{m_1+m_2}l^2 m1+m2m1m2l2\frac{m_1+m_2}{m_1m_2}l^2 (m1+m2)l2(m_1+m_2)l^2 m1m2l2m_1m_2l^2

Multiple Choice Options

  • A m1m2m1+m2l2{{{m_1}{m_2}} \over {{m_1} + {m_2}}}{l^2} m 1 _ + m 2 _ m 1 _ m 2 _ _ l 2
  • B m1+m2m1m2l2{{{m_1} + {m_2}} \over {{m_1}{m_2}}}{l^2} m 1 _ m 2 _ m 1 _ + m 2 _ _ l 2
  • C (m1+m2)l2\left( {{m_1} + {m_2}} \right){l^2} ( m 1 _ + m 2 _ ) l 2
  • D m1m2 l2\sqrt {{m_1}{m_2}} \,{l^2} m 1 _ m 2 _ _ l 2

Step-by-Step Verified Solution

Here, l 1 + l 2 = l Centre of mass of the system, l 1 = m 1 _ 0 + m 2 _ l m 1 + m 2 = m 2 l m 1 + m 2 l 2 = l _ l 1 = m 1 l m 1 + m 2 Required moment of inertia of the system, I = m 1 l 1 2 + m 2 l 2 2 = ( m 1 m 2 2 + m 2 m 1 2 ) l 2 ( m 1 + m 2 ) 2 = m 1 m 2 ( m 1 + m 2 ) l 2 ( m 1 + m 2 ) 2 = m 1 m 2 m 1 + m 2 l 2

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