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QAIPM-2015-Ph-1235: NEET 2015 Physics - Class 11 AIPM-2015-Ph-1235 Solution: Point masses \(m_1\) and \(m_2\) are placed at the opposite…

Point masses m1m_1 and m2m_2 are placed at the opposite ends of a rigid rod of length L and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0\omega_0 is minimum is given by x=m2m1Lx = \frac{m_2}{m_1}L x=m2m1+m2Lx = \frac{m_2}{m_1+m_2}L x=m1m1+m2Lx = \frac{m_1}{m_1+m_2}L x=m1m2Lx = \frac{m_1}{m_2}L

Multiple Choice Options

  • A x=m2m1Lx = {{{m_2}} \over {{m_1}}}L x = m 1 _ m 2 _ _ L
  • B x=m2Lm1+m2x = {{{m_2}L} \over {{m_1} + {m_2}}} x = m 1 _ + m 2 _ m 2 _ L _
  • C x=m1Lm1+m2x = {{{m_1}L} \over {{m_1} + {m_2}}} x = m 1 _ + m 2 _ m 1 _ L _
  • D x=m1m2Lx = {{{m_1}} \over {{m_2}}}L x = m 2 _ m 1 _ _ L

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