Rotational Motion MCQs for NEET — Physics Questions with Answers

Let mass per unit area of Plote = m Mass of whole Plote = M = $ \pi \left( { 56 \over 2} \right) ^2 m $ Mass of removed part = $M_1 = \pi \left( { 42 \over 2} \right) ^2 m $ Mass of remaining Portion $ M_2 = M – M_1 $ C.M of whole disc R = O at origin C.M of removed Plote = $r_1$ = 28 – 21 = 7cm C.M of remaining Portion $r_2$ = ? $M \times O = M_ir_i + M_2r_2 $

The Velocity of C.M. is given by $$ V_{cm} = { m_1 v_1 + m_2 v_2 \over m_1 m_2 } $$

$ E = {1 \over 2 } I \omega^2 = { 1 \over 2} I \omega . \omega = { 1 \over 2} L \omega $ $ \therefore L = { 2E \over \omega } $ $ \therefore L ' = { 2E' \over \omega' } $

Let m is the mass of unit area then mass of big disc = $ \pi (2R) ^2 m = M $ Let m is the mass of unit area then mass of small disc = $ \pi R^2 m = M_1 = { M \over 4 } $ Mass of remaining Portion = $ M_2 = M - M_1 $ $ M_2 = { 3M \over 4 } $ Let G be the C.M of remaining Portion $M_2(OG) = M_1(OO’)$ $ { 3M \over 4 } ( \alpha R ) = { M \over 4} R $ $ \therefore \alpha = {1 \over 3 } $

The moment of inertia about AD = ? $ I = m_1 (Perpendicular distance of m_1 from AD)^2 + m_2 (Perpendicular distance of m_2 from AD)^2 + m_3 (Perpendicular distance of m_3 from AD)^2 $ $ = 0 + m_2 \times \left( { 9 \over 2} \right) ^2 + m_3 \times \left( { 9 \over 2} \right) ^2 $ $ = ( m_2 + m_3 ) { a^2 \over 4 } $

$ I_A = m_a r_a^2 , I_B = m_b r_b^2 $ $ \therefore { I_B \over I_A} = { m_b \over m_a } \times \left( { r_b \over r_a } \right) ^2 $ Let K is the mass of unit length of the wire then $ m_a = ( 2 \pi r_a ) k and m_b = ( 2 \pi r_b ) K $ $ \therefore { m_b \over m_a } = { r_b \over r_a} $ $ \therefore { I_B \over I_A } = 8 = \left( { m_b \over m_a } \right) \left( { r_b \over r_a } \right) ^2 = \left( {r_b \over r_a } \right) ^3 $ $ \therefore { r_b \over r_a } = 2 $

Let M be the mass and $R_1$ the initial radius of the earth $ \omega_1 $ is the angular veloalty of the rotation of the earth, the duration $T_1$ of the day $ T_1 = { 2 \pi \over \omega_1 } $ and $ T_2 = { 2 \pi \over \omega_2 } $ According to law of conservation of angular momentum $ I_1 \omega_1 = I_2 \omega_2 $

$ m_1 = 1 , m_2 = 35.5 ,r_1 = 0 , r_2 = 1.27 \hat i $ $ \vec r_{cm} = { m_1 \vec r_1 + m_2 \vec r_2 \over m_1 + m_2} $

Theory [B] The centre of mass of lucky remains uncharged.

$ \omega_0 { G_0 \over r} = { 72 \times 1000 /3600 \over 0.25 } = 80 rad /sec $ $ \omega = 0 . \theta = 2 \pi n = 2 \pi \times 20 = 40 \pi rad $ $ As \;2 \alpha \theta = w^2 - w_0^2 $