Physics MCQs for NEET — Practice Questions with Answers

The potential energy of water molecules is more. The heat given to melt the ice at $ 0 ^\circ C$ used up in increasing the potentical energy of water molecules formed at $ 0 ^\circ C $

Water has maximum density at $ 4 ^\circ C$ on heating above $ 4 ^\circ C $ or cooling below $ 40 ^\circ C $ density of water decreases and its volume increases. Therefore water overflows in both the cases.

$ Let R_{cm } is at origin $ $ M \vec R_{cm} = m_1 \vec r_1 + m_2 \vec r_2 $ $ O = m_1 \vec r_1 + m_2 \vec r_2 $ $ - m_1 \vec r_1 = m_2 \vec r_2 $ $ { r_1 \over r _2 } = { m_2 \over m_1 } $ -ve sign ignore as distance

The x and y co.ordinates of centre of mass are $ x = { m_1 x_1 + m_2 x_2 + m_3 x_3 \over m_1 + m_2 + m_3 } $ $ as m_1 = m_2 =m_3 $ $ = { 1 \over 3 } (x_1 + x_2 + x_3 ) = 2 $ similarly for y =2 $ \therefore ( x,y) = (2,2) $

$m_1 x_1 = m_2 x_2 $ $ and m_1 ( x_1 - d ) = m_2 ( x_2 - d ) $ $ \therefore m_1 d = m_2 d ' $ $ \therefore d' = { m_1 d \over m_2 } $

Let mass per unit area of disc = m Mass of disc = M = $ \pi R^2 .m $ Mass of removed disc =M ' = $ \pi \left( { R \over 6 } \right) ^2 .m= { \pi R^2 m \over 36} $ from figure 00' = R/2 $ M \times 0 = M' \times { R \over 2} + ( M - M' ) x $ $ M' x = M' { R \over 2 } + Mx $ $ x = \left( { M' \over M - M' } . { R \over 2 } \right) $

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 } $