QJEEAdv18Ph3316: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph3316 Solution: A ring and disc are initially at rest, side by side, at the top …

Given, h = height of the top of inclined plane = ?, $\theta$ = 60 $^\circ$ , g = 10 m s ^{$-$ 2} and time difference between ring and disc reaching ground = ${{2 - \sqrt 3 } \over {\sqrt {10} }}$ s

We know that, $a = {{g\sin \theta } \over {1 + {I \over {M{R^2}}}}}$

For ring, $I = M{R^2}$ and for disc $I = {3 \over 2}M{R^2}$ .

So, ${a_{ring}} = {{g\sin \theta } \over 2}$ and ${a_{disk}} = {{2g\sin \theta } \over 3}$

Now using $s = {1 \over 2}a{t^2}$

${s_{ring}} = {1 \over 2}\left( {{{g\sin \theta } \over 2}} \right)t_1^2 \Rightarrow {h \over {\sin \theta }} = {1 \over 2}\left( {{{g\sin \theta } \over 2}} \right)t_1^2$ ...... (1)

${s_{disk}} = {1 \over 2}\left( {{{2g\sin \theta } \over 3}} \right)t_2^2 \Rightarrow {h \over {\sin \theta }} = {1 \over 2}\left( {{{2g\sin \theta } \over 3}} \right)t_2^2$ ...... (2)

Given, ${t_1} - {t_2} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$

From Eq. (1), we have ${t_1} = \sqrt {{{4h} \over {g{{\sin }^2}\theta }}}$

and from Eq. (2), we have ${t_2} = \sqrt {{{3h} \over {g{{\sin }^2}\theta }}}$

So, $\sqrt {{{4h} \over {g{{\sin }^2}\theta }}} - \sqrt {{{3h} \over {g{{\sin }^2}\theta }}} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$

$\Rightarrow \sqrt {{{4h} \over {10{{\sin }^2}60^\circ }}} - \sqrt {{{3h} \over {10{{\sin }^2}60^\circ }}} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$

$\Rightarrow \sqrt {{{16h} \over {3 \times 10}}} - \sqrt {{{4h} \over {10}}} = {{2 - \sqrt 3 } \over {\sqrt {10} }}$

$\Rightarrow {{4\sqrt h } \over {\sqrt 3 }} - 2\sqrt h = 2 - \sqrt 3$

$\Rightarrow \sqrt h \left( {{4 \over {\sqrt 3 }} - 2} \right) = (2 - \sqrt 3 )$

$\Rightarrow \sqrt h {{(4 - 2\sqrt 3 )} \over {\sqrt 3 }} = (2 - \sqrt 3 )$

$\Rightarrow \sqrt h \times {2 \over {\sqrt 3 }}(2 - \sqrt 3 ) = (2 - \sqrt 3 ) \Rightarrow \sqrt h = {{\sqrt 3 } \over 2}$

$\Rightarrow h = {3 \over 4} = 0.75$ m