JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 20)
A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $${1 \over 2}\sqrt {xgh} $$ m/s. The value of x is ___________.
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Explanation
For rolling wheel
$$[12g\sin \alpha - 3g\sin \alpha ] \times R = (2 \times 12{R^2} + 3{R^2}) \times {a \over R}$$
$$ \Rightarrow {{9g\sin \alpha } \over {27}} = a$$
$$ \Rightarrow a = {{g\sin \alpha } \over 3}$$
$$\therefore$$ $$v = \sqrt {2 \times {{g\sin \alpha } \over 3} \times {h \over {\sin \alpha }}} = \sqrt {{2 \over 3}gh} $$
$$ = {1 \over 2} \times \sqrt {{8 \over 3}gh} $$
$$\therefore$$ $$x = {8 \over 3} = 2.67$$
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