JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 23)
A solid sphere and solid cylinder of identical
radii approach an incline with the same linear
velocity (see figure). Both roll without slipping
all throughout. The two climb maximum
heights hsph and hcyl on the incline. The ratio
hsph/hcyl is given by :-
_8th_April_Evening_Slot_en_23_1.png)
1
14/15
4/5
2/$$\sqrt5$$
Explanation
For solid sphere
$${1 \over 2}m{v^2} + {1 \over 2}.{2 \over 5}m{R^2}.{{{v^2}} \over {{R^2}}} = mg{h_{sph}}$$
For solid cylinder
$${1 \over 2}m{v^2} + {1 \over 2}.{1 \over 2}m{R^2}.{{{v^2}} \over {{R^2}}} = mg{h_{cyl}}$$
$$ \Rightarrow {{{h_{sph}}} \over {{h_{cyl}}}} = {{7/5} \over {3/2}} = {{14} \over {15}}$$
$${1 \over 2}m{v^2} + {1 \over 2}.{2 \over 5}m{R^2}.{{{v^2}} \over {{R^2}}} = mg{h_{sph}}$$
For solid cylinder
$${1 \over 2}m{v^2} + {1 \over 2}.{1 \over 2}m{R^2}.{{{v^2}} \over {{R^2}}} = mg{h_{cyl}}$$
$$ \Rightarrow {{{h_{sph}}} \over {{h_{cyl}}}} = {{7/5} \over {3/2}} = {{14} \over {15}}$$
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