JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 25)
A small block slides down from the top of hemisphere of radius R = 3 m as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is __________ m.
(Assume there is no friction between the block and the hemisphere)
_27th_July_Evening_Shift_en_25_1.png)
(Assume there is no friction between the block and the hemisphere)
Answer
2
Explanation
_27th_July_Evening_Shift_en_25_2.png)
$$mg\cos \theta = {{m{v^2}} \over R}$$ .... (1)
$$\cos \theta = {h \over R}$$
Energy conservation
$$mg\{ R - h\} = {1 \over 2}m{v^2}$$ ..... (2)
from (1) & (2)
$$\Rightarrow$$ $$mg\left\{ {{h \over R}} \right\} = {{2mg\{ R - h\} } \over R}$$
$$h = {{2R} \over 3}$$ = 2m
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