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Question

A cylinder rolls down an inclined, plane of inclination 30, the acceleration of cylinder is

A
g3
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B
g
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C
g2
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D
g6
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Solution

The correct option is D g6
a=gsinθ1+ImR2
put I=mR22 (cylinder)
θ=30
we get a=g/21+2=g/6

1218544_434068_ans_6b6b4257a0f648fcb5092767fdf782f7.jpg

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