The ratio of the acceleration for a solid sphere mass -m- and radius -R- rolling down an incline of angle - - without slipping and slipping down the incline without rolling is-

The correct option is A 5:7 a_1=g sin θ1+I/MR^2 a_1=g sin θ1+2/5MR^2=57g sinθ a_1a_2=57 ...

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Standard XII

Physics

Pure Rolling

The ratio of ...

Question

The ratio of the acceleration for a solid sphere (mass $^{'} m^{'}$ and radius $^{'} R^{'}$ ) rolling down an incline of angle ' $θ$ ' without slipping and slipping down the incline without rolling is.

5 : 7

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2 : 3

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2 : 5

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7 : 5

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Solution

The correct option is A $5 : 7$
$a_{1} = \frac{g s i n θ}{1 + \frac{I}{M R^{2}}}$
$a_{1} = \frac{g s i n θ}{1 + \frac{2}{5} M R^{2}} = \frac{5}{7} g s i n θ$
$\frac{a_{1}}{a_{2}} = \frac{5}{7}$

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The ratio of the acceleration for a solid sphere (mass ′m′ and radius ′R′) rolling down an incline of angle 'θ' without slipping and slipping down the incline without rolling is.

The correct option is A 5:7a1=gsinθ1+IMR2a1=gsinθ1+25MR2=57gsinθa1a2=57

The ratio of the acceleration for a solid sphere (mass $^{'} m^{'}$ and radius $^{'} R^{'}$ ) rolling down an incline of angle ' $θ$ ' without slipping and slipping down the incline without rolling is.

The correct option is A $5 : 7$
$a_{1} = \frac{g s i n θ}{1 + \frac{I}{M R^{2}}}$
$a_{1} = \frac{g s i n θ}{1 + \frac{2}{5} M R^{2}} = \frac{5}{7} g s i n θ$
$\frac{a_{1}}{a_{2}} = \frac{5}{7}$