Question

139. A disc takes time t1 and t2 for sliding down and
rolling down an inclined plane of length L
respectively for reaching the bottom. The ratio of
t1 and t2 is :
(1) √2:1
(2) √2:3
(3) 2:3
(4) √3:1
​

Answer 1

Answer:

1:root2

Explanation:

forslidinga=gsinθt1=gsinθ2lforrollinga=2gsinθt2=2.gsinθ2l=2t1⇒t2t1=21

Answer 2

The ratio of time taken for sliding down and rolling down an inclined plane is √2 : √3. [mentioned options are incorrect ]

A disc takes time t₁ and t₂ for sliding down and rolling down an inclined plane of length, L respectively for the reaching bottom.

We have to find the ratio of t₁ and t₂.

Let Angle of inclination of plane is Ф.

then, acceleration along plane for sliding down , a = gsinФ

using formula,

$S=ut+\frac{1}{2}at^2$

Here, S = L  , a = gsinФ

$\therefore L=\frac{1}{2}sin\Phi t^2\\\\\\\implies t_1=\sqrt{\frac{2L}{gsin\Phi}}$

For rolling down,

$a=\frac{gsin\Phi}{1+\frac{I}{mr^2}}$

For disc, moment of inertia of disc about its centre is mr²/2.

$\therefore a=\frac{gsin\Phi}{1+\frac{1}{2}}=\frac{2}{3}sin\Phi$

$\text{time taken to rolling down},t_2=\sqrt{\frac{3L}{sin\Phi}}$

Therefore the the ratio of time taken to sling down and rolling down of a disc is

$\frac{t_1}{t_2}=\frac{\sqrt{2}}{\sqrt{3}}$

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