Question

If a disc slides from top to bottom in an inclined plane, it takes time $t_1$. If it rolls, it takes time $t_2$. Now, $\dfrac{{t_2^2}}{{t_1^2}}$ is:
A. $\dfrac{1}{2}$
B. $\dfrac{2}{3}$
C. $\dfrac{3}{2}$
D. $\dfrac{2}{5}$

Answer 1

Hint
In the first case the disc is sliding and in the second case the disc is rolling. So we have to calculate time for both cases which are taken by disc to reach the bottom. We will use Newton’s second equation of motion to find the time.
 $s = ut + \dfrac{1}{2}a{t^2}$.
Initial velocity in both cases is zero. Acceleration in sliding case is given by-
$a = g\sin \theta $,
and in rolling case acceleration is given by-
$a = \dfrac{{g\sin \theta }}{{1 + \dfrac{I}{{m{r^2}}}}}$

Complete Step By Step Solution
(1) Disc slides and time ($t_1$) is taken to reach the disc at the bottom of the inclined plane from top. So acceleration involve during this sliding motion is given by-
$a = g\sin \theta $
Acceleration due to gravity ($g$) divides into two components $g\sin \theta $ and $g\operatorname{Cos} \theta $ . $g\sin \theta $ is responsible for the body to slide in an inclined plane.
(2) From equation of motion
$s = ut + \dfrac{1}{2}a{t^2}$ -- (1)
Where s= length of inclined plane = l (say)
Initial velocity is zero, t=$t_1$ and $a = g\sin \theta $ so equation becomes
$l = \dfrac{1}{2}(g\sin \theta ){t_1}^2$
$t_1$, we get
${t_1}^2 = \dfrac{{2l}}{{g\sin \theta }}$ -- (2)
(3) When body rolls in an inclined plane its acceleration is given as
$a = \dfrac{{g\sin \theta }}{{1 + \dfrac{I}{{m{r^2}}}}}$
Moment of inertia of disc (I) = $\dfrac{1}{2}m{r^2}$
$a = \dfrac{{g\sin \theta }}{{1 + \dfrac{{m{r^2}}}{{2m{r^2}}}}}$
$a = \dfrac{{2g\sin \theta }}{3}$
Time taken to reach the disc at the bottom of the inclined plane from top is $t_2$.
Initial velocity is zero, put a, $t_2$ and s= l in equation (1) to get $t_2$.
$l = \dfrac{1}{2}(\dfrac{{2g\sin \theta }}{3}){t_2}^2$
${t_2}^2 = \dfrac{{3l}}{{g\sin \theta }}$ -- (3)
Divide equation (3) by equation (2) we get
$\dfrac{{{t_2}^2}}{{{t_1}^2}} = \dfrac{{3l}}{{g\sin \theta }}\dfrac{{g\sin \theta }}{{2l}}$
$ \Rightarrow \dfrac{{{t_2}^2}}{{{t_1}^2}} = \dfrac{3}{2}$
Correct option: $\dfrac{3}{2}$ (C)

Note
When the body rolls in an inclined plane it takes more time as compared to when the body slides in an inclined plane. So we can get the answer by using the above procedure.