JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 23)
A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it makes time 't1'. When it rolls down the plane, it takes time t2. The value of $${{{t_2}} \over {{t_1}}}$$ is $$\sqrt {{3 \over x}} $$. The value of x will be _______________.
Answer
2
Explanation
According to question, a circular disc reaches from top to bottom of an inclined plane of length L. This can be shown as
_20th_July_Morning_Shift_en_23_1.png)
When the disc slips down the inclined plane, it takes time t1. Therefore, in this case its acceleration, a1 = g sin$$\theta$$
$$\because$$ s = ut1 + $${1 \over 2}$$a1t$$_1^2$$
$$\Rightarrow$$ s = $${1 \over 2}$$ g sin$$\theta$$t$$_1^2$$ .... (i)
And when the disc rolls down the inclined plane, it takes time t2. Therefore in this case, its acceleration,
$${a_2} = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}} = {{g\sin \theta } \over {1 + {1 \over 2}}} = {2 \over 3}g\sin \theta $$ [$$\because$$ for disc, $${{{K^2}} \over {{R^2}}} = {1 \over 2}$$]
$$\because$$ s = ut2 + $${1 \over 2}$$ a2t$$_2^2$$ = $${1 \over 2}$$ . $${2 \over 3}$$ g sin$$\theta$$ t$$_2^2$$ .... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$${{{t_2}} \over {{t_1}}} = \sqrt {{3 \over 2}} $$ .... (iii)
According to question, value of $${{{t_2}} \over {{t_1}}}$$ is $$\sqrt {{3 \over x}} $$.
Comparing it with Eq. (iii), we get x = 2
When the disc slips down the inclined plane, it takes time t1. Therefore, in this case its acceleration, a1 = g sin$$\theta$$
$$\because$$ s = ut1 + $${1 \over 2}$$a1t$$_1^2$$
$$\Rightarrow$$ s = $${1 \over 2}$$ g sin$$\theta$$t$$_1^2$$ .... (i)
And when the disc rolls down the inclined plane, it takes time t2. Therefore in this case, its acceleration,
$${a_2} = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}} = {{g\sin \theta } \over {1 + {1 \over 2}}} = {2 \over 3}g\sin \theta $$ [$$\because$$ for disc, $${{{K^2}} \over {{R^2}}} = {1 \over 2}$$]
$$\because$$ s = ut2 + $${1 \over 2}$$ a2t$$_2^2$$ = $${1 \over 2}$$ . $${2 \over 3}$$ g sin$$\theta$$ t$$_2^2$$ .... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$${{{t_2}} \over {{t_1}}} = \sqrt {{3 \over 2}} $$ .... (iii)
According to question, value of $${{{t_2}} \over {{t_1}}}$$ is $$\sqrt {{3 \over x}} $$.
Comparing it with Eq. (iii), we get x = 2
Comments (0)
