JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 20)
A body of mass 'm' is launched up on a rough inclined plane making an angle of 30$$^\circ$$ with the horizontal. The coefficient of friction between the body and plane is $${{\sqrt x } \over 5}$$ if the time of ascent is half of the time of descent. The value of x is __________.
Answer
3
Explanation
$${t_a} = {1 \over 2}{t_d}$$
$$\sqrt {{{2s} \over {{a_a}}}} = {1 \over 2}\sqrt {{{2s} \over {{a_d}}}} $$ ..... (i)
$${a_a} = g\sin \theta + \mu g\cos \theta $$
$$ = {g \over 2} + {{\sqrt 3 } \over 2}\mu g$$
$${a_d} = g\sin \theta - \mu g\cos \theta $$
$$ = {g \over 2} - {{\sqrt 3 } \over 2}\mu g$$
using the above values of aa and ad and putting in equation (i) we will gate $$\mu = {{\sqrt 3 } \over 5}$$
$$\sqrt {{{2s} \over {{a_a}}}} = {1 \over 2}\sqrt {{{2s} \over {{a_d}}}} $$ ..... (i)
$${a_a} = g\sin \theta + \mu g\cos \theta $$
$$ = {g \over 2} + {{\sqrt 3 } \over 2}\mu g$$
$${a_d} = g\sin \theta - \mu g\cos \theta $$
$$ = {g \over 2} - {{\sqrt 3 } \over 2}\mu g$$
using the above values of aa and ad and putting in equation (i) we will gate $$\mu = {{\sqrt 3 } \over 5}$$
Comments (0)
