Question

Starting from rest, a body slides down a ${45}^{\circ }$ inclined plane in twice the time taken to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

• A. $0.25$
• B. $0.33$
• C. $0.75$
• D. $0.80$

Answer 1

The correct option is C $0.75$

if the length of incline plane is L, and time taken by body to reach at its bottom in absence of friction is 't'

then

$L=ut+\frac{1}{2}gsin\theta t^2(u=0)L=\frac{1}{2}gsin\theta t^{2}equation1$

In case 2nd

there is friction then acceleration becomes

$a=gsin\theta -\mu gcos\theta$

here friction force acts in upward direction as motion is in downward direction

let the time taken in this case is $t^{{}^{\prime }}$

then

$L=ut+\frac{1}{2}(gsin\theta -\mu gcos\theta ){t^{{}^{\prime }}}^{2}u=0L=\frac{1}{2}(gsin\theta -\mu gcos\theta ){t^{{}^{\prime }}}^{2}equation2$

here it is given that

$t^{{}^{\prime }}=2t$

then equating equation 1 and 2

we get,

$\frac{1}{2}gsin\theta t^2=\frac{1}{2}(gsin\theta -\mu gcos\theta ){t^{{}^{\prime }}}^{2}sin\theta t^2=(sin\theta -\mu cos\theta )\times {\left(2t\right)}^2\Rightarrow \frac{sin\theta }{sin\theta -\mu cos\theta }=4put\theta ={45}^{0}weget\mu =\frac{3}{4}=0.75$

Option C is correct