A body takes just twice the time as long to slide down a plane inclined at 30° to the horizontal as if the plane were frictionless. The coefficient of friction between the body and the plane is

The correct answer is √3/4.

We are given a body sliding down an inclined plane at angle θ = 30°. The time taken with friction is twice the time taken without friction, over the same distance. We need to find the coefficient of kinetic friction μ.

Step 1: Set up accelerations for both cases

Let the length of the incline be L.

Case 1 — Frictionless plane:
The only force along the incline is the component of gravity: mgsin\theta
So acceleration:

a_1=gsin30°=g\times \frac{1}{2}=\frac{g}{2}

Case 2 — Plane with friction:
The friction force opposes motion, so the net acceleration is:

a_2=g(sin\theta -\mu cos\theta )=g(sin30°-\mu cos30°)=g(\frac{1}{2}-\frac{\sqrt{3}}{2}\mu )

Step 2: Use the kinematic equation for same distance

Starting from rest, distance is: L=\frac{1}{2}a{t}^2

Let the frictionless time be t₁. Then the time with friction is t₂ = 2t₁.

Since both cover the same distance L:

\frac{1}{2}{a}_1{t}_1^2=\frac{1}{2}{a}_2{t}_2^2

a_1{t}_1^2=a_2(2t_1{)}^2=4a_2{t}_1^2

Dividing both sides by t_1^2:

a_1=4a_2

Step 3: Substitute the expressions for acceleration

\frac{g}{2}=4g(\frac{1}{2}-\frac{\sqrt{3}}{2}\mu )

Divide both sides by g:

\frac{1}{2}=4(\frac{1}{2}-\frac{\sqrt{3}}{2}\mu )

\frac{1}{2}=2-2\sqrt{3}\mu

Step 4: Solve for μ

2\sqrt{3}\mu =2-\frac{1}{2}=\frac{3}{2}

\mu =\frac{3}{2}\times \frac{1}{2\sqrt{3}}=\frac{3}{4\sqrt{3}}

Rationalize by multiplying numerator and denominator by \sqrt{3}:

\mu =\frac{3\sqrt{3}}{4\times 3}=\frac{3\sqrt{3}}{12}=\frac{\sqrt{3}}{4}

Therefore, the coefficient of friction is \mu =\frac{\sqrt{3}}{4}