Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

A block sliding down a rough $45 °$ inclined plane has half the velocity it would have had, the inclined plane been smooth. The coefficient of sliding friction between block and the inclined plane is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

\frac{3}{4}

\frac{1}{\sqrt{2}}

\frac{1}{4}

\frac{1}{2 \sqrt{2}}

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

v_{rough} = \frac{1}{2} v_{smooth}

$\sqrt{2 l g (\sin θ - μ \cos θ)} = \frac{1}{2} \sqrt{2 l g \sin θ}$

Substitute on both sides

$2 l g (\sin θ - μ \cos θ) = \frac{1}{4} 2 l g \sin θ$

$4 \sin θ - 4 μ \cos θ = \sin θ$

$3 \sin θ = 4 μ \cos θ$

$μ = \frac{3}{4} \tan θ$

$θ = 45 °$

$μ = \frac{3}{4}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE