A block moves down a smooth inclined plane of inclination θ . Its velocity on reaching the bottom is v . If it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is v / n , where n is a number greater than 1. The coefficient of friction is given by

Question

A block moves down a smooth inclined plane of inclination θ . Its velocity on reaching the bottom is v . If it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is v / n , where n is a number greater than 1. The coefficient of friction is given by (A) μ = tan θ (1 - (1/n2)) (B) μ = cot θ (1 - (1/n2)) (C) μ= tan θ(1-(1/n2))1 / 2 (D) μ = textcot θ (1 - (1/n2)) text1/2s

Tardigrade Admin · Accepted Answer

For smooth, v2=2(g sin θ ) s . . .(.1.) For rough, ((v/n))2= 2 g ( sin θ - μ cos θ )S. . .(.2.) on solving, μ =(1 - (1/n2))tanθ

$μ = tan θ (1 - \frac{1}{n ^{2}})$

$μ = cot θ (1 - \frac{1}{n ^{2}})$

$μ = tan θ (1 - \frac{1}{n ^{2}})^{1/2}$

$μ = cot θ (1 - \frac{1}{n ^{2}})^{1/2}$ s

For smooth, $v^{2} = 2 (g sin θ) s ... (1)$
For rough, $(\frac{v}{n})^{2} = 2 g (sin θ - μ cos θ) S ... (2)$
on solving, $μ = (1 - \frac{1}{n ^{2}}) t an θ$

Q. A block moves down a smooth inclined plane of inclination θ . Its velocity on reaching the bottom is v . If it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is v/n , where n is a number greater than 1. The coefficient of friction is given by

μ=tanθ(1−n21​)

μ=cotθ(1−n21​)

μ=tanθ(1−n21​)1/2

μ=cotθ(1−n21​)1/2s

For smooth, v2=2(gsinθ)s...(1) For rough, (nv​)2=2g(sinθ−μcosθ)S...(2) on solving, μ=(1−n21​)tanθ

$μ = tan θ (1 - \frac{1}{n ^{2}})$

$μ = cot θ (1 - \frac{1}{n ^{2}})$

$μ = tan θ (1 - \frac{1}{n ^{2}})^{1/2}$

$μ = cot θ (1 - \frac{1}{n ^{2}})^{1/2}$ s

For smooth, $v^{2} = 2 (g sin θ) s ... (1)$
For rough, $(\frac{v}{n})^{2} = 2 g (sin θ - μ cos θ) S ... (2)$
on solving, $μ = (1 - \frac{1}{n ^{2}}) t an θ$