A particle of mass 1kg and carrying charge 0.01C just fails to slide down on an inclined plane of angle 300 with horizontal when an electric field of 4903NC−1 is applied parallel to horizontal and pointing into the inclined plane. The coefficient of friction is g=9.8m/s2

When the particle just fails to slide down, limiting friction acts up the plane.mgsinθ=qEcosθ+f=qEcosθ+μN mgsinθ=qEcosθ+μ(mgcosθ+qEsinθ)μ=mgsinθ−qEcosθmgcosθ+qEsinθ,θ=300,g=9.8m/s2μ=19.8sin30°−0.014903cos30°19.8cos30°+0.014903sin30°=9.8−4.99.83+4.93=37μ=37

A particle of mass 1kg and carrying charge 0.01C just fails to slide down on an inclined plane of angle 300 with horizontal when an electric field of 4903NC−1 is applied parallel to horizontal and pointing into the inclined plane. The coefficient of friction is g=9.8m/s2

When the particle just fails to slide down, limiting friction acts up the plane.mgsinθ=qEcosθ+f=qEcosθ+μN mgsinθ=qEcosθ+μ(mgcosθ+qEsinθ)μ=mgsinθ−qEcosθmgcosθ+qEsinθ,θ=300,g=9.8m/s2μ=19.8sin30°−0.014903cos30°19.8cos30°+0.014903sin30°=9.8−4.99.83+4.93=37μ=37

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A particle of mass $1 k g$ and carrying charge $0.01 C$ just fails to slide down on an inclined plane of angle $30^{0}$ with horizontal when an electric field of $\frac{490}{\sqrt{3}} N C^{- 1}$ is applied parallel to horizontal and pointing into the inclined plane. The coefficient of friction is $(g = 9.8 m / s^{2})$

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0.5

$1 / \sqrt{3}$

$\sqrt{3} / 2$

$\sqrt{3} / 7$

answer is D.

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Detailed Solution

When the particle just fails to slide down, limiting friction acts up the plane.

$m g \sin θ = q E \cos θ + f = q E \cos θ + μ N$

$m g \sin θ = q E \cos θ + μ (m g \cos θ + q E \sin θ)$

$μ = \frac{m g \sin θ - q E \cos θ}{m g \cos θ + q E \sin θ}, θ = 30^{0}, g = 9.8 m / s^{2}$

$μ = \frac{(1) (9.8) \sin 30 ° - 0.01 (\frac{490}{\sqrt{3}}) \cos 30 °}{(1) (9.8) \cos 30 ° + 0.01 (\frac{490}{\sqrt{3}}) \sin 30 °} = \frac{9.8 - 4.9}{9.8 \sqrt{3} + \frac{4.9}{\sqrt{3}}} = \frac{\sqrt{3}}{7}$