A mass of $10$ kg is placed on an inclined plane and its inclination is gradually increased such that the mass just begins to slide down. The inclined plane makes an angle of $30^{o}$ at this time. The coefficient of friction between the mass and the plane is?

1 / 2

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1 / \sqrt{3}

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\sqrt{3} / 2

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1 / 3

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Solution

The correct option is B $1 / \sqrt{3}$
Coefficient of friction $μ = tan θ$
$θ =$ angle of inclined plane
$μ = tan 30^{\circ}$
$μ = 1 / \sqrt{3}$

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A mass of 10kg is placed on an inclined plane and its inclination is gradually increased such that the mass just begins to slide down. The inclined plane makes an angle of 30o at this time. The coefficient of friction between the mass and the plane is?

The correct option is B 1/√3Coefficient of friction μ=tanθθ=angle of inclined planeμ=tan30∘μ=1/√3

A mass of $10$ kg is placed on an inclined plane and its inclination is gradually increased such that the mass just begins to slide down. The inclined plane makes an angle of $30^{o}$ at this time. The coefficient of friction between the mass and the plane is?

The correct option is B $1 / \sqrt{3}$
Coefficient of friction $μ = tan θ$
$θ =$ angle of inclined plane
$μ = tan 30^{\circ}$
$μ = 1 / \sqrt{3}$