A body sliding down at an angle theta to the horizontal. then coefficient of friction is equal to

s i n θ

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c o s θ

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t a n θ

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c o t θ

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Solution

The correct option is C $t a n θ$
From free body diagram,
$N = m g c o s θ$ . . . . . . . .(1)
$f_{r} = m g s i n θ$
$μ . N = m g s i n θ$
$μ . m g c o s θ = m g s i n θ$ (from equation 1)
$μ . c o s θ = s i n θ$
$μ = \frac{s i n θ}{c o s θ}$
$μ = t a n θ$
The correct option is C.

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