A block released from rest from the top of a smooth inclined plane of inclination $45^{\circ}$ takes $t$ seconds to reach the bottom. The same block released from rest from the top of a rough inclined plane of the same inclination of $45^{\circ}$ takes $2 t$ seconds to reach the bottom. The coefficient of friction is

$\sqrt{0.5}$

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$\sqrt{0.75}$

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$0.5$

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$0.75$

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Solution

The correct option is D
$0.75$

The acceleration of the block sliding down the smooth inclined plane is $a_{1} = g s i n θ$ and down the rough inclined plane is $a_{2} = g s i n θ - μ g c o s θ$ .
Given $t_{1} = t a n d t_{2} = 2 t$
If the length of the inclined planes is s, we have
$s = \frac{1}{2} a_{1} t_{1}^{2} = \frac{1}{2} a_{2} t_{2}^{2} \Rightarrow a_{1} t_{1}^{2} = a_{2} t_{2}^{2} \Rightarrow g s i n θ \times t^{2} = (g s i n θ - μ g c o s θ) \times (2 t)^{2} \Rightarrow s i n 45^{\circ} = 4 (s i n 45^{\circ} - μ c o s 45^{\circ}) \Rightarrow μ = \frac{3}{4} = 0.75$
Hence, the correct choice is (d).

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A block released from rest from the top of a smooth inclined plane of inclination 45∘ takes t seconds to reach the bottom. The same block released from rest from the top of a rough inclined plane of the same inclination of 45∘ takes 2t seconds to reach the bottom. The coefficient of friction is