A wooden block sliding down from the top of a smooth inclined plane starting from rest takes $t_{1}$ seconds to reach the bottom of the plane and attains velocity $v_{1}$ . Another block of twice the mass falling freely in air from the same height takes $t_{2}$ seconds to reach the bottom of the plane and attains velocity $v_{2}$ . If the angle of inclination of the plane is $30^{0}$ , then

v_{1} = v_{2}

and

t_{1} = t_{2}

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v_{1} = v_{2}

and

t_{1} = 2 t_{2}

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v_{1} > v_{2}

and

t_{1} > t_{2}

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v_{1} = v_{2}

and

2 t_{1} = t_{2}

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Solution

The correct option is C $v_{1} = v_{2}$ and $t_{1} = 2 t_{2}$
The velocity attained will be same since the heights are same. But the acceleration for the vertical fall is $g$ , and for the sliding wooden block it will be $g s i n 30^{o} = 0.5 g$ .
$∴ v_{1} = v_{2}, \Rightarrow t_{1} = 2 t_{2}$

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