Question

A block, released from rest from the top of a smooth inclined plane of angle of inclination ${\theta }_1$, reaches the bottom in time $t_1$. The same block, released from rest from the top of another smooth inclined plane of angle of inclination ${\theta }_2$, reaches the bottom in time $t_2$. If the two inclined planes have the same height, the relation between $t_1$ and $t_2$ is

• A. $\frac{t_2}{t_1}={\left(\frac{sin{\theta }_1}{sin{\theta }_2}\right)}^{1/2}$
• B. $\frac{t_2}{t_1}=\frac{si{n}^2{\theta }_1}{si{n}^2{\theta }_2}$
• C. $\frac{t_2}{t_1}=\frac{sin{\theta }_1}{sin{\theta }_2}$
• D. $\frac{t_2}{t_1}=1$

Answer 1

The correct option is C

$\frac{t_2}{t_1}=\frac{sin{\theta }_1}{sin{\theta }_2}$

Let h be the height of each inclined plane.

Then, the distances along the plane are$S_1=\frac{h}{sin{\theta }_1}and{S}_2=\frac{h}{sin{\theta }_2}$ respectively.

The accelerations of the block are $a_1=gsin{\theta }_{1}and{a}_2=gsin{\theta }_2$ respectively.

Now, since the block is released from rest, the velocity of the block when it reaches the bottom of the planes is $v_1^2=2a_1{s}_{1}and{v}_2^2=2a_2{s}_2$ respectively.

But $v_1=a_1{t}_{1}and{v}_2=a_2{t}_{2}or{a}_1^2{t}_1^2=2a_1{s}_{1}and{a}_2^2{t}_2^2=2a_2{s}_2.$

These equations give

$\frac{t_2^2}{t_1^2}=\frac{a_1}{a_2}.\frac{s_2}{s_1}$

=$\frac{gsin{\theta }_1}{gsin{\theta }_2}.\frac{h}{sin{\theta }_2}.\frac{sin{\theta }_1}{h}$

which gives $\frac{t_2}{t_1}=\frac{sin{\theta }_1}{sin{\theta }_2}$