Question

A block released from rest from the top of a smooth inclined plane of angle ${\theta }_1$ reaches the bottom in time $t_1$. The same block released from rest from the top of another smooth inclined plane of angle ${\theta }_2$, 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. $\frac{t_2}{t_1}=1$
• C. $\frac{t_2}{t_1}=\left(\frac{sin{\theta }_1}{sin{\theta }_2}\right)$
• D. $\frac{t_2}{t_1}=\left(\frac{si{n}^2{\theta }_1}{si{n}^2{\theta }_2}\right)$

Answer 1

The correct option is C $\frac{t_2}{t_1}=\left(\frac{sin{\theta }_1}{sin{\theta }_2}\right)$

From the figure

we see that

down the incline acceleration us $a=g\sin \theta$

distance $s=\frac{h}{\sin {\theta }_1}$

using $s=ut+\frac{1}{2}a{t}^2;u=0$ (initially rest)

gives $t=\sqrt{\frac{2s}{a}}$

For ${\theta }_1{t}_1=\sqrt{\frac{2h}{\sin {\theta }_1\times g\sin {\theta }_1}}=\frac{1}{\sin {\theta }_1}\sqrt{\frac{2h}{g}}$

so for $O_2,t_2=\frac{1}{\sin {\theta }_2}\sqrt{\frac{2h}{g}}$

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