Question

A block at rest slides down a smooth inclined plane which makes an angle of ${30}^o$ with the vertical and it reaches the ground in $t_1$ second. Another block is dropped vertically from the same point and reaches the ground in $t_2$ second. What is the ratio $t_1/t_2$?

• A. $1:2$
• B. $2:1$
• C. $1:3$
• D. $1:\sqrt{2}$

Answer 1

Answer: (B)

$2:1$

Correct option is $\left(B\right)$

As we know ; $s=ut+1/2a{t}^2$

For wedge $\Rightarrow$

$\begin{array}{cc}r& =0+\frac{1}{2}g\cos \theta t_1^2\\ \text{(1)}-1-r& =\frac{g\cos \theta t_1^2\left(\theta \text{is vertical}\right)}{2}\end{array}$

For dropped block $\Rightarrow$

$r\cos \theta =\frac{1}{2}g{t}_2^2----\left(2\right)$

dividing eq. (1) and (2)

$\begin{array}{cc}\frac{1}{\cos \theta }& =\cos \theta +\frac{2}{t_2^2}\\ \sec \theta & =t_{1/t_2}\\ 2/1& =t_1/t_2\end{array}$