Question

A rod of length ${}^{\prime }{C}^{\prime }$ rests at a point $A$ against a smooth vertical wall while end $B$ is on the floor as shown in the figure. If the end $A$ move uniformly downwards, what will be the velocity of the end $B$ is $x$ is the distance of point $B$ from wall?

• A. $v_B=\left[\sqrt{\frac{t^2}{x^2}-1}\right]v_A$
• B. $[\sqrt{\frac{x^2}{{}^{t^2}}}-1]v_A$
• C. $\left[\frac{t^2-x^2}{x}\right]v_A$
• D. None of these

Answer 1

Answer: (A)

$v_B=\left[\sqrt{\frac{t^2}{x^2}-1}\right]v_A$

Option C

$\begin{array}{cc}\text{As length of rod remains constant:}& \\ \begin{array}{cc}{V}_A\cos \theta & =V_B\sin \theta \\ V_A\cot \theta & =V_B\end{array}& \\ {\left[\frac{x}{\sqrt{t^2-x^2}}\right]}^{-1}{V}_A=V_B& \\ & \left[\sqrt{\frac{t^2}{x^2}-1}\right]V_A=V_B\end{array}$