Question

A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity $2m{s}^{-1}$ towards right. The velocity of end B of rod at the instant shown in figure is:

• A. $\sqrt{3}m{s}^{-1}$
• B. $2m{s}^{-1}$
• C. $2\sqrt{3}m{s}^{-1}$
• D. $4m{s}^{-1}$

Answer 1

The correct option is C $2\sqrt{3}m{s}^{-1}$

Let L be the length of the rod AB. Let x be the distance from A to C and Y be the vertical distance from B to C.

Now, $x^2+y^2=L^2$

Differentiating with respect to time , we get

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0⟹\frac{dy}{dt}=V_B=\frac{-2x}{y}$

Here $\frac{dx}{dt}=V_A=2m/s,tan30=\frac{y}{x}=\frac{1}{\sqrt{3}}$

And hence the speed of the end B is $V_B=2\sqrt{3}m/s$