Question

A ball is held at rest in position $A$ by two light cords. The horizontal cord is now cut and the ball swings to position $B$. What is the ratio of the tension in the cord at position $B$ to that at position $A$ originally?

• A. $3$
• B. $3/4$
• C. $1/2$
• D. $1$

Answer 1

The correct option is B $3/4$

FBD of ball at position ${}^{\prime }{A}^{\prime }$:


For equilibrium of ball:
$T\cos {30}^{\circ }=mg$
$\therefore T=\frac{2}{\sqrt{3}}mg$ ... (i)
and $T_1=T\sin {30}^{\circ }$ ... (ii)

FBD of ball at position ${}^{\prime }{B}^{\prime }$:


Applying equilibrium condition for ball along the length of the string
$T_1=mg\cos {30}^{\circ }=\frac{\sqrt{3}}{2}mg$ ... (iii)

From eq (i) and (iii)
$\therefore \frac{T_1}{T}=\frac{\left(\frac{\sqrt{3}}{2}mg\right)}{\left(\frac{2}{\sqrt{3}}mg\right)}=\frac{3}{4}$