Question

If the surfaces shown in figure are frictionless, the ratio of $T_1$ and $T_2$ is

• A. $\sqrt{3}: 2$
• B. $1 : \sqrt{3}$
• C. $1 : 5$
• D. $5 : 1$

Answer 1

Answer: (D)

Acceleration of the system is
$a=\frac{F}{3+12+15}=\frac{F}{30}m\,s^{-2} ; T_{2}=3a=3\times\frac{F}{30}=\frac{F}{10}$
$T_{1}=\left(3+12\right)a=\left(3+12\right)\times\frac{F}{30}=\frac{F}{2}$
$\therefore \frac{T_{1}}{T_{2}}=\frac{5}{1}=5 : 1$