Question

In the following figures (i) and (ii), all the surfaces are smooth, string is massless and pulley is ideal. The mass of rope is $m$ in both the cases. If $T_1$ and $T_2$ be the tensions at the mid points of the ropes in figure (i) and figure (ii) respectively, then $T_1:T_2$ is

• A. $m:M$
• B. $m:(M+m)$
• C. $(M+m):m$
• D. $1:1$

Answer 1

The correct option is B $m:(M+m)$


Case I
$T=ma$
$Mg-T=Ma$
$\therefore Mg=(M+m)a$
$\therefore a=\frac{Mg}{(M+m)}$
$\therefore T=ma=\frac{Mm}{(M+m)}\times g$
$\therefore T_1=\frac{m}{2}\times \frac{M}{(M+m)}\times g$

Case II
Now $a_2=\frac{M}{m}g$
$\therefore T_2=\frac{m}{2}\times \frac{M}{m}g$
$\therefore T_1:T_2=\frac{1}{(M+m)}:\frac{1}{m}=m:(M+m)$