Question

In the given figure two identical particles each of mass $m$ are tied together with an inextensible string. This is pulled at its centre with a constant force $F$. If the whole system lies on a smooth horizontal plane, then the acceleration of each particle towards each other is:

• A. $\frac{\sqrt{3}}{2}\frac{F}{m}$
• B. $\frac{1}{\sqrt{3}}\frac{F}{m}$
• C. $\frac{2}{\sqrt{3}}\frac{F}{m}$
• D. $\sqrt{3}\frac{F}{m}$

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}\frac{F}{m}$

Let the tension developed in both the strings be $T$,as their masses are assumed to be equal.Now balancing components along x axis we get, $2Tcos30=F$,which gives $T=\frac{F}{\sqrt{3}}$.

Now,both the masses have acceleration $a=\frac{F}{2\sqrt{3}m}$,towards each other $i.e.$ in opposite direction.So,the acceleration with which both masses approach each other is $=\frac{F}{2\sqrt{3}m}$+$\frac{F}{2\sqrt{3}m}$.So, option $b$ is correct.