Question

In the given arrangement, $n$ number of equal masses are connected by strings of negligible masses. The tension in the string connected to the $n^{th}$ mass is

• A. $\frac{mMg}{nm+M}$
• B. $\frac{mMg}{nmM}$
• C. $mg$
• D. $nmg$

Answer 1

The correct option is A $\frac{mMg}{nm+M}$

Conidering all $n$ blocks as one mass, FBD of the system can be represented as follows. Let $a$ be the acceleration of the system.

From figure (a), we have
$Mg-T=Ma$..........(i) and
$T=nma$...........(ii)
From (i) & (ii) we have,
$a=\frac{Mg}{nm+M}$

FBD of the $n^{th}$ mass alone

From figure (b),
$T_n=ma⟹T_n=\frac{mMg}{nm+M}$