Question

$n$ blocks of different masses are placed on the frictionless inclined plane in contact. They are released at the same time. The force of interaction between $(n-1{)}^{th}$ and $n^{th}$ blocks is:

• A. $(m_{n-1}-m_n)gsin\theta$
• B. $zero$
• C. $nmgcos\theta$
• D. none of these

Answer 1

The correct option is B $zero$

Friction is absent on the incline surface So acceleration of each block is $g\sin \theta$. The FBD of $n^{th}$ block is shown in figure.
from Newton second law
$m_{n}g\sin \theta -T=m_{n}a$
$m_{n}g\sin \theta -T=m_n\left(g\sin \theta \right)\Rightarrow T=0$