Q. A block of mass $M$ placed on a frictionless horizontal table is pulled by another block of mass $m$ hanging vertically by a massless string passing over a frictionless pully. The tension in the string is
A) $\frac{m}{M + m}g$
B) $\frac{M}{M + m}g$
C) $\frac{M + m}{Mm}g$
D) $\frac{Mm}{M + m}g$
Explanation
(d) Force equation for ‘$M$’
[image:5_image154.png]
$Ma = T$ …(i)
Force equation for, $m$
$mg - T = ma$ …(ii)
On solving (i) and (ii)
$T = \left( \frac{Mm}{M + m} \right)g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.