Q. A block of mass $M$ placed on a frictionless horizontal table is pulled by an other block of mass $m$ hanging vertically by a massless string passing over a frictionless pulley. 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) $Tension\ in\ the\ string,\ \ T = \frac{m_{1}m_{2}}{m_{1} + m_{2}}g$
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$Here,\ \ tension\ T = \frac{Mm}{M + m}g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.