Question

Two particles of equal mass are connected to a rope $AB$ of negligible mass, such that one is at end $A$ and the other dividing the length of the rope in the ratio $1:2$ from $A$. The rope is rotated about end $B$ in a horizontal plane. Ratio of the tensions in the smaller part to the other is (ignore effect of gravity)

• A. $4:3$
• B. $1:4$
• C. $1:2$
• D. $1:3$

Answer 1

Answer: (C)

$1:2$

Answer 2

Outer part is the shorter part, $l =\frac{ L }{3}$
Mass connected to this part $= m$
$T _{1}=\frac{ mv ^{2}}{1}=\frac{3 mv ^{2}}{ L }$
Inner part is longer part, $l '=\frac{2 L }{3}$
Mass connected to this part $=2 m$
$T _{2}=\frac{6 mvv ^{2}}{2 L }+ T _{1}=\frac{6 muv ^{2}}{ L }$
$\frac{ T _{1}}{ T _{2}}=\frac{1}{2}$