Question

A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration ${}^{\prime }{a}^{\prime }$ directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration ${}^{\prime }{a}^{\prime }$ vertically. The tension in the string is equal to :

• A. $m\sqrt{g^2+a^2}$
• B. $m\sqrt{g^2+a^2}-ma$
• C. $m(g+a)$
• D. $m\sqrt{g^2+a^2}+ma$

Answer 1

Answer: (D)

$m\sqrt{g^2+a^2}+ma$

From NLM,

$T-m\sqrt{a^2+g^2}=ma$

$T=ma+m\sqrt{a^2+g^2}$

Hence, the option $D$ is the correct answer.