A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hands of a person standing in the car. The car is moving with constant acceleration `a` directed horizontally as shown in fig. The other end of the string is pulled with constant acceleration `a` vertically. The tension in the string is equal to

Options

A. `msqrt(g^(2)+a^(2))`
B. `msqrt(g^(2)+a^(2))-ma`
C. `msqrt(g^(2)+a^(2))+ma`
D. `m(g+a)`

Answer

C

Explanation

(Force diagram in the frame of the car)
Applying Newton's law perpendicular to string
`mgsintheta=macostheta`
`tantheta=(a)/(g)`
Applying Newton's law along string
implies `T-msqrt(g^(2)+a^(2))=ma` .