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. `mg(cos^(2) theta+sin theta)+ ma`
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 `mg sin theta=ma cos theta`
`tan theta=a/g`
Applying Newton's law along spring `rArr T-msqrt(g^(2)+a^(2))=ma T=msqrt(g^(2)+a^(2))+ma`