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

Applying Newton's law perpendicular to string
`mg sin theta =ma cos theta`
`implies tan theta =a/g`
Applying Newton's law along string,
`T-mg cos theta -ma sin theta =ma`
`implies T=msqrt(g^(2)+a^(2))+ma`.