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 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' (relative to car) vertically. the

Options

A. `msqrt(g^2+a^2)`
B. `msqrt(g^2+a^2)`
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 = mg cos theta = tan theta`
`=a/g`
Applying Newton.s law along string
`implies T – m sqrt(g^2 + a^2) = ma//T = m sqrt(g^2 + a^2) + ma`