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 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically (with respect to train). The tension in the string is equal to

• A.
• B.
• C.
• D. m(g + a)

Answer 1

The correct option is C

(From diagram in the frame of the car)

Applying Newton's law perpendicular to string

mg sin$\theta$ = ma cos$\theta$

$\Rightarrow tan\theta =\frac{a}{g}$

Applying Newton's law along string

T - mg cos$\theta$ - ma sin$\theta$ = ma

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