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A small sphere is suspended by a string from the ceiling of a car. If the car begins to move with constant acceleration \[a\], the inclination of string to the vertical is:
(A) \[{{\tan }^{-1}}\left( \frac{a}{g} \right)\] in direction of motion.
(B) \[{{\tan }^{-1}}\left( \frac{a}{g} \right)\] opposite to the direction of motion.
(C) \[{{\tan }^{-1}}\left( \frac{g}{a} \right)\] indirection of motion.
(D) \[{{\tan }^{-1}}\left( \frac{g}{a} \right)\] opposite to direction of motion.

Last updated date: 13th Jun 2024
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Hint By expanding components of tension \[\text{T}\] along \[x\] and \[y\] directions and their dividing these equations, value of \[\tan \theta \] and hence \[\theta \] can be found.

Complete Step by step solution
A is the position of the sphere before the car moves and B is the position afterwards. Components of \[\text{T}\] along \[x\] and \[y\] axis are:
\[\text{T sin}\theta \text{ = ma}\] …..(1)
\[\text{T cos}\theta \text{ = mg}\] …..(2)
Dividing equation (1) and (2), we get
  & \frac{\sin \theta }{\cos \theta }=\frac{a}{g} \\
 & \tan \theta =\frac{a}{g} \\
Hence, \[\theta ={{\tan }^{-1}}\left( \frac{a}{g} \right)\]
The direction will be opposite to the direction of motion of the car. This is due to the reason that when the car moves backwards due to inertia of motion.

Note Knowledge of tension, expansion of components along different axis is required beforehand.As the car moves with acceleration a, the pseudo force ma acts on the sphere in rear direction and gravitational force mg in the downward direction. The horizontal component of Tension becomes equal to the pseudo force and vertical component equals gravitational force.