Question

A bob is hanging over a pulley inside a car moving with constant acceleration \( a \) directed horizontally as shown. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration \( a \) horizontally as shown in figure. Other end of the string is pulled with constant acceleration \( a \) vertically. The tension in the string is equal to –
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• A. \( \frac{m g}{a^2} \)
• B. \( \frac{m g}{a} \)
• C. \( \frac{m g}{a + a^2} \)
• D. \( \frac{m g}{a + a^2} \)

Hint:

For a bob in motion inside a car with both vertical and horizontal acceleration, the tension in the string is a result of both accelerations and can be calculated using Newton's second law.

Answer 1

The Correct Option is C

Step 1: Understanding the forces on the bob.
The bob experiences two accelerations: one due to the horizontal motion of the car (\( a \)) and the other due to the vertical acceleration of the string. The net force on the bob is the vector sum of these two accelerations.
Step 2: Apply Newton's second law.
For the bob in equilibrium, the total acceleration vector is a result of both accelerations (horizontal and vertical). The net acceleration is given by: \[ a_{\text{net}} = \sqrt{a^2 + a^2} = \sqrt{2}a \] Now, the tension \( T \) in the string can be found using the equilibrium condition, considering the force balance along both axes: \[ T = \frac{m g}{a^2 + a} \] Final Answer: \[ \boxed{\frac{m g}{a + a^2}} \]