Question

The length of the seconds pendulum is decreased by 0.3 cm when it is shifted from place A to place B. If the acceleration due to gravity at place A is 981 cm/s\(^2\), the acceleration due to gravity at place B is (Take \( \pi^2 = 10 \))

• A. 975 cm/s\(^2\)
• B. 978 cm/s\(^2\)
• C. 984 cm/s\(^2\)
• D. 981 cm/s\(^2\)

Hint:

For pendulum problems involving changes in length, use the relationship between the period, length, and gravity to determine the change in gravity. The length of the pendulum is inversely proportional to the gravitational acceleration.

Answer 1

The Correct Option is B

Step 1: Formula for pendulum.
The period \( T \) of a seconds pendulum is given by: \[ T = 2 \pi \sqrt{\frac{L}{g}} \] where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. For the seconds pendulum, the length is related to the period by the above equation, and a change in length corresponds to a change in gravity. Since the length decreases, we use the relation: \[ L_2 = L_1 \left( \frac{g_1}{g_2} \right) \] where \( L_1 \) and \( L_2 \) are the lengths at places A and B, and \( g_1 \) and \( g_2 \) are the accelerations due to gravity at places A and B, respectively. Step 2: Solving for \( g_2 \).
The change in length \( \Delta L = 0.3 \) cm, and the initial gravity \( g_1 = 981 \, \text{cm/s}^2 \). Substituting into the equation, we find: \[ g_2 = 978 \, \text{cm/s}^2 \] Step 3: Conclusion.
Thus, the acceleration due to gravity at place B is \( 978 \, \text{cm/s}^2 \), which is option (B).