Answer
64.8k+ views
Hint:In this question, a pendulum is a weight suspended from a pivot so that it can swing freely. When a pendulum is displaced sideways from its resting equilibrium position it is subject to a restoring force due to gravity.
Complete step by step solution:
In this question, the job of a pendulum is positively charged. Another identical charge is placed at the point of suspension of the pendulum. The time period of pendulum
The force due to the charge kept at the suspension point always acts along the wire which is compensated by increasing the tension of the string. We have to increase the wire length to get the positive charge of the wire. This force never accounts for charge in a time period since it has no component perpendicular to the wire.
As we know that the time period of pendulum when job of a pendulum positively charged is
${T_1} = mg\sin \theta l......\left( {\text{I}} \right)$
Here, mass of the pendulum is $m$, the gravitational constant is $g$, the angle between the axis and the actual axis of the pendulum is $\theta $ and length of the pendulum string is $l$.
Similarly, the time period of pendulum when another identical charge is placed at the point of suspension of the pendulum is
${T_2} = mg\sin \theta l......\left( {{\text{II}}} \right)$
Now, we Compare equation (I) and (II).
$\therefore {T_1} = {T_2}$
Therefore, the time period will remain constant.
Hence, the correct option is (C).
Note:Do not confuse with the same formula of the time period in both cases as the angle $\alpha = - {\omega ^2}\theta $ is remains constant in both the cases, and due to which the time period is also remains constant for the both cases.
Complete step by step solution:
In this question, the job of a pendulum is positively charged. Another identical charge is placed at the point of suspension of the pendulum. The time period of pendulum
The force due to the charge kept at the suspension point always acts along the wire which is compensated by increasing the tension of the string. We have to increase the wire length to get the positive charge of the wire. This force never accounts for charge in a time period since it has no component perpendicular to the wire.
As we know that the time period of pendulum when job of a pendulum positively charged is
${T_1} = mg\sin \theta l......\left( {\text{I}} \right)$
Here, mass of the pendulum is $m$, the gravitational constant is $g$, the angle between the axis and the actual axis of the pendulum is $\theta $ and length of the pendulum string is $l$.
Similarly, the time period of pendulum when another identical charge is placed at the point of suspension of the pendulum is
${T_2} = mg\sin \theta l......\left( {{\text{II}}} \right)$
Now, we Compare equation (I) and (II).
$\therefore {T_1} = {T_2}$
Therefore, the time period will remain constant.
Hence, the correct option is (C).
Note:Do not confuse with the same formula of the time period in both cases as the angle $\alpha = - {\omega ^2}\theta $ is remains constant in both the cases, and due to which the time period is also remains constant for the both cases.
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