A mass falls from a height 'h' and its time of fall $'l'$ is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that I = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t' and T'.

Question

Two identical particles each of mass $ m $ go around a circle of radius $ a $ under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

To solve this problem, we need to assess the conditions under which a mass falls from a height 'h' and relate it to the time period of a simple pendulum on different planetary surfaces.

Let's break down the problem:

Initially, on the surface of earth, the time of fall 't' is related to the time period 'T' of a simple pendulum as: $t = 2T$.
The setup is taken to another planet with half the mass of Earth but the same radius. Both the time of fall (denoted as t') and pendulum period (denoted as T') change due to the new gravitational acceleration.

Now, consider the changes:

The gravitational acceleration 'g' on a planet is given by: $g_{planet} = \frac{G M_{planet}}{R^2}$. Here, M_{planet}\) is the mass of the planet, and R\) is the radius.
On Earth, gravitational acceleration g = \frac{G M_{earth}}{R_{earth}^2}\).
For the new planet, since the mass is half, g_{new} = \frac{1}{2}g\).

Time of fall of an object from a height 'h' is governed by the equation: $ h = \frac{1}{2}g t^2 $.

The time period of a simple pendulum is given by $ T = 2\pi \sqrt{\frac{L}{g}} $, and on the new planet, due to reduced gravity: $ T' = 2\pi \sqrt{\frac{L}{g_{new}}} = 2\pi \sqrt{\frac{L}{\frac{g}{2}}} = \sqrt{2} \times T $.

Since, on the surface of the new planet, both the pendulum period increases by a factor of \sqrt{2}, and because gravitational acceleration decreases, the fall time remains equivalent by the same factor due to the consistent ratio, leading to:

t' = \sqrt{2} \times t = \sqrt{2} \times 2T = 2 \times T', due to T' = \sqrt{2} T.

Thus, the correct option is:

t' = 2 T'

Answer 2

To solve this problem, we need to assess the conditions under which a mass falls from a height 'h' and relate it to the time period of a simple pendulum on different planetary surfaces.

Let's break down the problem:

Initially, on the surface of earth, the time of fall 't' is related to the time period 'T' of a simple pendulum as: $t = 2T$.
The setup is taken to another planet with half the mass of Earth but the same radius. Both the time of fall (denoted as t') and pendulum period (denoted as T') change due to the new gravitational acceleration.

Now, consider the changes:

The gravitational acceleration 'g' on a planet is given by: $g_{planet} = \frac{G M_{planet}}{R^2}$. Here, M_{planet}\) is the mass of the planet, and R\) is the radius.
On Earth, gravitational acceleration g = \frac{G M_{earth}}{R_{earth}^2}\).
For the new planet, since the mass is half, g_{new} = \frac{1}{2}g\).

Time of fall of an object from a height 'h' is governed by the equation: $ h = \frac{1}{2}g t^2 $.

The time period of a simple pendulum is given by $ T = 2\pi \sqrt{\frac{L}{g}} $, and on the new planet, due to reduced gravity: $ T' = 2\pi \sqrt{\frac{L}{g_{new}}} = 2\pi \sqrt{\frac{L}{\frac{g}{2}}} = \sqrt{2} \times T $.

Since, on the surface of the new planet, both the pendulum period increases by a factor of \sqrt{2}, and because gravitational acceleration decreases, the fall time remains equivalent by the same factor due to the consistent ratio, leading to:

t' = \sqrt{2} \times t = \sqrt{2} \times 2T = 2 \times T', due to T' = \sqrt{2} T.

Thus, the correct option is:

t' = 2 T'

The Correct Option is D

Solution and Explanation

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