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Chapter 2: Problem 234

Two bodies of masses \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) are dropped from heights \(\mathrm{H}\) and \(2 \mathrm{H}\) respectively. The ratio of time taken by the bodies to touch the ground is ... (A) \((1 / 2)\) (B) 2 (C) \((1 / \sqrt{2})\) (D) \((\sqrt{2} / 1)\)

Short Answer

Expert verified
The ratio of time taken by the bodies to touch the ground is \(\frac{1}{\sqrt{2}}\) (option C).

Step by step solution

01

Identify the given information and the equation to use

We are given that body 1 has mass \(m_1\) and is dropped from height \(H\), and body 2 has mass \(m_2\) and is dropped from height \(2H\). We need to find the ratio of time taken for both bodies to reach the ground. Since gravity is acting on both bodies, the acceleration due to gravity (\(g\)) can be assumed to be constant. Both bodies are dropped from rest, so their initial velocities are 0. We can use the second equation of motion to find the time taken to fall: \(s = ut + \frac{1}{2} at^2\) where \(s\) is the displacement, \(u\) is the initial velocity, \(t\) is the time taken, and \(a\) is the acceleration due to gravity.
02

Apply the equation of motion to both bodies

For body 1: \(H = 0 \times t_1 + \frac{1}{2} g t_1^2\) For body 2: \(2H = 0 \times t_2 + \frac{1}{2} g t_2^2\)
03

Solve for the time taken by each body

From the equations, we can solve for \(t_1\) and \(t_2\): For body 1: \(H = \frac{1}{2} g t_1^2\) => \(t_1^2 = \frac{2H}{g}\) => \(t_1 = \sqrt{\frac{2H}{g}}\) For body 2: \(2H = \frac{1}{2} g t_2^2\) => \(t_2^2 = \frac{4H}{g}\) => \(t_2 = \sqrt{\frac{4H}{g}}\)
04

Find the ratio of time taken by the bodies

We are asked to find the ratio of time taken by the bodies. Using the values of \(t_1\) and \(t_2\) found in the previous step, we can find this ratio: \(\frac{t_1}{t_2} = \frac{\sqrt{\frac{2H}{g}}}{\sqrt{\frac{4H}{g}}}\) => \(\frac{t_1}{t_2} = \frac{\sqrt{2H}}{\sqrt{4H}}\) Now we can simplify by noting that \(\sqrt{4H} = 2\sqrt{H}\): \(\frac{t_1}{t_2} = \frac{\sqrt{2H}}{2\sqrt{H}} = \frac{1}{\sqrt{2}}\) The ratio of time taken by the bodies to touch the ground is \(\frac{1}{\sqrt{2}}\), which corresponds to option (C).

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