There is a small hole in a table. A string of length 1m passes through it. Two bodies of masses 70g and 100g are attached at its ends. The 100g mass hangs freely at a depth of 60 cm from the table. If this mass is to be in equilibrium, the other mass should rotate in a circle with a frequency equal to

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$π / 140 Hz$

$π / \sqrt{140} Hz$

$4 π / 140 Hz$

$\sqrt{140} / 4 π Hz$

answer is D.

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Detailed Solution

$\begin{array}{l} Mg = mr ω^{2} \Rightarrow 100 \times 9.8 = 70 \times 0.4 \times ω^{2} \\ \Rightarrow 100 \times 9.8 = 70 \times 0.4 \times ω^{2} \\ 14 = \frac{4}{10} \times 4 π^{2} n^{2} \Rightarrow n^{2} = \frac{140}{16 π^{2}} \\ \Rightarrow n = \frac{\sqrt{140}}{4 π} \end{array}$

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