A mass of $M k g$ is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $45^{o}$ with the initial vertical direction is:

M g (\sqrt{2} + 1)

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M g \sqrt{2}

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\frac{M g}{\sqrt{2}}

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M g (\sqrt{2} - 1)

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Solution

The correct option is D $M g (\sqrt{2} - 1)$
By conservation of energy, work done by force $=$ increase in potential energy.
From the figure we have:
$\Rightarrow l = 2 R sin (22.5 °) \Rightarrow W = F l cos (22.5 °) \Rightarrow W = F R (2 sin (22.5 °) cos (22.5 °)) = F R sin (45 °) = \frac{F R}{\sqrt{2}}$
Change in potential energy $= M g h = M g R (1 - cos (45 °)) = M g R (1 - \frac{1}{\sqrt{2}})$
$\Rightarrow M g R (1 - \frac{1}{\sqrt{2}}) = \frac{F R}{\sqrt{2}} \Rightarrow F = M g (\sqrt{2} - 1)$

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