A mass of $M k g$ is suspended by a weightless string. The minimum horizontal force that is required to displace it slowly until the string makes an angle of $45^{0}$ 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 B $M g (\sqrt{2} - 1)$
Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using work energy theorem. It is given that body is taken slowly so that its speed doesn't change, then
$Δ K . E = 0$
$= W_{F} + W_{M g} + W_{t e n s i o n}$
[symbols have their usual meanings]
$W_{F} = F \times l s i n 45^{0},$
$W_{M g} = M_{g} (l - l c o s 45^{0}), W_{t e n s i o n} = 0$
$F = M g (\sqrt{2} - 1)$

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