A mass of M kg is suspended by a weightless string. The horizontal force required to displace it until the string makes an angel of $45^{\circ}$ with the initial vertical direction is

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

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$M g (\sqrt{2} - 1)$

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$M g (\sqrt{2} + 1)$

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$M g \sqrt{2}$

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Solution

The correct option is B
$M g (\sqrt{2} - 1)$

$W_{F} + W_{g r a v i t y} = (K_{f i n a l} - K_{i n i t i a l}) = (0 - 0) = 0$

Work done by F, $W_{F} = \times l s i n 45^{\circ} = \frac{F l}{\sqrt{2}}$

Work done by gravity,

$W_{g r a v i t y} = - M g (l - l c o s 45^{\circ}) = M g l (1 - \frac{1}{\sqrt{2}})$

$∴ \frac{F l}{\sqrt{2}} + (- \frac{M g l (\sqrt{2} - 1)}{\sqrt{2}}) = 0$

or $F = M g (\sqrt{2} - 1)$

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A mass of M kg is suspended by a weightless string. The horizontal force required to displace it until the string makes an angel of 45∘ with the initial vertical direction is

The correct option is B Mg(√2−1) WF+Wgravity=(Kfinal−Kinitial)=(0−0)=0 Work done by F, WF=×l sin45∘=Fl√2 Work done by gravity, Wgravity=−Mg(l−lcos45∘)=Mgl(1−1√2) ∴Fl√2+(−Mgl(√2−1)√2)=0 or F=Mg(√2−1)

A mass of M kg is suspended by a weightless string. The horizontal force required to displace it until the string makes an angel of $45^{\circ}$ with the initial vertical direction is