A mass of $\mathrm{M} \mathrm{kg}$ is suspended by a weightless string. The horizontal force that is…

A mass of $\mathrm{M} \mathrm{kg}$ is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $45^{\circ}$ with the initial vertical direction is
  1. $M g(\sqrt{2}-1)$
  2. $M g(\sqrt{2}+1)$
  3. $M g \sqrt{2}$
  4. $\frac{\mathrm{Mg}}{\sqrt{2}}$

Solution

$\mathrm{F} \ell \sin 45=M g(\ell-\ell \cos 45)$ $F=M g(\sqrt{2}-1)$

Asked in: JEE Main 2006

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