Q96 / 150 · JEE Main 2006 · Physics

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

  • Correct answer
    A
    $M g(\sqrt{2}-1)$
  • B
    $M g(\sqrt{2}+1)$
  • C
    $M g \sqrt{2}$
  • D
    $\frac{\mathrm{Mg}}{\sqrt{2}}$

Solution

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

Hence, the correct answer is option: 'A'.

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