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