Laws of Motion

JEE Main level, single correct, medium

106 of 210

Question 106 of 210

A mass of Mkg\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∘45^{\circ} with the initial vertical direction is

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Answer: (A) Mg(2−1)M g(\sqrt{2}-1)

Fℓsin⁡45=Mg(ℓ−ℓcos⁡45)\mathrm{F} \ell \sin 45=M g(\ell-\ell \cos 45)
F=Mg(2−1)F=M g(\sqrt{2}-1)

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