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JEE Main 2006
Exam
JEE Main
Year
2006
Subject
Physics
Chapter
Laws of Motion

JEE Main 2006 · Physics · Laws of Motion

A mass of M kg is suspended by a weightless string. The horizontal...

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

Options

  1. AMg(2−1)M g(\sqrt{2}-1)
  2. BMg(2+1)M g(\sqrt{2}+1)
  3. CMg2M g \sqrt{2}
  4. DMg2\frac{\mathrm{Mg}}{\sqrt{2}}

Correct answer

A. Mg(2−1)M g(\sqrt{2}-1)

Step-by-step solution

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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