A mass of $10\mathrm{ }\mathrm{kg}$ is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of $45°$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

• Correct answer
  A

  $100 N$
• B

  $200 N$
• C

  $140 N$
• D

  $70 N$

Solution

$T\cos 45°=mg$

$T\sin 45°=F$

$\mathrm{t}\mathrm{a}\mathrm{n} 45°=\frac{F}{mg}$

$F=mg=100\mathrm{ }\mathrm{N}$

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