Q73 / 90 · JEE Main 2024 (09 Apr Shift 2) · Physics

A $1 \mathrm{~kg}$ mass is suspended from the ceiling by a rope of length $4 \mathrm{~m}$. A horizontal force ' $F$ ' is applied at the mid point of the rope so that the rope makes an angle of $45^{\circ}$ with respect to the vertical axis as shown in figure. The magnitude of $F$ is : (Assume that the system is in equilibrium and $g=10 \mathrm{~m} / \mathrm{s}^2$ ) Figure

  • Correct answer
    A
    $10 \mathrm{~N}$
  • B
    $\frac{10}{\sqrt{2}} \mathrm{~N}$
  • C
    $1 \mathrm{~N}$
  • D
    $\frac{1}{10 \times \sqrt{2}} N$

Solution

$\begin{aligned} & \mathrm{T}_1 \sin 45^{\circ}=\mathrm{F} \\ & \mathrm{T}_1 \cos 45^{\circ}=\mathrm{T}_2=1 \times \mathrm{g} \\ & \therefore \tan 45^{\circ}=\frac{\mathrm{F}}{\mathrm{g}} \\ & \therefore \mathrm{F}=10 \mathrm{~N}\end{aligned}$

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

Relevant Questions