Question

One end of a stationary rope is fixed to a vertical wall and the other end pulled very slowly by a horizontal force of 40 N as shown in figure. Mass of rope is ($g=10m/s^2$)

• A.
  4 Kg
• B. $\frac{2}{\sqrt{3}}$ Kg
• C.
  $4\sqrt{3}$ Kg
• D.
  $2\sqrt{3}$ Kg

Answer 1

The correct option is C

$4\sqrt{3}$ Kg
Tcos$\theta$ = mg (1)
Tsin$\theta$ = F (2)
Here, $\theta$ = ${30}^{\circ }$ and F = 40 N
Equation 2 becomes,
$\therefore$ $T\sin {30}^{\circ }=40$
$\therefore$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{T}{2}$ = 40
$\therefore$ T = 80 N
From equation 1,
m = $\frac{T\cos {30}^{\circ }}{g}$
$\therefore$ m = $\frac{80\cos {30}^{\circ }}{10}$
$\therefore$ m = $\frac{80\sqrt{3}}{20}$
$\therefore$ m = $4\sqrt{3}$ kg