Question

A mass of $10\,kg$ is suspended from a spring balance. It is pulled aside by a horizontal string so that it makes an angle of $60^\circ$ with the vertical. The new reading of the balance is

• A. $20\,kg.wt$
• B. $10\,kg.wt$
• C. $10 \, \sqrt {3} \, kg.wt $
• D. $20 \, \sqrt {3} \, kg.wt $

Answer 1

Answer: (A)

The situation is shown in figure At an angle of $60^{\circ}$,
$T \cos \theta =m g $
$T =\frac{m g}{\cos \theta}=\frac{10 g}{\cos 60^{\circ}}$
$=\frac{10}{1 / 2} \,kg - wt . $
$=20\,kg - wt$