Q. A mass of 10 kg is suspended from a string 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
Explanation
(a) The situation is shown in figure.
At an angle of $60{^\circ}$.
$T\cos{\theta = mg}$
$T = \frac{mg}{\cos\theta}$
$= \frac{10g}{\cos{60{^\circ}}}$
$= \frac{10}{1/2}kg–wt$
$20\ kg–wt$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.