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Question

Physics Question on laws of motion

A mass of 10kg10\,kg is suspended from a spring balance. It is pulled aside by a horizontal string so that it makes an angle of 6060^\circ with the vertical. The new reading of the balance is

A

20kg.wt20\,kg.wt

B

10kg.wt10\,kg.wt

C

103kg.wt10 \, \sqrt {3} \, kg.wt

D

203kg.wt20 \, \sqrt {3} \, kg.wt

Answer

20kg.wt20\,kg.wt

Explanation

Solution

The situation is shown in figure At an angle of 6060^{\circ},
Tcosθ=mgT \cos \theta =m g
T=mgcosθ=10gcos60T =\frac{m g}{\cos \theta}=\frac{10 g}{\cos 60^{\circ}}
=101/2kgwt.=\frac{10}{1 / 2} \,kg - wt .
=20kgwt=20\,kg - wt