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A mass of $10 \mathrm{~kg}$ 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
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The correct answer is:
$20 \mathrm{~kg}-\mathrm{wt}$
The situation is shown in figure
At an angle of $60^{\circ}$,
$$
\begin{aligned}
\mathrm{T} \cos \theta &=\mathrm{mg} \\
\mathrm{T} &=\frac{\mathrm{mg}}{\cos \theta}=\frac{10 g}{\cos 60^{\circ}} \\
&=\frac{10}{1 / 2} \mathrm{~kg}-\mathrm{wt} . \\
&=20 \mathrm{~kg}-\mathrm{wt}
\end{aligned}
$$
At an angle of $60^{\circ}$,
$$
\begin{aligned}
\mathrm{T} \cos \theta &=\mathrm{mg} \\
\mathrm{T} &=\frac{\mathrm{mg}}{\cos \theta}=\frac{10 g}{\cos 60^{\circ}} \\
&=\frac{10}{1 / 2} \mathrm{~kg}-\mathrm{wt} . \\
&=20 \mathrm{~kg}-\mathrm{wt}
\end{aligned}
$$
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