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A mass of $10 \mathrm{~kg}$ is suspended from a spring balance. It is pulled a side 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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$20 \mathrm{~kg}$-wt
Given, mass suspended from spring balance, $m=10 \mathrm{~kg}$
$$
\theta=60^{\circ}
$$
The given situation is shown below
From the diagram, reading of spring balance,
$F_{s}=T$
Since, $T \cos \theta=m g$
$$
T=\frac{m g}{\cos \theta}
$$
From Eqs, (i) and (ii) we get
$$
\begin{aligned}
F_{s} &=\frac{m g}{\cos \theta}=\frac{m g}{\cos 60^{\circ}} \\
&=2 m g \\
&=2 \times 10 \times g \\
&=20 \mathrm{~kg}-\mathrm{wt}
\end{aligned}
$$
$$
\theta=60^{\circ}
$$
The given situation is shown below
From the diagram, reading of spring balance,
$F_{s}=T$
Since, $T \cos \theta=m g$
$$
T=\frac{m g}{\cos \theta}
$$
From Eqs, (i) and (ii) we get
$$
\begin{aligned}
F_{s} &=\frac{m g}{\cos \theta}=\frac{m g}{\cos 60^{\circ}} \\
&=2 m g \\
&=2 \times 10 \times g \\
&=20 \mathrm{~kg}-\mathrm{wt}
\end{aligned}
$$
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