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A system containing masses and pulleys connected on an inclined plane is shown in the figure. If the system is in equilibrium then the value of \(m\) is
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Verified Answer
The correct answer is:
\(0.5 \mathrm{~kg}\)
According to the question, the complete situation is shown in the following figure,
Since, system is in equilibrium, hence
\(\begin{aligned}
T & =m g \quad \ldots (i) \\
T & =\left(2 g-T_1\right) \sin 30^{\circ} \quad \ldots (ii) \\
T_1 & =1 g \quad \ldots (iii)
\end{aligned}\)
\(\therefore\) From Eqs. (ii) and (iii), we get
\(\begin{aligned}
& T=(2 g-\mathbf{l} g) \sin 30^{\circ}=g \sin 30^{\circ} \\
& T=\frac{g}{2} \quad \ldots (iv)
\end{aligned}\)
\(\therefore\) From Eqs. (i) and (iv), we get,
\(\begin{aligned}
m g & =\frac{g}{2} \\
\Rightarrow \quad m & =0.5 \mathrm{~kg}
\end{aligned}\)
Since, system is in equilibrium, hence
\(\begin{aligned}
T & =m g \quad \ldots (i) \\
T & =\left(2 g-T_1\right) \sin 30^{\circ} \quad \ldots (ii) \\
T_1 & =1 g \quad \ldots (iii)
\end{aligned}\)
\(\therefore\) From Eqs. (ii) and (iii), we get
\(\begin{aligned}
& T=(2 g-\mathbf{l} g) \sin 30^{\circ}=g \sin 30^{\circ} \\
& T=\frac{g}{2} \quad \ldots (iv)
\end{aligned}\)
\(\therefore\) From Eqs. (i) and (iv), we get,
\(\begin{aligned}
m g & =\frac{g}{2} \\
\Rightarrow \quad m & =0.5 \mathrm{~kg}
\end{aligned}\)
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