Q125 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Physics
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
- A\(1 \mathrm{~kg}\)
- Correct answerB\(0.5 \mathrm{~kg}\)
- C\(0.75 \mathrm{~kg}\)
- D\(0.25 \mathrm{~kg}\)
Solution
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, \(\begin{aligned} m g & =\frac{g}{2} \\ \Rightarrow \quad m & =0.5 \mathrm{~kg} \end{aligned}\)
Hence, the correct answer is option: 'B'.