JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 17)
Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $${{24} \over \pi } \times {10^2}$$ Nm-2. What is the minimum radius of the wire ? (Take g = 10 ms-2)
_26th_August_Evening_Shift_en_17_1.png)
125 cm
1250 cm
12.5 cm
1.25 cm
Explanation
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$$T = {{2{m_1}{m_2}g} \over {{m_1} + {m_2}}} = {{2 \times 3 \times 5 \times 10} \over 8}$$
$$ = {{75} \over 2}$$
Stress $$ = {T \over A}$$
$${{24} \over \pi } \times {10^2} = {{75} \over {2 \times \pi {R^2}}}$$
$${R^2} = {{75} \over {2 \times 24 \times 100}} = {3 \over {8 \times 24}}$$
$$\Rightarrow$$ R = 0.125 m
R = 12.5 cm
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