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A belt drives a pulley of 200 mm diameter such that the ratio of tensions in the tight side and the slack side is 1.2, the maximum tension in the belt is not to exceed 240 kN. The speed of pulley is 60 rpm. Find the safe power transmitted by the pulley. 
1. 50 kW
2. 20 kW
3. 35 kW
4. 25 kW

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Correct Answer - Option 4 : 25 kW

Concept:

Power transmitted by a belt:

P = (T1 – T2)V

T1 = Tension on the tight side, T2 = Tension on the slack side, V = velocity of the belt

Calculation:

Given:

D = 200 mm = 0.2 m, \(\frac{{T_1}}{{T_2}}=1.2\), Maximum tension = 240 kN, N = 60 rpm

Since, Between T1 and T2, T1 is always greater than T2.

Therefore, T1 = 240 kN, \(\frac{{T_1}}{{T_2}}=1.2 = T_2=\frac{{240}}{{1.2}}=200~ kN\)

\(V=\frac{{\pi DN}}{{60}}=\frac{{\pi ~×~0.2~×~60}}{{60}}=0.628 ~m/s\)

Power transmitted by the pulley, P = (T1 – T2)V = (240 - 200) × 0.628 = 25.12 kW

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