GATE 2026 AG – Question 38
A massless thin flat belt drives a 0.6 m diameter output pulley at 500 rpm without slip. Contact angles are 190° on this pulley and 170° on the smaller input pulley. Friction coefficient is 0.30 and slack tension 250 N. Ignore other losses, use π=3.14,and find the nearest transmitted power in kW.
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Correct answer: (B) 5.6
Explanation
The smaller contact angle limits traction:$\theta=170(3.14)/180=2.96556$ rad. At the limiting tension ratio,$T_1/T_2=e^{\mu\theta}=e^{0.889667}$,so $T_1\approx608.6$ N. Belt speed is $v=\pi DN/60=15.7$ m/s. Hence $P=(T_1-T_2)v\approx5.63$ kW, nearest **5.6,B**. This is the maximum transmissible power at impending slip;without that implicit limit, slack tension alone does not fix actual power.