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NEET PHYSICSEasy

An automobile engine develops 100 kW100\text{ kW} when rotating at a speed of 1800 rev/min1800\text{ rev/min}. What torque does it deliver?

A

350 N m350\text{ N m}

B

440 N m440\text{ N m}

C

531 N m531\text{ N m}

D

628 N m628\text{ N m}

Step-by-Step Solution

Given, Power (PP) = 100 kW=100×103 W100\text{ kW} = 100 \times 10^3\text{ W} Angular speed (ω\omega) = 1800 rev/min1800\text{ rev/min} First, convert angular speed from rev/min to rad/s: ω=1800×2π60 rad/s=60π rad/s\omega = 1800 \times \frac{2\pi}{60}\text{ rad/s} = 60\pi\text{ rad/s} We know the relation between power, torque (τ\tau), and angular speed is: P=τωP = \tau\omega τ=Pω\Rightarrow \tau = \frac{P}{\omega} τ=100×10360π=100006π=50003π\tau = \frac{100 \times 10^3}{60\pi} = \frac{10000}{6\pi} = \frac{5000}{3\pi} τ50003×3.1450009.42530.7 N m531 N m\tau \approx \frac{5000}{3 \times 3.14} \approx \frac{5000}{9.42} \approx 530.7\text{ N m} \approx 531\text{ N m}

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