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An automobile engine develops 10⁵ W power when rotating at an angular speed of 900 rad/minute. The instantaneous torque delivered by the engine is

Dynamics of rotational motion about a fixed axis
NEET
1

10⁴ N-m

2

\frac{1}{3} \times 10⁴ N-m

3

\frac{2}{3} \times 10⁴ N-m

4

2 \times 10⁴ N-m

Solution:

The power developed by a rotating engine is given by the formula P = \tau \omega, where P is the power, \tau is the torque, and \omega is the angular speed.

Given:

Power P = 10⁵ W

Angular speed \omega = 900 rad/minute

First, convert the angular speed from rad/minute to rad/second:

\omega = 900 \frac{rad}{minute} \times \frac{1 minute}{60 seconds} = \frac{900}{60} rad/s = 15 rad/s

Now, use the power formula to find the torque \tau:

P = \tau \omega

\tau = \frac{P}{\omega}

\tau = \frac{10⁵ W}{15 rad/s}

\tau = \frac{100000}{15} N-m

\tau = \frac{20000}{3} N-m

\tau = \frac{2}{3} \times 10000 N-m

\tau = \frac{2}{3} \times 10⁴ N-m

Comparing this result with the given options, we find that it matches option (3).