For a watt governor, what will be the angular speed corresponding to the height of 10 cm, if g = 10 m/sec² ?

Question

Question

This question was previously asked in

UKPSC JE Mechanical 2013 Official Paper I

Answer 1

Concept:

The height of a Watt governor is given by

\(h = \frac{{895}}{{{N^2}}}\;m\)

N is revolution per minute

Calculation:

Given,

h = 10 cm = 0.1 m, g = 10 m/sec2

So, \(N = \sqrt {\frac{{895}}{{0.1}}} = 94.60\;rpm\)

Angular velocity, \(\omega = \frac{2 \pi N}{60}\)

\(\omega = \frac{2 \pi \times 94.60}{60}=10~rad/sec\)