The height of Watt’s governor is proportional to

(where N : rpm of balls)

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

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

\(h \propto \frac{{1}}{{{N^2}}}\;m\)

N is revolution per minute

Additional Information

Height of a porter governor is given by,

Gov2

\({N^2} = \frac{{895}}{h}\left( {\frac{{\left( {2mg + \left( {Mg \pm f} \right)\left( {1 + k} \right)} \right)}}{{2mg}}} \right)\)

Here \(k = \frac{{tan\beta }}{{tan\theta }}\)