SSC JE Mechanical Engineering 2018 Set 1 Paper

Question : 112 of 200

Marks: +1, -0

What will be the change in the vertical height (in m) of a watt governor, when the speed is decreased from 50 rpm to 25 rpm?

Solution:

decrease in height

=895/N_1^2-895/N_2^2

=895/50^2-895/ 25^2

–1.074 M

(–ve sign means height will be decrease)

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