SSC JE Mechanical Engineering 2013 Paper

Question : 101 of 200

Marks: +1, -0

MECHANICAL

In wheel and differential axle, the velocity ratio is given by

Solution:

Total unwounding = Unwounding (1) – wounding (2)

= πd_1 - πd_2

Distance moved by weigh

W=1/2π(d_1-d_2)

Distance moved by effort wheel

= πD

Velocity ratio

= {\Velocity \effort \wheel}/{\Velocity \of \weight}

={πD}/{(1/2)π(d_1-d_2)}={2D}/{d_1-d_2

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