SSC JE Mechanical Engineering 2012 Paper

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Question : 128 of 200

Marks: +1, -0

The kinetic energy associated with a jet through a cross-sectional area ‘a’ having density ‘p’ and velocity ‘v’ is

$1/2\pav^ 3$
$1/2\pa^2v^ 3$
$1/2\p v^ 2$
$1/2\pa v^ 2$

Solution:

We know K.E

= 1 /2 mv^2

=1/2(\mass/s×(\velocity)^2)

∵

m/v=p

=1/2(pav).v^2\;\;\;m=p.v=pa.v

=1/2pav^3

Volume

V = a. l

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