A liquid of specific gravity 0.8 is flowing in a pipeline with a speed of 2 m/ s. The K.E. per cubic meter of it is

160 J

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1600 J

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160.5 J

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1.6 J

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Solution

The correct option is B 1600 J
$1 m^{3} = 100 c m \times 100 c m \times 100 c m$
[B]
$\begin{matrix} V o l u m e i n c m^{3} = 1000000 c m^{3} 1 l i t r e = 1000 c m^{3} V o l u m e i n l i t r e = \frac{1000000}{1000} = 1000 l i t r e s S p e c i f i c g r a v i t y = 0.8 m a s s o f w a t e r = 1000 \times 0.8 = 800 k g m a s s o f 1 m^{3} = 800 k g k i n e t i c e n e r g y = \frac{1}{2} m V^{2}, V = 2 m / s = \frac{1}{2} \times 800 \times 4 = 1600 J \end{matrix}$

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A liquid of specific gravity 0.8 is flowing in a pipeline with a speed of 2 m/ s. The K.E. per cubic meter of it is

The correct option is B 1600 J1m3=100cm×100cm×100cm[B] Volumeincm3=1000000cm31litre=1000cm3Volumeinlitre=10000001000=1000litresSpecificgravity=0.8massofwater=1000×0.8=800kgmassof1m3=800kgkineticenergy=12mV2,V=2m/s=12×800×4=1600J