A) \[n\text{-}\]times
B) \[{{n}^{2}}\text{-}\]times
C) \[{{n}^{3}}\text{-}\]times
D) \[{{n}^{4}}\text{-}\]times
Correct Answer: C
Solution :
If the motor pumps water (density =\[\rho \]] continuously through a pipe of area of cross-section A with velocity v, then mass flowing out per second. \[m=A\nu \rho \] ?(i) Rate of increase of kinetic energy \[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(A\nu \rho ){{\nu }^{2}}\] ?(ii) Mass m, flowing out per sec, can be increased to m by increasing \[\nu \] to \[\nu ,\]then power increases from P to P. \[\frac{P}{P}=\frac{\frac{1}{2}A\rho \nu {{}^{3}}}{\frac{1}{2}A\rho {{\nu }^{3}}}\] or \[\frac{P}{P}={{\left( \frac{\nu }{\nu } \right)}^{3}}\] Now, \[\frac{m}{m}=\frac{A\rho \nu }{A\rho \nu }=\frac{\nu }{\nu }\] As \[m=nm,\]\[\nu =n\nu \] \[\therefore \] \[\frac{P}{P}={{n}^{3}}\] \[\Rightarrow \] \[P={{n}^{3}}P\]
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