A) \[n-\]times
B) \[{{n}^{2}}-\]times
C) \[{{n}^{3}}-\]times
D) \[{{n}^{4}}-\]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. \[m=Av\rho \] ... (i) Rate of increase of kinetic energy \[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(Av\rho ){{v}^{2}}\] ? (ii) Mass\[m\], flowing out per sec, can be increased to \[m'\] by increasing \[v\] to\[v'\], then power increases from \[P\] to\[P'\]. \[\frac{P'}{P}=\frac{\frac{1}{2}A\rho v{{'}^{3}}}{\frac{1}{2}A\rho {{v}^{3}}}or\frac{P'}{P}={{\left( \frac{v'}{v} \right)}^{3}}\] Now, \[\frac{m'}{m}=\frac{A\rho v'}{A\rho v}=\frac{v'}{v}\] As \[m'=nm,\,\,v'=nv\] \[\therefore \] \[\frac{P'}{P}={{n}^{3}}\Rightarrow P'={{n}^{3}}p\]
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