A) \[4{{r}_{1}}\]
B) \[{{r}_{1}}\]
C) \[2{{r}_{1}}\]
D) \[\frac{{{r}_{1}}}{2}\]
Correct Answer: D
Solution :
\[\frac{\phi v}{dt}=\frac{\pi }{8}\,\frac{p{{r}^{4}}}{nL}\] \[\frac{{{p}_{1}}r_{1}^{4}}{{{L}_{1}}}\,=\frac{{{p}_{2}}r_{2}^{4}}{{{L}_{2}}}\] \[\frac{{{p}_{1}}r_{1}^{4}}{{{I}_{2}}}\,=\frac{4{{p}_{1}}r_{2}^{4}}{{{I}_{1/4}}}\,=r_{2}^{4}\,=\frac{r_{1}^{4}}{16}\] \[{{r}_{2}}=\frac{{{r}_{1}}}{2}\]You need to login to perform this action.
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