A) \[1:4\]
B) \[1:2\]
C) \[1:8\]
D) \[2:1\]
Correct Answer: C
Solution :
\[V\alpha \frac{1}{\sqrt{r}},{{V}_{A}}\alpha \frac{1}{\sqrt{r}},{{V}_{B}}\alpha \frac{1}{\sqrt{4r}}\] \[\frac{{{V}_{A}}}{{{V}_{B}}}=2\] \[{{t}_{A}}/{{t}_{B}}=\frac{{\scriptstyle{}^{2\pi r}/{}_{{{V}_{A}}}}}{{\scriptstyle{}^{2\pi 4r}/{}_{{{V}_{B}}}}}\] \[=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}\] \[\therefore \] \[\frac{{{t}_{A}}}{{{t}_{B}}}=1:8\]You need to login to perform this action.
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