A) \[200%\]
B) \[100%\]
C) \[50%\]
D) \[300%\]
Correct Answer: D
Solution :
Given: \[l'=l+100%\,l=2l\] Initial volume = Final volume i.e., \[\pi {{r}^{2}}l=\pi {{r}^{'2}}l'\] \[\Rightarrow \] \[r{{'}^{2}}=\frac{{{l}^{2}}l}{l'}={{r}^{2}}\times \frac{l}{2\,l}\,\,\,\,\Rightarrow \,\,\,{{r}^{2}}=\frac{{{r}^{2}}}{2}\] \[\therefore \] \[R'=\rho \frac{l'}{A'}=\rho \frac{2\,l}{4\pi {{'}^{2}}}\] \[\left( \because R=\frac{\rho l}{A} \right)\] \[=\frac{\rho .\,4l}{\pi {{r}^{2}}}=4R\] Thus, \[\Delta R=R'-R=4R=3R\] \[\therefore \] \[%\Delta R=\frac{3R}{R}\times 100%=300%\]
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