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