A) \[\frac{{{(8.02)}^{4}}-{{(7.98)}^{4}}}{{{(0.8)}^{4}}}\]
B) \[\frac{{{(8.02)}^{2}}-{{(7.98)}^{2}}}{{{(0.8)}^{2}}}\]
C) \[\frac{{{(0.8)}^{2}}}{{{(8.02)}^{4}}-{{(7.98)}^{4}}}\]
D) \[\frac{{{(0.8)}^{2}}}{{{(8.02)}^{3}}-{{(7.98)}^{2}}}\]
Correct Answer: A
Solution :
[a] \[{{C}_{1}}=\frac{\pi \eta (r_{2}^{4}-r_{1}^{4})}{2\ell },{{C}_{2}}=\frac{\pi \eta {{r}^{4}}}{2\ell }\] Initial volume = Final volume \[\therefore \,\,\,\pi [r_{2}^{2}-r_{1}^{2}]\ell \rho =\pi {{r}^{2}}\ell \rho \] \[\Rightarrow \,\,{{r}^{2}}=r_{2}^{2}-r_{1}^{2}\Rightarrow {{r}^{2}}=({{r}_{2}}+{{r}_{1}})({{r}_{2}}-{{r}_{1}})\] \[\Rightarrow \,\,{{r}^{2}}=(8.02+7.98)(8.02-7.98)\] \[\Rightarrow \,\,{{r}^{2}}=16\times 0.04=0.64cm\Rightarrow r=0.8cm\] \[\therefore \,\,\,\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{r_{2}^{4}-r_{1}^{4}}{{{r}^{4}}}=\frac{{{[8.02]}^{4}}-{{[7.98]}^{4}}}{{{[0.8]}^{4}}}\]
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