A) \[2\times {{10}^{9}}N/{{m}^{2}}\]
B) \[2\times {{10}^{8}}N/{{m}^{2}}\]
C) \[2\times {{10}^{6}}N/{{m}^{2}}\]
D) \[2\times {{10}^{4}}N/{{m}^{2}}\]
Correct Answer: C
Solution :
\[Mass=volume\times density=Vd=cons\tan t\] \[V\Delta d+d\Delta V=0\] \[\Rightarrow \] \[\frac{\Delta V}{V}=-\frac{\Delta d}{d}=-\frac{0.1}{100}={{10}^{-3}}\] \[\therefore \] \[K=\frac{\Delta p}{\Delta V/V}\] \[\Rightarrow \] \[\Delta p=-K\frac{\Delta V}{V}\] \[\Delta p=2\times {{10}^{9}}\times {{10}^{-3}}\] \[=2\times {{10}^{6}}N/{{m}^{2}}\]You need to login to perform this action.
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