A) 18 cm
B) 0.9 cm
C) 9 cm
D) 1.8 cm
Correct Answer: C
Solution :
By cubical expansion relation. \[\Delta V=V\times \gamma \times \Delta T\]where \[\gamma \] is coefficient of cubical expansion and \[V={{10}^{-6}}{{m}^{3}}=\]initial volume \[\gamma =18\times {{10}^{-5}}/{{\,}^{o}}C\] \[\Delta T=100{{\,}^{o}}C\] \[\therefore \] \[\Delta V={{10}^{-6}}\times 18\times {{10}^{-5}}\times {{10}^{2}}\] \[=18\times {{10}^{-9}}\] Since, \[\Delta V=A\times \Delta l\] \[\therefore \] \[18\times {{10}^{-9}}=2\times {{10}^{-7}}\times \Delta l\] or \[9\times {{10}^{-2}}=\Delta l\] or \[\Delta l=9\,cm\]You need to login to perform this action.
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