A) 88 N
B) UN
C) 44 N
D) 22 N
Correct Answer: C
Solution :
The change in tension, when wire is cooled is given by \[F=yA\,\,\alpha \,\Delta \theta \] Given, \[\Delta \theta ={{40}^{o}}-{{20}^{o}}={{20}^{o}}C\], \[A=1\,m{{m}^{2}}=1\times {{10}^{-6}}{{m}^{2}}\], \[\alpha =1.1\times {{10}^{-5}}{{/}^{o}}C\], \[Y=2\times {{10}^{11}}N/{{m}^{2}}\] \[\therefore \] \[F=2\times {{10}^{11}}\times {{10}^{-6}}\times 1.1\times {{10}^{5}}\times 20\] \[\Rightarrow \] F = 44 NYou need to login to perform this action.
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