A) \[\text{8}\text{.7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
B) \[\text{12}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
C) \[\text{8}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-1}}}\text{J}\]
D) \[\text{12}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-1}}}\text{J}\]
Correct Answer: A
Solution :
Force exerted on spring is given by \[F=(-5x-16{{x}^{3}})N\] or \[F=-(5+16{{x}^{2}})x\,N\] ?(i) Comparing Eq. (i) with \[F=-kx\] where k is a force constant. We have \[k=5+16{{x}^{2}}\] Work done in stretching the spring from \[{{x}_{1}}=0.1\,m\,\] to\[{{x}_{2}}=0.2\,m\,\] is \[W=\frac{1}{2}{{k}_{2}}x_{2}^{2}-\frac{1}{2}{{k}_{1}}x_{1}^{2}\] \[W=\frac{1}{2}[5+16x_{2}^{2}]x_{2}^{2}-\frac{1}{2}[5+16x_{1}^{2}]x_{1}^{2}\] Substituting the given values, we obtain \[W=\frac{1}{2}[5+16{{(0.2)}^{2}}]{{(0.2)}^{2}}\] \[-\frac{1}{2}[5+16{{(0.1)}^{2}}]{{(0.1)}^{2}}\] \[=2.82\times 4\times {{10}^{-2}}-2.58\times 1\times {{10}^{-2}}\] \[=8.7\times {{10}^{-2}}\,J\]
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