A) \[a=\frac{F}{m+{{C}^{2}}{{B}^{2}}l}\]
B) \[\left( \text{take g }=\text{ 1}0\text{ m}/{{\text{s}}^{\text{2}}} \right)\]
C) \[1.4\times {{10}^{-4}}\text{J}\]
D) \[0.75\times {{10}^{-3}}\text{J}\]
Correct Answer: C
Solution :
As, Young's modulus of elasticity \[\text{mol }{{\text{L}}^{\text{-1}}}\] \[\text{mi}{{\text{n}}^{\text{-1}}}\] \[\text{2}.\text{3}\times \text{1}{{\text{0}}^{\text{-5}}}\text{ and 1}\text{.2 }\times \text{1}{{0}^{-\text{5}}}\] ?.. (i) where, L is the original length of the wire and also \[\text{3}.\text{8}\times \text{1}{{\text{0}}^{-4}}\text{ and }0.\text{6}\times \text{1}{{0}^{-\text{4}}}\text{ respectively}\] ...(ii) From Eqs. (i) and (ii), we get \[\frac{{{T}_{1}}L}{A({{l}_{1}}-L)}=\frac{{{T}_{2}}L}{A({{l}_{2}}-L)}\] \[\text{4}.\text{8}\times \text{1}{{0}^{-\text{4}}}\text{ and 1}.\text{2}\times \text{1}{{0}^{-\text{4}}}\text{ respectively}\]
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