A) 0.5%
B) 1%
C) 2%
D) 4%
Correct Answer: C
Solution :
[c] : Length of the wire at temperature\[{{T}_{2}}\] is \[{{L}_{t}}=L\left( 1+\frac{1}{100} \right)\therefore 2L_{t}^{2}=2{{L}^{2}}{{\left( 1+\frac{1}{100} \right)}^{2}}\] Now \[2L_{t}^{2}=\]area of the plate at temperature \[{{T}_{2}}({{A}_{t}})\]and \[2L_{{}}^{2}=\]area of the plate at temperature\[{{T}_{1}}(A)\]. \[\therefore \]\[{{A}_{t}}=A{{\left( 1+\frac{1}{100} \right)}^{2}}=A\left( 1+\frac{2}{100} \right)=\frac{102A}{100}\] Thus, the area increases by 2%.
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