A) \[{{15.24}^{o}}C\]
B) 15.00°C
C) \[{{24.15}^{o}}C\]
D) None of the above
Correct Answer: A
Solution :
Heat developed by the heater \[H=\frac{{{V}^{2}}}{R}.\frac{t}{J}=\frac{{{(200)}^{2}}\times t}{20\times 4.2}\] Heat conducted by the glass \[H=\frac{0.2\times 1\times (20-\theta )t}{0.002}\] Hence\[\frac{{{(200)}^{2}}\times t}{20\times 4.2}=\frac{0.2\times (20-\theta )t}{0.002}\]\[\Rightarrow \theta ={{15.24}^{o}}C\]
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