A) 100 W bulb
B) 25 W bulb
C) None of them
D) Both of them
Correct Answer: B
Solution :
Resistance of 25 W bulb, \[K=\frac{233.33\times 4\times {{10}^{-3}}}{5\times {{10}^{-4}}\times 32}\] Resistance of 100 W bulb, \[=58.33W/m{{\text{-}}^{o}}C\] Series resistance, \[v=\sqrt{2\left( \frac{g}{8} \right)h}\] Current through series combination, \[v=\frac{\sqrt{gh}}{2}\] \[t=\frac{v}{g}\left[ 1+\sqrt{1+\frac{2gh}{{{v}^{2}}}} \right]\] Potential difference across 100 W bulb, \[=\frac{\sqrt{gh}/2}{g}\left[ 1+\sqrt{1+\frac{2gh}{gh/4}} \right]\] Thus, the bulb of 25 W will be fused because it can tolerate only 220 V while, the voltage across it is 352 V.
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