A) \[30\,\,kJ\]
B) \[20\,\,kJ\]
C) \[15\,\,kJ\]
D) \[10\,\,kJ\]
Correct Answer: C
Solution :
From Joule's law, the total energy produced by bulb is \[H=\frac{{{V}^{2}}}{R}t\] Also, \[P=\frac{{{V}^{2}}}{R}\] Given,\[P=100\,\,W,\,\,V=230\,\,volt\] \[\therefore \] \[R=\frac{{{(230)}^{2}}}{100}=529\,\,\Omega \] Hence, heat produced\[=\frac{{{V}^{2}}}{R}\times t\] or \[H=\frac{115\times 115}{529}\times 10\times 60\] \[=15\,\,kJ\]
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