A) 30cal
B) 60cal
C) 100 cal
D) 120 cal
Correct Answer: D
Solution :
Heat produced \[H={{I}^{2}}Rt\] In both the parts, current is in the ratio of \[2:1\] \[\therefore \] \[{{I}_{1}}=\frac{2I}{3},\]\[{{I}_{2}}=\frac{I}{3}\] So, \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{I_{1}^{2}{{R}_{1}}t}{I_{2}^{2}{{R}_{2}}t}\] or \[\frac{{{H}_{1}}}{60}={{\left( \frac{21/3}{1/3} \right)}^{2}}\times \frac{3}{6}\] or \[{{H}_{1}}=120cal.\]You need to login to perform this action.
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