A) \[8\,%\]
B) \[6\,%\]
C) \[18\,%\]
D) \[12\,%\]
Correct Answer: B
Solution :
In a conductor of resistance \[R\] and current flowing is \[i\] for time\[t\], then the quantity of heat generated is \[H={{i}^{2}}R\,t\] Taking log on both sides, we have \[\log H=2\log \,i+\log \,R+\log \,\,t\] By partial differentiation, we have \[\frac{\Delta H}{H}=\frac{2\Delta \,i}{i}+\frac{\Delta R}{R}+\frac{\Delta t}{t}\] Given, \[\frac{\Delta \,i}{i}=2%,\frac{\Delta R}{R}=1%,\frac{\Delta t}{t}=1%\] \[\therefore \] \[\frac{\Delta H}{H}=\frac{2\times 2}{100}+\frac{1}{100}+\frac{1}{100}\] \[\frac{\Delta H}{H}=6%\] Note: Higher the power of a quantity in the formula of an experiment, greater care should be taken in the measurement of that quantity.
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