NEET Solved Paper 2016 Question 14
Question: The charge flowing through a resistance R varies with time t as $ Q=at-bt^{2}, $ where a and b are positive constants. The total heat produced in R is:
Options:
A) $ \frac{a^{3}R}{6b} $
B) $ \frac{a^{3}R}{3b} $
C) $ \frac{a^{3}R}{2b} $
D) $ \frac{a^{3}R}{b} $
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Answer:
Correct Answer: A
Solution:
$ Q=at-bt^{2} $ $ i=a-2bt $ {for $ i=0\Rightarrow ,t=\frac{a}{2b} $ }
From joule’s law of heating $ dH=i^{2}Rdt $ $ H=\int\limits_0^{a/2b}{{{(a-2b)}^{2}}}Rdt $ $ H=\frac{{{(a-2b)}^{3}}R}{-3\times 2b}| _0^{\frac{a}{2b}} .=\frac{a^{3}R}{6b} $
BETA
