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JEE MAIN 2021
16-03-2021 S2
Question
The energy dissipated by a resistor is 10 mJ in 1 s when an electric current of 2 mA flows through it. The resistance is $\_\_\_\_$ $\Omega$. (Round off to the Nearest Integer)
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Solution
$\begin{aligned} & Q=i^2 R T \\ & R=\frac{Q}{i^2 t}=\frac{10 \times 10^{-3}}{4 \times 10^{-6} \times 1}=2500 \Omega\end{aligned}$
Question Tags
JEE Main
Physics
Easy
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