CUET Preparation Today
A wire of resistance 5Ω is drawn out so that its length is increased by thrice its original length. The new resistance would be: |
10 Ω 20 Ω 40 Ω 45 Ω |
45 Ω |
The correct answer is Option (4) → 45 Ω The resistance of a wire is - $R=ρ\frac{L}{A}=5Ω$ [Given] $L'=3L$ and $A'=A/3$ $R'=ρ\frac{3L}{A/3}=9\frac{ρL}{A}=9×5=45 Ω$ |
