First slide

Ohm's law

Question

A cylindrical copper rod is reformed to twice its original length with no change in volume. The resistance between its ends before the change was R Now its resistance will be

Moderate

A

8 R
B

6 R
C

4 R
D

2 R

Solution

$R = \frac{{ρl}_{1}}{{πr}_{1}^{2}}$ ……….(1)
Now the rod is reformed such that $I_{2} = 2 I_{1}$ and. ${πr}_{1}^{2} l_{1} = {πr}_{2}^{2} l_{2}$ (No change in volume)
or $(r_{1}^{2} / r_{2}^{2}) = (l_{2} / h)$ ……..(2)
Now $R_{2} = \frac{{ρl}_{2}}{{πr}_{2}^{2}}$ …………(3)
From eqs. [1) and (3), we get
$\frac{R}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{r_{2}^{2}}{r_{2}^{2}} = \frac{l_{1}}{l_{2}} \times \frac{l_{1}}{l_{2}} = \frac{1}{2} \times \frac{1}{2}$ [using eq. (2)]
$∴ R_{2} = 4 R$

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