A wire of resistance 5 ohm is drawn out so that its new length is three times

Discussion Forum : Current Electricity

Question -

A wire of resistance 5 ohm is drawn out so that its new length is three times its original length. What is the resistance of the new wire ?

Options:

A .

\frac{5}{3} Ω

B .

5 Ω

C .

15 Ω

D .

45 Ω

Answer: Option D
:
D

We know that $V o l u m e = \frac{M a s s}{d e n s i t y}$
Let m be the mass of the given wire, d its density and $ρ$ its resistivity.
These quantities remain unchanged when the wire is drawn out.
Let $l_{1}$ be the length of the smaller wire and $r_{1}$ its corresponding radius of cross section. If $l_{2}$ and $r_{2}$ represent the length and radius of cross section of the elongated wire,
$⟹ m = π r_{1}^{2} l_{2} d$ for smaller wire and for elongated wire, $m = π r_{2}^{2} l_{2} d$ .
Since mass and density are constant
$⟹ r_{1}^{2} l_{1} = r_{2}^{2} l_{2}$
If $R_{1}$ is the resistance of the smaller wire and $R_{2}$ is the resistance of the elongated wire, we have,
$R_{1} = \frac{ρ l_{1}}{π r_{1}^{2}}$
$R_{2} = \frac{ρ l_{2}}{π r_{2}^{2}}$
Now, $\frac{R_{2}}{R_{1}} = \frac{l_{2}}{l_{1}} \times \frac{r_{1}^{2}}{r_{2}^{2}}$
It is given that, $l_{2} = 3 l_{1} .$
Therefore $r_{2}^{2} = \frac{r_{1}^{2}}{3}$
Substituting the values,
$\frac{R_{2}}{R_{1}} = 9$
Since, $R_{1} = 5 Ω$
$R_{2} = 45 Ω$ .

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