A wire of resistance $3$ ohm and length $10$ cm is stretched to length $30$ cm. Assuming that it has a uniform cross section, what will be its new resistance?

Open in App

Solution

For constant volume

$A_{1} L_{1} = A_{2} L_{2}$

$A_{2} = \frac{A_{1} L_{1}}{L_{2}} = \frac{A_{1} \times 10}{30} = \frac{A_{1}}{3} . . . . (1)$

Formula for resistance,

$R = \frac{ρ L}{A}$ (Where, resistivity $ρ = constant$ )

$R_{1} = \frac{ρ L_{1}}{A_{1}} a n d R_{2} = \frac{ρ L_{2}}{A_{2}}$

$\frac{R_{2}}{R_{1}} = \frac{(\frac{ρ L_{2}}{A_{2}})}{(\frac{ρ L_{1}}{A_{1}})} = \frac{L_{2} A_{1}}{L_{1} A_{2}} = 3 \times 3 = 9$

$R_{2} = 9 R_{1} = 9 \times 3 = 27 Ω$

Hence, resistance become is $27 Ω$

Suggest Corrections

Similar questions

3 Ω

A wire of 3 ohms resistance and 15 cm in length is stretched to 45 cm in length. Calculate the following:

i) New cross sectional area

ii) New resistance

iii) Resistivity.

Assume that wire has a uniform cross sectional area.

Join BYJU'S Learning Program