Question

When a resistane 5 ohm twice its length wha will its new length?

Answer 1

GIVEN

Resistance of a wire is 5 ohm

Let the length of the wire be L

Area of cross section be A

Since,

Resistance is directly proportional to resistance and inversely proportional to area  and when we remove a proportianality sign we put a constant so let the constant be $\rho$.

∴RESISTANCE = $\rho\dfrac{length}{Area}$

So,

5 = $\rho\dfrac{L}{A}$-----(2)

According to question ,

Length is made twice , so new length $L_2= 2L$

And area becomes $A_2=\dfrac{A}{2}$

∴RESISTANCE$R_2 = \rho\dfrac{2L}{\dfrac{A}{2}}$

So,

$R_2 =\rho\dfrac{4L}{A}$----(1)

On dividing eq(2) by (1)

$\dfrac{R}{R_2}= \rho\dfrac{L}{A} \div \rho\dfrac{4L}{A}$

$\dfrac{R}{R_2}= \dfrac{L}{A} \times \dfrac{A}{4L}$

on solving

$\dfrac{R}{R_2}= \dfrac{1}{4}$

$\dfrac{5}{R_2}= \dfrac{1}{4}$

$R_2 = 5 \times 4$

so,new resistance is 20 ohms.