Name: Resistivity and Its Temperature Dependence
Uploaded: 2019-07-19
Duration: 5 min 51 s
Description: In this video, we will discuss the resistivity of different materials and the variation of resistivity with temperature. Let us watch this video and learn.

Question

A cylindrical wire having

10 Ω

resistance is stretched to increase its length by

10 %

. Calculate its new resistance.

Accepted Answer

Let the initial length of the wire be $l$ , and the new length be $l'$ .

Now, the increase in length is $10 %$ ,

So, $l' = l + (\frac{10}{100} \times l) = (\frac{11}{10}) l$

Now, let the initial area of wire be $A$ , the area after the wire is stretched be $A'$ , the initial resistance be $R$ , and new resistance after the wire is stretched be $R'$ .

As we know that the resistance in terms of resistivity, length and area is given by the expression $R = \frac{ρ L}{A}$ .

The volume of the wire will remain constant as $l A = l' A'$ or, $A' = \frac{l A}{l'}$

Now, using the above expression $\frac{R}{R'} = \frac{l A'}{l' A}$ ,

Now,substituting the value of $A'$ in above equation $\frac{R}{R'} = \frac{l^{2}}{l'^{2}} = \frac{100}{121}$

$R' = \frac{121}{100} R$

Now, the value of $R = 10 Ω$ , The value of new resistance $R' = \frac{121}{100} \times 10 = 12.1 Ω$ .

Now, the percentage change in resistance $= \frac{12.1 - 10}{10} \times 100 = 21 %$ .

Hence, the new resistance is $12.1 Ω$ .

A cylindrical wire having $10 Ω$ resistance is stretched to increase its length by $10 %$ . Calculate its new resistance.

Important Questions on Current Electricity

Learn and Prepare for any exam you want !

A cylindrical wire having 10 Ω resistance is stretched to increase its length by 10%. Calculate its new resistance.

Important Questions on Current Electricity

Learn and Prepare for any exam you want !

A cylindrical wire having $10 Ω$ resistance is stretched to increase its length by $10 %$ . Calculate its new resistance.