Question

A 2m long wire with a cross sectional area of 1mm*2 has a resistance of 16 ohm what is the resistance of the wire if it has a cross sectional area of 2mm*2 A. 4ohm B. 8ohm C. 32ohm D. 64ohm​

Answer 1

Given,

Length of wire = 2m

Area of cross-section = 1mm²

Area of the cross-section of other wire = 2mm²

To Find,

Resistance of wire

Solution,

In first case

ρ = RA/l

ρ = 16*1/2

ρ = 8

Now, for the other wire

R = ρL/A

  = 8*2/2

   = 8 ohm

Hence, the resistance of the wire is 8 ohm.

Answer 2

Answer:

The resistance of the wire will be 4Ω.i.e.option(A).      

Explanation:

The volume of the wire remains constant i.e.

$V_1=V_2$              (1)

Where,

V₁ and V₂ are the volumes of the wire for different areas A₁ and A₂.

$A_1\times l_1=A_2\times l_2$

$\frac{l_1}l_2}=\frac{A_2}{A_1}$     (2)

From the questions we have,

l₁=2m

R=16Ω

A₁=1mm²

A₂=2mm²

By substituting the value of l₁, A₁, and A₂ in equation (2) we get;

$\frac{2}l_2}=\frac{2}{1}$

$l_2=1m$

The resistance of the wire having different areas and lengths is given as;

$R_1=\rho\frac{l_1}{A_1}$                   (3)

$R_2=\rho\frac{l_2}{A_2}$                  (4)

By taking the ratio of R₁ and R₂ we get;

$\frac{R_1}{R_2}=\frac{\rho\frac{l_1}{A_1}}{\rho\frac{l_2}{A_2}}=\frac{l_1\time A_2}{l_2\times A_1}$        (5)

By placing the required values  in equation (5) we get;

$\frac{16}{R_2}=\frac{l_1\time A_2}{l_2\times A_1}=\frac{2\times 2}{1\times 1}=4$    

$R_2=\frac{16}{4}=4\Omega$

Hence, the resistance of the wire will be 4Ω.i.e.option(A).