Question

4. Two wires of the same metal have their lengths in the ratio of 3.5 but radii in the rate of
2 3. Find the ratio of their resistances​

Answer 1

Question

Two wires of the same metal have their lengths in the ratio of 3:5 but radii in the rate of 2:3. Find the ratio of their resistances.

Solution

We know that-

R = p l/A

Resistance of a wire is directly proportional to it's length and inversely proportional to Area of cross-section.

A = πr²

So,

R = p l/(πr²) ..................(1st equation)

As per given condition,

Two wires of the same metal have their lengths in the ratio of 3:5 but radii in the rate of 2:3.

For first wire:

Length = 3l and radius = 2r

For second wire:

Length = 5l and radius = 3r

Substitute value of length and radius of first wire in (1st equation)

R' = p 3l/(π(2r)²)

R' = p 3l/(π4r²) ...................(A)

Similarly for second wire,

R" = p 5l/(π(3r)²)

R" = p 5l/(π9r²) ...................(B)

{ p = rho = resistivity (constant) }

Divide (A) and (B)

→ R'/R" = [p 3l/(π4r²)]/[p 5l/(π9r²)]

→ R'/R" = (3/4)/(5/9)

→ R'/R" = (3 × 9)/(4 × 5)

→ R'/R" = 27/20

Therefore, the ratio of their resistances is 27:20.

Answer 2

GIVEN:

• Ratio of the wires = 3 : 5
• Rate of radii = 2 : 3

TO FIND:

• Ratio of resistances

SOLUTION:

R = ρI/A

We know that

If length of wire increases resistance also increases & if resistance of wire increases area of cross section is decreases . That means,

Resistance is direct proportional to length

Resistance is inversly proportional to Area

Area = πr²

R = ρI/πr²

Assuming as equation 1

Here , given that 3I & 2r is the length & radius of the first wire ann 5I & 3r is the length and radius of 2nd wire

Substituting the values in Eq 1 we have

→ R = ρ(3I)/[π(2r)²]

→ R = 3ρI/4πr²

________________

→ R' = ρ(5I)/[π(3r)²]

→ R' = 5ρI/9πr²

Now, R : R'

♦ R/R' = [3ρI/4πr²]/[5ρI/9πr²]

♦ R/R' = (3/4)/(5/9)

♦ R/R' = 27/20. [.Answer]