Question

When a potential difference is applied across the ends of a uniform wire of length / and radius r, a current / flows in the wire. If some potential difference is applied to the ends of another wire of the same material but of length 27 and radius 2r, the current in the wire is St

C)
D) 1/2
A)7/4
B) 27

Answer 1

The current in the second wire is 4 times the current in the first wire,

The correct answer is C) 4.

To compare the currents in the two wires, we can use the formula for the electrical resistance of a wire:

Resistance (R) = (resistivity * length) / cross-sectional area

Since both wires are made of the same material, they have the same resistivity (ρ).

For the first wire:

Length (L1) = L (given as "/")

Radius (r1) = r

For the second wire:

Length (L2) = 27

Radius (r2) = 2r

Let's calculate the resistance of each wire and then compare the currents:

Resistance of the first wire (R1):

R1 = (ρ * L1) / (π * r1^2)

Resistance of the second wire (R2):

R2 = (ρ * L2) / (π * r2^2)

Now, to compare the currents, we can use Ohm's Law:

Current (I) = Voltage (V) / Resistance (R)

Assume the potential difference applied to both wires is the same, so V is constant.

Now, the ratio of currents (I1 and I2) can be expressed as:

I2 / I1 = (V / R2) / (V / R1) = R1 / R2

Substitute the values of R1 and R2:

I2 / I1 = [(ρ * L1) / (π * r1^2)] / [(ρ * L2) / (π * r2^2)]

Since resistivity (ρ) and π (pi) are common to both sides of the equation, they cancel out:

I2 / I1 = (L1 / r1^2) / (L2 / r2^2)

Now, substitute the given values:

I2 / I1 = (L / r^2) / (27 / (2r)^2)

I2 / I1 = (L / r^2) / (27 / 4r^2)

I2 / I1 = (4r^2 * L) / (27 * r^2)

I2 / I1 = (4 * L) / 27

Finally, the current ratio (I2 / I1) is:

I2 / I1 = 4L / 27

Thus, the current in the second wire is 4 times the current in the first wire.