Question

a rod has a resistivity of 2×10^-3 ohm metre . if its Length is doubled and area is halved what is the new resistivity.​

Answer 1

A rod of resistivity 2 × 10³- ohm meter. Iy's length is doubled and area is halved.

We know that, R = p l/A

RA/l = p

Resistance of a wire is directly proportional to length of the wire and inversely proportional to the area of cross-section. Where rho (p = resistivity) is a constant.

As per given condition, length is doubled mean resistance will also become double. But as said in question, length is doubled and area of cross-section is halved.

So, new length = 2l and new area of cross-section = A/2

Let's denote the new resistivity by p'.

p' = R(A/2)/2l

p' = RA/4l

p' = 1/4 RA/l

p' = 1/4 p

Value of p is 2 × 10³- ohm meter (given in question). So,

p' = (2 × 10³-)/4

p' = 0.5 × 10³- ohm meter

Therefore, the value of new resistivity is 0.5 × 10³- ohm meter.

Answer 2

◙ Given ◙

A rod has a resistivity of 2 × 10⁻³ Ω-m. Its length is doubled and area is halved.

• ρ = 2 × 10⁻³ Ω-m
• l = 2l
• A = A/2

◙ To Find ◙

The new resistivity of the rod.

◙ Solution ◙

We know,

$\boxed{\sf{\color{red}{\rho=\dfrac{RA}{l} }}}$

After the length is doubled and area is halved,

$\sf{\hookrightarrow \rho'=\dfrac{R \times \frac{A}{2} }{2l}}\\\\\sf{\hookrightarrow \rho' = \dfrac{\frac{1}{2}RA}{2l}}\\\\\sf{\hookrightarrow \rho'=\dfrac{1}{4}\times \dfrac{RA}{l}}\\\\\sf{\hookrightarrow \rho'=\dfrac{1}{4}\times \rho}$

→ 1st resistivity = 2 × 10⁻³

→ New resistivity(ρ') = 1/4 × 2 × 10⁻³

⇒ρ' = 10⁻³/2

⇒ρ' = 0.0005

⇒ρ' = 0.5 × 10⁻³

Hence, the new resistivity of the rod is 0.5 × 10⁻³ Ω-m.