This question was previously asked in

DSSSB TGT Natural Science Female Subject Concerned - 27 Sept 2018 Shift 3

Option 3 : 20 ohm

CT 1: Indian History

20365

10 Questions
10 Marks
6 Mins

**Concept:**

**Resistance: **

- Resistance is a measure of the
**opposition to current flow**by molecules of the conductor. It is denoted by R. - The resistance R of a conductor
**depends**on its**length**l and constant cross-sectional**area**A through the relation,

\(R = \frac{ρ L}{A}\)

where ρ, called **resistivity**, is a property of the material and depends on temperature and pressure.

L = length of the conductor.

**Calculation:**

Given that,

Initial length of conductor = 20 cm

Final length of conductor = 40 cm

The initial resistance of conductor wire = 5 Ω

Let initial and final cross-section area of the conductor are A and A' respectively.

We know that, on stretching the wire, the volume of material remains the same. Hence

Initial volume = final volume

⇒ A × L = A' × L'

⇒ A × 20 = A' × 40

\(⇒ A' = \frac{A}{2}\) ..........(1)

Now, initial resistance

\(R = \frac{ρ L}{A}\) .....(2)

Final resistance

\(R' = \frac{ρ L'}{A'}\) .......(3)

(3) ÷ (2)

\(\frac{R'}{R} = \frac{\rho L'}{A'}\ × \frac{A}{\rho L}\)

\(⇒ \frac{R'}{R} = \frac{40 \ × 2\ × A}{20\ × A}\)

⇒ R' = 4 × 5 (∵ R = 5 Ω)

⇒ R' = 20 Ω

Hence, resistance of a stretched wire is 20 Ω.

__Alternate Method__

If the Initial length of resistance wire is R and it is stretched n times of initial; length, then the final resistance

R' = (n^{2})R

According to question, Wire stretched up to 2 times its initial length. Hence, final resistance

R' = (2^{2}) × 5 (∵ Initial resistance = 5 Ω)

⇒ R' = 20 Ω