A wire 1 m long has a resistances of 1Ω. It is uniformly stretched, so that its length increases by 25%.Therefore its resistance will increase by:

25%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

50%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

56.25%

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

77.33%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
56.25%

As R = ρ $\frac{L}{A}$ and volume = L x A = Constant.

Final resistance = ${1.25}^{2}$ R = 1.5625R

Therefore, Increase in Resistance = 56.25%

Suggest Corrections

Similar questions

A wire 1 m long has a resistances of 1Ω. If it is uniformly stretched, so that its length increases by 25%, then its resistance will increase by:

Q. The resistance of a 10 m long wire is

10 Ω

. Its length is increased by 25% by streatching the wire uniformly . the resistance of wire will change to (approximately)

An increase in the length of a copper wire due to stretching is 25%, $R_{1}$ is its resistance before stretching and $R_{2}$ is its resistance after stretching, then $R_{1} : R_{2}$ ?

Q. A wire of

1 Ω

has a length of

1 m

. It is stretched till its length increases by

25

%. The percentage change in the resistance to the nearest integer is -

Join BYJU'S Learning Program

A wire 1 m long has a resistances of 1Ω. It is uniformly stretched, so that its length increases by 25%.Therefore its resistance will increase by:

The correct option is C 56.25% As R = ρLA and volume = L x A = Constant. Final resistance = 1.252R = 1.5625R Therefore, Increase in Resistance = 56.25%

The correct option is C
56.25%

As R = ρ $\frac{L}{A}$ and volume = L x A = Constant.

Final resistance = ${1.25}^{2}$ R = 1.5625R

Therefore, Increase in Resistance = 56.25%