Question

A wire of length = l and resistance = R , if it streched such that its length becomes 3 times more of its initial one. Find the ratio of initial resistance to the newly obtained resistance.​

Answer 1

Answer :-

Ratio of initial resistance to the newly obtained resistance is 1 : 9 .

Explanation :-

For the initial case :-

l = length

A₁ = Cross sectional area

R = Resistance

For the new case :-

3l = length

A₂ = Cross sectional area

R' = Resistance

Since the same wire is stretched to make it's length triple, so the volume of the wire is same in both cases.

∴ lA₁ = 3lA₂

⇒ A₁ = 3A₂

_____________________________

Initial case :-

⇒ R = ρl/A₁

⇒ R = ρl/3A₂ -----(1)

New case :-

⇒ R' = (ρ × 3l)/A₂

⇒ R' = 3ρl/A₂ -----(2)

On dividing eq.1 by eq.2, we get :-

⇒ R/R' = ρl/3A₂ × A₂/3ρl

⇒ R/R' = ρlA₂/9ρlA₂

⇒ R/R' = 1/9

⇒ R : R' = 1 : 9

Answer 2

Question:-

• A wire of length = l and resistance = R , if it streched such that its length becomes 3 times more of its initial one. Find the ratio of initial resistance to the newly obtained resistance.

To Find:-

• Find the ratio.

Solution:-

Case 1:-

$\sf\dashrightarrow \: R = \dfrac { ρl } { A_1 }$

$\sf\dashrightarrow \: R = \dfrac { ρl } { 3A_2 }$ . . . . ( 1 )

Case 2:-

$\sf\dashrightarrow \: R' = \dfrac { ρ \times 3l } { A_2 }$

$\sf\dashrightarrow \: R' = \dfrac { 3ρl } { A_2 }$ . . . . ( 2 )

Now ,

We have do to divide ( 1 )/( 2 ) then we get:-

$\sf\dashrightarrow \: \dfrac { R } { R' } = \dfrac { \: \: \dfrac { ρl } { 3A_2 } \: \: } { \dfrac { A_2 } { 3ρl } }$

$\sf\dashrightarrow \: \dfrac { R } { R' } = \dfrac { ρl } { 3A_2 } \times \dfrac { A_2 } { 3ρl }$

$\sf\dashrightarrow \: \dfrac { R } { R' } = \dfrac {1 } { 9 }$

$\sf\dashrightarrow \: R : R' = 1 : 9$