Question

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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.




hiiii people ​

Answer 1

Answer:

Heya Here's the required Answer mate

• 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

Hope it helps dear

Hola✌️

Answer 2

Speed of car = 3 whole $\dfrac{1}{9}$ km/hr

=$\dfrac{28}{9}$ km/hr

time taken to travel x km=5 whole $\dfrac{1}{7}$ hours

=$\dfrac{36}{7}$ hr

we know that, distance = speed ${×}$ time

or, x = $\dfrac{28}{9}$ ${×}$ $\dfrac{36}{7}$ km

or, x = 16 km

So. the car will travel 16 km in $\dfrac{36}{7}$ hours

$\huge\bold{\textbf{\textsf{{\color{turquoise}{Hope \ it \ helps}}}}}$