Question

the resistance of a wire is R. if the diameter of the wire is halved by stretching it then resistance of the wire will become
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Answer 1

Answer:

R* = 4R

Explanation:

Given the resistance of the wire is R

The diameter of the wire is halved.

Let the new resistance be R*.

We know that

R  ∝  1/A       [where A is the area of cross-section of the wire]

Let the area of the wire unstretched be A₁.

Let the diameter of wire unstretched be 'd'.

⇒A₁ = $\pi$r² = $\pi$ (d/2)² = $\pi$*d²/4 ________(1)    [∵ r = d/2]

Now the wire is stretched so the diameter is halved.

Let the new diameter be d' and the area is A₂.

⇒d' = d/2

⇒A₂ = $\pi$*(r')² = $\pi$*(d'/2)² =$\pi$ *(d/4)²    [∵d' = d/2 ⇒ d'/2 = d/4]

⇒A₂ = $\pi$*d²/16 ________(2)

∴ R/R* = A₂/A₁          [∵  R  ∝  1/A]

          =($\pi$*d²/16) / ( $\pi$*d²/4)

          = (1/16) / (1/4)

          = 4 / 16 = 1/4

⇒ R/R* = 1/4

⇒R* = 4R

∴ R* = 4R

∴ New resistance is 4 times the old resistance

Answer 2

Answer: The correct answer is 4R; the resistance of the wire will become 4R.

Explanation:

Given that the wire's resistance is R. The wire's diameter is cut in half. Let R* represent the new resistance.
The fact that,

∵ $R\alpha 1/A$ ( here A is the cross section area of the given wire)

Let A₁ represent the un-stretched region of the wire. Unstretched wire should have a diameter of "d."

⇒ A₁ = $\pi r^{2}$ = $\pi (d/2)^{2}$ = $\pi d^{2} /4$ =$\pi$* $d^{2} /4$------------- (1)  (r=d/2)

The wire is now stretched till the diameter is cut in half. Let's say the area is A₂, and the new diameter is d'.

Here d' = d/2

⇒ A₂ =$\pi$*(r')² = $\pi$*(d/4)²

⇒ A₂ = $\pi$*d²/16 ---------- (2)

Solving R/R* which is equal to A₂/ A₁ equals 1/4

Thus R/R* = 1/4

R* = 4R so the new resistance will be 4 times of the old resistance.

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