Question

A peice of wire of resistance 20 ohm
is drawn out so that its length is
increased to twice its origional
length. Calculate the resistance of
the a wire in the new situation​

Answer 1

Solution :-

Given :

Initial length of wire = L

Final length of wire = 2L

Initial resistance of wire = 20Ω

To Find :

Final resistance of wire

Concept :

This question is completely based on concept of volume conservation.

We also know that, resistivity is metalistic property of metal.

Resistivity doesn't depend upon length and area of conductor.

Calculation :

As per volume conservation

$\Rightarrow\bf\:V_{initial}=V_{final}\\ \\ \Rightarrow\sf\:A_1L_1=A_2L_2\\ \\ \Rightarrow\sf\:A_1\cancel{L_1}=A_2(2\:\cancel{L_1})\\ \\ \Rightarrow\sf\:A_1=2A_2$

Formula of resistance in terms of resistivity, length of conductor and area of cross section is given by...

$\implies\bf\:R=\rho\times \dfrac{L}{A}\\ \\ \therefore\underline{\purple{\sf{R\propto \dfrac{L}{A}}}}\\ \\ \mapsto\bf\:\dfrac{R_1}{R_2}=\dfrac{L_1A_2}{A_1L_2}\\ \\ \mapsto\sf\:\dfrac{20}{R_2}=\dfrac{(\cancel{L_1})(\cancel{A_2})}{(2 \cancel{A_2})\:(2\cancel{L_1})}\\ \\ \mapsto\sf\:\dfrac{R_2}{20}=4\\ \\ \mapsto\underline{\boxed{\pink{\bf{R_2=80\Omega}}}}$

Answer 2

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QUESTION :

A peice of wire of resistance 20 ohm is drawn out so that its length is increased to twice its origional length.

Calculate the resistance of the wire in the new situation.

SOLUTION :

The following Information is given :

The wire innitially has a resistance of 20 Ohm.

Now this wire is drawn out so that its length is increased to twice its origional length.

So the new length becomes half of it's original length.

Cross sectional Area is decreased to the square of the ratio of the original length to the new length.

So the cross Sectional Area becomes - { 1 / 2 } ^ 2 = 1 / 4

R = \rho { L / A }

R 1 / R 2 => Increase in cross Sectional Area.

=> R 1 / R 2 = 1 / 4

=> 20 / R 2 = 1 / 4

=> R 2 is 20 × 4 = 80 ohm.

Therefore the resistance of the a wire in the new situation is 80 Ohm.

ANSWER :

The resistance of the a wire in the new situation is 80 Ohm.