Question

A wire has a resistance of 10 ohm .It is melted and drawn into a wire of half of its length. Calculate the resistance of the new wire.What is the percentage change in its resistance?
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Answer 1

It is melted and drawn into a wire of half its length. Calculate the resistance of the new wire. What is the percentage change in its resistance?

Answer 2

Solution

Here the concept of Ohms Law is used. Its given that the resistance of the wire is 10 Ω. When melted the length becomes the half of intital length. So we have to find that resistance of new wire and change in its percentage%.

Let's proceed!!

First Let us assume few things.

$\sf{ \to Initial \: length= L_1} \\ \\\sf{ \to Final\: length= L_2} \\ \\ \sf{ \to Initial \: Area = A_1 }\\ \\ \sf{ \to Final \: area = A_2 }\\ \\ \sf{ \to Initial \: Resistance= R_1} \\ \\ \sf{ \to Final\:Resistance=R_2}\\ \\ \sf { \to initial\: volume= V_1} \\ \\ \sf{ \to final \: volume = V_2}$

~When the wire is melted the final length becomes half of intital.

$\sf{ \bull \: L_2 =\frac{ L_1}{2}}$

~When the wire is melted the final area becomes double of the inital area.

$\sf{ \bull \: A_2 = 2A_1}$

~But the volume after melting and before melting will remain same.

$\sf{ \implies V_1 = V_2}\\ \\\sf{ \implies L_1×A_1 = L_2×A_2}\\ \\\sf { \implies\frac{L_1}{L_2} = \frac{A_2}{A_1}}\\ \\ \sf{ \implies \frac{L_1}{ \frac{L_1}{2} } =\frac{A_2}{A_1} } \\ \\ \sf{ \implies 2 =\frac{A_2}{A_1} }$

Resistance of a wire is given as

${\underline{\boxed{\sf{ \bull \: R=\rho \frac{L}{A}}}}}$

Ratio

$\sf{ \implies \frac{R_2}{R_1} = \frac{\frac{ \rho L_2}{A_2}} {\frac{ \rho \: L_1}{A_1}}} \\ \\ \sf{ \implies \frac{R_2}{R_1} = \frac{L_2}{L_1} \times \frac{A_1}{A_2} } \\ \\ \sf{ \implies \frac{R_2}{R_1} = \frac{1}{2} \times \frac{1}{2} } \\ \\ \sf{ \implies \frac{R_2}{R_1} = \frac{1}{4}} \\ \\ \sf{ \implies R_2= \frac{R_1}{4} } \\ \\ \sf{ \implies R_2= \frac{10}{4} = 2.5 Ω}$

Hence the new resistance is of 2.5 Ω.

Now we have to calculate the percentage change

$\sf{ \implies \% \: change = \frac{R_1-R_2}{R_1}×100\%} \\ \\ \sf{ \implies \% \: change= \frac{10-2.5 }{10} ×100\%} \\ \\ \sf{ \implies \% \: change= 75 \%}$

So the % change in its resistance is 75%

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Knowledge Bytes !!

■ Ohm's law - the current passing through a metallic element is directly proportional to the potential difference across its end provided the temperature remains constant.

■ Resistance is the property of a conductor by which it obstructs the flow of current through it.

→R= V/I

• Higher is a resistance the lower is the current.
• good conductors have low resistance.

■ factors on which resistance depends

• material
• cross section area
• length
• temperature.

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