Physics, asked by sangwankailash2733, 1 year ago

A wire of resistance 16 ohm is melted and drawn into a wire of half of its original length calculate the resistance of new wire what is its percentage change in resistance

Answers

Answered by mapsvanshi3741
109

ANSWER IS-: So the new resistance is 75% less than old resistance.

                      NEW RESISTANCE IS 4 Ohms


We know that

Resistance = Rho (Resistivity) * Length of Conductor /(Area of Cross-Section)

SO given wire has a resistance of 16 Ohms.

16 = Rho * Length / Area

If the length is halved after melting wire, it means Area of cross section would have doubled..

New resistance = Rho * 1/2 Length / 2*Area of Cross Section

New resistance = Rho * Length / 4* AREA OF Cross Section


16/ New Resistance = (Length of Conductor / Area of Cross - Section) / Length / 4* Area of Cross Section.     {rho gets cancelled by rho}

1/New Resistance = 4/16                        {Length of conductor and Area of                                .                                                               cross Section also get cancelled}

New Resistance = 4 Ohms...

Percent of Change of Resistance = New Resistance / Old Resistance * 100

4/16*(100) =100/4= 25%

So the new resistance is 75% less than old resistance


I request you to please once verify the calculations as I am not good at calculations...

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Answered by munirsheikh2006
0

Answer:

Explanation: new resistance is 75%

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