1.

A wire of length 9 cm and resistance 9 ohm is tripled on itself. Calculate it's final resistance.​

Answer»

Answer:

Let Initial RESISTANCE be R1=9 Ω.

Initial length be L1=30 cm/0.3m.

Initial AREA of cross section of wire be A1=a m2 .

THE WIRE IS TRIPLED ON ITSELF i.e. it is folded on itself 3 TIMES.So:-

Let Final resistance be R2.

Length would become 1/3 times the initial length so,

Final length= 1/3*0.3

As same wire is folded on itself so volume of the wire will remain same i.e.:-

L1*A1=L2*A2,

A2=L1*A1/L2,

A2=L1*A1/(1/3*L1),

A2=3A1.

Our new area is 3A1.

R1=ρ*(0.3/a)=9Ω

R2=ρ(1/3*0.3)/3a=1/9(ρ*0.3/a) and this[ρ*0.3/a]=R1=9Ω,so:-

R2=1/9*9=1Ω.


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