Question

a wire of length l and uniform cross section of resistance 16 ohm. it is cut into 4 equal parts.each wire length is increased to l find the resistance in parallel. plzzz solve. priperly no shortcuts. it is 100 point ques

Answer 1

Length of wire = L
resistance of wire = 16 Ohm = R ( Let)
now, wire cut four equal parts
e.g Length of each part = L/4

we know,
R = ρL/A
where ρ is resistivity ( constant )
when we cut the wire , cross section area won't be change .
hence, R is directly proportional to L
so, each part have resistance = R/4

again,
A/C to question ,
each part stretched and length increases L/4 to L

[ when , we stretched the wire , volume of wire is constant ]
R = ρL²/AL
AL = volume of each wire ( constant)
hence, R is directly proportional to L²
L is increased to four times so,
R' = 16( R/4) = 4R
hence, each of wire have 4× 16 = 64 ohm resistance .

now,
all wire connected in parallel .
e.g 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ = 1/Req
1/Req = 1/64 + 1/64 + 1/64 + 1/64
1/Req = 4/64
1/Req = 1/16
Req = 16 ohm

Answer 2

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Resistance of a material depends upon the nature of material, length of the wire and area of cross section.

Let the resistivity of the material is ρ, length of wire initially l and the area of cross section is A.

Then the resistance of the wire will be :

R = ​ρl/A = 16 Ω

Now when the wire is cut into 4 equal pieces and each part is stretched to the initial length l. when we stretch each part, the new length will be same of each part and the area of each part get reduced to one fourth.

i.e., l' = l

A' = A/4

Then resistance of each piece will be :

R' = ​ ​ρl'/A'

or R' = ​ ​ρl/A/4 = 4R = 64 Ω

Now connect these four pieces in parallel.

then the equivalent resistance will be :

Rp = 1/R' + 1/R' + 1/R' + 1/R' = 16 ​Ω