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Question

Masses of three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical resistances are ___ .


A

1 : 3 : 5

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B

5 : 3 : 1

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C

1 : 15 : 125

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D

125 : 15 : 1

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Solution

Step 1: Given that:

The ratio of the masses of three wires if copper = 1:3:5

The ratio of the lengths of the wires = 5:3:1

Let, l1,l2 , l3 and m1,m2,m3 be the lengths and masses of the three wires resppectively.

Then, m1:m2:m3=1:3:5

and, l1:l2:l3=5:3:1

Step 2: Formula and concept used:

The volume of wire= Area× length

If the volume of a wire is V, the area of cross-section is A and the length of the wire is l, then

V=Al

And, Mass(m)= Volume(V)× density(d)

Thus,

m=Al×d

Al×d=m

A=mld

The electric resistance of a wire is given by

R=ρlA

Where ρ is the resistivity of the material of the wire.

Putting the value of the area of cross-section , we get

R=ρlmld

R=ρl2dm

Step 3: Calculation of the ratio of their resistances:

Let R1,R2 and R3 be the resistances of the three wires of copper respectively.

As the wires are made of copper the value of resistivity(ρ) and the density of all three wires will be the same.

Thus,

R1=ρ(l1)2dm1

R2=ρ(l2)2dm2

And,

R3=ρ(l3)2dm3

Now,

R1R2=ρ(l1)2dm1ρ(l2)2dm2

R1R2=(l1)2(l1)2×m2m1

R1R2=(l1l2)2×m2m1R1R2=259×31

R1R2=253

R1R2=25×53×5

R1R2=12515.............(1)

And,

R2R3=ρ(l2)2dm2ρ(l3)2dm3

R2R3=(l2)2(l3)2×m3m2

R2R3=91×53

R2R3=151..............(2)

From equation (1) and (2), we get

R1:R2:R3=125:15:1

Thus,

The ratio of the resistances of the wires is 125:15:1 .

Hence option D) 125:15:1 is the correct option.


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