Question

Masses of three pieces of
wires made of the same metal
are in the ratio 1:3:5 & their
lengths are in the ratio 5:3:1.
The ratios of their resistances
are...
1:3:5
5:3:1
1:15:125
125:15:1​

Answer 1

Answer:

25 : 3: 1/5

or

125 : 15 : 1

Explanation:

mass ratio of 3 wires

1 : 3 : 5

lengths are in ratio

5 : 3: 1

V = A * L

V = pi*r^2 * L

density = d

m = V*d

m = A*L * d

A = m/(L *d)

R = p*L/A

p is constant for all copper so now we just have to find the ratio of L to area

since d = density of copper it is constant too .. so the only variables are

A varies m/L

R varies L/A

R varies by L/ (m/L)

R varies by L*L/m

R varies by L^2/m

first wire

R1 = 5^2/1 = 25

R2 = 3^2/3 = 3

R3 = 1^2/5 = 1/5

the resistance vary by

25 : 3: 1/5

or

125 : 15 : 1

I multiplied by 5 through the whole thing

Assuming that the cross-sectional areas are not equal.

We know

R=ρlA

R=ρl2Al

R=ρl2V

Where V is the volume of each wire. Let d be the density.

d=mV⟹V=md

R=ρl2dm

Therefore,

R∝l2m

From the given ratios,

R1:R2:R3=521:323:125

R1:R2:R3=125:15:1

M1=m, M2=3m,M3=5m

So their volume will be directly proportional to the mass…

V1=v, V2=3v,V3=5v

R=resistivity(length) /(area)

As length*Area=volume…

A1=V1/L1 and so on..

So..

A1=v/5l,A2=3v/3l=v/l,A3=5v/l..

Resistance is directly proportional to (length/area)..

Answer =>Ratio->

25:3:1/5

=125:15:1

P. S. :just convey in comment if it is correct or not.. :)