Question

1.
Three copper wires have their lengths in the ratio
5:3:1 and their masses are in the ratio 1:3
5. Their electrical resistance will be in the ratio

pls with explanation​

Answer 1

Answer:

mass ratio of the three wire

1:3:5

length are in the ratio

5:3:1

v=A*L

v=pi*r2*l

density =d

m=v*d

m=a*l*d

A=m/(l*d)

R=p*l/a

p is constant to 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 variable are

a varies m/l

R varies l/a

R varies by l/(m/l)

r varies by l*l/m b

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 verify

25:3:1/5

or

125:15:1

Explanation:

i multiplied by 5 to the whole thing