Question

Physics question :- Three uniform copper wires have their lengths in ratio 1:4:5 and their mass in ratio 5:2:1 their electrical resistance will be in the ratio..

Answer 1

Please no qualified helper will help in this

Answer 2

Sorry for thid

Answer 3

OK, I won't.

Answer 4

Thanks

Answer 5

Why though? lol

Answer 6

hint:
If such wires have the same cross sectional area, then their electrical resistance is proportional to the respectively length of such wires

Answer 7

Because I want to know the views of my friends like me...

Answer 8

nevertheless that is not the case of your exercise

Answer 9

But I'm in 11th grade.

Answer 10

OK, let's get to it. We're given that they're made out of the same material. Their mass density is the same.\[\rho = \frac{m}{V}=\frac{m}{A \ell }\]\[\Rightarrow \rho_1 = \rho_2 = \rho_3 = \rho_{copper}\]\[\Rightarrow\frac{m_1}{A_1\ell_1}=\frac{m_2}{A_2\ell 2} = \frac{m_3}{A_3\ell_3 }\]Since you're given the relation between masses and lengths, you can obtain the ratio of cross-sectional areas. Then use \(R \propto \frac{\ell }{A}\)

Answer 11

we can express the electrical resistance \(R\) of a conductor, like a function of its mass \(m\) as below:
\[\huge R = \frac{k}{{\delta \cdot {S^2}}}m\]
where \(k\) is the resistivity, \(S\) is the cross sectional area, and \(\delta\) is the density of copper

Answer 12

R~L/A
Since resistivity of all the three wires are same as all are of same material
So, where do we found the area?

Answer 13

L/A

Answer 14

We obtained the ratio of areas by using the fact that the volume-density (mass/volume) is the same across all wires. That's a vital assumption.

Answer 15

Can you find me pls

Answer 16

being:
\[\Large \frac{m}{{{S^2}}} = \frac{{{\delta ^2}{l^2}}}{m}\]
after a simplification, we can write:

\[\Large R = k\delta \frac{{{l^2}}}{m}\]

Answer 17

OK done

Answer 18

For L/A
For all wire
Multiplying numerator and denominator by L
We get (L)(L)/(AL)=L^2/V
Also the ratio of mass is the ratio of volume for every wire because density is constant (all wire are of copper)
We need to find the ratio of (L)^2/M
Now, 1/5:16/2:25/1
1/5:8:25
Now we are left with 5 in one of the denominator so multiply every ratio with 5
And we get
1×5/5=1
8×5=40
25×5=125
So the ratio of resistance is
1:40:125

Answer 19

heyyyy that's what I said