Question

three metals rods of same length and same area of cross sections having conductivities 1,2,3 units are connected in series then their effective conductivity will be 1
1) 2 units
2) 1.6 units
3) 2.4 units
4) 2.8 units​

Answer 1

Answer:

1.6

Explanation:

compare to resistive circuit

R = L/KA

in series R equivalent = R1 + R2 ...

3L/k(eq)A = L/1A + L/2A + L/3A

here 3L is taken because they are connected on series ie. equivalent length

solve to get aprox 1.6

Answer 2

Answer:

Given:

3 rods of same length and area of cross-section have been attached in series. Their conductivities are 1,2,3 units.

To find:

Effective conductivity of the combination

Calculation:

Since the pipes have been connected in series, we can say that :

$R \: eq. = R1 + R2 + R3$

$= > \dfrac{3l}{(k \: eq.)a} = \dfrac{l}{(1)a} + \dfrac{l}{(2)a} + \dfrac{l}{(3)a}$

$= > \dfrac{3l}{(k \: eq.)a} = \dfrac{l}{a} \bigg \{ 1 + \dfrac{1}{2} + \dfrac{1}{3} \bigg \}$

$= > \dfrac{3l}{(k \: eq.)a} = \dfrac{l}{a} \bigg \{ \dfrac{11}{6} \bigg \}$

$= > k \: eq. = \dfrac{18}{11}$

So final answer :

$\boxed{ \red{ \huge{ \bold{k \: eq. = \dfrac{18}{11} }}}}$