Question

Q7:
7. A conductor of uniform cross-section, its length 20 m and its resistance 1080 Another
conductor of the same material its length 5 m, and of cross-sectional area equals three
times the cross-sectional area of the first. The resistance of the second conductor equals
a. 9
b. 27
c. 84​

Answer 1

Explanation:

Let the elementary ring of width dx and radius r.

From figure, y=mx+c (straight line equation)

Using y=0 at x=0 ⟹c=0

Also y=b−a at x=l ⟹ m=

l

b−a

⟹ y=

l

b−a

∴ Radius of elementary ring r=a+mx=a+

l

b−a

x

Now resistance of elementary ring dR=

πr

2

ρdx

=

π

ρl

2

((b−a)x+al)

2

dx

∴ Total resistance R=

π

ρl

2

∫

o

l

((b−a)x+al)

2

dx

=

(b−a)π

−ρl

2

×

(b−a)x+al

1

∣

∣

∣

∣

∣

o

l

⟹ R=

π(b−a)

−ρl

2

[

bl

1

−

al

1

]=

πab

ρl

solution

Answer 2

Answer:

your answer is b .27. hope you like it