Question

a conductor of length L and area a has a resistance of 10 ohm what will be the resistance of same material if the length is doubled and the radius of its cross-section is increased by twice​

Answer 1

Answer:

Explanation:

Given:

Length = 2L

Radius = 2r

Resistance,R = 10 ohm

To find out:

Resistance =?

Solution:

As it is known:

R  = ρ * L / A

R = ρ * L /πr^2

Putting the values that have been changed:

R' = ρ * (2L)/π(2r)^2

Rearranging:

R' = [ρ*L/πr^2] * 2/4    (As square of 2 is 4)

The expression in bracket is for resistance.

R'= R * 1/2

R'= 10 * 1/2

R'= 5 ohm

So,the new resistance will be 5 ohm.

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Answer 2

$\huge{\star}{\underline{\boxed{\red{\mathfrak{Answer :}}}}}{\star}$

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$\Large{\sf{ R \: = \: \rho \: \frac{L}{A}}}$

Where,

R is resistance

L is length

A is area of cross section.

And

$\large{\sf{10 \: = \: \rho \: \frac{L}{A}----(1)}}$

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Now, come to the question,

If ,

length is doubled and radius of cross section is increased twice, then

R = R' (new resistance)

L' = 2L

R = R'

____[Put Values in formula]

$\huge{\boxed{\boxed{\sf{R' \: = \: \rho \: \frac{L'}{A'}}}}}$

Substite value of L' and R'

We, get

$\large{\sf{R' \: = \: \rho \: \frac{2L}{\pi R'^2 h}}}$

As, radius is twiced then

$\large{\sf{R' \: = \: \rho \: \frac{2L}{\pi 2R^2 h}}}$

$\large{\sf{R' \: = \: \rho \: \frac{L}{\pi R^2h} \times \frac{2}{4}}}$

Substite value of 1

$\large{\sf{R' \: = \: R \ times \frac{2}{4}}}$

$\large{\sf{R' \: = \: 10 \times \frac{2}{4}}}$

$\huge{\boxed{\boxed{\sf{R' \: = \: 5 \: Ω }}}}$