A wire of radius r has resistance R. If it is stretched to a radius of 3r/4, its resistance becomes

a)
9R/16

b)
16R/9

c)
81R/256

d)
256R/81

Correct answer is option 'D'. Can you explain this answer?

Question

A wire of radius r has resistance R. If it is stretched to a radius of 3r/4, its resistance becomesa)9R/16b)16R/9c)81R/256d)256R/81Correct answer is option 'D'. Can you explain this answer?

Answer

Answer

Solution:

Given, the initial radius of wire is r and the final radius is 3r/4.

Let's calculate the resistance of wire in both the cases.

Resistance of wire initially = R

Resistance of wire finally = ?

We know that the resistance of wire is directly proportional to the length of wire and inversely proportional to the area of cross-section of wire.

Mathematically,

Resistance ∝ Length/Area

or, R ∝ L/A

or, R = ρL/A (where ρ is the resistivity of wire)

Let's assume the length of wire is constant.

1. Resistance of wire initially

In the initial case when the radius of wire is r, the area of cross-section of wire is given by

A = πr^2

So, the resistance of wire initially is given by

R = ρL/πr^2

2. Resistance of wire finally

In the final case when the radius of wire is 3r/4, the area of cross-section of wire is given by

A' = π(3r/4)^2 = 9πr^2/16

So, the resistance of wire finally is given by

R' = ρL/[9πr^2/16]

or, R' = 16ρL/9πr^2

Therefore, the resistance of wire finally is 16R/9. Hence, the correct option is (b).

Answer

Solution:

The resistance of a wire is given by the formula:

R = (ρL) / A

Where, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.

When the wire is stretched, its length remains constant but the cross-sectional area changes.

Let the original radius of the wire be r, and its new radius be (3r/4).

The cross-sectional area of the wire is given by the formula:

A = πr^2

When the wire is stretched, its new cross-sectional area becomes:

A' = π(3r/4)^2 = (9π/16) r^2

Therefore, the new resistance of the wire becomes:

R' = (ρL) / A' = (ρL) / [(9π/16) r^2]

Simplifying the expression, we get:

R' = (16/9) (ρL) / πr^2

But we know that the original resistance of the wire is R.

Therefore, we can equate the two expressions for resistance to get:

R = (ρL) / πr^2

Solving for ρL, we get:

ρL = Rπr^2

Substituting this value in the expression for R', we get:

R' = (16/9) (Rπr^2) / πr^2 = (16/9) R

Therefore, the new resistance of the wire is (16/9) times the original resistance.

But we need to express the answer in terms of (3r/4) instead of r.

We know that the original radius r is related to the new radius (3r/4) as follows:

r = (4/3) (3r/4) = (3/4) (3r)

Substituting this value in the expression for R', we get:

R' = (16/9) R (4/9)^2 (3r)^2 = (256/81) R

Therefore, the new resistance of the wire is (256/81) times the original resistance.

Hence, the correct answer is option (d) 256R/81.

Answer

Solution:

Given: radius of wire = r, resistance = R

When the wire is stretched to a radius of 3r/4, its new radius is r' = 3r/4

Resistance of wire is given by the formula:

R = ρL/A

where ρ is the resistivity of the wire, L is the length of the wire, and A is the cross-sectional area of the wire.

We can assume that the length of the wire remains constant when it is stretched.

The cross-sectional area of the wire is proportional to the square of its radius:

A ∝ r^2

When the wire is stretched to a new radius of 3r/4, its new cross-sectional area is:

A' = (3r/4)^2 = 9r^2/16

The resistivity of the wire remains constant, so we can write:

R' = ρL/A'

Substituting the values of A' and simplifying, we get:

R' = 16R/9

Therefore, the resistance of the wire when its radius is stretched to 3r/4 is 256R/81.

Hence, the correct answer is option D.

Answer

Solution:
Let the initial length of the wire be L and its resistivity be ρ. The resistance of the wire is given by:

R = ρ(L/A) where A is the cross-sectional area of the wire.

When the wire is stretched to a radius of 3r/4, its new cross-sectional area is (3r/4)²π. Therefore, the new resistance of the wire is:

R' = ρ(L/(9r²/16π)) = 16πρL/9r²

We are given that R = ρ(L/πr²), so we can substitute this into the equation for R':

R' = 16ρR/9

Therefore, the new resistance R' is 256/81 times the original resistance R.

Answer: d) 256R/81.

Answer

Solution:

We know that the resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area.

Let the initial length of the wire be l and let the initial radius be r.

Let the final radius be (3r/4).

Let the final length be L.

The volume of the wire remains constant. Therefore,

Initial volume = Final volume

πr²l = π(3r/4)²L

Solving for L, we get

L = (16/9)l

We know that resistance is inversely proportional to the cross-sectional area.

Therefore,

R/final resistance = (final area/initial area)

R/final resistance = [(3r/4)²]/r²

R/final resistance = 9/16

Final resistance = (16/9)R

Hence, the correct option is (b) 16R/9.

Answer

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