Question

Two wires of the same material and having the same volume, are fixed

from one end. A mass (m1 = 3 kg) is hanged to the first wire and a mass

(m2) is hanged to the other wire. If the radius of the first wire is half that

of the second wire and the elongation in the two wires are the same, find m2​

Answer 1

Answer:

48 kg

Explanation:

Given that the two wires are of same material, so their value of young's modulus will be same

Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × ($r_{1}$)² × $l_{1}$ = π × ($r_{2}$)² × $l_{2}$

Here

$r_{1}$ is the radius of the first wire

$r_{2}$ is the radius of the second wire

$l_{1}$ is the length of the first wire

$l_{2}$ is the length of the second wire

⇒ π × (($r_{2}$)² ÷ 4) × $l_{1}$ = π × ($r_{2}$)² × $l_{2}$ (∵ radius of first wire is half that of the second wire)

By cancelling the same terms on both sides

we get

$l_{1}$ = 4 × $l_{2}$

⇒ Length of first wire will be four times of the length of second wire

Strain is defined as the elongation per unit length

Strain in first wire = ΔL ÷ $l_{1}$ = ΔL ÷ (4 × $l_{2}$)

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ $l_{2}$

∴ Strain in second wire is four times of strain in first wire

Stress = F ÷ A

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = $m_{1}$ × g

where $m_{1}$ is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = $m_{2}$ × g

where $m_{2}$ is the mass hanged to the second wire

g is the acceleration due to gravity

Let $A_{1}$ be the cross-sectional area of first wire

$A_{2}$ be the cross-sectional area of second wire

$A_{2}$ = 4 × $A_{1}$ (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = ($m_{1}$ × g) ÷ ($A_{1}$)

Stress in second wire = ($m_{2}$ × g) ÷ ($A_{2}$) = ($m_{2}$ × g) ÷ (4 × $A_{1}$)

Young's modulus is defined as Stress per unit strain

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = (($m_{1}$ × g) ÷ ($A_{1}$)) ÷ (ΔL ÷ (4 × $l_{2}$))

Stress per unit strain of second wire = (($m_{2}$ × g) ÷ (4 × $A_{1}$)) ÷ (ΔL ÷ $l_{2}$)

By equating them we get

$m_{2}$ = 16 × $m_{1}$

⇒ $m_{2}$ = 16 × 3 = 48 kg

∴ $m_{2}$ = 48 kg