The stress experienced by a wire of diameter {{y}}{π}m loaded by a weight y²N is

Correct Answer is : 4πππNm−2^{-2}−2

Correct Answer is : 4πππNm−2^{-2}−2

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JAMB Physics 2024 Paper

Question : 1

Total: 125

\frac{\text{y}}{\pi}

^2

N is

Solution:

Concept:

Stress is defined as the force applied per unit cross-sectional area of a wire.

Explanation:

Given diameter

D = \frac{y}{\pi}\,m

and the load is the weight

F = y^2\,N

Radius is half the diameter:

r = \frac{D}{2} = \frac{y}{2\pi}\,m

Cross-sectional area of the wire:

A = \pi r^2 = \pi \left(\frac{y}{2\pi}\right)^2 = \frac{y^2}{4\pi}\,m^2

Now, stress =

\frac{F}{A}

Substitute the values:

\text{Stress} = \frac{y^2}{y^2/(4\pi)} = 4\pi\,Nm^{-2}

Thus, the wire experiences a stress of

4\pi\,Nm^{-2}

Answer:

4\pi\,Nm^{-2}

(Option C).

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