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CGPET 2016 Solved Paper

Section: Physics

Question : 5 of 150

Marks: +1, -0

Solution:

Let the Young’s modulus of copper wire is

Y

. So,

Y = \frac{\frac{F}{A}}{\frac{\Delta l}{l}} = \frac{F \cdot L}{A \cdot \Delta l}

(i)

When the arrangement is stretched by applying forces at two ends, the same force will experienced by the wires.

From Eq. (i),

\Delta l = \frac{F \cdot L}{A \cdot Y}

If the increase in length of the wires are

\Delta l_1

and

\Delta l_2

, then

\Delta l_1 = \frac{F \cdot L}{A \cdot Y}

\Delta l_2 = \frac{F \cdot L'}{A' \cdot Y}

[ ∵

A = \pi R^2

]

\frac{\Delta l_1}{\Delta l_2} = \frac{L}{R^2} \times \frac{2R^2}{2L} = 2

[∴

A' = \pi (2R)^2

]

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