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VITEEE 2013 Paper

Section: Physics

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Question : 7 of 120

Marks: +1, -0

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm are connected end to end. When stretched by a force, the elongation in length 0.50 mm is produced in the copper wire. The stretching force is

Y_{\text{cu}}

10^{11}

\mathrm{m}^2

Y_{\text{steel}}

10^{11}

\mathrm{m}^2

)

5.4 x $10^2$ N
3.6 x $10^2$ N
2.4 x $10^2$ N
1.8 x $10^2$ N

Solution:

For Cu wire , ,

l_1

= 2.2 m ,

r_1

= 1.5 mm

= 1.5 ×

10^{-3}

n

Y_1

= 1.1 ×

10^{11}

N/

\mathrm{m}^2

For steel wire

l_2

= 1.6 m ,

r_2

= 1.5 mm

= 1.5 ×

10^{-3}

m

Y_2

= 2.0 ×

10^{11}

N/

\mathrm{m}^2

Let F be the stretching force in both the wires the

For Cu wire,

Y_1

=

\frac{F}{\pi r_1^2} \times \frac{l_1}{\Delta l_1}

⇒ F =

\frac{Y_1 \pi r_1^2 \times \Delta l_1}{l_1}

=

\frac{1.1 \times 10^{-11}}{2.2}

×

\frac{22}{7}

×

(1.5 \times 10^{-3})^2

× 0.5 ×

10^3

= 1.8 x

10^2

N

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