A 5 m long aluminium wire $(Y = 7 \times 10^{10} N / m^{2})$ of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10} N / m^{2})$ of the same length under the same weight, the diameter should now be, in mm

1.75

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2.3

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Solution

The correct option is C 2.3
$l = \frac{F L}{π r^{2} Y} = \frac{4 F L}{π d^{2} Y} [A s r = \frac{d}{2}]$
If the elongation in both wires (of same length) are same under the same weight then $d^{2} Y$ = constant
$(\frac{d_{C u}}{d_{A l}})$ = $\frac{Y_{A l}}{Y_{C u}} \Rightarrow d_{C u} = d_{A l} \times \sqrt{\frac{Y_{A l}}{Y_{C u}}} = 3 \times \sqrt{\frac{7 \times 10^{10}}{12 \times 10^{10}}} = 2.29 m m$

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Q. A 5 m long aluminium wire

(Y = 7 \times 10^{10} N / m^{2})

of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire

(Y = 12 \times 10^{10} N / m^{2})

of the same length under the same weight, the diameter should now be, in mm

Q. A 5 m long aluminium wire

(Y = 7 \times 10^{10} N / m^{2})

of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire

(Y = 12 \times 10^{10} N / m^{2})

of the same length under the same weight, the diameter should now be, in mm

A $5$ m long aluminium wire $(Y = 7 \times 10^{10} N m^{- 2})$ of diameter $3$ mm supports a $40$ kg mass. In order to have the same elongation in the copper wire $(Y = 12 \times 10^{10} N m^{- 2})$ of the same length under the same weight, the diameter of the copper wire should now be (in mm).