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An aluminum rod, Young's modulus 7.0x10⁹ newton/metre² has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in m² in order to support a load of 10⁴ newton is

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a

1x 10^-2

b

1.0x10^-3

c

1.4 x 10^-2

d

7.1x10^-4

answer is D.

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Detailed Solution

$Y = \frac{FL}{Al} or A = \frac{E}{Y} \times (\frac{L}{l})$
Given $(l / L) = 0 \cdot 2 % = 0 \cdot 2 \times 10^{- 2}$
$F = 10^{4} N$
and $Y = 7 \cdot 0 \times 10^{9} N / m^{2}$
$\begin{array}{l} ∴ A = \frac{10^{4}}{7 \cdot 0 \times 10^{9}} \times (0 \cdot 2 \times 10^{- 2}) \\ \approx 7 \cdot 1 \times 10^{- 4} m^{2} \end{array}$

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An aluminum rod, Young's modulus 7.0x109 newton/metre2 has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in m2 in order to support a load of 104 newton is