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Q.

A brass bar having a cross sectional area $10 {cm}^{2}$ is subjected to axial force as shown in the figure. Take $Y = 8 \times 10^{2} N / {cm}^{2}$ . The total elongation of the bar is

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a

0.050 cm

b

0.095 cm

c

0.0775 cm

d

–0.0.25 cm

answer is A.

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Detailed Solution

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$Δ ℓ = \frac{F_{1} ℓ_{1}}{A Y} + \frac{F_{2} ℓ_{2}}{A Y} + \frac{F_{3} ℓ_{3}}{A Y}$

$= \frac{5 \times 60 + 2 \times 100 + 1 \times 120}{10 \times 8 \times 10^{2}}$

$= \frac{6 .2}{80} = 0 .0775 cm$

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A brass bar having a cross sectional area 10 cm2 is subjected to axial force as shown in the figure. Take Y=8×102 N/cm2 . The total elongation of the bar is