An aluminium rod, Young's modulus 7.0 × 109 Nm -2, has a breaking strain of 0.2 %. The minimum cross-sectional area of the rod in m 2 in order to support a load of 104 N is

Question

An aluminium rod, Young's modulus 7.0 × 109 Nm -2, has a breaking strain of 0.2 %. The minimum cross-sectional area of the rod in m 2 in order to support a load of 104 N is (A) 1 × 10-2 (B) 1.4 × 10-3 (C) 1.0 × 10-3 (D) 7.1 × 10-4

Tardigrade Admin · Accepted Answer

A s, Y=(F / A/ text Breaking strain ) or A =(F/Y × text Breaking strain )=(104 × 100/7 × 10 × 0.2) =0.71 × 10-3=7.1 × 10-4

Q. An aluminium rod, Young's modulus $7.0 \times 1 0^{9} N m^{- 2}$ , has a breaking strain of $0.2%$ . The minimum cross-sectional area of the rod in $m^{2}$ in order to support a load of $1 0^{4} N$ is

$1 \times 1 0^{- 2}$

$1.4 \times 1 0^{- 3}$

$1.0 \times 1 0^{- 3}$

$7.1 \times 1 0^{- 4}$

$A s, Y = \frac{F / A}{Breaking strain}$
or $A = \frac{F}{Y \times Breaking strain} = \frac{1 0 ^{4} \times 100}{7 \times 10 \times 0.2}$
$= 0.71 \times 1 0^{- 3} = 7.1 \times 1 0^{- 4}$

Q. An aluminium rod, Young's modulus 7.0×109Nm−2, has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in m2 in order to support a load of 104N is

1×10−2

1.4×10−3

1.0×10−3

7.1×10−4