A load of 10 kg is suspended by a metal wire 3m long and having a cross-sectional area 4mm^2. Find. 1. The stress. 2. The strain and. 3. The elongation. Young's modulus of the metal is 2.0 10N^ 11 N/ m^2.

Length of wire L = 3m, Load F = 10 10N = 100N, Taking g = 10m/s^2 Area of cross-section A = 4mm^2 =4 10^ -6 m^2 1. The stress = Load (force) on unit area of cross-section = F A =100 4 10^ -6 N/m^2 =25 10^6N/m^2 =2.5 10^7N/m^2 2. Let the elongation of the wire under this stress bel, The strain = l L , Young's modulus of the metal Y = 2.0 10^ 11 N/m^2 We have Stress Strain =Y (constant) ( F A ) Strain = Y = ( F A ) Y = 2.5 10^7 2.0 10^ 11 =1.25 10^ -4 m 3. strain = l L =1.25 10^ -4 =3.0 1.25 10^ -4 m =3.75 10^ -4 m

A load of 10 kg is suspended by a metal wire 3m long and having a…

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Question

A load of $10\ kg$ is suspended by a metal wire $3m$ long and having a cross-sectional area $4mm^2$. Find.

The stress.
The strain and.
The elongation. Young's modulus of the metal is $2.0 \times 10N^{11}N/ m^2$.

✓

Answer

Length of wire $L = 3m,$
Load $F = 10 \times 10N = 100N, \{$Taking $g = 10m/s^2\}$
Area of cross$-$section $A = 4mm^2$
$\Rightarrow\text{A}=4\times10^{-6}\text{m}^2$

The stress $=$ Load $($force$)$ on unit area of cross$-$section $=\frac{\text{F}}{\text{A}}$

$=\frac{100}{4}\times10^{-6}\text{N}/\text{m}^2$
$=25\times10^6\text{N}/\text{m}^2$
$=2.5\times10^7\text{N}/\text{m}^2$

Let the elongation of the wire under this stress bel, The strain $=\frac{\text{l}}{\text{L}},$ Young's modulus of the metal $Y = 2.0 \times 10^{11}N/m^2$ We

have $\frac{\text{Stress}}{\text{Strain}}=\text{Y} ($constant$)$
$\Rightarrow\frac{\Big(\frac{\text{F}}{\text{A}}\Big)}{\text{Strain}}= \text{Y}$
$\Rightarrow\text{Strain}= \frac{\Big(\frac{\text{F}}{\text{A}}\Big)}{\text{Y}}= 2.5 \times\frac{10^7}{2.0}\times10^{11}$
$\Rightarrow\text{strain=1.25}\times10^{-4}\text{m}$