Question

Calculate the stress applied to a wire having mass
per unit length 0.2 g/cm and density 8 g/cm
when 15 g-wt is attached to the free end of the
wire.​

Answer 1

Not the correct answer but for reference
ANSWER
A wire having mass per unit
length =10g/cm
density =800kg/m
3

which is suspended rigid support
The stress in the wire if 10kg−wt is attrached to free its end is
800kg=1m
3

10kg=
800
10
​
m
3

80
1
​
m
3

1000
10
​
kg=
100
1
​
m
∴
100
1
​
kg=
100
1
​
m
=
100
1
​

80
1
​

​

=
80
100
​

=
4
5
​
m
2

Stress=
Area
Force
​

=Pressure
=
4
5
​

100
1
​

​

=
5
4
​
×
100
1
​

=
100
1
​
N/m
2

Answer 2

$\sigma=0.0588\ MPa$

Explanation:

Given:

mass per unit length, $\bar m=0.2\ g.cm^{-1}$

density of the wire, $\rho=8\ g.cm^{-3}$

mass attached to the end of wire, $m=15\ g$

So, force due to the mass:

$w=\frac{15}{1000} \times 9.8$

$w=0.147\ N$

Now the cross sectional area of the wire:

$A=\bar m \div\rho$

$A=0.2\div 8$

$A=0.025\ cm^2=2.5\ mm^2$

• Now stress:

$\sigma=\frac{w}{A}$

$\sigma=\frac{0.147}{2.5}$

$\sigma=0.0588\ MPa$

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TOPIC: stress

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