
A wire having mass per unit length 10 g/cm and density \[800\,{\text{kg/}}{{\text{m}}^3}\] suspended rigid support. The stress in the wire if 10 kg-wt is attached to its free end is?
Answer
458.4k+ views
Hint:Use the relation between density, mass and volume to express the volume of the wire. Divide the relation by the length of wire to get the area of the wire. The stress is the ratio of applied force and area of cross-section.
Formula used:
Stress, \[\sigma = \dfrac{F}{A}\]
Here, F is the applied force and A is the area.
Complete step by step answer:
We have given the mass per unit length of the wire, \[\dfrac{m}{l} = \left( {10\,\dfrac{{\text{g}}}{{{\text{cm}}}}} \right)\left( {\dfrac{{1\,{\text{kg}}}}{{1000\,{\text{g}}}}} \right)\left( {\dfrac{{100\,{\text{cm}}}}{{1\,{\text{m}}}}} \right) = 1\,{\text{kg/m}}\].
The density of the wire is, \[\rho = 800\,{\text{kg/}}{{\text{m}}^3}\].
We have the relation,
\[\rho = \dfrac{m}{V}\]
\[ \Rightarrow V = \dfrac{m}{\rho }\]
Here, m is the mass of the wire and V is the volume.
Dividing the above equation by \[l\], we get,
\[\dfrac{V}{l} = \dfrac{m}{{\rho l}}\]
We know that the volume is given as, \[V = {l^3}\]. Therefore, the above expression becomes,
\[\dfrac{{{l^3}}}{l} = \dfrac{m}{{\rho l}}\]
\[ \Rightarrow {l^2} = \dfrac{m}{{\rho l}}\]
\[ \Rightarrow A = \dfrac{m}{{\rho l}}\] ……. (Since \[Area = {l^2}\]) …… (1)
We know that the stress is the ratio of applied force per unit area of cross-section. Therefore,
\[\sigma = \dfrac{F}{A}\]
Here, F is the applied force and A is the area.
In this case, the applied force is the weight of the block attached to the free end of the wire. Therefore, we can write the above equation as,
\[\sigma = \dfrac{W}{A}\]
Using equation (1) in the above equation, we get,
\[\sigma = \dfrac{W}{{\dfrac{m}{{\rho l}}}}\]
\[ \Rightarrow \sigma = \dfrac{{W\rho }}{{m/l}}\]
Substituting\[W = 10\,{\text{N}}\], \[\rho = 800\,{\text{kg/}}{{\text{m}}^3}\] and \[\dfrac{m}{l} = 1\,{\text{kg/m}}\] in the above equation, we get,
\[\sigma = \dfrac{{\left( {10} \right)\left( {800} \right)}}{1}\]
\[\therefore\sigma = 8000\,{\text{N/}}{{\text{m}}^2}\]
Thus,the stress in the wire if 10 kg-wt is attached to its free end is \[8000\,{\text{N/}}{{\text{m}}^2}\].
Note: Always convert the mass per unit length from \[{\text{gm/c}}{{\text{m}}^3}\] to \[{\text{kg/}}{{\text{m}}^3}\]. We have taken the volume of the wire irrespective of its shape to express it in terms of area of the wire. The stress is also the pressure because the pressure is also the force per unit area of cross-section.
Formula used:
Stress, \[\sigma = \dfrac{F}{A}\]
Here, F is the applied force and A is the area.
Complete step by step answer:
We have given the mass per unit length of the wire, \[\dfrac{m}{l} = \left( {10\,\dfrac{{\text{g}}}{{{\text{cm}}}}} \right)\left( {\dfrac{{1\,{\text{kg}}}}{{1000\,{\text{g}}}}} \right)\left( {\dfrac{{100\,{\text{cm}}}}{{1\,{\text{m}}}}} \right) = 1\,{\text{kg/m}}\].
The density of the wire is, \[\rho = 800\,{\text{kg/}}{{\text{m}}^3}\].
We have the relation,
\[\rho = \dfrac{m}{V}\]
\[ \Rightarrow V = \dfrac{m}{\rho }\]
Here, m is the mass of the wire and V is the volume.
Dividing the above equation by \[l\], we get,
\[\dfrac{V}{l} = \dfrac{m}{{\rho l}}\]
We know that the volume is given as, \[V = {l^3}\]. Therefore, the above expression becomes,
\[\dfrac{{{l^3}}}{l} = \dfrac{m}{{\rho l}}\]
\[ \Rightarrow {l^2} = \dfrac{m}{{\rho l}}\]
\[ \Rightarrow A = \dfrac{m}{{\rho l}}\] ……. (Since \[Area = {l^2}\]) …… (1)
We know that the stress is the ratio of applied force per unit area of cross-section. Therefore,
\[\sigma = \dfrac{F}{A}\]
Here, F is the applied force and A is the area.
In this case, the applied force is the weight of the block attached to the free end of the wire. Therefore, we can write the above equation as,
\[\sigma = \dfrac{W}{A}\]
Using equation (1) in the above equation, we get,
\[\sigma = \dfrac{W}{{\dfrac{m}{{\rho l}}}}\]
\[ \Rightarrow \sigma = \dfrac{{W\rho }}{{m/l}}\]
Substituting\[W = 10\,{\text{N}}\], \[\rho = 800\,{\text{kg/}}{{\text{m}}^3}\] and \[\dfrac{m}{l} = 1\,{\text{kg/m}}\] in the above equation, we get,
\[\sigma = \dfrac{{\left( {10} \right)\left( {800} \right)}}{1}\]
\[\therefore\sigma = 8000\,{\text{N/}}{{\text{m}}^2}\]
Thus,the stress in the wire if 10 kg-wt is attached to its free end is \[8000\,{\text{N/}}{{\text{m}}^2}\].
Note: Always convert the mass per unit length from \[{\text{gm/c}}{{\text{m}}^3}\] to \[{\text{kg/}}{{\text{m}}^3}\]. We have taken the volume of the wire irrespective of its shape to express it in terms of area of the wire. The stress is also the pressure because the pressure is also the force per unit area of cross-section.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
