An ideal gas of bulk modulus K is enclosed in a container of fixed walls. Find the stress set up in the gas when heated through temperature θ. Let γ be the temperature coefficient of cubical expansion of gas. 

Given, an ideal gas of bulk modulus 'K' and is heated to a temperature θ

Increase of temperature of gas at constant volume is equivalent if the gas is first heated at constant pressure and then compressed isothermally to original volume.

Let V be the volume of gas. When it is heated through a temperature θ at constant pressure, then the increase in volume of gas is,

∆V = Vγθ

Now compress the gas isothermally so that the volume decreases by ∆V = Vγθ.

If ∆P be the stress set up in the gas, then

K=stressstrain=ΔPΔV/V=ΔPγθ         ΔP = Kγθ 

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Derive an expression for the radius of rope used in a crane to lift the mass M kg. The elastic limit of the material of rope is σ N/m2, and safety factor of crane is K.

Given mass = M kg
Elastic limit of the material = σ N/m2
Safety factor for crane = K

The design of the rope is such that it should tolerate a load of KM kg.
Let r be the radius of rope required to be used in the crane.
So, using the formula, 

   Elastic limit =Maximum loadArea of cross sectioni.e.            σ=KMgπr2            r = KMgπσ 
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Derive an expression for the increase in the length of a uniform rope hanging from the ceiling of a room, due to its own weight.

Let,

Mass of the wire = M

Area of cross section = A

Length of the rope = L 

Let, be the increase in the length of rope due to its own weight.

Since the rope is uniform and whole of the weight of rope can supposed to be concentrated at the centre of gravity, therefore the effective length of the rope is L/2 loaded by weight Mg.

Thus, Young's modulus is given by, 

   
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A load W is hanging from the string of length l and area of cross section A. The string suddenly breaks. Find the change in the temperature of the string. Take heat capacity of string as Q and Young’s modulus as Y.

Length of the string =l
Heat capacity of the string = Q
Young's modulus = Y

The potential energy stored in the loaded string is,

                 U=12×stress×strain×volume   = 12(Stress)2YxVolume   =12W2A2YA   =12W2AY

When the string breaks, the potential energy stored in the string is converted into heat.

Let θ be the rise in temperature.
Therefore,

         =12W2AY      θ=12W2QAY 
is the required change in temperature of the string. 
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A rod of metal is clamped between two rigid supports and is heated by temperature θ. If wire is not allowed to bend then what type of stress will be set up in the wire and how does stress developed in wire depend on length of the wire? 


When a rod of metal is heated, compressive stress will be set up in the rod. The wire is not allowed to bend and the force is applied normal to it's cross sectional area.  
Let 'L' be the length of wire, A be the area of cross section, Y be Young's modulus and 'α' be the coefficient of linear expansion, heated to a temperature 'θ'.
If the rod was free then increase in length of wire due to rise in the temperature will be Lαθ.

Here, since rod is not allowed to expand, therefore, compression in rod is equal to increase in the length of rod i.e. Lαθ. 

If S is the compression stress in rod then,

Y = SLLαθ=Sαθor     S=Yαθ

Hence stress is independent of the length of the wire.

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