Answers to Problems on (Some Mechanical Properties Of Matter) HC Verma's Questions for Short Answer (1-10)

  Q#1

The ratio stress/strain remains constant for small deformation of a metal wire. When the deformation is made larger, will this ratio increase or decrease?

Answer:

For large deformations beyond the elastic limit, a small increase in the stress makes the strain larger. So the ratio Stress/Strain decreases.

Q#2

When a block of mass M is suspended by a long wire of length L, the elastic potential energy stored in the wire is ½ (stress)(strain)(volume). Show that it is equal to ½Mgl, where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining ½Mgl energy go? 

Answer:
Stress = Mg/A, {Where A is the cross-sectional area of the wire}

Strain = l/L.
Since the elastic potential energy

U = ½(Stress)(Strain)(Volume)
→U = ½(Mg/A)(l/L)(AL) = ½Mgl

The loss in gravitational potential energy will be Mgl only when the load is suddenly released from the initial position i.e. when elongation is zero. The elastic potential energy stored in the wire =½Mgl when the elongation is l, the rest of the gravitational potential energy Mgl-½Mgl =½Mgl is converted to the kinetic energy. At the time elongation is l, the mass M will have a certain speed say v. So,

 ½Mv² = ½Mgl
v² = gl
v =√(gl).

This point as a mean position the mass will be in a simple harmonic motion vertically. Slowly it will come to rest at the mean position by dissipating this energy in the form of heat.

Q#3
When the skeleton of an elephant and the skeleton of a mouse are prepared in the same size, the bones of the elephant is shown thicker than those of the mouse. Explain why the bones of an elephant are thicker than proportionate. The bones are expected to withstand the stress due to the weight of the animal.

Answer:
The weight of an elephant is so much that, to bring the stress on the bones under permissible limit the cross-sectional areas of the bones are increased by nature because of stress = Force/area. So the bones of an elephant are thicker than proportionate.

Q#4
The yield point of a typical solid is about 1%. Suppose you are lying horizontally and two persons are pulling your hands and two persons are pulling your legs along your own length. How much will be the increase in your length if the strain is 1%? Do you think your yield point is 1% or, much less than that?

Answer:
If the height is H and the increase in height/length is h due to pulling then the strain = h/H. If it is 1% then

h/H = 1/100

h = H/100
If the height H = 180 cm (say)
then the increase in length h = 1.80 cm.

Since our body has joints which are weaker than the bones so this increase in length will be located to the weakest joint. Any joint having 1.80 cm separation will be much damaging. So the yield point (or the bearable pain) due to pulling will be much less than 1%.

Q#5
When rubber sheets are used in a shock absorber, what happens to the energy of vibration?

Answer:
The energy of shock due to vibration absorbed by the rubber sheets is converted to heat. It is also due to a property of rubber which is called "elastic hysteresis".

Q#6
If a compressed spring is dissolved in acid what happens to the elastic potential energy of the spring?

Answer:
In a compressed spring the molecules are under stress due to the elastic potential energy stored in it. When the acid dissolves the spring molecules the P.E. stored in it is converted to K.E. and the new molecule it forms with acid ion has this extra K.E. Due to this the temperature of the acid increases slightly. It means that the elastic potential energy is converted to heat energy.

Q#7
A steel blade placed gently on the surface of water floats on it. If the same blade is kept well inside the water, it sinks. Explain.

Answer:
Due to the surface tension the surface of the water acts like a stretched membrane. The surface tension along the edges of the blade is more than the weight of the blade thus it floats.
Now the surface tension is a surface phenomenon, so when the same blade is kept well inside the water there is no surface tension to counter its weight. There is only the weight of the blade and buoyancy force on it. Since the density of the blade is more than the water, the weight is greater than the buoyancy force, thus it sinks.

Q#8
When some wax is rubbed on a cloth, it becomes waterproof. Explain.

Answer:
The cloth is made of fine fibers and the spaces between the fine fibers act as capillary tubes. We know that when the angle of contact between the water and the side of the capillary tube is less than 90°, the water level in the tube rises. On the other hand, if the angle is more than 90°, the level of water in the tube goes down. In the case of water and the cloth, the angle of contact is nearly zero, so the water goes readily inside to wet it. The angle of contact between the water and the wax is about 107°. So when some wax is rubbed on the cloth, the water first comes in contact with the wax and the angle of contact being more than 90° it does not enter the cloth. So the cloth becomes waterproof.

Q#9
The contact angle between pure water and pure silver is 90°. If a capillary tube made of silver is dipped at one end in pure water, will the water rise in the capillary?

Answer:
No. The rise of liquid in a capillary tube is given as,

h = 2Scosθ/(rρg)

Where S = Surface tension, θ = the angle of contact between the liquid and the wall of the tube, r = radius of the radius of the capillary tube, ρ = density of the liquid and g = acceleration due to gravity.
Here given that θ = 90°, hence cosθ = 0.

∴ h = 0.

So the water will not rise in the pure silver capillary tube.

Q#10
It is said that a liquid rises or is depressed in a capillary due to the surface tension. If a liquid neither rises nor depresses in a capillary, can we conclude that the surface tension of the liquid is zero?

Answer:
No. Because the rise of a liquid in a capillary tube depends on other factors also as we can see in the expression for height h,

h = 2Scosθ/(rρg)

The rise h depends on S, θ, r and ρ. Since it is a capillary tube, r is small and it can not make h zero. Even the density of a liquid ρ will not be so high to make h zero. But the cosθ can be zero for θ =90°, thus making the rise h =0.

 So if a liquid neither rises nor depresses in a capillary we can not conclude that the surface tension of the liquid is zero. The angle of contact between the liquid and the wall of the tube may be about 90°.

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