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Q. 290. What pressure has to be applied to the ends of a steel cylinder to keep its length constant on raising its temperature by 100°C? 

Solution. 290. Variation of length with temperature is given by

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

But     Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus σ = αΔtE, which is the sought stress of pressure.
Putting the value of a and E from Appendix and taking Δt = 100°C, we get

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 291. What internal pressure (in the absence of an external pressure) can be sustained
 (a) by a glass tube;
 (b) by a glass spherical flask, if in both cases the wall thickness is equal to Δr = 1.0 mm and the radius of the tube and the flask equals r = 25 mm?

Solution. 291. (a) Consider a transverse section of the tube and concentrate on an element which subtends an angle Δφ at the centre. The forces acting on a portion of length A/ on the element are
(1) tensile forces side ways of magnitude σΔrΔl.
The resultant of these is

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

radially towards the cente.

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(2) The force due to fluid pressure  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEETSince these balance, we get Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET where σm is the maximum tensile force. Putting the values we get  atmos. Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(b) Consider an element of area  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET about z -a x is chosen arbitrarily. There are tangential tensile forces all around the ring of the cap. Their resultant is

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 292. A horizontally oriented copper rod of length l = 1.0 m is rotated about a vertical axis passing through its middle. What is the number of rps at which this rod ruptures?

Solution. 292. Let us consider an element of rod at a distance x from its rotation axis (Fig.). From Newton’s second law in projection form directed towards the rotation axis

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

On integrating

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

But    Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus     Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Hence     Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus     Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Condition renuired for the. problem

  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Hence the sought number of r p s

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 293. A ring of radius r = 25 cm made of lead wire is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring. What is the number of rps at which the ring ruptures? 

Solution. 293. Let us consider an element of the ring (Fig.). From Newton’s law Fn = mwn for this element, we get,

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

So,     Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Condition for the problem is :

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus sought number of rps

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Using the table of appendices n = 23rps


Q. 294. A steel wire of diameter d = 1.0 mm is stretched horizontally between two clamps located at the distance l = 2.0 m from each other. A weight of mass m = 0.25 kg is suspended from the midpoint O of the wire. What will the resulting descent of the point O be in centimetres? 

Solution. 294.  Let the point O desend by the distance x (Fig.). From the condition of equilibrium of point O.

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET  (1)

Now,   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET      (2)

(σ here is stress and ε is strain.)

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 295. A uniform elastic plank moves over a smooth horizontal plane due to a constant force F0 distributed uniformly over the end face. The surface of the end face is equal to S, and Young's modulus of the material to E. Find the compressive strain of the plank in the direction of the acting force.

Solution. 295. Let us consider an element of the rod at a distance x from the free end (Fig.). For the considered element ‘T - T’ are internal restoring forces which produce elongation and dT provides the acceleration to the element. For the element from Newton’s law :

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

As free end has zero tension, on integrating the above expression,

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Elongation in the considered element of lenght dx :

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus total elengation  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Hence the sought strain

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 296. A thin uniform copper rod of length l and mass m rotates uniformly with an angular velocity ω in a horizontal plane about a vertical axis passing through one of its ends. Determine the tension in the rod as a function of the distance r from the rotation axis. Find the elongation of the rod.

Solution. 296. Let us consider an element of the rod at a distance r from it’s rotation axis. As the element rotates in a horizontal circle of radius r, we have from Newton’s second law in projection form directed toward the axis of rotation :

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

or,   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

At the free end tension becomes zero. Integrating the above experession we get, thus

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Elongation in elemental length dr is given by :

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(where S is the cross sectional area of the rod and T is the tension in the rod at the considered element)

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET  (where p is the density of the copper.)


Q. 297. A solid copper cylinder of length l = 65 cm is placed on a horizontal surface and subjected to a vertical compressive force F = 1000 N directed downward and distributed uniformly over the end face. What will be the resulting change of the volume of the cylinder in cubic millimetres?

Solution. 297. Volume of a solid cylinder

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

So,   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET     (1)
But longitudinal strain Δl/l and accompanying lateral strain Δ r/r are related as

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET    (2)

Using (2) in (1), we get :

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET    (3)

But  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(Because the increment in the length of cylinder Δl is negative)

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus,    Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Negative sign means that the volume of the cylinder has decreased.


Q. 298. A copper rod of length l is suspended from the ceiling by one of its ends. Find:
 (a) the elongation Δl of the rod due to its own weight;
 (b) the relative increment of its volume ΔVIV. 

Solution. 298. (a) As free end has zero tension, thus the tension in the rod at a vestical distance y from its lower end

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET be the elongation of the element of length dy, then

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET (where p is the density of the copper) 

Thus the sought elongation

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET   (2)

(b) If the longitudinal (tensile) strain is   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET the accompanying lateral (compressive) strain is given by

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET    (3)

Then since  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET we have

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

where  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET is given in part (a), μ is the Poisson ratio for copper.


Q. 299. A bar made of material whose Young's modulus is equal to E and Poisson's ratio to μ, is subjected to the hydrostatic pressure p. Find:
 (a) the fractional decrement of its volume;
 (b) the relationship between the compressibility β and the elastic constants E and μ. 

Solution. 299.  Consider a cube of unit length before pressure is applied. The pressure acts on each face. The pressures on the opposite faces constitute a tensile stress producing longitudianl compression and lateral extension. The compressions is P/E and the lateral extension is  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

The net result is a compression  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Hence   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET because from symmetry Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(b) Let us consider a cube under an equal compressive stress σ, acting on all its faces.

Then,   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET   (1)

where k is the bulk modulus of elasticity.

So  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

or,  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET are both to remain positive.


Q. 300. One end of a steel rectangular girder is embedded into a wall (Fig. 1.74). Due to gravity it sags slightly. Find the radius of curvature of the neutral layer (see the dotted line in the figure) in the vicinity of the point O if the length of the protruding section of the girder is equal to l = 6.0 m and the thickness of the girder equals h = 10 cm. 

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Solution. 300. A beam clamped at one end and supporting an applied load at the free end is called a cantilever. The theory of cantilevers is discussed in advanced text book on mechanics. The key result is that elastic forces in the beam generate a couple, whose moment, called the moment of resistances, balances the external bending moment due to weight of the beam, load etc. The moment of resistance, also called internal bending moment (I.B.M) is given by

I.B.M. = EI/R

Here R is the radius of curvature of the beam at the representative point (x, y). I is called the geometrical moment of inertia

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

of the cross section relative to the axis passing through the netural layer which remains unstretched. (Fig.l.). The section of the beam beyond P exerts the bending moment N (x) and we have,

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

If there is no load other than that due to the weight of the beam, then

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

where p = density of steel. Hence, at x = 0

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Here b = width of the beam perpendicular to paper.

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 301. The bending of an elastic rod is described by the elastic curve passing through centres of gravity of rod's cross-sections. At small bendings the equation of this curve takes the form Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

where N (x) is the bending moment of the elastic forces in the cross- section corresponding to the x coordinate, E is Young's modulus, I is the moment of inertia of the cross-section relative to the axis pass- ing through the neutral laye Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET 

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Suppose one end of a steel rod of a square cross-section with side a is embedded into a wall, the protruding section being of length l (Fig. 1.76). Assuming the mass of the rod to be negligible, find the shape of the elastic curve and the deflection of the rod X, if its end A experiences 
 (a) the bending moment of the couple N0;
 (b) a force F oriented along the y axis

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Solution. 301. We use the equation given above and use the result that when y is small

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(a) Here N (x) = N0 is a constant. Then integration gives,

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

But   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

where we have used y = 0 for x = 0 to set the constant of integration at zero. This is the equation of a parabola. The sag of the free end is

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(b) In this case N (x) - F (l - x) because the load F at the extremity is balanced by a similar force at F directed upward and they constitute a couple. Then

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Integrating,  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

As before C1 = 0. Integrating again, using y = 0 for x = 0

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Here for a square cross section

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET



Q. 302.  A steel girder of length l rests freely on two supports (Fig. 1.77). The moment of inertia of its cross-section is equal to I (see the foregoing problem). Neglecting the mass of the girder and assuming the sagging to he slight, find the deflection λ due to the force F applied to the middle of the girder.

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Solution. 302. One can think of it as analogous to the previous case but with a beam of length l/2 loaded upward by a force F/2.

Thus Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

On using the last result of the previous problem.


Q. 303. The thickness of a rectangular steel girder equals h. Using the equation of Problem 1.301, find the deflection λ caused by the weight of the girder in two cases:
 (a) one end of the girder is embedded into a wall with the length of the protruding section being equal to l (Fig. 1.78a);
 (b) the girder of length 2l rests freely on two supports (Fig. 1.78b). 

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Solution. 303.  (a)   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET  where b = width of the girder.

Also   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Integrating,   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

using   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(b) As before,  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET is the bending moment due to section PB.

This bending moment is clearly

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(Here w = p g b h is weight o f the beam per unit length)

Now integrating,  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

or since  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEETIntegrating again,  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

As y = 0 for x = 0, c1 = 0. From this we find

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 304. A steel plate of thickness h has the shape of a square whose side equals l, with h <<  1. The plate is rigidly fixed to a vertical axle OO which is rotated with a constant angular acceleration 13 (Fig. 1.79). Find the deflection λ, assuming the sagging to be small. 

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Solution. 304. The deflection of the plate can be noticed by going to a co- rotating frame. In this frame each element of the plate experiences a pseudo force proportional to its mass. These forces have a moment which constitutes the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET from the axis is Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET and the moment of the forces exerted by the section between x and l is

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

From the fundamental equation   

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

The moment of inertia  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Vofe that the neutral surface (i.e. the surface which contains lines which are neither stretched nor compressed) is a vertical plane here and z is perpendicular to it.

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Since Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 305. Determine the relationship between the torque N and the torsion angle φ for
 (a) the tube whose wall thickness Δr is considerably less than the tube radius;
 (b) for the solid rod of circular cross-section. Their length l, radius r, and shear modulus G are supposed to be known. 

Solution. 305. (a) Consider a hollow cylinder of length l, outer radius r + Δr inner radius r, fixed at one end and twisted at the other by means of a couple of moment N. The angular displacement φ, at a distance l from the fixed end, is proportional to both l and N. Consider an element of length dx at the twisted end. It is moved by an angle φ as shown. A vertical section is also shown and the twisting of the parallelopipe of length l and area Δr dx under the action of the twisting couple can be discussed by elementary means. If f is the tangential force generated then shearing stress is f/Δr dx and this must equal

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Hence ,   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

The force / has moment f r about the axis and so the total moment is

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

(b) For a solid cylinder we must integrate over r. Thus

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEETIrodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 306. Calculate the torque N twisting a steel tube of length l = 3.0 m through an angle φ = 2.0° about its axis, if the inside and outside diameters of the tube are equal to d1 =  30 mm and d2  = 50 mm. 

Solution. 306.  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

using   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 307. Find the maximum power which can be transmitted by means of a steel shaft rotating about its axis with an angular velocity ω = 120 rad/s, if its length l = 200 cm, radius r = 1.50 cm, and the permissible torsion angle φ = 2.5°. 

Solution. 307. The maximum power that can be transmitted by means of a shaft rotating about its axis is clearly N ω where N is the moment of the couple producing the maximum permissible torsion, φ. Thus

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 308. A uniform ring of mass m, with the outside radius r2, is fitted tightly on a shaft of radius r1. The shaft is rotated about its axis with a constant angular acceleration β. Find the moment of elastic forces in the ring as a function of the distance r from the rotation axis. 

Solution. 308. Consider an elementary ring of width dr at a distant r from the axis. The part outside exerts a couple  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET dr on this ring while the part inside exerts a couple N in the opposite direction. We have for equilibrium

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

where dl is the moment of inertia o f the elementary ring, β is the angular acceleration and minus sign is needed because the couple N (r) decreases, with distance vanshing at the outer radius, N (r2) = 0. Now

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Thus  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

or,   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEETIrodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 309. Find the elastic deformation energy of a steel rod of mass m = 3.1 kg stretched to a tensile strain ε = 1.0-10-3

Solution. 309. We assume that the deformation is wholly due to external load, neglecting the effect of the weight of the rod (see next problem). Then a well known formula says,
elastic energy per unit volume

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

This gives   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET for the total deformation energy.


Q. 310. A steel cylindrical rod of length l and radius r is suspended by its end from the ceiling.
 (a) Find the elastic deformation energy U of the rod.
 (b) Define U in terms of tensile strain Δl/l of the rod.

Solution. 310. When a rod is deformed by its own weight the stress increases as one moves up, the stretching force being the weight of the portion below the element considered.
The stress on the element dx is   

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Integrating   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET  is the extension of the whole rod. The elastic energy of the element is

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Integrating 

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 311. What work has to be performed to make a hoop out of a steel band of length l = 2.0 m, width h = 6.0 cm, and thickness δ = 2.0 mm? The process is assumed to proceed within the elasticity range of the material. 

Solution. 311. The work done to make a loop out of a steel band appears as the elastic energy of the loop and may be calculated from the same.

If the length of the band is l, the radius of the loop  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET Now consider an element ABCD of the loop. The elastic energy of this element can be calculated by the same sort of arguments as used to derive the formula for internal bending moment. Consider a fibre at a distance z from the neutral surface PQ. This fibre experiences a forcep and undergoes an extension  Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET If α is the cross sectional area of the fibre, the elastic energy associated with it is

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Summing over all the fibres we get

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

For the whole loop this gives,

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Now   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

So the energy is    Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 312. Find the elastic deformation energy of a steel rod whose one end is fixed and the other is twisted through an angle φ = 6.0°. The length of the rod is equal to l = 1.0 m, and the radius to r = 10 mm.

Solution. 312. When the rod is twisted through an angle θ a couple 

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET appears to resist this. Work done in twisting the rod by an angle φ is then

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 313. Find how the volume density of the elastic deformation energy is distributed in a steel rod depending on the distance r from its axis. The length of the rod is equal to l, the torsion angle to φ. 

Solution. 313. The energy between radii r and r + dr is, by differentiation,   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

Its density is   Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET


Q. 314. Find the volume density of the elastic deformation energy in fresh water at the depth of h = 1000 m. 

Solution. 314. The energy density is as usual 1/2 stress x strain. Stress is the pressure p g h. Strain is β x p gh by defination of β. Thus

Irodov Solutions: Elastic Deformations of a Solid Body | Physics Class 11 - NEET

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FAQs on Irodov Solutions: Elastic Deformations of a Solid Body - Physics Class 11 - NEET

1. What is elastic deformation of a solid body?
Ans. Elastic deformation refers to the temporary change in shape or size of a solid body when an external force is applied to it. In this process, the body is able to return to its original shape and size once the force is removed. It occurs due to the rearrangement of atoms or molecules within the material without any permanent damage.
2. What causes elastic deformation in a solid body?
Ans. Elastic deformation in a solid body is caused by the application of an external force. When a force is applied, it induces stress within the material, which leads to the distortion of its shape or size. The elastic properties of the material allow it to undergo temporary deformation without permanent damage, as long as the stress applied is within its elastic limit.
3. How can we measure elastic deformation in a solid body?
Ans. Elastic deformation in a solid body can be measured using various techniques. One common method is the use of strain gauges or extensometers, which are devices that can accurately measure the change in length or shape of a material under stress. These measurements can then be used to calculate the strain, which represents the extent of elastic deformation.
4. What is the relationship between stress and elastic deformation?
Ans. The relationship between stress and elastic deformation in a solid body is described by Hooke's Law. According to this law, the amount of elastic deformation (strain) is directly proportional to the applied stress, as long as the material remains within its elastic limit. This relationship is represented by the equation: stress = elastic modulus × strain.
5. What are some real-life examples of elastic deformation in solid bodies?
Ans. Elastic deformation can be observed in various real-life scenarios. For example, when a spring is stretched or compressed, it undergoes elastic deformation and returns to its original shape once the force is removed. Similarly, rubber bands, trampoline mats, and diving boards exhibit elastic deformation when subjected to external forces. These materials can stretch or deform temporarily and then regain their original shape.
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