Elasticity and Fluid Machanics-03-SUBJECTIVE SOLVED

SOLVED SUBJECTIVE PROBLEMS Problem 1. A wire of radius r stretched without tension along a straight line is lightly fixed at A and B (figure). What is the tension in the wire when it is pulled into the shape ACB. Assume Young’s modulus of the material of wire to be Y. Solution: Increase in length = (AC + CB) – 2 l Longitudinal stress = , (where is radius of wire). Longitudinal strain Now, Young’s modulus or, If , then . Problem 2. Two rods of different metals having the same area of cross-section A, are placed between the two massive walls as shown in figure. The first rod has a length , coefficient of linear expansion and Young’s modulus . The corresponding quantities for second rod are and . The temperature of both the rods is now raised by T degrees. (i) Find the force with which the rods act on each other (at higher temperature) in terms of given quantities. (ii) Also find the length of the each rod at higher temperature. Solution: (i) When the temperature is raised by T, then Increase in length of first rod = and increase in length of second rod = Total increase in length …(i) As the walls are rigid, the above increase will not be possible. This will be compensated by the force producing decrease in the rods due to elasticity. decrease in length of first rod decrease in length of second rod Total decrease in length due to F …(ii) From eq. (i) and (ii), we have or …(iii) (ii) Length of the first rod = original length + increase in length due to temp. - decrease in length due to force F …(iv) Length of the second rod …(v) The total length will remain unaltered. Problem 3. A composite rod is made by joining a copper rod end to end, with a second rod of different material, but of the same cross-section. At 25ºC, the composite rod is 1 m in length, of which the length of the copper rod is 30 cm. At 125ºC the length of the composite rod increases by 1.91 mm. When the composite rod is not allowed to expand by holding it between two rigid walls, it is found that the length of the two constituents do not change with rise in temperature. Find the Young’s modulus and the coefficient of Linear expansion of the second rod. . Solution: Length of the composite rod = 100 cm Length of copper rod at 25ºC = 30 cm Length of the second rod at 25ºC = 100 – 30 = 70 cm Length of the copper rod at 125ºC, where is the coefficient of linear expansion of copper and  is change in temperature. cm For second rod, where = coefficient of linear expansion increase in length of Cu rod = 0.051 cm increase in length of second rod = 7000 cm Total increase in length = 0.051 + 7000 = 0.191 Solving we get = The force required to maintain the Cu rod at its initial length, when the rod is heated, is given by dyne The forced required to maintain the other rod at its initial length, when the rod is heated Now if the composite rod is not allowed to expand by holding it between two rigid walls, . Problem 4. A cylinder tank of height 0.4 m is open at the top and has a diameter 0.16 m. Water is filled in it upto a height of 0.16 meter. Calculate how long will it take to empty the tank through a hole of radius m in its bottom. Solution: Let rate of drop of water is - , then or Integrating this expression, we get or second]. Problem 5. A conical glass capillary tube of length 0.1 m has diameters m and m at the ends. When it is just immersed in a liquid at 0ºC with larger radius in contact with it, the liquid rises to m in the tube. In another cylindrical glass capillary tube , when immersed in the same liquid at 0ºC, the liquid rises to m height. The rise of liquid in tube is only m when the liquid is at 50ºC. Find the rate at which the surface tension changes with temperature considering the change to be linear. The density of liquid is (1/14) and the angle of contact is zero. Effect of temperature on the density of liquid and glass is negligible. Solution: The situation is shown in figure. Let and be radii of upper and lower ends of the conical capillary tube. The radius at the meniscus is given by m The surface tension at 0ºC is given by For tube or Considering the change in surface tension as linear, the change in surface tension with temperature is given by . Problem 6. A non-viscous liquid of constant density 1000 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross-section of the tube at two points and at heights of 2 meter and 5 meter are respectively and . The velocity of the liquid at point is 1 . Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point and . Solution: As gravitational field is conservative. i.e., so, So work done by the force of gravity per unit volume …(i) Now in case of ideal fluid motion by conservation of mass, i.e., and or [as =constant (given)] so … (ii) Now as work done per unit volume by pressure, [as ] But by Bernoulli’s theorem, so … (iii) Using (i), (ii) and (iii) we get . Problem 7. A cylindrical tank 1 m in radius rests on a platform 5 m high. initially the tank is filled with water up to a height of 5 m. A plug whose area is is removed from an orifice on the side of the tank at the bottom. Calculate (a) initial speed with which the water flows from the orifice (b) initial speed with which the water strikes the ground (c) time taken to empty the tank to half its original value (d) Does the time to empty the tank depend upon the height of stand. Solution: (a) As speed of efflux is given by so here (b) As vertical speed with which water strikes the ground, So the initial speed with which water strikes the ground, (c) When the height of water level above the hole is , velocity of flow will be and so rate of flow or [as ] Which on integration gives So (d) No, as expression of is independent of height of stand. Problem 8. A rod of length 6 m has a mass 12 kg. It is hinged at one end at a distance of 3 m below water surface. (a) What weight must be attached to the other end of the rod so that 5 m of the rod are submerged ? (b) Find the magnitude and direction of the force exerted by the hinge on the rod. (Specific gravity of rod is 0.5). Solution: As shown in figure, the forces acting on the rod are : (1) The weight of rod 12 g N acting downwards through the CG of the rod, i.e., at a distance of 3 m from the hinge. (2) Force of buoyancy through the CG of displaced liquid vertically upwards. As Force of buoyancy = and acts at a distance 2.5 m from the hinge. (3) Extra weight at the other end of the rod at a distance 6 m from acting vertically downwards. (4) Reaction at the hinge at will be vertical (as here all the forces are vertical, so for horizontal equilibrium of the rod ) So for translatory equilibrium of rod, i.e., …(i) And for rotational equilibrium of rod (taking moments about ) or …(ii) Substituting the value of from Eqn. (ii) in (i) and solving for , we get Negative sign implies that is directed vertically downwards. Problem 9. A block of wood weighs 12 kg and has a relative density 0.6. It is to be in water with 0.9 of its volume immersed. What weight of a metal is needed (a) If the metal is on the top of wood, (b) If the metal is attached below the wood ? [RD of metal = 14]. Solution: (a) When the metal is on the top of wood, (b) When the metal is attached at the bottom of wood, or or or or, . Problem 10. A body of mass 3.14 kg is suspended from one end of a wire of length 10.0 m. The radius of the wire is changing uniformly from m at one end to m at the other end. Find the change in length of the wire. What will be the change in length if the ends are interchanged? Young’s modulus of the material of the wire is . Solution: Consider an element of length at a distance from the fixed end; then by definition of , change in the length of the element will be [as here ] But here , So total change in length of wire To integrate is let , so that the above equation becomes or [as ] … (i) So Further on interchanging a and b in Eqn. (i), , i.e., change in length remains same. Another Method : The situation is shown in figure. Let be the length of the wire. Suppose and be the radii of the upper and lower ends of the wire respectively. Now, the radius of the wire at a distance from the upper end is given by Stress at a distance is given by Corresponding strain Consider a very small part of the wire. Extension in this part is given by Total extension in the length of the wire ’ m. On interchanging the ends, the change in length remains the same. Problem 11. A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its end. One of the wires is made of steel and is of cross-section 0.1 cm2 and the other of brass of cross-section 0.2 cm2. Along the rod at which distance a weight be hung to produce (a) equal stresses in both the wires (b) equal strains in both the wires? Y for brass and steel are respectively. Solution: (a) As stresses are equal, …(i) Now for translatory equilibrium of the rod, which in the light of Eqn. (i) gives and …(ii) Now if is the distance of weight from steel wire, for rotational equilibrium of rod, or i.e., (b) As strains are equal, So So for translatory equilibrium of rod, =W in the light of Eqn. (iii) yields And for rotational equilibrium of rod Problem 12. Under isothermal condition two soap bubbles of radii a and b coalesce to form a single bubble of radius c. If the external pressure is show that surface tension, . Solution: As excess pressure for a soap bubble is and external pressure , so …(i) and …(ii) Now as mass is conserved, i.e., As temp. is constant, i.e., the above expression reduces to which in the light of Eqn. (i) and (ii) becomes i.e., i.e., . Problem 13. The fresh water behind a reservoir dam is 15 m deep. A horizontal pipe 4.0 cm in diameter passed through the dam 6.0 m below the water surface as shown in figure. A plug secures the pipe opening. (a) Find the friction force between the plug and pipe wall. (b) The plug is removed. What volume of water flows out of the pipe in 3.0 hour ? Solution: (a) As the plug secures the pipe opening, the force of friction between plug and pipe wall. But so i.e., (b) As the velocity of efflux, so assuming the level of water in the tank to be constant [(i.e., area = ) as it is not given] the volume coming out per second will be So the volume of the water flowing through the pipe in 3 hours . Problem 14. A cylindrical vessel of base area has a small hole of cross-section ‘ ’ punched near its base. At time , water is supplied into the vessel at a constant rate ‘ ’ . Find (a) The maximum water level in the vessel (b) The time ‘ ’ when water level becomes . Solution: (a) Water level will have maximum height when inflow rate = outflow rate and there will be no further change in level. or, or, (b) Let the water level be at time . Here is positive as increases with time and instantaneous efflux velocity . Rearranging the above equation . Integrating under the given limits, we get the required time, This gives the time as a function of . For any value , the corresponding time can be evaluated. Problem 15. A cylindrical vessel of (radius ) containing a liquid spins continuously with constant angular velocity as shown in the figure. Show that the pressure at a radial distance from the axis is , where = atmospheric pressure. Solution : Consider particle of the fluid at a point w.r.t. the coordinate axes as shown in the figure. The forces acting on this particle are (the centrifugal force) and the weight . The net force acting at should be perpendicular to the free surface, so that or, or, . This equation represents a parabola; for which the elevation from the origin at will be . Pressure .