Elasticity and Fluid Machanics -10-SOLUTION-II

Problem 16. A container of large uniform cross-section area resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities and 2 each of height ( ) as shown in figure. The lower density liquid is open to the atmosphere having pressure . (a) A homogeneous solid cylinder of length , cross-sectional area is immersed such that it floats with its axis vertical at the liquid-liquid interface with length in the denser liquid. Determine (i) The density of solid and (ii) The total pressure at the bottom of the container. (b) The cylinder is removed and original arrangement is restored. A tiny hole of area is punched on the vertical side of the container at a height . Determine (i) the initial speed of efflux of the liquid at the hole (ii) the horizontal distance traveled by the liquid initially and (iii) the height at which the hole should be punched so that the liquid travels the maximum distance initially. Also calculate . Solution: (a) (i) As for floating, or i.e., (ii) Total pressure = + (weight of liquid + weight of solid) / i.e., i.e., (b) (i) By Bernoulli’s theorem for a point just inside and outside the hole i.e., or or (ii) As at the hole vertical velocity of liquid is zero so time taken by it to reach the ground, so that (iii) For to be maximum must be maximum , i.e., or or and . Problem 17. A wooden stick of length , radius and density has a small metal piece of mass (of negligible volume) attached to its one end. Find the minimum value for the mass (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density . Solution: For the stick to be vertical for rotational equilibrium, center of gravity should be below in a vertical line through the center of buoyancy. For minimum , the two will coincide. Let be the length of immersed portion. For translatory equilibrium, Wt. of rod + mass attached = force of buoyancy …(i) where . The height of center of mass from bottom For rotatory equilibrium and for minimum , this should be equal to . …(ii) Substituting for in Equn. (i), we get . 12. When the stretching force is increased upto the proportional limit. Hooke’s law is valid. Beyond the proportional limit and upto the elastic limit, Hooke’s law is not valid but the elastic property is maintained. Beyond the elastic limit the wire does not regain the original length and remains slightly elongated. Ans. (c) 13. The force developed at the mid-point of the rod perpendicular to the cross-section (area = A) Stress Ans. (a) 14. If be the natural length, and Simplifying, we get . Ans. (c) 15. Viscous force develop when different layers of the fluid move with different velocities. No such motion exists within the oil drop. Hence the motion in the viscous medium depends only on . Ans. (b) 13. If be the angle of contact between the liquid and the glass, the capillary radius and meniscus radius are expressed as and . Ans. (d) 14. The pressure inside the water column at the uppermost & lowermost points are and , where is the atmospheric pressure. or, . Ans. (a) 15. Linear mass density Tension at section is weight of part Elongation of an element dy at Total elongation . Ans. (b) 9. The fluid pressure at a distance below the hinge is If be the width of the barrier, the force acting on a thin area is The torque of about is Total deflecting torque due to fluid pressure is . The restoring torque produced by the spring force is . Hence, , N/m.