Q: A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 ~N ~m ^ -1 , then the excess force required to take it away from the surface

aligned & Excess force =T 2 R \\ & = 7 100 2 3.14 4.5 100 \\ & =197.82 10^ -4 \\ & =19.8 10^ -3 ~N \\ & =19.8 mN aligned

Question 1

In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity (v) with t

Accepted Answer

The terminal velocity v of a spherical ball of radius r and density falling through a liquid of density and viscosity is given by Stokes' law: v = 2r^2g 9 ( - ) Rearranging the terms to express v as a function of the ratio : v = 2r^2g 9 ( - 1 ) Let y = v and x = . The equation becomes: y = k(x - 1) where k = 2r^2g 9 is a positive constant. This represents the equation of a straight line with a positive slope k and an x-intercept at x = 1 (i.e., v = 0 when = 1). Looking at the given graphs, only

Question 2

A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface ? (Consider the density of water = 1000 kg m^ -3 , 1 atm = 1 10^ 5 Pa and gravitational

Accepted Answer

The absolute pressure at a depth h below the water surface is given by P = P_0 + g h, where P_0 is the atmospheric pressure. Given: P = 100 atm = 100 10^5 Pa P_0 = 1 atm = 1 10^5 Pa = 1000 kg m^ -3 g = 10 m/s^2 Substituting the values into the equation: 100 10^5 = 1 10^5 + 1000 10 h 99 10^5 = 10^4 h h = 99 10^5 10^4 h = 990 m Answer: 990 m

Question 3

Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between P and Q is 15 Nm^ -2 . The area of cross-section

Accepted Answer

From the equation of continuity, the rate of flow is constant: A_P v_P = A_Q v_Q Given A_P = 40 cm ^2 and A_Q = 20 cm ^2, we have: 40 v_P = 20 v_Q v_Q = 2 v_P Applying Bernoulli's equation for a horizontal pipe: P_P + 1 2 v_P^2 = P_Q + 1 2 v_Q^2 P_P - P_Q = 1 2 (v_Q^2 - v_P^2) Substitute the given values (P_P - P_Q = 15 N m ^ -2 , = 1000 kg m ^ -3 ) and v_Q = 2 v_P: 15 = 1 2 1000 ((2 v_P)^2 - v_P^2) 15 = 500 (4 v_P^2 - v_P^2) 15 = 500 3 v_P^2 15 = 1500 v_P^2 v_P^2 = 1 100 v_P = 0.1 m s ^ -1 = 10

Question 4

Consider a water tank shown in the figure. It has one wall at x=L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density , the liquid surface m

Accepted Answer

For the given element, (consider length d in Z direction) Net force in upward direction = Weight (S ( +d )-S ) d=m g Angle is small d y d x = g S ...(1) aligned & = d y d x Differentiating wrt x \ & ^2 d d x = d^2 y d x^2 ...(2) aligned Put d from (2) in (1) d take 1, we get d^2 y d^2 = g y S Alternative Solution : aligned & P_A=P_B=P_0 \ & P_C=P_0- g y \ & P_C=P_0- S R \ & g y= S R \ & g y=S d^2 y d x^2 \ & d^2 y d^2 = g y S aligned R = \ 1+ ( dy dx )^2 \ ^ 3 / 2 ~d ^2 y dx ^2 dy / dx is

Question 5

A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 ~N ~m ^ -1 , then the excess force required to take it away from the surface

Accepted Answer

aligned & Excess force =T 2 R \ & = 7 100 2 3.14 4.5 100 \ & =197.82 10^ -4 \ & =19.8 10^ -3 ~N \ & =19.8 mN aligned

Question 6

An ideal fluid is flowing in a non-uniform cross-sectional tube X Y (as shown in the figure) from end X to end Y. If K_1 and K_2 are the kinetic energy per unit volume of the fluid at X and Y respecti

Accepted Answer

According to Bernoulli's principle, Kinetic energy per unit volume + Potential energy per unit volume + Pressure = Constant 1 2 V^2+ g h+P= constant Apply Bernoulli's principle at point X and Y, P+K_1+ g(0)=P+K_2+ g( ~h ) K_1=K_2+ g h K_1 _2

Mechanical Properties of Fluids — Previous Year Questions & Solutions