Q: A spherical soap bubble inside an air chamber at pressure P_0=10^5 ~Pa has a certain radius so that the excess pressure inside the bubble is P=144 ~Pa . Now, the chamber pressure is reduced to 8 P_0 /

Case-1 aligned & P - P _0= P = 4 ~T R \\ & P = ( P _0+ 4 ~T R ) aligned Case-2 aligned & P _1- 8 P _0 27 = P _1= 4 ~T R _1 \\ & P _1= 4 ~T R _1 + 8 P _0 27 aligned Constant temperature process aligned & PV = P _1 ~V _1 \\ & ( P _0+ 4 ~T R ) 4 3 R ^3= ( 4 ~T R _1 + 8 P _0 27 ) 4 3 R _1^3 ; ( 4 ~T R ), ( 4 ~T R _1 ) (Neglected) \\ & R = 2 3 R _1 R _1= 3 2 R \\ & P _1= 4 ~T R _1 = 4 ~T 3 R 2= 2 3 (144)=96 ~Pa aligned

Q: Two large, identical water tanks, 1 and 2, kept on the top of a building of height H, are filled with water up to height h in each tank. Both the tanks contain an identical hole of small radius on the

aligned & Av = av _1 \\ & ~A (- dy dt )= a 2 gy ; dt = A a 2 ~g -d y y \\ & _0^ t_1 dt = A a 2 ~g _ h ^0- d y y \\ & t_1= A a 2 ~g 2 h ; t _1= A a 2 h g aligned A v^ = av _2 aligned & A (- d y d t )=a 2 g(H+y) \\ & d t=- A a 2 g d y H+y \\ & _0^ t_2 d t=- A a 2 g _H^0 d y H+y \\ & t_2= A a 2 g (2)( H+h - H ) \& H= 16 h 9 \\ & = A a 2 h g ( 5 3 - 4 3 ) \\ & t_2= A a 2 h g ( 1 3 ) aligned ratio t _1 t _2 =3

Q: A cylindrical furnace has height H and diameter D both 1 m . It is maintained at temperature 360 K . The air gets heated inside the furnace at constant pressure P a and

Applying Bernoulli’s theorem between top and bottom points of furnace : P a + 1 2 ρ a V 2 = P a + ρ g H + 1 2 ρ V 2 . . . 1 Also, since P M = ρ R T ⇒ ρ a × 300 = ρ × 360 ⇒ ρ = 1 kg m - 3 . . .

Q: A cylindrical furnace has height H and diameter D both 1 m . It is maintained at temperature 360 K . The air gets heated inside the furnace at constant pressure P a and

Applying Bernoulli’s theorem between top and bottom points of furnace : P a + 1 2 ρ a V 2 = P a + ρ g H + 1 2 ρ V 2 . . . 1 Also, since P M = ρ R T ⇒ ρ a × 300 = ρ × 360 ⇒ ρ = 1 kg m - 3 . . .

Q: A cylindrical tube, with its base as shown in the figure., is filled with water. It is moving down with a constant acceleration a along a fixed inclined plane with angle θ = 45

Given, β = P 1 − P 2 ρ g d . . . 1 Here we can write now, P 3 = P 2 + ρ g − a 2 d P 1 = P 3 − ρ a 2 d P 1 = P 2 + ρ g d − 2 ρ a d P 1 = P 2 + ρ g − a 2 d − ρ a 2 d P 1 − P 2 ρ d g = 1 − 2 a g Compare it with given equation 1 , β = 1 − 2 a g Now check the options, If a = g 2 , β = 0 . If a = g 2 , β = 2 - 1

Question 1

Passage: A container of height 2 m, length 2 m and breadth 1 m is made of insulating vertical walls and two large area horizontal metal plates (M_1 and M_2) which extend far beyond the vertical walls

Accepted Answer

Let A be the cross-sectional area of each chamber and a be the cross-sectional area of the hole. From the given dimensions, the area of each chamber is A = 1 m 1 m = 1 m ^2. The area of the hole is a = 10 cm ^2 = 10 10^ -4 m ^2. Let h_1 and h_2 be the heights of the liquid in the left and right chambers at time t, respectively. By conservation of volume, A h_1 + A h_2 = A H, where H = 2 m is the initial height of the liquid in the left chamber. h_1 + h_2 = 2 The velocity of efflux through the ho

Question 2

A table tennis ball has radius (3 / 2) 10^ -2 ~m and mass (22 / 7) 10^ -3 ~kg . It is slowly pushed down into a swimming pool to a depth of d=0.7 ~m below the water surface and then released from rest

Accepted Answer

Work done in pushing the ball W=(v g) d-(v g) d Where, Density of water Density of ball aligned & W= 4 3 R^3 10 0.7 [1000- 3 4 10^ -3 R^3 ] \ & W=0.077 ~J aligned [1 is correct] When ball is released at bottom same work (i.e. 0.077 ~J ) is done on ball. aligned & 1 2 m v^2=0.077 \ & v= 0.077 2 22 7 10^ -3 \ & =7 ~m / s aligned [2 is correct] also, H= v^2 2 g = 7 7 2 10 =2.45 ~m [3 is incorrect] Net force F_ net =v g-v g=0.11 ~N Also, viscous force is maximum when v=7 ~m / s aligned & (F_v )_

Question 3

A spherical soap bubble inside an air chamber at pressure P_0=10^5 ~Pa has a certain radius so that the excess pressure inside the bubble is P=144 ~Pa . Now, the chamber pressure is reduced to 8 P_0 /

Accepted Answer

Case-1 aligned & P - P _0= P = 4 ~T R \ & P = ( P _0+ 4 ~T R ) aligned Case-2 aligned & P _1- 8 P _0 27 = P _1= 4 ~T R _1 \ & P _1= 4 ~T R _1 + 8 P _0 27 aligned Constant temperature process aligned & PV = P _1 ~V _1 \ & ( P _0+ 4 ~T R ) 4 3 R ^3= ( 4 ~T R _1 + 8 P _0 27 ) 4 3 R _1^3 ; ( 4 ~T R ), ( 4 ~T R _1 ) (Neglected) \ & R = 2 3 R _1 R _1= 3 2 R \ & P _1= 4 ~T R _1 = 4 ~T 3 R 2= 2 3 (144)=96 ~Pa aligned

Question 4

Two large, identical water tanks, 1 and 2, kept on the top of a building of height H, are filled with water up to height h in each tank. Both the tanks contain an identical hole of small radius on the

Accepted Answer

aligned & Av = av _1 \ & ~A (- dy dt )= a 2 gy ; dt = A a 2 ~g -d y y \ & _0^ t_1 dt = A a 2 ~g _ h ^0- d y y \ & t_1= A a 2 ~g 2 h ; t _1= A a 2 h g aligned A v^ = av _2 aligned & A (- d y d t )=a 2 g(H+y) \ & d t=- A a 2 g d y H+y \ & _0^ t_2 d t=- A a 2 g _H^0 d y H+y \ & t_2= A a 2 g (2)( H+h - H ) \& H= 16 h 9 \ & = A a 2 h g ( 5 3 - 4 3 ) \ & t_2= A a 2 h g ( 1 3 ) aligned ratio t _1 t _2 =3

Question 5

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0 . 1 &#160; mm is dipped vertically into the liquid through the airtight pisto

Accepted Answer

Let P f  be the air pressure. As the process is isothermal, we can write P 0 V 0 = P f V f &#8658; P 0 V 0 = P f 100 101 V 0 &#8658; P f = 101 &#215; 10 3 &#160; Pa &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#8757; &#160; P 0 = 10 5 &#160; N &#160; m - 2 Now, consider the 4 points shown in diagram P d - P c = 2 T R &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#8757; &#160; P d = P 0 &#8756; &#160; P c = P 0 - 2 T R Now, P a = P b  (

Question 6

A cylindrical furnace has height H and diameter D both 1 &#160; m . It is maintained at temperature 360 &#160; K . The air gets heated inside the furnace at constant pressure  P a  and

Accepted Answer

Applying Bernoulli&rsquo;s theorem between top and bottom points of furnace : P a + 1 2 &#961; a V 2 = P a + &#961; g H + 1 2 &#961; V 2 &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . &#160; 1 Also, since  P M = &#961; R T &#8658; &#160; &#961; a &#215; 300 = &#961; &#215; 360 &#8658; &#160; &#961; = 1 &#160; kg &#160; m - 3 &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . .

Question 7

A cylindrical furnace has height H and diameter D both 1 &#160; m . It is maintained at temperature 360 &#160; K . The air gets heated inside the furnace at constant pressure  P a  and

Accepted Answer

Applying Bernoulli&rsquo;s theorem between top and bottom points of furnace : P a + 1 2 &#961; a V 2 = P a + &#961; g H + 1 2 &#961; V 2 &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . &#160; 1 Also, since  P M = &#961; R T &#8658; &#160; &#961; a &#215; 300 = &#961; &#215; 360 &#8658; &#160; &#961; = 1 &#160; kg &#160; m - 3 &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . .

Question 8

A soft plastic bottle, filled with water of density 1 gm / cc , carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm , and it

Accepted Answer

When all the forces are balanced on tube, it will start sinking. F u p w a r d = F d o w n w a r d &#961; w V tube &#160; + V air &#160; g = m tube &#160; g 2 &#160; cm 3 + V air &#160; g = ( 5 &#160; gm ) g V air = 3 &#160; cm , while the initial volume of the air was  3 . 3 &#160; cm 3  so the decrease in the volume  &#916; V = 3 . 3 - 3 = 0 . 3 &#160; cm 3

Question 9

A soft plastic bottle, filled with water of density 1 gm / cc , carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm , and it

Accepted Answer

When all the forces on tube are balanced then the tube will start sinking. F u p w a r d = F d o w n w a r d &#961; w V tube &#160; + V air &#160; g = m tube &#160; g 2 &#160; cm 3 + V air &#160; g = ( 5 &#160; gm ) g V air = 3 &#160; cm , while the initial volume of the air was  3 . 3 &#160; cm 3  so the decrease in the volume  &#916; V = 3 . 3 - 3 = 0 . 3 &#160; cm 3 V tube &#160; = m tube &#160; &#961; tube &#160; = 5 &#160; gm 2 . 5 &#160; gm &#160; cm - 3 V tube &#160; =

Question 10

A cylindrical tube, with its base as shown in the figure., is filled with water. It is moving down with a constant acceleration  a  along a fixed inclined plane with angle  &#952; = 45

Accepted Answer

Given, &#946; = P 1 &#8722; P 2 &#961; g d &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . 1 Here we can write now, P 3 = P 2 + &#961; g &#8722; a 2 d P 1 = P 3 &#8722; &#961; a 2 d P 1 = P 2 + &#961; g d &#8722; 2 &#961; a d P 1 = P 2 + &#961; g &#8722; a 2 d &#8722; &#961; a 2 d P 1 &#8722; P 2 &#961; d g = 1 &#8722; 2 a g Compare it with given equation 1 , &#946; = 1 &#8722; 2 a g Now check the options, If  a = g 2 ,  &#946; = 0 . If  a = g 2 ,  &#946; = 2 - 1

Question 11

A train with cross-sectional area  S t is moving with speed  v t inside a long tunnel of cross-sectional area  S 0 S 0 = 4 S t . Assume that almost all the air (density &#961; ) in fron

Accepted Answer

4 S t v t = 3 S t v ⇒ v = 4 3 v t Apply Bernoulli's theorem, P 0 + 1 2 ρ v t 2 = P + 1 2 ρ 4 3 v t 2 P 0 - P = 7 18 ρ v t 2 ⇒ N = 9

Question 12

A hot air balloon is carrying some passengers, and a few sandbags of mass  1 &#160; kg each so that its total mass is  480 &#160; kg . Its effective volume giving the balloon its buoyancy is

Accepted Answer

Initially m g = F b 480 &#215; g = V &#961; 0 e - 100 6000 g       ....(1) Finally ( 480 - m ) g = V &#961; 0 e - 150 6000 g      .....(2) dividing 480 - m 480 = e - 1 40 + 1 60 m = 480 1 - e - 1 / 120 Hence e - 1 / 120 &#8776; 1 - 1 120 1 - e - 1 / 110 = 1 120 m = 4 &#160; kg Hence, four sandbags are thrown.

Question 13

An open-ended U-tube of uniform cross-sectional area contains water (density 10 3 kg m - 3 ). Initially the water level stands at 0 . 29 m from the bottom in each arm. Kerosene oil (a water-immiscible

Accepted Answer

p 1 = p 2 ⇒ p 0 + 800 g × 0 . 1 + 1000 g ( 0 . 29 - x ) = 1000 g ( 0 . 29 + x ) + p 0 ⇒ 0 . 8 + 2 . 9 - 10 x = 2 . 9 + 10 x ⇒ x = 0 . 8 20 = 0 . 04 h 2 = 0 . 1 + 0 . 29 - x = 0 . 35 h 1 = 0 . 29 + x = 0 . 33 h 2 h 1 = 35 33 = 1 . 06

Question 14

As shown schematically in the figure, two vessels contain water solutions (at temperature T ) of potassium permanganate KMnO 4 of different concentrations n 1 and n 2 n 1 > n 2 molecules per unit volu

Accepted Answer

(a) F = Δ P . A = P 1 - P 2 A = n 1 × RTA - n 2 RT . A F = Δ n R T A (b) β v × n 1 × A × ℓ = Δ n k B T A β v n 1 ℓ = Δ n k B T (c) dN dt = S v n 1 = s n 1 × Δ n k B T β n 1 ℓ = S × Δ n k B T S β ℓ (d) Δ n → decreasing then d N d t → decreasing

Question 15

When water is filled carefully in a glass, one can fill it to a height h above the rim of the glass due to the surface tension of water. To calculate h just before water starts flowing, model the shap

Accepted Answer

Pressure at the bottom of disc $=$ pressure due to surface tension $ aligned & g h=T ( 1 R _1 + 1 R _2 ) \ & R _1 > R _2 \ & So 1 R _1 < 1 R _2 and R _2= h / 2 \ & gh = T [ 1 R _1 + 1 R _2 ]= T [0+ 1 ~h / 2 ] \ & h ^2= 2 ~T g \ & h = 2 ~T g = 2 0.07 10^3 10 = 14 100 10^4 100 \ & ~h = 14 ~mm =3.741 aligned $

Mechanical Properties of Fluids — Previous Year Questions & Solutions