Q: A barometer is constructed using a liquid (density = 760 kg/m^3). What would be the height of the liquid column, when a mercury barometer reads 76 cm? (Density of mercury = 13600 kg/m^3)

13.6 m. \rho_l = 760 kg/m^3, \rho_m = 13600 kg/m^3, h_m = 76 cm = 0.76 m. Equating pressures p_{liquid} = p_{mercury} \Rightarrow h_l\rho_l g = h_m\rho_m g \Rightarrow h_l = \dfrac{h_m\rho_m}{\rho_l} = \dfrac{0.76\times13600}{760} = 13.6 m.

Q: Two non-mixing liquids of densities \rho and n\rho (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to

\{1 + (n-1)p\}\rho. By Archimedes principle, weight of cylinder = (upthrust)_1 + (upthrust)_2, i.e. ALdg = (1-p)LA\rho g + (pLA)n\rho g \Rightarrow d = (1-p)\rho + pn\rho = \rho - p\rho + np\rho = \rho + (n-1)p\rho = \rho[1 + (n-1)p].

Q: The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4\times10^{-11} Pa^{-1} and density of water is 10^3 kg/m^3. What fractional compression of water will be obtained at the bottom of the ocean?

1.2\times10^{-2}. Pressure at the bottom of the ocean p = \rho g d = 10^3\times10\times2700 = 27\times10^6 Pa. Fractional compression = compressibility \times pressure = 45.4\times10^{-11}\times27\times10^6 = 1.2\times10^{-2}.

Q: The velocity of a small ball of mass M and density d when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be

\dfrac{Mg}{2}. At constant (terminal) velocity, viscous force = weight - buoyant force = Vd_1 g - Vd_2 g, with d_1 = d and d_2 = d/2. So viscous force = Vdg - V\dfrac{d}{2}g = \dfrac{Vdg}{2} = \dfrac{Mg}{2} (since M = d\times V).

Q: Two small spherical metal balls, having equal masses, are made from materials of densities \rho_1 and \rho_2 (\rho_1 = 8\rho_2) and have radii of 1 mm and 2 mm, respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals \eta and whose density is 0.1\rho_2. The ratio of their terminal velocities would be

\dfrac{79}{36}. Terminal velocity v_t = \dfrac{2(\rho-\sigma)r^2 g}{9\eta}, so \dfrac{v_{t_1}}{v_{t_2}} = \dfrac{(8\rho_2-\sigma)}{(\rho_2-\sigma)}\left(\dfrac{r_1}{r_2}\right)^2. With \sigma = 0.1\rho_2, r_1 = 1 mm and r_2 = 2 mm, = \dfrac{7.9\rho_2}{0.9\rho_2}\left(\dfrac{1}{2}\right)^2 = \dfrac{79}{36}.

Q: A small hole of area of cross-section 2 mm^2 is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s^2, the rate of flow of water through the open hole would be nearly

12.6\times10^{-6} m^3/s. Rate R = a\times v with v = \sqrt{2gh}, so R = a\sqrt{2gh}. With a = 2\times10^{-6} m^2 and h = 2 m, R = 2\times10^{-6}\times\sqrt{2\times10\times2} = 2\times10^{-6}\times6.32 = 12.64\times10^{-6} \approx 12.6\times10^{-6} m^3/s.

Q: A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

r^5. Rate of heat produced \dfrac{dQ}{dt} = F\times v_T = 6\pi\eta r v_T\times v_T = 6\pi\eta r v_T^2. Since v_T = \dfrac{2}{9}\dfrac{r^2(\rho-\sigma)}{\eta}g, v_T \propto r^2. Hence \dfrac{dQ}{dt} \propto r\cdot(r^2)^2 \propto r^5.

Q: A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m^2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (\rho_{air} = 1.2 kg/m^3)

2.4\times10^5 N, upwards. From Bernoulli's theorem, p_1 - p_2 = \dfrac{1}{2}\rho(v_2^2 - v_1^2) = \dfrac{1}{2}\times1.2(40^2 - 0) = 960 N/m^2. Force = (p_1 - p_2)A = 960\times250 = 24\times10^4 = 2.4\times10^5 N. As the pressure inside is greater than outside, the force acts upwards.

Question 1

A barometer is constructed using a liquid (density = 760 kg/m^3). What would be the height of the liquid column, when a mercury barometer reads 76 cm? (Density of mercury = 13600 kg/m^3)

Accepted Answer

13.6 m. ho_l = 760 kg/m^3, ho_m = 13600 kg/m^3, h_m = 76 cm = 0.76 m. Equating pressures p_{liquid} = p_{mercury} \Rightarrow h_lho_l g = h_mho_m g \Rightarrow h_l = \dfrac{h_mho_m}{ho_l} = \dfrac{0.76	imes13600}{760} = 13.6 m.

Question 2

In a U-tube as shown in a figure, water and oil are in the left side and right side of the tube respectively. The heights from the bottom for water and oil columns are 15 cm and 20 cm respectively. The density of the oil is [take \rho_{water} = 1000 kg/m^3]

Accepted Answer

750 kg/m^3. Pressure due to water column of height 15 cm = pressure due to oil column of height 20 cm. h_who_w g = h_0ho_0 g \Rightarrow 15ho_w = 20ho_0 \Rightarrow ho_0 = \dfrac{15}{20}ho_w = \dfrac{15}{20}	imes1000 = 750 kg/m^3.

Question 3

A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is

Accepted Answer

928 kg/m^3. Both ends are open, so pressures at the same horizontal level are equal, p_1 = p_2 \Rightarrow h_{oil}\cdot S_{oil}\cdot g = h_{water}\cdot S_{water}\cdot g. From the figure S_{oil} = \dfrac{(65+65)	imes1000}{(65+65+10)} = 928 kg/m^3.

Question 4

Two non-mixing liquids of densities \rho and n\rho (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to

Accepted Answer

\{1 + (n-1)p\}\rho. By Archimedes principle, weight of cylinder = (upthrust)_1 + (upthrust)_2, i.e. ALdg = (1-p)LA\rho g + (pLA)n\rho g \Rightarrow d = (1-p)\rho + pn\rho = \rho - p\rho + np\rho = \rho + (n-1)p\rho = \rho[1 + (n-1)p].

Question 5

The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4	imes10^{-11} Pa^{-1} and density of water is 10^3 kg/m^3. What fractional compression of water will be obtained at the bottom of the ocean?

Accepted Answer

1.2	imes10^{-2}. Pressure at the bottom of the ocean p = ho g d = 10^3	imes10	imes2700 = 27	imes10^6 Pa. Fractional compression = compressibility 	imes pressure = 45.4	imes10^{-11}	imes27	imes10^6 = 1.2	imes10^{-2}.

Question 6

The velocity of a small ball of mass M and density d when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be

Accepted Answer

\dfrac{Mg}{2}. At constant (terminal) velocity, viscous force = weight - buoyant force = Vd_1 g - Vd_2 g, with d_1 = d and d_2 = d/2. So viscous force = Vdg - V\dfrac{d}{2}g = \dfrac{Vdg}{2} = \dfrac{Mg}{2} (since M = d	imes V).

Question 7

Two small spherical metal balls, having equal masses, are made from materials of densities \rho_1 and \rho_2 (\rho_1 = 8\rho_2) and have radii of 1 mm and 2 mm, respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals \eta and whose density is 0.1\rho_2. The ratio of their terminal velocities would be

Accepted Answer

\dfrac{79}{36}. Terminal velocity v_t = \dfrac{2(\rho-\sigma)r^2 g}{9\eta}, so \dfrac{v_{t_1}}{v_{t_2}} = \dfrac{(8\rho_2-\sigma)}{(\rho_2-\sigma)}\left(\dfrac{r_1}{r_2}\right)^2. With \sigma = 0.1\rho_2, r_1 = 1 mm and r_2 = 2 mm, = \dfrac{7.9\rho_2}{0.9\rho_2}\left(\dfrac{1}{2}\right)^2 = \dfrac{79}{36}.

Question 8

A small hole of area of cross-section 2 mm^2 is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s^2, the rate of flow of water through the open hole would be nearly

Accepted Answer

12.6	imes10^{-6} m^3/s. Rate R = a	imes v with v = \sqrt{2gh}, so R = a\sqrt{2gh}. With a = 2	imes10^{-6} m^2 and h = 2 m, R = 2	imes10^{-6}	imes\sqrt{2	imes10	imes2} = 2	imes10^{-6}	imes6.32 = 12.64	imes10^{-6} \approx 12.6	imes10^{-6} m^3/s.

Question 9

A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

Accepted Answer

r^5. Rate of heat produced \dfrac{dQ}{dt} = F	imes v_T = 6\pi\eta r v_T	imes v_T = 6\pi\eta r v_T^2. Since v_T = \dfrac{2}{9}\dfrac{r^2(ho-\sigma)}{\eta}g, v_T \propto r^2. Hence \dfrac{dQ}{dt} \propto r\cdot(r^2)^2 \propto r^5.

Question 10

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m^2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (\rho_{air} = 1.2 kg/m^3)

Accepted Answer

2.4	imes10^5 N, upwards. From Bernoulli's theorem, p_1 - p_2 = \dfrac{1}{2}ho(v_2^2 - v_1^2) = \dfrac{1}{2}	imes1.2(40^2 - 0) = 960 N/m^2. Force = (p_1 - p_2)A = 960	imes250 = 24	imes10^4 = 2.4	imes10^5 N. As the pressure inside is greater than outside, the force acts upwards.

Question 11

The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is v, the speed of the ejection of the liquid through the holes is

Accepted Answer

\dfrac{vR^2}{nr^2}. By the equation of continuity, Av= constant, so the volume flow rate in the tube equals the total volume flow rate out of the holes: \pi R^2 v = n\pi r^2 v'. \Rightarrow v' = \dfrac{R^2 v}{nr^2}. Hence the speed of ejection through the holes is \dfrac{vR^2}{nr^2}.

Question 12

A liquid does not wet the solid surface if angle of contact is

Accepted Answer

greater than 90^\circ. A liquid does not wet a solid surface when the angle of contact is obtuse, i.e. 	heta > 90^\circ.

Question 13

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary tube is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is

Accepted Answer

2.5 g. Height of water in a capillary tube, h=\dfrac{2S\cos	heta}{ho g r}\Rightarrow h\propto\dfrac{1}{r}, so \dfrac{h_1}{h_2}=\dfrac{r_2}{r_1}=\dfrac{2r}{r}=2. As mass m=A\cdot h\cdotho, \dfrac{m_2}{m_1}=\dfrac{h_2}{h_1}=\dfrac{1}{2}. Therefore m_2=\dfrac{m_1}{2}=\dfrac{5}{2}=2.5 g.

Question 14

A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5	imes10^{-2} N/m. The pressure inside the bubble equals at a point Z_0 below the free surface of water in a container. Taking g=10 m/s^2, density of water =10^3 kg/m^3, the value of Z_0 is

Accepted Answer

1 cm. Excess pressure inside a soap bubble is p=\dfrac{4T}{r}, so the total pressure inside is p_1=p_0+\dfrac{4T}{r}. The pressure at depth Z_0 is p_2=p_0+Z_0ho g. Equating gives Z_0=\dfrac{4T}{rho g}=\dfrac{4	imes2.5	imes10^{-2}}{1	imes10^{-3}	imes10^3	imes10}=10	imes10^{-3} m =1 cm.

Question 15

A rectangular film of liquid is extended from (4 cm 	imes 2 cm) to (5 cm 	imes 4 cm). If the work done is 3	imes10^{-4} J, the value of the surface tension of the liquid is

Accepted Answer

0.125 Nm^{-1}. Increase in surface energy = increase in area 	imes surface tension. As the film has two surfaces, \Delta A=(5	imes4-4	imes2)	imes2=24 cm^2=24	imes10^{-4} m^2. From W=T\cdot\Delta A: 3	imes10^{-4}=T	imes24	imes10^{-4}\Rightarrow T=\dfrac{1}{8}=0.125 N/m.

Question 16

Three liquids of densities ho_1,ho_2 and ho_3 (with ho_1>ho_2>ho_3), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact 	heta_1,	heta_2 and 	heta_3 obey

Accepted Answer

0\leq heta_1< heta_2< heta_3<\dfrac{\pi}{2}. By the ascent formula for a capillary tube, h=\dfrac{2T\cos heta}{ ho g r}. Since h, T and r are the same, \dfrac{\cos heta_1}{ ho_1}=\dfrac{\cos heta_2}{ ho_2}=\dfrac{\cos heta_3}{ ho_3}, so \cos heta\propto ho. As ho_1> ho_2> ho_3, \cos heta_1>\cos heta_2>\cos heta_3, giving 0\leq heta_1< heta_2< heta_3<\dfrac{\pi}{2}.

Question 17

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

Accepted Answer

energy =3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right) is released. Energy released =(A_f-A_i)T, where the single big drop has area A_i=4\pi R^2=\dfrac{3V}{R} and the small drops together have area A_f=n\cdot4\pi r^2=\dfrac{3V}{r} (using V=\dfrac{4}{3}\pi R^3). Hence energy released =3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right).

Question 18

The wettability of a surface by a liquid depends primarily on

Accepted Answer

angle of contact between the surface and the liquid. The wettability of a surface by a liquid depends primarily on the angle of contact between the surface and the liquid.

Mechanical Properties of Fluids — NEET Physics PYQ MCQs with Solutions

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