Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 40

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Question 1
1 / -0

A small spherical drop fall from rest in viscous liquid. Due to friction, heat is produced. The correct relation between the rate of production of heat and the radius of the spherical drop at terminal velocity will be

A
$$\dfrac { dH }{ dt } \propto \dfrac { 1 }{ { r }^{ 5 } } $$

B
$$\dfrac { dH }{ dt } \propto { r }^{ 4 }$$

C
$$\dfrac { dH }{ dt } \propto \dfrac { 1 }{ { r }^{ 4 } } $$

D
$$\dfrac { dH }{ dt } \propto { r }^{ 5 }$$

Solution

Viscous force acting on spherical drop
$${ F }_{ v }=6\pi \eta v$$
$$\therefore $$ Terminal velocity, $$v=\dfrac { 2 }{ 9 } \dfrac { { r }^{ 2 }\left( \sigma -\rho \right) g }{ \eta }$$
where, $$ \eta =$$ coefficient of viscosity of liquid
$$\sigma =$$ density of material of spherical drop
$$\rho =$$ density of liquid
Power imparted by viscous force $$=$$ Rate of production of heat
$$\rho =\dfrac { dH }{ dt } ={ F }_{ v }\cdot v=6\pi \eta { v }^{ 2 }$$
$$=6\pi \eta \cdot { \left( \dfrac { 2 }{ 9 } \dfrac { { r }^{ 2 }\left( \sigma -\rho \right) g }{ \eta } \right) }^{ 2 }$$
$$\Rightarrow \dfrac { dH }{ dt } \propto { r }^{ 5 }$$
Question 2
1 / -0

When one end of the capillary is dipped in water, the height of water column is $$'h'$$. The upward force of $$105$$ dyne due to surface tension balanced by the force due to the weight of water column. The inner circumference of the capillary is
(Surface tension of water $$= 7\times 10^{-2}N/m)$$

A
$$1.5\ cm$$

B
$$2\ cm$$

C
$$2.5\ cm$$

D
$$3\ cm$$

Solution

Upward force acting $$F = 105 $$ dyne $$ = 105\times 10^{-5}$$ $$N$$
Surface tension of water $$T =7\times 10^{-2} N/m$$
Surface tension acts at the circumference $$C$$ of the capillary tube.
.$$\therefore$$ $$F = C T$$
Or $$105\times 10^{-5} = C\times 7\times 10^{-2}$$
Or $$C = 15\times 10^{-3} m$$
$$\implies$$ $$C = 15\times 10^{-1} cm = 1.5 cm$$
Question 3
1 / -0

A tank is filled with water of density 1 g $$cm^{-3}$$ and oil of density 0.9 g $$cm^{-3}$$. The height of water layer is 100 cm and of the oil layer is 400 cm. If g = 980 cm $$s^{-2}$$, then the velocity of efflux from an opening in the bottom of the tank is :

A
$$\sqrt{920\times 980}cms^{-1}$$

B
$$\sqrt{900\times 980}cms^{-1}$$

C
$$\sqrt{1000\times 980}cms^{-1}$$

D
$$\sqrt{92\times 980}cms^{-1}$$

Solution

The pressure at the bottom of the tank must be equal to the pressure due to water of height h.

If $$d_w$$ and $$d_o$$ be the densities of water and oil respectively, then the pressure at the bottom of the tank
$$= h_wd_wg + h_od_og$$
If this pressure is equivalent to pressure due to water of height h.

Then,
$$hd_wg=h_wd_wg+h_od_og$$
$$h=h_w+\dfrac{h_od_o}{d_w}=100+\dfrac{400\times 0.9}{1}$$
$$=100+360=460$$

According to Torricelli's theorem
$$v=\sqrt{2gh}=\sqrt{2\times 980\times 460}$$
$$=\sqrt{920\times 980}cms^{-1}$$
Question 4
1 / -0

The potential energy of a molecules on the surface of a liquid compared to the one inside the liquid is :

A
Zero

B
Lesser

C
Equal

D
Greater

Solution

The Surface tension acts on the molecules on the surface of a liquid.
while there is no such effect on the molecules in side the bulk of the liquid.

So potential energy of a molecules on the surface of a liquid compared to the one inside the liquid is greater.
Question 5
1 / -0

A sphere of radius R and density $$\rho_1$$ is dropped in a liquid of density $$\sigma$$. Its terminal velocity is $$v_1$$. If another sphere of radius R and density $$\rho_2$$ is dropped in the same liquid, its terminal velocity will be

A
$$\displaystyle \left( \frac{\rho_2 - \sigma}{\rho_1 - \sigma} \right) v_1$$

B
$$\displaystyle \left( \frac{\rho_1 - \sigma}{\rho_2 - \sigma} \right) v_1$$

C
$$\displaystyle \left( \frac{\rho_1}{\rho_2} \right) v_1$$

D
$$\displaystyle \left( \frac{\rho_2 }{\rho_1} \right) v_1$$

Solution

Here, $$\displaystyle 6 \pi \eta Rv_1 =\frac{4}{3} \pi R^3 (\rho_1 - \sigma)$$ ...(i)
and $$\displaystyle 6 \pi \eta Rv_2 =\frac{4}{3} \pi R^3 (\rho_2 - \sigma)$$ ....(ii)
Eq. (ii) dividing by Eq. (i) we get
$$\displaystyle \frac{v_2}{v_1} = \frac{(\rho_2 - \sigma)}{(\rho_1 - \sigma)}$$
$$\displaystyle u_2 = \left( \frac{P_2 - \sigma}{P_1 - \sigma} \right) v_1$$
Question 6
1 / -0

A capillary tube of radius $$r$$ is immersed in water and water rises in it to a height $$h$$. Mass of water in the capillary tube is $$m$$. If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be

A
$$m$$

B
$$2m$$

C
$$\cfrac { m }{ 2 } $$

D
$$4m$$

Solution

$$\textbf{Step 1 - Calculation of new height}$$
Let initial and final mass of water be $$m_{1}$$ and $$m_{2}$$
Let initial and final height be $$h_{1}$$ and $$h_{2}$$
Capillary height is given by : $$(h) = \dfrac {2T\cos \theta}{\rho rg}$$
Where $$T\rightarrow \text {Surface Tension}$$
$$\theta \rightarrow \text {contact angle}$$
$$\rho \rightarrow \text{density}$$
$$r \rightarrow \text{radius of tube}$$
Here, $$T , \theta, \rho, g$$ are constant
$$\therefore r_{1}h_{1} = r_{2}h_{2}$$
$$\Rightarrow h_{2} = \dfrac {r_{1}\cdot h_{2}}{r_{2}}$$
$$h_{2} = \dfrac {r\cdot h}{2r} = \dfrac {h}{2}$$

$$\textbf{Step 2 - Calculation of new mass of water}$$
Now, mass of water in capillary tube
$$\therefore \dfrac {m_{2}}{m_{1}} = \dfrac {V_{2}}{V_{1}}$$
$$\Rightarrow m_{2} = \dfrac {V_{2}}{V_{1}}\times m_{1}$$
$$\Rightarrow m_{2} = \dfrac {\pi \cdot (r_{2})^{2} \cdot h_{2}}{\pi \cdot (r_{1})^{2} \cdot h_{1}} \times m$$
$$\Rightarrow m_{2} = \dfrac {\pi \cdot 4r^{2} \cdot (h/2)}{\pi\cdot r^{2} \cdot h} \times m$$
$$\Rightarrow m_{2} = 2m$$

Therefore, mass of water will be $$2m$$.
Question 7
1 / -0

Two spherical rain drops with radii in the ratio $$1:2$$ fall from a great height through the atmosphere. The ratio of their momenta after they have attained terminal velocity is

A
1 : 8

B
2 : 1

C
1 : 32

D
1 : 2

E
1 : 16

Solution

Given, $$r_1:r_2=1:2$$
We know that,
$$v\propto r^3$$ and $$m\propto r^3$$
$$\dfrac{v_1}{v_2}=\begin{pmatrix}\dfrac{r_1}{r_2}\end{pmatrix}^3$$ and $$\dfrac{m_1}{m_2}=\begin{pmatrix}\dfrac{r_1}{r_2}\end{pmatrix}^3$$
We also know that momentum
$$p\propto mv$$
Hence, $$\dfrac{p_1}{p_2}=\dfrac{m_1v_1}{m_2v_2}=\begin{pmatrix}\dfrac{r_1}{r_2}\end{pmatrix}^5$$
$$\dfrac{p_1}{p_2}=\begin{pmatrix}\dfrac{1}{2}\end{pmatrix}^5$$
$$p_1:p_2=1:32$$
Question 8
1 / -0

Equal volume of two immiscible liquids of densities $$\rho$$ and $$2\rho$$ are filled in a vessel as shown in figure. Two small holes are made at depth $$\dfrac {h}{2}$$ and $$\dfrac {3h}{2}$$ from the surface of lighter liquid. If $$v_{1}$$ and $$v_{2}$$ are the velocities of efflux at these two holes, then $$\dfrac{v_{1}}{v_{2}}$$ will be

A
$$\dfrac {1}{\sqrt {2}}$$

B
$$\dfrac {1}{2\sqrt {2}}$$

C
$$\dfrac {1}{2}$$

D
$$\dfrac {1}{4}$$

Solution

We have $$v_{1} = \sqrt {2gh(h/2)} = \sqrt {gh} .... (i)$$
and by using Bernoulli's theorem
$$\rho gh + 2\rho g\left (\dfrac {h}{2}\right ) = \dfrac {1}{2}(2\rho) v_{2}^{2}$$
$$\Rightarrow v_{2} = \sqrt {2}gh .... (ii)$$
From Eqs. (i) and (ii)
$$\dfrac {v_{1}}{v_{2}} = \dfrac {1}{\sqrt {2}}$$.
Question 9
1 / -0

A large tank is filled with water to a height H. A small hole is made at the base of the tank. It takes 71 time to decrease the height of water to $$\frac {H}{\eta }(\eta > 1)$$ and $$T_2$$ time to take out the rest of 11 water. If $$T_1 = T_2$$ then the value of $$\eta$$ is

A
2

B
4

C
6

D
8

Solution

$$t=\dfrac{A}{a}\sqrt{\dfrac{2}{g}}[\sqrt{H_1}-\sqrt{H_2}]$$
$$T_1 =\dfrac{A}{a}\sqrt{\dfrac{2}{g}} \left [ \sqrt{H_1} \sqrt{\dfrac{H}{\eta }}\right ]$$
$$T_2 =\dfrac{A}{a}\sqrt{\dfrac{2}{g}} \left [ \sqrt{\dfrac{H}{\eta }-0}\right ]$$
Given, $$T_1=T_2$$
$$\sqrt{H}-\sqrt{\dfrac{H}{\eta }}=\sqrt{\dfrac{H}{\eta }}-0$$
$$\Rightarrow \sqrt{H}=2 \sqrt{\dfrac{H}{\eta }}$$
$$\Rightarrow \eta =4$$
Question 10
1 / -0

A cylindrical tank has a hole of $$1$$ $$cm^2$$ m bottom. If the water is allowed to flow into the tank from a tube above it at the rate of $$70cm^2/s$$, then the maximum height up to which water can rise in the tank is?

A
$$2.5$$cm

B
$$5$$cm

C
$$10$$cm

D
$$0.25$$cm

Solution

The height of water in the tank becomes maximum when the volume of water flows into the tank per second becomes equal to the volume flowing out per second.
Volume of water flowing out per second $$=A\sqrt{2gh}$$
Volume of water flowing in per second $$=70cm^3/s$$
$$\therefore A\sqrt{2gh}=70$$
$$1\sqrt{2\times 980\times h}=70$$
$$h=\displaystyle\frac{4900}{1960}=2.5$$cm.

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Mechanical Properties of Fluids Test - 40

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Mechanical Properties of Fluids Test - 40

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Question 1

$$\dfrac { dH }{ dt } \propto \dfrac { 1 }{ { r }^{ 5 } } $$

$$\dfrac { dH }{ dt } \propto { r }^{ 4 }$$

$$\dfrac { dH }{ dt } \propto \dfrac { 1 }{ { r }^{ 4 } } $$

$$\dfrac { dH }{ dt } \propto { r }^{ 5 }$$

Solution

Question 2

$$1.5\ cm$$

$$2\ cm$$

$$2.5\ cm$$

$$3\ cm$$

Solution

Question 3

$$\sqrt{920\times 980}cms^{-1}$$

$$\sqrt{900\times 980}cms^{-1}$$

$$\sqrt{1000\times 980}cms^{-1}$$

$$\sqrt{92\times 980}cms^{-1}$$

Solution

Question 4

Zero

Lesser

Equal

Greater

Solution

Question 5

$$\displaystyle \left( \frac{\rho_2 - \sigma}{\rho_1 - \sigma} \right) v_1$$

$$\displaystyle \left( \frac{\rho_1 - \sigma}{\rho_2 - \sigma} \right) v_1$$

$$\displaystyle \left( \frac{\rho_1}{\rho_2} \right) v_1$$

$$\displaystyle \left( \frac{\rho_2 }{\rho_1} \right) v_1$$

Solution

Question 6

$$m$$

$$2m$$

$$\cfrac { m }{ 2 } $$

$$4m$$

Solution

Question 7

1 : 8

2 : 1

1 : 32

1 : 2

1 : 16

Solution

Question 8

$$\dfrac {1}{\sqrt {2}}$$

$$\dfrac {1}{2\sqrt {2}}$$

$$\dfrac {1}{2}$$

$$\dfrac {1}{4}$$

Solution

Question 9

2

4

6

8

Solution

Question 10

$$2.5$$cm

$$5$$cm

$$10$$cm

$$0.25$$cm

Solution

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