Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 30

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Question 1
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A leakage begins in water tank at position $$P$$ as shown in the figure. The initial gauge pressure pressure above that of the atmosphere) at $$P$$ was $$ 5 \times 10^{5} N/m^{2}$$ .If the density of water is $$ 1000 kg/m^{3}$$ the initial velocity with which water gushes out is:

A
$$3.2 ms^{-1}$$

B
$$32 ms^{-1}$$

C
$$28 ms^{-1}$$

D
$$2.8 ms^{-1}$$

Solution

$$\displaystyle \Delta p=\dfrac{1}{2}\rho v^{2}$$
$$\therefore $$ $$\displaystyle v=\sqrt{\dfrac{2\Delta P}{\rho }}=\sqrt{\dfrac{2\times 5\times 10^{5}}{1000}}=31.5 m/s$$
Question 2
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Directions For Questions

The spouting can is something used to demonstrate the variation of pressure with depth. When the corks are removed from the tubes in the side of the can, water flows out with a speed that depends on the depth. In a certain can, three tubes $$T_{1},T_{2}$$ and $$T_{3}$$ are set at equal distances a above the base of the can. When water contained in this can is allowed to come out of the tubes the distances on the horizontal surface are measured as $$x_{1},x_{2}$$ and $$x_{3}$$

...view full instructions

Distance $$x_{3}$$ is given by

A
$$\sqrt{3} a$$

B
$$\sqrt{2} a$$

C
$$\displaystyle \frac{1}{2}\sqrt{3} a$$

D
$$\displaystyle 2\sqrt{3} a$$

Solution

$$x_{3}=2\sqrt{h_{Top}\times h_{bottom}}$$
$$=2\sqrt{3a\times a}$$
$$=2a\sqrt{3}$$
Question 3
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Directions For Questions

The spouting can is something used to demonstrate the variation of pressure with depth. When the corks are removed from the tubes in the side of the can, water flows out with a speed that depends on the depth. In a certain can, three tubes $$T_{1},T_{2}$$ and $$T_{3}$$ are set at equal distances a above the base of the can. When water contained in this can is allowed to come out of the tubes the distances on the horizontal surface are measured as $$x_{1},x_{2}$$ and $$x_{3}$$

...view full instructions

The correct sketch is

A

B

C

D
None of these

Solution

$$x_{1}=2\sqrt{a\times 3a}=2\sqrt{3a}$$
$$x_{2}=2\sqrt{2a\times 2a}=2\sqrt{4a}$$
$$x_{1}=2\sqrt{3a}$$
$$x_{1}=x_{3}< x_{2}$$
Question 4
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Two holes $$1$$ and $$2$$ are made at depths $$h$$ and $$16 h$$ respectively. Both the holes are circular but radius of hole-$$1 $$ is two times

A
Initially equal volumes of liquid will flow from both the holes in unit time

B
Initially more volume of liquid will flow from hole-2 per unit time

C
After some time more volume of liquid will flow from hole-1 per unit time

D
After some time more volume of liquid will flow from hole-2 per unit time

Solution

Initially
$$\displaystyle \frac{dV_{1}}{dt}=v_{1}a_{1}=\left (

\sqrt{2gh} \right )\left ( \pi \right )\left ( 2R \right

)^{2}$$ (i)
$$\displaystyle

\frac{dV_{2}}{dt}=v_{2}a_{2}=\left ( \sqrt{2g\left ( 16h \right )}

\right )\left ( \pi \right )\left ( R \right )^{2}$$ (ii)
From Eqs. (i) and (ii) we can see that,
$$\displaystyle \frac{dV_{1}}{dt}=\frac{dV_{2}}{dt}$$
After some time $$v_{1}$$ and $$v_{2}$$ both will decrease, but decrease in the value of $$v_{1}$$ is more dominating. So,
$$a_{1}v_{1}$$ or $$\displaystyle \frac{dV_{1}}{dt}< a_{2}v_{2}$$ or $$\displaystyle \frac{dV_{2}}{dt}$$
Question 5
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A viscous liquid flows through a horizontal pipe of varying cross-sectional area, Identify the option which correctly represent,, the variation of heighl of rise of liquid in each vertical tube:

A

B

C

D
None of these

Solution

Answer is C
As the cross section of pipe decreases the velocity of flow increases, by continuity equation.
As per Bernoulli's theorem, when the velocity increases, the pressure will decrease since the mean height of flow is same. Hence the level of liquid will keep on decreasing as the velocity increases. Along the velocity direction, height of liquid will decrease.
Question 6
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A steel ball of mass $$m$$ falls in a viscous liquid with terminal velocity $$v$$, then the steel ball of mass $$8\ m$$ will fall in the same liquid with terminal velocity:

A
$$v$$

B
$$4\ v$$

C
$$8\ v$$

D
$$16\sqrt{2}\ v$$

Solution

The terminal velocity $$V$$ of the body of radius $$r$$, density $$\rho$$ falling through a medium of density $$\rho_o$$ is given by
$$V=\displaystyle\frac{2r^2(\rho-\rho_o)g}{9 \eta}$$ where $$\eta$$ is the coefficient of viscosity of medium.
clearly $$V \propto r^2 $$
$$\Rightarrow V \propto (volume)^{2/3}$$
$$\Rightarrow V \propto (m)^{2/3}$$
$$\Rightarrow \dfrac{V_2}{V_1} =\left (\frac{m_2}{m_1} \right)^{2/3} $$
$$\Rightarrow \dfrac{V_2}{V} =\left (\frac{8m}{m} \right)^{2/3} $$
$$\Rightarrow V_2=4V$$
Question 7
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Uniform speed of $$2 cm$$ diameter ball is $$20 cm/s$$ in a viscous liquid. Then, the speed of $$1 cm$$ diameter ball in the same liquid is:

A
$$5cms^{-1}$$

B
$$10cms^{-1}$$

C
$$40cms^{-1}$$

D
$$80cms^{-1}$$

Solution

The terminal velocity $$V$$ of the body of radius $$r$$, density $$\rho$$ falling through a medium of density $$\rho_o$$ is given by
$$V=\displaystyle\dfrac{2r^2(\rho-\rho_o)g}{9 \eta}$$ where $$\eta$$ is the coefficient of viscosity of medium.
clearly $$V \propto r^2 $$, So we have
$$\dfrac{V_2}{V_1}=\left(\dfrac{r_2}{r_1}\right)^2$$
$${V_2}=\left(\dfrac{1}{2}\right)^2 V_1=\dfrac{20}{4}m/s=5m/s$$
Question 8
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There is a small hole at the bottom of tank filled with water. If total pressure at the bottom is 3 atm (1 atm $$= 10^{5} Nm^{-2})$$, then velocity of water flowing from hole is

A
$$\sqrt{400}ms^{-1}$$

B
$$\sqrt{600}ms^{-1}$$

C
$$\sqrt{60}ms^{-1}$$

D
None of these

Solution

$$\displaystyle \Delta p=\frac{1}{2}\rho v^{2}$$
$$\therefore $$   $$\displaystyle v=\sqrt{\frac{2\Delta p}{\rho }}$$
   $$\displaystyle =\sqrt{\frac{2\left ( 3\: atm-1\: atm \right )}{\rho }}$$
   $$\displaystyle =\sqrt{\frac{2\times 2\times 10^{5}}{10^{3}}}=\sqrt{400}$$ m/s
Question 9
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A box exerts a force of $$420N$$ on a floor. The bottom of the box has an area of $$0.7m^2$$. What is the pressure exerted by the box on the floor?

A
$$600 Pa$$

B
$$500 Pa$$

C
$$400 Pa$$

D
$$700 Pa$$

Solution

Given, force, $$F=420N$$; area$$=0.7m^2$$;
Pressure, $$P=?$$
Pressure, $$P=\displaystyle\frac{Force}{Area}=\frac{F}{A}=\frac{420}{0.7}=600Pa$$
Question 10
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For the determination of the coefficient of viscosity of a given liquid, a graph between square of the radius of the spherical steel balls and their terminal velocity is plotted. The slope of the graph is given by

A
$$\cfrac { { r }^{ 2 } }{ v } $$

B
$$\cfrac { { v }^{ 2 } }{ r } $$

C
$$\cfrac{v}{r}$$

D
$$\cfrac{v}{{r}^{2}}$$

Solution

According to Stoke's law, terminal velocity $$v = \dfrac{2}{9\eta } gr^2 (\rho - \sigma)$$
$$\implies$$ $$\dfrac{v}{r^2} =\dfrac{2g (\rho -\sigma)}{ 9\eta} =constant$$
Thus the $$v$$ Vs $$r^2$$ graph is a straight line having slope $$ =\dfrac{v}{r^2}=\dfrac{2g (\rho - \sigma)}{9 \eta}$$