Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 20

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Question 1
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In the experimental arrangement shown in figure, the areas of cross-section of the wide and narrow portions of the tube are 5 cm$$^{2}$$ and 2 cm$$^{2}$$ respectively. The rate of flow of water through the tube is 500 cm$$^{3}$$ s$$^{-1}$$. The difference of mercury levels in the U-tube is:

A
$$0.97 \ cm$$

B
$$1.97 \ cm$$

C
$$0.67 \ cm$$

D
$$4.67 \ cm$$

Solution

$$Q = 500 cm^{3}/s$$
Using continuity equation we have $$Q=A_1v_1=A_2v_2$$
or
$$500=5(v_1)=2(v_2)$$
Thus we get
$$v_1=100 cm/s=1 m/s$$ and $$v_2=250 cm/s=2.5 m/s$$

Applying Bernoullis therom, In point (1) and (2),
$$\Rightarrow P_{1}-P_{2}=\dfrac{1}{2}\rho(v_2^2-v_1^2)=\dfrac{1}{2}1000(2.5^2-1^2)=2625 N/m^{2}$$
$$\Rightarrow P_{1}-P_{2}=2625= \rho gh=13.6\times 1000\times 9.8\times h$$
$$\Rightarrow h = 1.97 \ cm$$
Question 2
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A square hole of side a is made at a depth $$h$$ below water surface and to the side of a water container another circular hole of radius $$r$$ is made to the same container at a depth of $$4h$$. It is found that volume flow rate of water through both the holes is found to be same then:

A
$$r=\dfrac{a}{2\sqrt{\pi }}$$

B
$$a=\dfrac{r}{2\sqrt{\pi }}$$

C
$$a=\dfrac{r}{\sqrt{2\pi }}$$

D
$$r=\dfrac{a}{\sqrt{2\pi }}$$

Solution

$$\rho ^{1}A_{1}v_{1} = \rho ^{2}A_{2}v_{2}$$ as volume rate is constant
   velocity by toricellis therom is $$\sqrt{2gh}$$
$$\therefore a^{2}\left ( \sqrt{2gh} \right ) = \pi r^{2}\left ( \sqrt{2g\left ( 4h \right )} \right )$$
     $$\Rightarrow    2\pi r^{2} = a^{2}$$
              $$r = \dfrac{a}{\sqrt{2\pi }}$$
Question 3
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A cylinder of height $$20m$$ is completely filled with water. The velocity of efflux of water (in $$ms^{-1}$$) through a small hole on the side wall of the cylinder near its bottom is:

A
$$10 ms^{-1}$$

B
$$20 ms^{-1}$$

C
$$25.5 ms^{-1}$$

D
$$5 ms^{-1}$$

Solution

Given that
Pressure at the top surface $$=P_1 =p_{atm} $$
Pressure at the hole, $$P_2= p_{atm} $$
Here height of the cylinder , $$h=20\ m$$
From Bernoulli's Theorem we have:
Sum of all mechanical energies remains constant at a point over all points.
$$\implies P +\rho g h+ \dfrac12 \rho v^2=constant$$
where $$\rho $$ is the density of the fluid and $$v$$ is the velocity at that point .

So, applying Bernoulli Theorem at top and bottom of the cylinder we get :

$$P_{atm}+ \rho gh = P_{atm} +\dfrac12 \rho v^2$$

$$v=\sqrt{2gh}=\sqrt{2\times 20\times 10 }= 20 ms^{-1}$$
Question 4
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Two drops of small radius are falling in air with constant velocity $$5 cms^{-1}$$. If they coalesce, then the terminal velocity will be

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

B
$$2.5 cms^{-1}$$

C
$$5 \times \sqrt[3]{4}cms^{-1}$$

D
$$5.\sqrt{2}cms^{-1}$$

Solution

Let the radius of smaller drops =r $$r^{3}=V$$
radius of larger coslesce drop= $$2^{1/3}r $$
$$V$$ of smaller drops =$$=\dfrac{2g(T-s)r^{2}}{9 \eta}=5$$ ----------(1)
$$V$$ of larger drops= $$=\dfrac{2g(T-s)r^{2}}{9 \eta} 2^{2/3}$$[from(1)]
$$=5 \times 2 ^{2/3} cm/s$$
$$= 5 \times \sqrt[3]{4} cm/s$$
Question 5
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The velocity head of a stream of water is 40 cm. Then, the velocity of flow of water is :

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

B
$$5.6$$ $$ms^{-1}$$

C
$$7.2$$ $$ms^{-1}$$

D
$$9.8$$ $$ms^{-1}$$

Solution

Velocity head of water $$= 40\ cm $$ $$=\dfrac{V^{2}}{2g}$$
$$\Rightarrow V^{2} = \dfrac{40}{100} \times 2 \times 10$$
$$\Rightarrow V^{2}= 8\\ \Rightarrow V= \sqrt{8}\ m/s$$
$$\Rightarrow V= 2.8\ m/s$$
Question 6
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A tank with vertical walls is mounted so that its base is at a height $$H$$ above the horizontal ground. The tank is filled with water to a depth '$$h$$' . A hole is punched in the side wall of the tank at a depth ' $$x$$ ' below the water surface. To have maximum range of the emerging stream,the value of $$x$$ is

A
$$\dfrac{H+h}{4}$$

B
$$\dfrac{H+h}{2}$$

C
$$\dfrac{H+h}{3}$$

D
$$\dfrac{3(H+h)}{4}$$

Solution

Horizontal velocity of ejection, when a vessel is filled with
water to height h is $$\sqrt{2gh}$$
$$\therefore$$ velocity of ejection $$= \sqrt{2g(h-x)}$$
Time taken will be $$\sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2(H+h-x)}{g}}$$
$$\therefore range = 2\sqrt{(h-x)(H+h-x)} [\because R = ut]$$
range is max , when $$\dfrac{d R}{d x} = 0$$
$$ \Rightarrow x = \dfrac{H+h}{2}$$
Question 7
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Two equal drops of water are falling through air with a steady velocity of $$10\ cm/s$$. If the drops recombine to form a single drop then the terminal velocity is:

A
$$2^{\frac{2}{3}}$$ x $$5\ cm/s$$

B
$$2^{\frac{2}{3}}$$ x $$10\ cm/s$$

C
$$2^{\frac{2}{3}}$$ x $$15\ cm/s$$

D
$$2^{\frac{2}{3}}$$ x $$4\ cm/s$$

Solution

From the initial Terminal velocity given,
Let radius of drop = $$r$$.
So, $$V= 10 cm/s= \dfrac{2 g (T-\sigma )r^{2}}{9 \eta}$$
So,If two drops combine
As, $$\dfrac{4}{3}\pi r^{3}= V \Rightarrow r =\left ( \dfrac{3V}{4 \pi} \right )^{1/3}$$
So $$\dfrac{4}{3} \pi r_{1}^{3}= 2V$$
$$\Rightarrow r_{1}= 2^{1/3}\left ( \dfrac{3V}{4 \pi} \right )^{1/3}= 2^{1/3}r$$
Now,
New Terminal velocity =$$\dfrac{2 2^{2/3} r^{2}(T-\sigma )g}{9 \eta}$$
$$= 2^{2/3} \times 10 cm/s$$
Question 8
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Two hail stones with radii in the ratio of $$1:2$$ fall from a great height through the atmosphere. Then the ratio of their momenta after they have attained terminal velocity is

A
$$1:1$$

B
$$1:4$$

C
$$1:16$$

D
$$1:32$$

Solution

As the density of hailstones are same,
Mass of first = $$d \times \dfrac{4}{3} \pi r^{3}= M$$
Velocity of first =$$\dfrac{2 r^{2}(T_{h}-T_{a})g}{9 \eta}= V$$
For $$2^{nd}$$ hail with radius $$= 2r$$,
Mass of $$2^{nd}$$= $$d \times \dfrac{4}{3} \pi r^{3} 2^{3}= 8M$$
Velocit of $$2^{nd}$$=$$\dfrac{2 g r^{2}(T_{h}-T_{a})4}{9 \eta}=4V$$
Ratio of Momenta =$$MV : 8M \times 4V$$
$$MV : 32MV = 1:32$$
Question 9
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A vessel is filled with water and kerosene oil. The vessel has a small hole in the bottom. Neglecting viscosity if the thickness of water layer is $$h_{1}$$ and kerosene layer is $$h_{2}$$ then the velocity $$v$$ of flow of water will be (Given : density of water is $$\rho _{1}$$ g/ cc and that of kerosene is $$\rho _{2}$$ g/cc, neglecting viscosity):

A
$$v=\sqrt{2g(h_{1}+h_{2})}$$

B
$$v=\sqrt{2g\left[ h_{1}+h_{2}\left(\dfrac{\rho _{2}}{\rho _{1}} \right ) \right ]}$$

C
$$v=\sqrt{2g(h_{1}\rho _{1}+h_{2}\rho _{2})}$$

D
$$v=\sqrt{2g \left[h_{1}\left(\dfrac{\rho _{1}}{\rho _{2}}\right)+h_{2}\right]}$$

Solution

Berneoulli's Theorm : $$\rho gh + \dfrac{1}{2}\rho v^2 + P_o= constant $$
Pressure at A and B, $$P_A=P_B=P_o$$ and velocity at B be $$v$$ and at A is zero.
Applying Berneoulli's Theorem at points A and B,
$$P_o + \rho_1 g h_1 +\rho_2 g h_2 +0= P_o + 0 +\dfrac{1}{2}\rho_1 v^2 $$
$$\implies \dfrac{1}{2}\rho_1 v^2= \rho_1 g h_1+ \rho_2 g h_2 $$
$$v=\sqrt{\bigg[2g(h_1+h_2\dfrac{\rho_2}{\rho_1})\bigg]}$$
Question 10
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The area of cross-section of the two arms of a hydraulic press are 1 cm$$^{2}$$ and 10 cm$$^{2}$$ respectively(fig.). A force of 5 N is applied on the water in the thinner arm. What force should be applied on the water in the thicker arm so that the water may remain in equilibrium?

A
50 N

B
60 N

C
70 N

D
80N

Solution

For the water to remain in equilibrium, the pressure at same height i.e.,A and B should be same according to Pascals law,
$$\dfrac{5N}{1cm^{2}}=\dfrac{xN}{10cm^{2}}$$

$$x=50N$$