Mechanical Properties of Fluids Test

Result

Mechanical Properties of Fluids Test - 54

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Question 1
1 / -0

A ball of mass $$m$$ and radius $$r$$ is gently released in a viscous liquid. The mass of the liquid displaced by it is $$'m$$ such that $$m > m'$$. The terminal velocity is proportional to :

A
$$\dfrac{m-m'}{r}$$

B
$$\dfrac{m+m'}{r}$$

C
$$\dfrac{(m+m')}{r^2}$$

D
$$(m-m')r^2$$

Solution
Question 2
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A jet of water having velocity $$=10\ m/s$$ and stream cross-section $$=2\ cm^2$$ hits a flat plate perpendicularly, with the water splashing out parallel to plate. The plate experiences a force of :

A
$$40\ N$$

B
$$20\ N$$

C
$$8\ N$$

D
$$10\ N$$

Solution
Question 3
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The atmospheric pressure at earth surface is$$ P_{1}$$ and inside mine is $$ P_{2}.$$ They are related as:

A
$$ P_{1} = P_{2}.$$

B
$$ P_{1} > P_{2}.$$

C
$$ P_{1} < P_{2}.$$

D
$$ P_{2} $$ = 0

Solution
Question 4
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From the adjacent figure, the correct observation is

[ KCET 2005 ]

A
The pressure on the bottom of tank (a) is greater than at the bottom of (b).

B
The pressure on the bottom of the tank (a) is smaller than at the bottom of (b).

C
The pressure depend on the shape of the container

D
The pressure on the bottom of (a) and (b) is the same

Solution

Pressure = $$h\rho g$$ i.e. pressure at the bottom is independent of the area of the bottom of the tank. It depends on the height of water upto which the tank is filled with water. As in both the tanks, the levels of water are the same, pressure at the bottom is also the same.
Question 5
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A ball whose density is 0.4 × 103 kg/m falls into water from a height of 9 cm . To what depth does the ball sink

A
9 cm

B
6 cm

C
4.5 cm

D
2.25 cm

Solution

The velocity of ball before entering the water surface
$$v = \sqrt{2gh} = \sqrt{2g \times 9}$$
When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded)
The retardation, $$\alpha = \frac{apparent weight}{mass of ball}$$
$$= \frac{V(\rho - \sigma) g}{V \rho} $$ $$= \left ( \frac{\rho - \sigma}{\rho} \right ) g = \left ( \frac{0.4 - 1}{0.4} \right ) \times g = - \frac{3}{2} g$$
If h be the depth upto which ball sink, then,
$$0 - v^2 = 2 \times \left ( - \frac{3}{2} g \right ) \times h \Rightarrow 2g \times 9 = 3gh \therefore h = 6 cm$$
Question 6
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Radius of an air bubble at the bottom of the lake is $$r$$ and it becomes $$2r$$ when the air bubbles rises to the top surface of the lake. If $$P$$ cm of water be the atmospheric pressure, then the depth of the lake is

A
$$2p$$

B
$$8p$$

C
$$4p$$

D
$$7p$$

Solution

Initially the radius of the bubble is $$r$$. After reaching surface it becomes $$2r$$. The atmospheric pressure is given as,
$$P_{atm}= P \space cm \space of \space water$$
What we can conclude from this process is that the volume is changing in the air bubble but the temperature remains unchanged.
For isothermal process,
$$P_1V_1=P_2V_2$$
Let the height of water surface be $$h$$.
$$(P+h)(\frac{4}{3}\pi r^3)=P(\frac{4}{3}\pi 8r^3)$$
$$\Rightarrow h+P=8P$$
$$\Rightarrow h=7P$$ is our required answer.
Question 7
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A triangular lamina of area $$ A $$and height $$ h $$ is immersed in a liquid of density $$\rho$$ in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is

A
$$\dfrac{1}{2}A\rho gh$$

B
$$\dfrac{1}{3}A\rho gh$$

C
$$\dfrac{1}{6}A \rho gh$$

D
$$\dfrac{2}{3}A \rho gh$$

Solution

The Area of the lamina is A and the height is h which is immersed in a liquid of the density $$\rho $$ in a vertical plane with its base on the surface of the liquid.
The thrust on the lamina = pressure at centroid $$\times$$ Area
$$=\dfrac{h\rho g}{3}\times A = \dfrac{1}{3} A \rho gh$$
Question 8
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A hydraulic lift is designed to lift heavy objects of maximum mass $$2000 kg$$. The area of cross-section of piston carrying the load is $$2.25 \times 10^{-2} m^2$$. What is the maximum pressure the smaller piston would have to bear?

A
$$0.8711 \times 106 \ N/m^2$$

B
$$0.5862 \times 107 \ N/m^2$$

C
$$0.4869 \times 105 \ N/m^2$$

D
$$0.3271 \times 104 \ N/m^2$$

Solution

Pressure on the piston=$$\dfrac{F}{A}$$

Force $$F=m×a$$

=$$2000×9.8$$

Area of cross section $$A=2.25×10^{-2}m^2$$

Therefore the pressure $$P=\dfrac{2000×9.8}{2.25×10^(-2)}$$
$$P=0.8711×10^6\dfrac{N}{m^2}$$
Question 9
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A water supply maintains a constant rate of flow for water in a hose. You want to change the opening of the nozzle so that water leaving the nozzle will reach a height that is four times the current maximum height the water reaches with the nozzle vertical. To do so, should you

A
decrease the area of the opening by a factor of $$16$$

B
decrease the area by a factor of $$8$$

C
decrease the
area by a factor of $$4$$

D
decrease the area by a factor of $$2$$

E
give up because it cannot be done?

Solution
Question 10
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Figure shows aerial views from directly above two dams. Both dams are equally wide (the vertical dimension in the diagram) and equally high (into the page in the diagram). The dam on the left holds back a very large lake, and the dam on the right holds back a narrow river. Which dam has to be built more strongly?

A
the dam on the left

B
the dam on the right

C
both the same

D
cannot be predicted

Solution