Mechanical Properties of Fluids Test

Result

Mechanical Properties of Fluids Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

A small rain droplet is falling with terminal speed $$\upsilon$$. Terminal speed of another droplet of radius twice the radius of first droplet will be :

A
$$\upsilon$$

B
$$2 \upsilon$$

C
$$8 \upsilon$$

D
$$4 \upsilon$$.

Solution
Question 2
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The mercury columns in a manometer are given as 2$$\mathrm { cm }$$ and 12$$\mathrm { cm }$$ respectively. Specific gravity of mercury is13.6. The gauge pressure is pressure is $$\left( g = 10 \mathrm { m } / \mathrm { s } ^ { 2 } \right)$$

A
$$13.6 \times 10 ^ { 3 } N / m ^ { 2 }$$

B
$$13.6N / m ^ { 2 }$$

C
$$1.36 \times 10 ^ { 3 } N / m ^ { 2 }$$

D
$136$N / m ^ { 2 }$$

Solution
Question 3
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Two tubes of radii $$r_1$$ and $$r_2$$, and length $$l_1$$ and $$l_2$$ respectively, are connected in series and a liquid flows through each of them in streamline conditions. $$P_1$$ and $$P_2$$ are pressure differences across the two tubes. If $$P_2$$ is $$4P_1$$ and $$l_2$$ is $$\frac {l_1}{4}$$, then the radius $$r_2$$ will be equal to :

A
$$2{r_1}$$

B
$$4{r_1}$$

C
$${r_1}$$

D
$$\frac {r_1}{2}$$
Question 4
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The terminal speed $$\upsilon$$ of a rain water drop in air is :

A
$$\upsilon = k r \eta$$

B
$$\upsilon = k r^2 \eta$$

C
$$\upsilon = k r \eta^2$$

D
$$\upsilon = k r^2/\eta$$.
Question 5
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The area of cross section of the wider tube shown in figure is 900 $$
\mathrm{cm}^{2}
$$. If the boy standing on the piston weighs 45 kg, the difference in the levels of water in the two tubes.

A
50 cm

B
25cm

C
75 cm

D
100 cm

Solution

R.E.F image
$$A_{1}=900 cm^{2}$$
$$F_{1}= 45 g$$ newton
difference of water level = l
pressure at right end P =$$\frac{F_{1}}{A_{1}}=\frac{45g}{.09m^{2}}$$
so pressure in the right arm for the difference l scaled maintain the pressure
otherwise liquid will have either towards right or left.
so $$\frac{45\times 10}{.09} = pgh=\frac{1000kg}{m^{3}}\times 10\times l$$
$$\Rightarrow l= \frac{45}{.09\times 1000} m=\frac{4500}{90}cm=50 cm$$
Question 6
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Consider a wooden block immersed in a beaker containing water. If the beaker now starts accelerating upwards then :

A
Tension in the string will increase

B
Tension in the string will decrease

C
Tension in the string will remain unchanged

D
Data insufficient

Solution
Question 7
1 / -0

A block of wood of length $$50\ cm$$ and area of cross-section $$10\ cm^{2}$$, floats in water with $$\dfrac{3}{5}$$ of its lengths above water. Calculate weight of wood

A
$$200\ gf$$

B
$$300\ gf$$

C
$$400\ gf$$

D
$$500\ gf$$

Solution
Question 8
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A $$0.50\ L$$ container is occupied by nitrogen at a pressure of $$800\ torr$$ and a temperature of $$0^{o}C$$. The container can only withstand a pressure of $$3.0\ atm$$. What is the highest temperature $$(^{o}C)$$ to which the container may be heated?

A
$$505$$

B
$$450$$

C
$$625$$

D
$$500$$

Solution

As we know the

$$\frac{P V}{T}=\text { Canstant }$$

Given

$$P_{i}=800 \text { torr }=\frac{800}{760} \text { atm } $$

$$f_{f}=3 \text { atm } $$

$$V_{i}=0.5, \quad V_{f}=0.5 $$

$$T_{i}=(273+0)=273 \mathrm{~K}$$

From the formula

$$\frac{P_{i} V_{i}}{T_{i}}=\frac{P_{j} V_{f}}{T_{f}}$$

$$\frac{800 \times 0.5}{760 \times 273}=\frac{3 \times 0.5}{T} $$

$$T_{f}=778 \mathrm{~K} $$

$$T_{f}=505^{\circ} \mathrm{C}$$
Question 9
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Two vessels A and B of different shapes have the same base area and are filled with water up to the same height h (see figure). The force exerted by water on the base is $$F_{A}$$ for vessel A and $$F_{B}$$ for vessel B. The respective weights of the water filled in vessels are $$W_{A}$$ and $$W_{B}$$ . Then

A
$$F_{A}>F_{B} ;\ W_{A}>W_{B}$$

B
$$F_{A}=F_{B} ;\ W_{A}>W_{B}$$

C
$$F_{A}=F_{B} ;\ W_{A}<W_{B}$$

D
$$F_{A}>F_{B} ;\ W_{A}=W_{B}$$
Question 10
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Water enters a horizontal pipe of non uniform cross section with a velocity
of $$0.4 \mathrm { ms } ^ { - 1 }$$ and leaves the other end with a velocity of 0.6$$\mathrm { ms } ^ { - 1 } .$$ Pressure of water at the first end is $$1500 \mathrm { Nm } ^ { - 2 } ,$$ then pressure at the other end is

A
1000$$\mathrm { Nm } ^ { - 2 }$$

B
1200$$\mathrm { Nm } ^ { - 2 }$$

C
1400$$\mathrm { Nm } ^ { - 2 }$$

D
1600$$\mathrm { Nm } ^ { - 2 }$$

Solution