QBManager

Mechanical Properties of Fluids

142688 A liquid $X$ of density $3.36 g / {cm}^{3}$ is poured in a U-tube in right arm with height $10 cm$ , which contains Hg. Another liquid $Y$ is poured in left arm with height $8 cm$ . Upper levels of $X$ and $Y$ are same. What is the density of $Y$ ?

1

0.8 g / cc

2

1.2 g / cc

3

1.4 / cc

4

1.6 g / cc

Explanation:

A
Let, $P_{atm} =$ atmospheric pressuring
$ρ_{1} =$ density of liquid $Y$
$ρ_{2} =$ density of $Hg$
$ρ_{3} =$ density of liquid $X$
Its shown in figure, in the two arms of tube pressure remains same on surface. (i.e. pressure in the left limb equal to pressure in right limb).
Therefore, $P_{a t m} + ρ_{1} g h_{1} + ρ_{2} g h_{2} = ρ_{3} g h_{3} + P_{a t m}$
$ρ_{1} h_{1} + ρ_{2} h_{2} = ρ_{3} h_{3}$
$ρ_{1} \times 8 + 13.6 \times 2 = 3.36 \times 10$
$8 ρ_{1} + 27.2 = 33.6$
$8 ρ_{1} = 33.6 - 27.2 = 6.4$
or $ρ_{1} = \frac{6.4}{8} = 0.8 g / cc$

JCECE-2014

Mechanical Properties of Fluids

142689 An air bubble of radius $1 cm$ rises from the bottom portion through a liquid of density 1.5 $g / cc$ at a constant speed of $0.25 cm s^{- 1}$ . If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in Pas)

1 13000

2 1300

3 130

4 13

Explanation:

C Given, Radius of air bubble $(r) = 0.01 m$ Terminal velocity $(v_{T}) = 0.25 \times 10^{- 2} m / s$
Density of bubble $(ρ) = 1.5 \times 10^{3} kg / m^{3}$
Density of air $(σ_{air}) = 0, g = 9.8 m / s^{2}, η =$ ?
We know that, terminal velocity $(v_{T}) = \frac{2}{9} \frac{r^{2} (ρ - σ_{air}) g}{η}$
or $η = \frac{2 r^{2} (ρ - σ_{air}) g}{9 v_{T}}$
$η = \frac{2}{9} \times \frac{(0.01)^{2} (1.5 \times 10^{3} - 0) 9.8}{0.25 \times 10^{- 2}} = 130.667$
$η = 130.667 ≃ 130 Pa - s$

JCECE-2009

Mechanical Properties of Fluids

142691 An adulterated sample of milk has a density of $1032 kg - m^{- 3}$ , while pure milk has a density of $1080 kg - m^{- 3}$ . The volume of pure milk in a sample of $10 L$ of adulterer milk is :

1

0.5 L

2

1.0 L

3

2.0 L

4

4.0 L

Explanation:

D Given that, $ρ_{a} = 1032 kg / m^{3}, ρ_{p} = 1080 kg / m^{3}$
We know that, $ρ = \frac{m}{V}$
Mass of adulterated milk $(m_{a}) = ρ V_{a}$
$= 1032 \times 10 \times 10^{- 3} = 10.32 kg$
and mass of pure milk $(m_{p}) = 1080 \times V_{p}$
Therefore, volume of pure milk-
$(ρ_{w} V_{w}) = m_{a} - m_{p}$
$ρ_{w} (V_{a} - V_{p}) = m_{a} - m_{p}$
$10^{3} (10 \times 10^{- 3} - V_{p}) = 10.32 - 1080 V_{p}$
$80 V_{p} = 10.32 - 10$
$V_{p} = \frac{0.32}{80} \times 1000 = 4 L$

JCECE-2005

Mechanical Properties of Fluids

142693 Assertion: Machine parts are jammed in winter.
Reason: The viscosity of lubricant used in machine parts increase at low temperature.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

A Viscosity is an internal fluid friction at between two adjacent layers of same liquid which opposes their relative motion.
Viscosity increases with decrease in temperature in case of liquid. Due to low temperature in winter, the coefficient of viscosity of engine oil and the lubricant increases.
Hence, machine parts jammed in winter.
Viscosity $(v) \propto \frac{1}{T}$

AIIMS-2007

Mechanical Properties of Fluids

142688 A liquid $X$ of density $3.36 g / {cm}^{3}$ is poured in a U-tube in right arm with height $10 cm$ , which contains Hg. Another liquid $Y$ is poured in left arm with height $8 cm$ . Upper levels of $X$ and $Y$ are same. What is the density of $Y$ ?

1

0.8 g / cc

2

1.2 g / cc

3

1.4 / cc

4

1.6 g / cc

Explanation:

A
Let, $P_{atm} =$ atmospheric pressuring
$ρ_{1} =$ density of liquid $Y$
$ρ_{2} =$ density of $Hg$
$ρ_{3} =$ density of liquid $X$
Its shown in figure, in the two arms of tube pressure remains same on surface. (i.e. pressure in the left limb equal to pressure in right limb).
Therefore, $P_{a t m} + ρ_{1} g h_{1} + ρ_{2} g h_{2} = ρ_{3} g h_{3} + P_{a t m}$
$ρ_{1} h_{1} + ρ_{2} h_{2} = ρ_{3} h_{3}$
$ρ_{1} \times 8 + 13.6 \times 2 = 3.36 \times 10$
$8 ρ_{1} + 27.2 = 33.6$
$8 ρ_{1} = 33.6 - 27.2 = 6.4$
or $ρ_{1} = \frac{6.4}{8} = 0.8 g / cc$

JCECE-2014

Mechanical Properties of Fluids

142689 An air bubble of radius $1 cm$ rises from the bottom portion through a liquid of density 1.5 $g / cc$ at a constant speed of $0.25 cm s^{- 1}$ . If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in Pas)

1 13000

2 1300

3 130

4 13

Explanation:

C Given, Radius of air bubble $(r) = 0.01 m$ Terminal velocity $(v_{T}) = 0.25 \times 10^{- 2} m / s$
Density of bubble $(ρ) = 1.5 \times 10^{3} kg / m^{3}$
Density of air $(σ_{air}) = 0, g = 9.8 m / s^{2}, η =$ ?
We know that, terminal velocity $(v_{T}) = \frac{2}{9} \frac{r^{2} (ρ - σ_{air}) g}{η}$
or $η = \frac{2 r^{2} (ρ - σ_{air}) g}{9 v_{T}}$
$η = \frac{2}{9} \times \frac{(0.01)^{2} (1.5 \times 10^{3} - 0) 9.8}{0.25 \times 10^{- 2}} = 130.667$
$η = 130.667 ≃ 130 Pa - s$

JCECE-2009

Mechanical Properties of Fluids

142691 An adulterated sample of milk has a density of $1032 kg - m^{- 3}$ , while pure milk has a density of $1080 kg - m^{- 3}$ . The volume of pure milk in a sample of $10 L$ of adulterer milk is :

1

0.5 L

2

1.0 L

3

2.0 L

4

4.0 L

Explanation:

D Given that, $ρ_{a} = 1032 kg / m^{3}, ρ_{p} = 1080 kg / m^{3}$
We know that, $ρ = \frac{m}{V}$
Mass of adulterated milk $(m_{a}) = ρ V_{a}$
$= 1032 \times 10 \times 10^{- 3} = 10.32 kg$
and mass of pure milk $(m_{p}) = 1080 \times V_{p}$
Therefore, volume of pure milk-
$(ρ_{w} V_{w}) = m_{a} - m_{p}$
$ρ_{w} (V_{a} - V_{p}) = m_{a} - m_{p}$
$10^{3} (10 \times 10^{- 3} - V_{p}) = 10.32 - 1080 V_{p}$
$80 V_{p} = 10.32 - 10$
$V_{p} = \frac{0.32}{80} \times 1000 = 4 L$

JCECE-2005

Mechanical Properties of Fluids

142693 Assertion: Machine parts are jammed in winter.
Reason: The viscosity of lubricant used in machine parts increase at low temperature.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

A Viscosity is an internal fluid friction at between two adjacent layers of same liquid which opposes their relative motion.
Viscosity increases with decrease in temperature in case of liquid. Due to low temperature in winter, the coefficient of viscosity of engine oil and the lubricant increases.
Hence, machine parts jammed in winter.
Viscosity $(v) \propto \frac{1}{T}$

AIIMS-2007

Mechanical Properties of Fluids

142688 A liquid $X$ of density $3.36 g / {cm}^{3}$ is poured in a U-tube in right arm with height $10 cm$ , which contains Hg. Another liquid $Y$ is poured in left arm with height $8 cm$ . Upper levels of $X$ and $Y$ are same. What is the density of $Y$ ?

1

0.8 g / cc

2

1.2 g / cc

3

1.4 / cc

4

1.6 g / cc

Explanation:

A
Let, $P_{atm} =$ atmospheric pressuring
$ρ_{1} =$ density of liquid $Y$
$ρ_{2} =$ density of $Hg$
$ρ_{3} =$ density of liquid $X$
Its shown in figure, in the two arms of tube pressure remains same on surface. (i.e. pressure in the left limb equal to pressure in right limb).
Therefore, $P_{a t m} + ρ_{1} g h_{1} + ρ_{2} g h_{2} = ρ_{3} g h_{3} + P_{a t m}$
$ρ_{1} h_{1} + ρ_{2} h_{2} = ρ_{3} h_{3}$
$ρ_{1} \times 8 + 13.6 \times 2 = 3.36 \times 10$
$8 ρ_{1} + 27.2 = 33.6$
$8 ρ_{1} = 33.6 - 27.2 = 6.4$
or $ρ_{1} = \frac{6.4}{8} = 0.8 g / cc$

JCECE-2014

Mechanical Properties of Fluids

142689 An air bubble of radius $1 cm$ rises from the bottom portion through a liquid of density 1.5 $g / cc$ at a constant speed of $0.25 cm s^{- 1}$ . If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in Pas)

1 13000

2 1300

3 130

4 13

Explanation:

C Given, Radius of air bubble $(r) = 0.01 m$ Terminal velocity $(v_{T}) = 0.25 \times 10^{- 2} m / s$
Density of bubble $(ρ) = 1.5 \times 10^{3} kg / m^{3}$
Density of air $(σ_{air}) = 0, g = 9.8 m / s^{2}, η =$ ?
We know that, terminal velocity $(v_{T}) = \frac{2}{9} \frac{r^{2} (ρ - σ_{air}) g}{η}$
or $η = \frac{2 r^{2} (ρ - σ_{air}) g}{9 v_{T}}$
$η = \frac{2}{9} \times \frac{(0.01)^{2} (1.5 \times 10^{3} - 0) 9.8}{0.25 \times 10^{- 2}} = 130.667$
$η = 130.667 ≃ 130 Pa - s$

JCECE-2009

Mechanical Properties of Fluids

142691 An adulterated sample of milk has a density of $1032 kg - m^{- 3}$ , while pure milk has a density of $1080 kg - m^{- 3}$ . The volume of pure milk in a sample of $10 L$ of adulterer milk is :

1

0.5 L

2

1.0 L

3

2.0 L

4

4.0 L

Explanation:

D Given that, $ρ_{a} = 1032 kg / m^{3}, ρ_{p} = 1080 kg / m^{3}$
We know that, $ρ = \frac{m}{V}$
Mass of adulterated milk $(m_{a}) = ρ V_{a}$
$= 1032 \times 10 \times 10^{- 3} = 10.32 kg$
and mass of pure milk $(m_{p}) = 1080 \times V_{p}$
Therefore, volume of pure milk-
$(ρ_{w} V_{w}) = m_{a} - m_{p}$
$ρ_{w} (V_{a} - V_{p}) = m_{a} - m_{p}$
$10^{3} (10 \times 10^{- 3} - V_{p}) = 10.32 - 1080 V_{p}$
$80 V_{p} = 10.32 - 10$
$V_{p} = \frac{0.32}{80} \times 1000 = 4 L$

JCECE-2005

Mechanical Properties of Fluids

142693 Assertion: Machine parts are jammed in winter.
Reason: The viscosity of lubricant used in machine parts increase at low temperature.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

A Viscosity is an internal fluid friction at between two adjacent layers of same liquid which opposes their relative motion.
Viscosity increases with decrease in temperature in case of liquid. Due to low temperature in winter, the coefficient of viscosity of engine oil and the lubricant increases.
Hence, machine parts jammed in winter.
Viscosity $(v) \propto \frac{1}{T}$

AIIMS-2007

Mechanical Properties of Fluids

142688 A liquid $X$ of density $3.36 g / {cm}^{3}$ is poured in a U-tube in right arm with height $10 cm$ , which contains Hg. Another liquid $Y$ is poured in left arm with height $8 cm$ . Upper levels of $X$ and $Y$ are same. What is the density of $Y$ ?

1

0.8 g / cc

2

1.2 g / cc

3

1.4 / cc

4

1.6 g / cc

Explanation:

A
Let, $P_{atm} =$ atmospheric pressuring
$ρ_{1} =$ density of liquid $Y$
$ρ_{2} =$ density of $Hg$
$ρ_{3} =$ density of liquid $X$
Its shown in figure, in the two arms of tube pressure remains same on surface. (i.e. pressure in the left limb equal to pressure in right limb).
Therefore, $P_{a t m} + ρ_{1} g h_{1} + ρ_{2} g h_{2} = ρ_{3} g h_{3} + P_{a t m}$
$ρ_{1} h_{1} + ρ_{2} h_{2} = ρ_{3} h_{3}$
$ρ_{1} \times 8 + 13.6 \times 2 = 3.36 \times 10$
$8 ρ_{1} + 27.2 = 33.6$
$8 ρ_{1} = 33.6 - 27.2 = 6.4$
or $ρ_{1} = \frac{6.4}{8} = 0.8 g / cc$

JCECE-2014

Mechanical Properties of Fluids

142689 An air bubble of radius $1 cm$ rises from the bottom portion through a liquid of density 1.5 $g / cc$ at a constant speed of $0.25 cm s^{- 1}$ . If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in Pas)

1 13000

2 1300

3 130

4 13

Explanation:

C Given, Radius of air bubble $(r) = 0.01 m$ Terminal velocity $(v_{T}) = 0.25 \times 10^{- 2} m / s$
Density of bubble $(ρ) = 1.5 \times 10^{3} kg / m^{3}$
Density of air $(σ_{air}) = 0, g = 9.8 m / s^{2}, η =$ ?
We know that, terminal velocity $(v_{T}) = \frac{2}{9} \frac{r^{2} (ρ - σ_{air}) g}{η}$
or $η = \frac{2 r^{2} (ρ - σ_{air}) g}{9 v_{T}}$
$η = \frac{2}{9} \times \frac{(0.01)^{2} (1.5 \times 10^{3} - 0) 9.8}{0.25 \times 10^{- 2}} = 130.667$
$η = 130.667 ≃ 130 Pa - s$

JCECE-2009

Mechanical Properties of Fluids

142691 An adulterated sample of milk has a density of $1032 kg - m^{- 3}$ , while pure milk has a density of $1080 kg - m^{- 3}$ . The volume of pure milk in a sample of $10 L$ of adulterer milk is :

1

0.5 L

2

1.0 L

3

2.0 L

4

4.0 L

Explanation:

D Given that, $ρ_{a} = 1032 kg / m^{3}, ρ_{p} = 1080 kg / m^{3}$
We know that, $ρ = \frac{m}{V}$
Mass of adulterated milk $(m_{a}) = ρ V_{a}$
$= 1032 \times 10 \times 10^{- 3} = 10.32 kg$
and mass of pure milk $(m_{p}) = 1080 \times V_{p}$
Therefore, volume of pure milk-
$(ρ_{w} V_{w}) = m_{a} - m_{p}$
$ρ_{w} (V_{a} - V_{p}) = m_{a} - m_{p}$
$10^{3} (10 \times 10^{- 3} - V_{p}) = 10.32 - 1080 V_{p}$
$80 V_{p} = 10.32 - 10$
$V_{p} = \frac{0.32}{80} \times 1000 = 4 L$

JCECE-2005

Mechanical Properties of Fluids

142693 Assertion: Machine parts are jammed in winter.
Reason: The viscosity of lubricant used in machine parts increase at low temperature.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

A Viscosity is an internal fluid friction at between two adjacent layers of same liquid which opposes their relative motion.
Viscosity increases with decrease in temperature in case of liquid. Due to low temperature in winter, the coefficient of viscosity of engine oil and the lubricant increases.
Hence, machine parts jammed in winter.
Viscosity $(v) \propto \frac{1}{T}$

AIIMS-2007

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