QBManager

Mechanical Properties of Fluids

143193 A block of iron contains a hollow cavity as shown below. The block weighs $6000 N$ in air and $4000 N$ in water. If the density of iron and water are $6 g / {cm}^{3}$ and $1 g / {cm}^{3}$ , then the volume of the cavity is

1

0.05 m

2

0.5 m^{3}

3

0.25 m^{3}

4

0.1 m^{3}

Explanation:

D Given,
$ρ_{iron} = 6 g / {cm}^{3} = 6 \times 10^{3} kg / m^{3}$
$ρ_{Water} = 1 g / {cm}^{3} = 1 \times 10^{3} kg / m^{3}$
$W_{air} = 6000 N$
$w_{water} = 4000 N$
Volume of cavity
$V_{cavity} = \frac{w_{air} - w_{water}}{ρ_{w} \times g} - \frac{w_{air}}{ρ_{iron} \times g}$
$= \frac{6000 - 4000}{1 \times 10^{3} \times 10} - \frac{6000}{6 \times 10^{3} \times 10}$
$= 0.2 - 0.1$
$= 0.1 m^{3}$

TS- EAMCET-11.09.2020

Mechanical Properties of Fluids

143194 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 cm$ of $Hg$ , where the velocity is $35 cm - s^{- 1}$ . At a point where the velocity is $65 cm - s^{- 1}$ , the pressure will be

1

0.89 cm

of

Hg

2

8.9 cm

of

Hg

3

0.5 cm

of

Hg

4

1 cm

of

Hg

Explanation:

A Given,
$v_{1} = 35 cm / \sec = 35 \times 10^{- 2} m / \sec$
$v_{2} = 65 cm / \sec = 65 \times 10^{- 2} m / \sec$
$P_{1} = ρ {gh}_{1} = 13600 \times 9.8 \times 10^{- 2}$
${∵ h_{1} = 1 cm = 1 \times 10^{- 2} m}$
$P_{2} = {ggh}_{2} = 13600 \times 9.8 \times h$
$z_{1} = z_{2}$
According to Bernoulli equation
$P_{1} + \frac{1}{2} ρ v_{1}^{2} + z_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + z_{2}$
$13600 \times 9.8 \times 10^{- 2} + 1 / 2 \times 1000 (0.35)^{2} = 13600 \times 9.8$
$\times h + \frac{1}{2} \times 1000 (0.65)^{2}$
$1394.05 = 133280 h + 221.25$
$\Rightarrow h = 0.00887 m$
$h = 0.89 cm of Hg$

JCECE-2014

Mechanical Properties of Fluids

143195 Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is $V$ . If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

1

V / 4

2

V / 8

3

16 V

4

V / 32

Explanation:

D Given,
$l_{2} = 2 l_{1}$
$d_{2} = \frac{d_{1}}{2} \Rightarrow r_{2} = \frac{r_{1}}{2}$
According to Poiseuille's formula-
$V = \frac{π \Pr^{4}}{8 n l}$
$V \propto \frac{r^{4}}{l}$
$\frac{V_{2}}{V_{1}} = \frac{r_{1}^{4}}{r_{2}^{4}} \times \frac{l_{2}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{r_{1}^{4}}{{(\frac{r_{1}}{2})}^{4}} \times \frac{2 l_{1}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = (2)^{4} \times 2$
$V_{2} = \frac{V}{32} (∵ V_{1} = V)$

COMEDK 2015

Mechanical Properties of Fluids

143197 A boy has $60 kg$ weight. He wants to swim in a river with the help of a wooden log. If relative density of wood is 0.6 , what is the minimum volume of wooden log?
(density of river water is $1000 kg / m^{3}$ )

1

0.66 m^{3}

2

150 m^{3}

3

\frac{3}{1} m^{3}

4

\frac{3}{20} m^{3}

Explanation:

D Given,
$m = 60 kg$
Relative density of wood $= 0.6$
$ρ_{wood} = 600 kg / m^{3}$
$ρ_{water} = 1000 kg / m^{3}$
According to Archimedes' principle,
body displaced by liquid $=$ up thrust
$mg + V ρ_{wood} g = V ρ_{water} g$
$60 \times g + V \times 0.6 \times 10^{3} g = V \times 1000 g$
$60 + 600 V = 1000 V$
$V = 60 / 400$
$V = \frac{3}{20} m^{3}$

AIIMS-2010

Mechanical Properties of Fluids

143199 A piece of wax weighs $18.03 g$ in air. A piece of metal is found to weigh $17.03 g$ in water. It is tied to the wax and both together weight $15.23 g$ in water. Then, the specific gravity of wax is :

1

\frac{18.03}{17.03}

2

\frac{17.03}{18.03}

3

\frac{18.03}{19.83}

4

\frac{15.03}{17.03}

Explanation:

C Given,
$W_{w_{1}} = 18.03 g$
$W_{m} = 17.03 g$
$W = 15.23 g$
$W_{w_{2}} = 15.23 - 17.03 = - 1.8 g$
We know that,
Specific gravity of wax $= \frac{W_{w_{1}}}{W_{w_{1}} - W_{w_{2}}}$
$g = \frac{18.03}{18.03 - (- 1.8)}$
$g = \frac{18.03}{19.83}$
Total weight
$W = W_{m} + (W_{wax} - v \frac{ρ_{wax}}{ρ_{water}} \times ρ_{w} g)$
$15.23 = 17.03 + (18.03 - \frac{18.03}{S.G.})$
${∵ S.G. specific gravity}$
$S.G. = \frac{18.03}{19.83}$

BCECE-2003

Mechanical Properties of Fluids

143193 A block of iron contains a hollow cavity as shown below. The block weighs $6000 N$ in air and $4000 N$ in water. If the density of iron and water are $6 g / {cm}^{3}$ and $1 g / {cm}^{3}$ , then the volume of the cavity is

1

0.05 m

2

0.5 m^{3}

3

0.25 m^{3}

4

0.1 m^{3}

Explanation:

D Given,
$ρ_{iron} = 6 g / {cm}^{3} = 6 \times 10^{3} kg / m^{3}$
$ρ_{Water} = 1 g / {cm}^{3} = 1 \times 10^{3} kg / m^{3}$
$W_{air} = 6000 N$
$w_{water} = 4000 N$
Volume of cavity
$V_{cavity} = \frac{w_{air} - w_{water}}{ρ_{w} \times g} - \frac{w_{air}}{ρ_{iron} \times g}$
$= \frac{6000 - 4000}{1 \times 10^{3} \times 10} - \frac{6000}{6 \times 10^{3} \times 10}$
$= 0.2 - 0.1$
$= 0.1 m^{3}$

TS- EAMCET-11.09.2020

Mechanical Properties of Fluids

143194 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 cm$ of $Hg$ , where the velocity is $35 cm - s^{- 1}$ . At a point where the velocity is $65 cm - s^{- 1}$ , the pressure will be

1

0.89 cm

of

Hg

2

8.9 cm

of

Hg

3

0.5 cm

of

Hg

4

1 cm

of

Hg

Explanation:

A Given,
$v_{1} = 35 cm / \sec = 35 \times 10^{- 2} m / \sec$
$v_{2} = 65 cm / \sec = 65 \times 10^{- 2} m / \sec$
$P_{1} = ρ {gh}_{1} = 13600 \times 9.8 \times 10^{- 2}$
${∵ h_{1} = 1 cm = 1 \times 10^{- 2} m}$
$P_{2} = {ggh}_{2} = 13600 \times 9.8 \times h$
$z_{1} = z_{2}$
According to Bernoulli equation
$P_{1} + \frac{1}{2} ρ v_{1}^{2} + z_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + z_{2}$
$13600 \times 9.8 \times 10^{- 2} + 1 / 2 \times 1000 (0.35)^{2} = 13600 \times 9.8$
$\times h + \frac{1}{2} \times 1000 (0.65)^{2}$
$1394.05 = 133280 h + 221.25$
$\Rightarrow h = 0.00887 m$
$h = 0.89 cm of Hg$

JCECE-2014

Mechanical Properties of Fluids

143195 Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is $V$ . If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

1

V / 4

2

V / 8

3

16 V

4

V / 32

Explanation:

D Given,
$l_{2} = 2 l_{1}$
$d_{2} = \frac{d_{1}}{2} \Rightarrow r_{2} = \frac{r_{1}}{2}$
According to Poiseuille's formula-
$V = \frac{π \Pr^{4}}{8 n l}$
$V \propto \frac{r^{4}}{l}$
$\frac{V_{2}}{V_{1}} = \frac{r_{1}^{4}}{r_{2}^{4}} \times \frac{l_{2}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{r_{1}^{4}}{{(\frac{r_{1}}{2})}^{4}} \times \frac{2 l_{1}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = (2)^{4} \times 2$
$V_{2} = \frac{V}{32} (∵ V_{1} = V)$

COMEDK 2015

Mechanical Properties of Fluids

143197 A boy has $60 kg$ weight. He wants to swim in a river with the help of a wooden log. If relative density of wood is 0.6 , what is the minimum volume of wooden log?
(density of river water is $1000 kg / m^{3}$ )

1

0.66 m^{3}

2

150 m^{3}

3

\frac{3}{1} m^{3}

4

\frac{3}{20} m^{3}

Explanation:

D Given,
$m = 60 kg$
Relative density of wood $= 0.6$
$ρ_{wood} = 600 kg / m^{3}$
$ρ_{water} = 1000 kg / m^{3}$
According to Archimedes' principle,
body displaced by liquid $=$ up thrust
$mg + V ρ_{wood} g = V ρ_{water} g$
$60 \times g + V \times 0.6 \times 10^{3} g = V \times 1000 g$
$60 + 600 V = 1000 V$
$V = 60 / 400$
$V = \frac{3}{20} m^{3}$

AIIMS-2010

Mechanical Properties of Fluids

143199 A piece of wax weighs $18.03 g$ in air. A piece of metal is found to weigh $17.03 g$ in water. It is tied to the wax and both together weight $15.23 g$ in water. Then, the specific gravity of wax is :

1

\frac{18.03}{17.03}

2

\frac{17.03}{18.03}

3

\frac{18.03}{19.83}

4

\frac{15.03}{17.03}

Explanation:

C Given,
$W_{w_{1}} = 18.03 g$
$W_{m} = 17.03 g$
$W = 15.23 g$
$W_{w_{2}} = 15.23 - 17.03 = - 1.8 g$
We know that,
Specific gravity of wax $= \frac{W_{w_{1}}}{W_{w_{1}} - W_{w_{2}}}$
$g = \frac{18.03}{18.03 - (- 1.8)}$
$g = \frac{18.03}{19.83}$
Total weight
$W = W_{m} + (W_{wax} - v \frac{ρ_{wax}}{ρ_{water}} \times ρ_{w} g)$
$15.23 = 17.03 + (18.03 - \frac{18.03}{S.G.})$
${∵ S.G. specific gravity}$
$S.G. = \frac{18.03}{19.83}$

BCECE-2003

Mechanical Properties of Fluids

143193 A block of iron contains a hollow cavity as shown below. The block weighs $6000 N$ in air and $4000 N$ in water. If the density of iron and water are $6 g / {cm}^{3}$ and $1 g / {cm}^{3}$ , then the volume of the cavity is

1

0.05 m

2

0.5 m^{3}

3

0.25 m^{3}

4

0.1 m^{3}

Explanation:

D Given,
$ρ_{iron} = 6 g / {cm}^{3} = 6 \times 10^{3} kg / m^{3}$
$ρ_{Water} = 1 g / {cm}^{3} = 1 \times 10^{3} kg / m^{3}$
$W_{air} = 6000 N$
$w_{water} = 4000 N$
Volume of cavity
$V_{cavity} = \frac{w_{air} - w_{water}}{ρ_{w} \times g} - \frac{w_{air}}{ρ_{iron} \times g}$
$= \frac{6000 - 4000}{1 \times 10^{3} \times 10} - \frac{6000}{6 \times 10^{3} \times 10}$
$= 0.2 - 0.1$
$= 0.1 m^{3}$

TS- EAMCET-11.09.2020

Mechanical Properties of Fluids

143194 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 cm$ of $Hg$ , where the velocity is $35 cm - s^{- 1}$ . At a point where the velocity is $65 cm - s^{- 1}$ , the pressure will be

1

0.89 cm

of

Hg

2

8.9 cm

of

Hg

3

0.5 cm

of

Hg

4

1 cm

of

Hg

Explanation:

A Given,
$v_{1} = 35 cm / \sec = 35 \times 10^{- 2} m / \sec$
$v_{2} = 65 cm / \sec = 65 \times 10^{- 2} m / \sec$
$P_{1} = ρ {gh}_{1} = 13600 \times 9.8 \times 10^{- 2}$
${∵ h_{1} = 1 cm = 1 \times 10^{- 2} m}$
$P_{2} = {ggh}_{2} = 13600 \times 9.8 \times h$
$z_{1} = z_{2}$
According to Bernoulli equation
$P_{1} + \frac{1}{2} ρ v_{1}^{2} + z_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + z_{2}$
$13600 \times 9.8 \times 10^{- 2} + 1 / 2 \times 1000 (0.35)^{2} = 13600 \times 9.8$
$\times h + \frac{1}{2} \times 1000 (0.65)^{2}$
$1394.05 = 133280 h + 221.25$
$\Rightarrow h = 0.00887 m$
$h = 0.89 cm of Hg$

JCECE-2014

Mechanical Properties of Fluids

143195 Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is $V$ . If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

1

V / 4

2

V / 8

3

16 V

4

V / 32

Explanation:

D Given,
$l_{2} = 2 l_{1}$
$d_{2} = \frac{d_{1}}{2} \Rightarrow r_{2} = \frac{r_{1}}{2}$
According to Poiseuille's formula-
$V = \frac{π \Pr^{4}}{8 n l}$
$V \propto \frac{r^{4}}{l}$
$\frac{V_{2}}{V_{1}} = \frac{r_{1}^{4}}{r_{2}^{4}} \times \frac{l_{2}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{r_{1}^{4}}{{(\frac{r_{1}}{2})}^{4}} \times \frac{2 l_{1}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = (2)^{4} \times 2$
$V_{2} = \frac{V}{32} (∵ V_{1} = V)$

COMEDK 2015

Mechanical Properties of Fluids

143197 A boy has $60 kg$ weight. He wants to swim in a river with the help of a wooden log. If relative density of wood is 0.6 , what is the minimum volume of wooden log?
(density of river water is $1000 kg / m^{3}$ )

1

0.66 m^{3}

2

150 m^{3}

3

\frac{3}{1} m^{3}

4

\frac{3}{20} m^{3}

Explanation:

D Given,
$m = 60 kg$
Relative density of wood $= 0.6$
$ρ_{wood} = 600 kg / m^{3}$
$ρ_{water} = 1000 kg / m^{3}$
According to Archimedes' principle,
body displaced by liquid $=$ up thrust
$mg + V ρ_{wood} g = V ρ_{water} g$
$60 \times g + V \times 0.6 \times 10^{3} g = V \times 1000 g$
$60 + 600 V = 1000 V$
$V = 60 / 400$
$V = \frac{3}{20} m^{3}$

AIIMS-2010

Mechanical Properties of Fluids

143199 A piece of wax weighs $18.03 g$ in air. A piece of metal is found to weigh $17.03 g$ in water. It is tied to the wax and both together weight $15.23 g$ in water. Then, the specific gravity of wax is :

1

\frac{18.03}{17.03}

2

\frac{17.03}{18.03}

3

\frac{18.03}{19.83}

4

\frac{15.03}{17.03}

Explanation:

C Given,
$W_{w_{1}} = 18.03 g$
$W_{m} = 17.03 g$
$W = 15.23 g$
$W_{w_{2}} = 15.23 - 17.03 = - 1.8 g$
We know that,
Specific gravity of wax $= \frac{W_{w_{1}}}{W_{w_{1}} - W_{w_{2}}}$
$g = \frac{18.03}{18.03 - (- 1.8)}$
$g = \frac{18.03}{19.83}$
Total weight
$W = W_{m} + (W_{wax} - v \frac{ρ_{wax}}{ρ_{water}} \times ρ_{w} g)$
$15.23 = 17.03 + (18.03 - \frac{18.03}{S.G.})$
${∵ S.G. specific gravity}$
$S.G. = \frac{18.03}{19.83}$

BCECE-2003

Mechanical Properties of Fluids

143193 A block of iron contains a hollow cavity as shown below. The block weighs $6000 N$ in air and $4000 N$ in water. If the density of iron and water are $6 g / {cm}^{3}$ and $1 g / {cm}^{3}$ , then the volume of the cavity is

1

0.05 m

2

0.5 m^{3}

3

0.25 m^{3}

4

0.1 m^{3}

Explanation:

D Given,
$ρ_{iron} = 6 g / {cm}^{3} = 6 \times 10^{3} kg / m^{3}$
$ρ_{Water} = 1 g / {cm}^{3} = 1 \times 10^{3} kg / m^{3}$
$W_{air} = 6000 N$
$w_{water} = 4000 N$
Volume of cavity
$V_{cavity} = \frac{w_{air} - w_{water}}{ρ_{w} \times g} - \frac{w_{air}}{ρ_{iron} \times g}$
$= \frac{6000 - 4000}{1 \times 10^{3} \times 10} - \frac{6000}{6 \times 10^{3} \times 10}$
$= 0.2 - 0.1$
$= 0.1 m^{3}$

TS- EAMCET-11.09.2020

Mechanical Properties of Fluids

143194 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 cm$ of $Hg$ , where the velocity is $35 cm - s^{- 1}$ . At a point where the velocity is $65 cm - s^{- 1}$ , the pressure will be

1

0.89 cm

of

Hg

2

8.9 cm

of

Hg

3

0.5 cm

of

Hg

4

1 cm

of

Hg

Explanation:

A Given,
$v_{1} = 35 cm / \sec = 35 \times 10^{- 2} m / \sec$
$v_{2} = 65 cm / \sec = 65 \times 10^{- 2} m / \sec$
$P_{1} = ρ {gh}_{1} = 13600 \times 9.8 \times 10^{- 2}$
${∵ h_{1} = 1 cm = 1 \times 10^{- 2} m}$
$P_{2} = {ggh}_{2} = 13600 \times 9.8 \times h$
$z_{1} = z_{2}$
According to Bernoulli equation
$P_{1} + \frac{1}{2} ρ v_{1}^{2} + z_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + z_{2}$
$13600 \times 9.8 \times 10^{- 2} + 1 / 2 \times 1000 (0.35)^{2} = 13600 \times 9.8$
$\times h + \frac{1}{2} \times 1000 (0.65)^{2}$
$1394.05 = 133280 h + 221.25$
$\Rightarrow h = 0.00887 m$
$h = 0.89 cm of Hg$

JCECE-2014

Mechanical Properties of Fluids

143195 Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is $V$ . If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

1

V / 4

2

V / 8

3

16 V

4

V / 32

Explanation:

D Given,
$l_{2} = 2 l_{1}$
$d_{2} = \frac{d_{1}}{2} \Rightarrow r_{2} = \frac{r_{1}}{2}$
According to Poiseuille's formula-
$V = \frac{π \Pr^{4}}{8 n l}$
$V \propto \frac{r^{4}}{l}$
$\frac{V_{2}}{V_{1}} = \frac{r_{1}^{4}}{r_{2}^{4}} \times \frac{l_{2}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{r_{1}^{4}}{{(\frac{r_{1}}{2})}^{4}} \times \frac{2 l_{1}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = (2)^{4} \times 2$
$V_{2} = \frac{V}{32} (∵ V_{1} = V)$

COMEDK 2015

Mechanical Properties of Fluids

143197 A boy has $60 kg$ weight. He wants to swim in a river with the help of a wooden log. If relative density of wood is 0.6 , what is the minimum volume of wooden log?
(density of river water is $1000 kg / m^{3}$ )

1

0.66 m^{3}

2

150 m^{3}

3

\frac{3}{1} m^{3}

4

\frac{3}{20} m^{3}

Explanation:

D Given,
$m = 60 kg$
Relative density of wood $= 0.6$
$ρ_{wood} = 600 kg / m^{3}$
$ρ_{water} = 1000 kg / m^{3}$
According to Archimedes' principle,
body displaced by liquid $=$ up thrust
$mg + V ρ_{wood} g = V ρ_{water} g$
$60 \times g + V \times 0.6 \times 10^{3} g = V \times 1000 g$
$60 + 600 V = 1000 V$
$V = 60 / 400$
$V = \frac{3}{20} m^{3}$

AIIMS-2010

Mechanical Properties of Fluids

143199 A piece of wax weighs $18.03 g$ in air. A piece of metal is found to weigh $17.03 g$ in water. It is tied to the wax and both together weight $15.23 g$ in water. Then, the specific gravity of wax is :

1

\frac{18.03}{17.03}

2

\frac{17.03}{18.03}

3

\frac{18.03}{19.83}

4

\frac{15.03}{17.03}

Explanation:

C Given,
$W_{w_{1}} = 18.03 g$
$W_{m} = 17.03 g$
$W = 15.23 g$
$W_{w_{2}} = 15.23 - 17.03 = - 1.8 g$
We know that,
Specific gravity of wax $= \frac{W_{w_{1}}}{W_{w_{1}} - W_{w_{2}}}$
$g = \frac{18.03}{18.03 - (- 1.8)}$
$g = \frac{18.03}{19.83}$
Total weight
$W = W_{m} + (W_{wax} - v \frac{ρ_{wax}}{ρ_{water}} \times ρ_{w} g)$
$15.23 = 17.03 + (18.03 - \frac{18.03}{S.G.})$
${∵ S.G. specific gravity}$
$S.G. = \frac{18.03}{19.83}$

BCECE-2003

Mechanical Properties of Fluids

143193 A block of iron contains a hollow cavity as shown below. The block weighs $6000 N$ in air and $4000 N$ in water. If the density of iron and water are $6 g / {cm}^{3}$ and $1 g / {cm}^{3}$ , then the volume of the cavity is

1

0.05 m

2

0.5 m^{3}

3

0.25 m^{3}

4

0.1 m^{3}

Explanation:

D Given,
$ρ_{iron} = 6 g / {cm}^{3} = 6 \times 10^{3} kg / m^{3}$
$ρ_{Water} = 1 g / {cm}^{3} = 1 \times 10^{3} kg / m^{3}$
$W_{air} = 6000 N$
$w_{water} = 4000 N$
Volume of cavity
$V_{cavity} = \frac{w_{air} - w_{water}}{ρ_{w} \times g} - \frac{w_{air}}{ρ_{iron} \times g}$
$= \frac{6000 - 4000}{1 \times 10^{3} \times 10} - \frac{6000}{6 \times 10^{3} \times 10}$
$= 0.2 - 0.1$
$= 0.1 m^{3}$

TS- EAMCET-11.09.2020

Mechanical Properties of Fluids

143194 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 cm$ of $Hg$ , where the velocity is $35 cm - s^{- 1}$ . At a point where the velocity is $65 cm - s^{- 1}$ , the pressure will be

1

0.89 cm

of

Hg

2

8.9 cm

of

Hg

3

0.5 cm

of

Hg

4

1 cm

of

Hg

Explanation:

A Given,
$v_{1} = 35 cm / \sec = 35 \times 10^{- 2} m / \sec$
$v_{2} = 65 cm / \sec = 65 \times 10^{- 2} m / \sec$
$P_{1} = ρ {gh}_{1} = 13600 \times 9.8 \times 10^{- 2}$
${∵ h_{1} = 1 cm = 1 \times 10^{- 2} m}$
$P_{2} = {ggh}_{2} = 13600 \times 9.8 \times h$
$z_{1} = z_{2}$
According to Bernoulli equation
$P_{1} + \frac{1}{2} ρ v_{1}^{2} + z_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + z_{2}$
$13600 \times 9.8 \times 10^{- 2} + 1 / 2 \times 1000 (0.35)^{2} = 13600 \times 9.8$
$\times h + \frac{1}{2} \times 1000 (0.65)^{2}$
$1394.05 = 133280 h + 221.25$
$\Rightarrow h = 0.00887 m$
$h = 0.89 cm of Hg$

JCECE-2014

Mechanical Properties of Fluids

143195 Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is $V$ . If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

1

V / 4

2

V / 8

3

16 V

4

V / 32

Explanation:

D Given,
$l_{2} = 2 l_{1}$
$d_{2} = \frac{d_{1}}{2} \Rightarrow r_{2} = \frac{r_{1}}{2}$
According to Poiseuille's formula-
$V = \frac{π \Pr^{4}}{8 n l}$
$V \propto \frac{r^{4}}{l}$
$\frac{V_{2}}{V_{1}} = \frac{r_{1}^{4}}{r_{2}^{4}} \times \frac{l_{2}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{r_{1}^{4}}{{(\frac{r_{1}}{2})}^{4}} \times \frac{2 l_{1}}{l_{1}}$
$\frac{V_{1}}{V_{2}} = (2)^{4} \times 2$
$V_{2} = \frac{V}{32} (∵ V_{1} = V)$

COMEDK 2015

Mechanical Properties of Fluids

143197 A boy has $60 kg$ weight. He wants to swim in a river with the help of a wooden log. If relative density of wood is 0.6 , what is the minimum volume of wooden log?
(density of river water is $1000 kg / m^{3}$ )

1

0.66 m^{3}

2

150 m^{3}

3

\frac{3}{1} m^{3}

4

\frac{3}{20} m^{3}

Explanation:

D Given,
$m = 60 kg$
Relative density of wood $= 0.6$
$ρ_{wood} = 600 kg / m^{3}$
$ρ_{water} = 1000 kg / m^{3}$
According to Archimedes' principle,
body displaced by liquid $=$ up thrust
$mg + V ρ_{wood} g = V ρ_{water} g$
$60 \times g + V \times 0.6 \times 10^{3} g = V \times 1000 g$
$60 + 600 V = 1000 V$
$V = 60 / 400$
$V = \frac{3}{20} m^{3}$

AIIMS-2010

Mechanical Properties of Fluids

143199 A piece of wax weighs $18.03 g$ in air. A piece of metal is found to weigh $17.03 g$ in water. It is tied to the wax and both together weight $15.23 g$ in water. Then, the specific gravity of wax is :

1

\frac{18.03}{17.03}

2

\frac{17.03}{18.03}

3

\frac{18.03}{19.83}

4

\frac{15.03}{17.03}

Explanation:

C Given,
$W_{w_{1}} = 18.03 g$
$W_{m} = 17.03 g$
$W = 15.23 g$
$W_{w_{2}} = 15.23 - 17.03 = - 1.8 g$
We know that,
Specific gravity of wax $= \frac{W_{w_{1}}}{W_{w_{1}} - W_{w_{2}}}$
$g = \frac{18.03}{18.03 - (- 1.8)}$
$g = \frac{18.03}{19.83}$
Total weight
$W = W_{m} + (W_{wax} - v \frac{ρ_{wax}}{ρ_{water}} \times ρ_{w} g)$
$15.23 = 17.03 + (18.03 - \frac{18.03}{S.G.})$
${∵ S.G. specific gravity}$
$S.G. = \frac{18.03}{19.83}$

BCECE-2003

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