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Mechanical Properties of Fluids

142978 If a capillary tube of radius $R$ is immersed in water, the mass of water risen in capillary is $M$ . If the radius of capillary be doubled, the mass of water risen in capillary will be

1

\frac{M}{2}

2

M

3

2 M

4

4 M

Explanation:

C Given,
Initial mass, $m_{1} = M$
Initial radius, $r_{1} = r$
Final radius, $r_{2} = 2 r$
We know that,
Mass of water in capillary tube $(m) =$ volume $\times$ density $= A \times h \times ρ$
Where $A =$ Area of cross-section $= π r^{2}$
Height of water raised $(h) = \frac{2 σ \cos θ}{r ρ g}$
$ρ =$ density of water $m = (π r^{2}) \times h \times ρ$
$m = π r^{2} \times \frac{2 σ \cos θ}{r ρ g} \times ρ$
$m = 2 π r \times \frac{\cos θ}{g}$
$m \propto r$
Then,
$\frac{m_{1}}{m_{2}} = \frac{r_{1}}{r_{2}}$
When radius is double,
$\frac{M}{M_{2}} = \frac{r}{2 r}$
Hence, $M_{2} = 2 M$

UP CPMT-2002

Mechanical Properties of Fluids

142979 In a capillary tube, water rises by $2 mm$ . The height of water that will rise in another capillary tube of half to radius, will be

1

1 mm

2

2 mm

3

4 mm

4

8 mm

Explanation:

C Given,
Initial Height, $h_{1} = 2 mm$
Final Height, $h_{2} =$ ?
Initial radius, $r_{1} = r$
Final radius, $r_{2} = r / 2$
We know that,
Capillary rise in a tube $(h) = \frac{2 T \cos θ}{ρ g r}$
$h \propto \frac{1}{r}$
So, $\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}$
$\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}$
$\frac{2}{h_{2}} = \frac{r}{2 \times r}$
Hence, $h_{2} = 4 mm$

UP CPMT-2001

Mechanical Properties of Fluids

142980 Water rises to a height $H$ in a capillary tube of area of cross section A. To what height will water rise in a capillary tube of area of cross section $4 A$ .

1

\frac{H}{4}

2

\frac{H}{2}

3

2 H

4

4 H

Explanation:

B Given, $H_{1} = H, H_{2} =$ ?, $A_{1} = A, A_{2} = 4 A$ We know that,
$H = \frac{2 T \cos θ}{ρ gr}$
Since, $T, ρ, g, \cos θ$ are constant in both cases.
Then,
$H \propto \frac{1}{r}$
And, $A = π r^{2}$
$r = {(\frac{A}{π})}^{1 / 2}$
$r \propto (A)^{1 / 2}$
Hence, $H \propto \frac{1}{(A)^{1 / 2}}$
Now, $\frac{H_{2}}{H_{1}} = \sqrt{\frac{A_{1}}{A_{2}}}$
$H_{2} = \sqrt{\frac{1}{4}} H$
$H_{2} = \frac{H}{2}$

TS EAMCET 31.07.2022

Mechanical Properties of Fluids

142981 Liquid $A$ rises to a height of $10 cm$ in a capillary tube and liquid $B$ falls to a depth of 2 $cm$ in the same tube. The density of $A$ and $B$ are $1 g / {cm}^{3}$ and $10 g / {cm}^{3}$ respectively. The contact angle of $A$ and $B$ with the tube is $0^{\circ}$ and $135^{\circ}$ respectively. If the surface tension of $A$ and $B$ are $S_{A}$ and $S_{B}$ then the ratio $\frac{S_{B}}{S_{A}}$ is:

1

\sqrt{2}

2

2 \sqrt{2}

3

\frac{1}{\sqrt{2}}

4

\frac{1}{2 \sqrt{2}}

Explanation:

B Given that, $h_{A} = 10 cm, h_{B} = 2 cm$ , $ρ_{A} = 1 g / {cm}^{3}, ρ_{B} = 10 g / {cm}^{3}, θ_{A} = 0^{\circ}, θ_{B} = 135^{\circ}$
We know,
$h = \frac{2 T \cos θ}{r ρ g}$
Now, $\frac{h_{A}}{h_{B}} = \frac{2 S_{A} \cos θ_{A}}{r ρ_{1} g} \times \frac{r ρ_{2} g}{2 S_{B} \cos θ_{B}}$
$\frac{h_{A}}{h_{B}} = \frac{S_{A} \cos θ_{A}}{ρ_{1}} \times \frac{ρ_{2}}{S_{B} \cos θ_{B}}$
Then, $\frac{S_{A}}{S_{B}} = \frac{h_{A}}{h_{B}} \frac{ρ_{1}}{ρ_{2}} \frac{\cos θ_{B}}{\cos θ_{A}}$
Putting the value in above equation,
$= \frac{10}{- 2} \times \frac{1}{10} \times \frac{\cos 135^{\circ}}{\cos 0^{\circ}}$
$= \frac{1}{- 2} \times \frac{\cos (180 - 45^{\circ})}{1}$
$= \frac{1}{- 2} \times - \cos 45^{\circ}$
$\frac{S_{A}}{S_{B}} = \frac{1}{- 2} \times (\frac{- 1}{\sqrt{2}}) = \frac{1}{2 \sqrt{2}} m / \sec$
$\frac{S_{B}}{S_{A}} = 2 \sqrt{2}$

TS EAMCET 05.08.2021

Mechanical Properties of Fluids

142978 If a capillary tube of radius $R$ is immersed in water, the mass of water risen in capillary is $M$ . If the radius of capillary be doubled, the mass of water risen in capillary will be

1

\frac{M}{2}

2

M

3

2 M

4

4 M

Explanation:

C Given,
Initial mass, $m_{1} = M$
Initial radius, $r_{1} = r$
Final radius, $r_{2} = 2 r$
We know that,
Mass of water in capillary tube $(m) =$ volume $\times$ density $= A \times h \times ρ$
Where $A =$ Area of cross-section $= π r^{2}$
Height of water raised $(h) = \frac{2 σ \cos θ}{r ρ g}$
$ρ =$ density of water $m = (π r^{2}) \times h \times ρ$
$m = π r^{2} \times \frac{2 σ \cos θ}{r ρ g} \times ρ$
$m = 2 π r \times \frac{\cos θ}{g}$
$m \propto r$
Then,
$\frac{m_{1}}{m_{2}} = \frac{r_{1}}{r_{2}}$
When radius is double,
$\frac{M}{M_{2}} = \frac{r}{2 r}$
Hence, $M_{2} = 2 M$

UP CPMT-2002

Mechanical Properties of Fluids

142979 In a capillary tube, water rises by $2 mm$ . The height of water that will rise in another capillary tube of half to radius, will be

1

1 mm

2

2 mm

3

4 mm

4

8 mm

Explanation:

C Given,
Initial Height, $h_{1} = 2 mm$
Final Height, $h_{2} =$ ?
Initial radius, $r_{1} = r$
Final radius, $r_{2} = r / 2$
We know that,
Capillary rise in a tube $(h) = \frac{2 T \cos θ}{ρ g r}$
$h \propto \frac{1}{r}$
So, $\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}$
$\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}$
$\frac{2}{h_{2}} = \frac{r}{2 \times r}$
Hence, $h_{2} = 4 mm$

UP CPMT-2001

Mechanical Properties of Fluids

142980 Water rises to a height $H$ in a capillary tube of area of cross section A. To what height will water rise in a capillary tube of area of cross section $4 A$ .

1

\frac{H}{4}

2

\frac{H}{2}

3

2 H

4

4 H

Explanation:

B Given, $H_{1} = H, H_{2} =$ ?, $A_{1} = A, A_{2} = 4 A$ We know that,
$H = \frac{2 T \cos θ}{ρ gr}$
Since, $T, ρ, g, \cos θ$ are constant in both cases.
Then,
$H \propto \frac{1}{r}$
And, $A = π r^{2}$
$r = {(\frac{A}{π})}^{1 / 2}$
$r \propto (A)^{1 / 2}$
Hence, $H \propto \frac{1}{(A)^{1 / 2}}$
Now, $\frac{H_{2}}{H_{1}} = \sqrt{\frac{A_{1}}{A_{2}}}$
$H_{2} = \sqrt{\frac{1}{4}} H$
$H_{2} = \frac{H}{2}$

TS EAMCET 31.07.2022

Mechanical Properties of Fluids

142981 Liquid $A$ rises to a height of $10 cm$ in a capillary tube and liquid $B$ falls to a depth of 2 $cm$ in the same tube. The density of $A$ and $B$ are $1 g / {cm}^{3}$ and $10 g / {cm}^{3}$ respectively. The contact angle of $A$ and $B$ with the tube is $0^{\circ}$ and $135^{\circ}$ respectively. If the surface tension of $A$ and $B$ are $S_{A}$ and $S_{B}$ then the ratio $\frac{S_{B}}{S_{A}}$ is:

1

\sqrt{2}

2

2 \sqrt{2}

3

\frac{1}{\sqrt{2}}

4

\frac{1}{2 \sqrt{2}}

Explanation:

B Given that, $h_{A} = 10 cm, h_{B} = 2 cm$ , $ρ_{A} = 1 g / {cm}^{3}, ρ_{B} = 10 g / {cm}^{3}, θ_{A} = 0^{\circ}, θ_{B} = 135^{\circ}$
We know,
$h = \frac{2 T \cos θ}{r ρ g}$
Now, $\frac{h_{A}}{h_{B}} = \frac{2 S_{A} \cos θ_{A}}{r ρ_{1} g} \times \frac{r ρ_{2} g}{2 S_{B} \cos θ_{B}}$
$\frac{h_{A}}{h_{B}} = \frac{S_{A} \cos θ_{A}}{ρ_{1}} \times \frac{ρ_{2}}{S_{B} \cos θ_{B}}$
Then, $\frac{S_{A}}{S_{B}} = \frac{h_{A}}{h_{B}} \frac{ρ_{1}}{ρ_{2}} \frac{\cos θ_{B}}{\cos θ_{A}}$
Putting the value in above equation,
$= \frac{10}{- 2} \times \frac{1}{10} \times \frac{\cos 135^{\circ}}{\cos 0^{\circ}}$
$= \frac{1}{- 2} \times \frac{\cos (180 - 45^{\circ})}{1}$
$= \frac{1}{- 2} \times - \cos 45^{\circ}$
$\frac{S_{A}}{S_{B}} = \frac{1}{- 2} \times (\frac{- 1}{\sqrt{2}}) = \frac{1}{2 \sqrt{2}} m / \sec$
$\frac{S_{B}}{S_{A}} = 2 \sqrt{2}$

TS EAMCET 05.08.2021

Mechanical Properties of Fluids

142978 If a capillary tube of radius $R$ is immersed in water, the mass of water risen in capillary is $M$ . If the radius of capillary be doubled, the mass of water risen in capillary will be

1

\frac{M}{2}

2

M

3

2 M

4

4 M

Explanation:

C Given,
Initial mass, $m_{1} = M$
Initial radius, $r_{1} = r$
Final radius, $r_{2} = 2 r$
We know that,
Mass of water in capillary tube $(m) =$ volume $\times$ density $= A \times h \times ρ$
Where $A =$ Area of cross-section $= π r^{2}$
Height of water raised $(h) = \frac{2 σ \cos θ}{r ρ g}$
$ρ =$ density of water $m = (π r^{2}) \times h \times ρ$
$m = π r^{2} \times \frac{2 σ \cos θ}{r ρ g} \times ρ$
$m = 2 π r \times \frac{\cos θ}{g}$
$m \propto r$
Then,
$\frac{m_{1}}{m_{2}} = \frac{r_{1}}{r_{2}}$
When radius is double,
$\frac{M}{M_{2}} = \frac{r}{2 r}$
Hence, $M_{2} = 2 M$

UP CPMT-2002

Mechanical Properties of Fluids

142979 In a capillary tube, water rises by $2 mm$ . The height of water that will rise in another capillary tube of half to radius, will be

1

1 mm

2

2 mm

3

4 mm

4

8 mm

Explanation:

C Given,
Initial Height, $h_{1} = 2 mm$
Final Height, $h_{2} =$ ?
Initial radius, $r_{1} = r$
Final radius, $r_{2} = r / 2$
We know that,
Capillary rise in a tube $(h) = \frac{2 T \cos θ}{ρ g r}$
$h \propto \frac{1}{r}$
So, $\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}$
$\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}$
$\frac{2}{h_{2}} = \frac{r}{2 \times r}$
Hence, $h_{2} = 4 mm$

UP CPMT-2001

Mechanical Properties of Fluids

142980 Water rises to a height $H$ in a capillary tube of area of cross section A. To what height will water rise in a capillary tube of area of cross section $4 A$ .

1

\frac{H}{4}

2

\frac{H}{2}

3

2 H

4

4 H

Explanation:

B Given, $H_{1} = H, H_{2} =$ ?, $A_{1} = A, A_{2} = 4 A$ We know that,
$H = \frac{2 T \cos θ}{ρ gr}$
Since, $T, ρ, g, \cos θ$ are constant in both cases.
Then,
$H \propto \frac{1}{r}$
And, $A = π r^{2}$
$r = {(\frac{A}{π})}^{1 / 2}$
$r \propto (A)^{1 / 2}$
Hence, $H \propto \frac{1}{(A)^{1 / 2}}$
Now, $\frac{H_{2}}{H_{1}} = \sqrt{\frac{A_{1}}{A_{2}}}$
$H_{2} = \sqrt{\frac{1}{4}} H$
$H_{2} = \frac{H}{2}$

TS EAMCET 31.07.2022

Mechanical Properties of Fluids

142981 Liquid $A$ rises to a height of $10 cm$ in a capillary tube and liquid $B$ falls to a depth of 2 $cm$ in the same tube. The density of $A$ and $B$ are $1 g / {cm}^{3}$ and $10 g / {cm}^{3}$ respectively. The contact angle of $A$ and $B$ with the tube is $0^{\circ}$ and $135^{\circ}$ respectively. If the surface tension of $A$ and $B$ are $S_{A}$ and $S_{B}$ then the ratio $\frac{S_{B}}{S_{A}}$ is:

1

\sqrt{2}

2

2 \sqrt{2}

3

\frac{1}{\sqrt{2}}

4

\frac{1}{2 \sqrt{2}}

Explanation:

B Given that, $h_{A} = 10 cm, h_{B} = 2 cm$ , $ρ_{A} = 1 g / {cm}^{3}, ρ_{B} = 10 g / {cm}^{3}, θ_{A} = 0^{\circ}, θ_{B} = 135^{\circ}$
We know,
$h = \frac{2 T \cos θ}{r ρ g}$
Now, $\frac{h_{A}}{h_{B}} = \frac{2 S_{A} \cos θ_{A}}{r ρ_{1} g} \times \frac{r ρ_{2} g}{2 S_{B} \cos θ_{B}}$
$\frac{h_{A}}{h_{B}} = \frac{S_{A} \cos θ_{A}}{ρ_{1}} \times \frac{ρ_{2}}{S_{B} \cos θ_{B}}$
Then, $\frac{S_{A}}{S_{B}} = \frac{h_{A}}{h_{B}} \frac{ρ_{1}}{ρ_{2}} \frac{\cos θ_{B}}{\cos θ_{A}}$
Putting the value in above equation,
$= \frac{10}{- 2} \times \frac{1}{10} \times \frac{\cos 135^{\circ}}{\cos 0^{\circ}}$
$= \frac{1}{- 2} \times \frac{\cos (180 - 45^{\circ})}{1}$
$= \frac{1}{- 2} \times - \cos 45^{\circ}$
$\frac{S_{A}}{S_{B}} = \frac{1}{- 2} \times (\frac{- 1}{\sqrt{2}}) = \frac{1}{2 \sqrt{2}} m / \sec$
$\frac{S_{B}}{S_{A}} = 2 \sqrt{2}$

TS EAMCET 05.08.2021

Mechanical Properties of Fluids

142978 If a capillary tube of radius $R$ is immersed in water, the mass of water risen in capillary is $M$ . If the radius of capillary be doubled, the mass of water risen in capillary will be

1

\frac{M}{2}

2

M

3

2 M

4

4 M

Explanation:

C Given,
Initial mass, $m_{1} = M$
Initial radius, $r_{1} = r$
Final radius, $r_{2} = 2 r$
We know that,
Mass of water in capillary tube $(m) =$ volume $\times$ density $= A \times h \times ρ$
Where $A =$ Area of cross-section $= π r^{2}$
Height of water raised $(h) = \frac{2 σ \cos θ}{r ρ g}$
$ρ =$ density of water $m = (π r^{2}) \times h \times ρ$
$m = π r^{2} \times \frac{2 σ \cos θ}{r ρ g} \times ρ$
$m = 2 π r \times \frac{\cos θ}{g}$
$m \propto r$
Then,
$\frac{m_{1}}{m_{2}} = \frac{r_{1}}{r_{2}}$
When radius is double,
$\frac{M}{M_{2}} = \frac{r}{2 r}$
Hence, $M_{2} = 2 M$

UP CPMT-2002

Mechanical Properties of Fluids

142979 In a capillary tube, water rises by $2 mm$ . The height of water that will rise in another capillary tube of half to radius, will be

1

1 mm

2

2 mm

3

4 mm

4

8 mm

Explanation:

C Given,
Initial Height, $h_{1} = 2 mm$
Final Height, $h_{2} =$ ?
Initial radius, $r_{1} = r$
Final radius, $r_{2} = r / 2$
We know that,
Capillary rise in a tube $(h) = \frac{2 T \cos θ}{ρ g r}$
$h \propto \frac{1}{r}$
So, $\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}$
$\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}$
$\frac{2}{h_{2}} = \frac{r}{2 \times r}$
Hence, $h_{2} = 4 mm$

UP CPMT-2001

Mechanical Properties of Fluids

142980 Water rises to a height $H$ in a capillary tube of area of cross section A. To what height will water rise in a capillary tube of area of cross section $4 A$ .

1

\frac{H}{4}

2

\frac{H}{2}

3

2 H

4

4 H

Explanation:

B Given, $H_{1} = H, H_{2} =$ ?, $A_{1} = A, A_{2} = 4 A$ We know that,
$H = \frac{2 T \cos θ}{ρ gr}$
Since, $T, ρ, g, \cos θ$ are constant in both cases.
Then,
$H \propto \frac{1}{r}$
And, $A = π r^{2}$
$r = {(\frac{A}{π})}^{1 / 2}$
$r \propto (A)^{1 / 2}$
Hence, $H \propto \frac{1}{(A)^{1 / 2}}$
Now, $\frac{H_{2}}{H_{1}} = \sqrt{\frac{A_{1}}{A_{2}}}$
$H_{2} = \sqrt{\frac{1}{4}} H$
$H_{2} = \frac{H}{2}$

TS EAMCET 31.07.2022

Mechanical Properties of Fluids

142981 Liquid $A$ rises to a height of $10 cm$ in a capillary tube and liquid $B$ falls to a depth of 2 $cm$ in the same tube. The density of $A$ and $B$ are $1 g / {cm}^{3}$ and $10 g / {cm}^{3}$ respectively. The contact angle of $A$ and $B$ with the tube is $0^{\circ}$ and $135^{\circ}$ respectively. If the surface tension of $A$ and $B$ are $S_{A}$ and $S_{B}$ then the ratio $\frac{S_{B}}{S_{A}}$ is:

1

\sqrt{2}

2

2 \sqrt{2}

3

\frac{1}{\sqrt{2}}

4

\frac{1}{2 \sqrt{2}}

Explanation:

B Given that, $h_{A} = 10 cm, h_{B} = 2 cm$ , $ρ_{A} = 1 g / {cm}^{3}, ρ_{B} = 10 g / {cm}^{3}, θ_{A} = 0^{\circ}, θ_{B} = 135^{\circ}$
We know,
$h = \frac{2 T \cos θ}{r ρ g}$
Now, $\frac{h_{A}}{h_{B}} = \frac{2 S_{A} \cos θ_{A}}{r ρ_{1} g} \times \frac{r ρ_{2} g}{2 S_{B} \cos θ_{B}}$
$\frac{h_{A}}{h_{B}} = \frac{S_{A} \cos θ_{A}}{ρ_{1}} \times \frac{ρ_{2}}{S_{B} \cos θ_{B}}$
Then, $\frac{S_{A}}{S_{B}} = \frac{h_{A}}{h_{B}} \frac{ρ_{1}}{ρ_{2}} \frac{\cos θ_{B}}{\cos θ_{A}}$
Putting the value in above equation,
$= \frac{10}{- 2} \times \frac{1}{10} \times \frac{\cos 135^{\circ}}{\cos 0^{\circ}}$
$= \frac{1}{- 2} \times \frac{\cos (180 - 45^{\circ})}{1}$
$= \frac{1}{- 2} \times - \cos 45^{\circ}$
$\frac{S_{A}}{S_{B}} = \frac{1}{- 2} \times (\frac{- 1}{\sqrt{2}}) = \frac{1}{2 \sqrt{2}} m / \sec$
$\frac{S_{B}}{S_{A}} = 2 \sqrt{2}$

TS EAMCET 05.08.2021