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Mechanical Properties of Fluids

143097 In a barometer, the mercury level is $76 cm$ at sea level. On a hill of height $3 km$ . If the ratio of density of $Hg$ to that of air is $10^{4}$ , the atmospheric pressure on the hill is

1

26 cm

of

Hg

2

46 cm

of

Hg

3

36 cm

of Hg

4

56 cm

of

Hg

Explanation:

B Given,
Height of mercury level $= 76 cm$
Pressure at sea level $(P_{1}) = ρ_{Hg} \times g \times 0.76$
Height of hill from sea level $= 3000 m$
Pressure at hill $(P_{2}) = ρ_{air} \times g \times 3000$
Now,
Atmospheric pressure at hill $= P_{3}$
$∴ P_{3} + P_{2} = P_{1}$
$P_{3} + ρ_{air} \times g \times 3000 = ρ_{Hg} \times g \times 0.76$
$P_{3} + 1.36 \times g \times 3000 = 13.6 \times 10^{3} \times g \times 0.76$
$P_{3} + 40.02 \times 10^{3} = 101.396 \times 10^{3}$
Now, $P_{3} = 61.37 \times 10^{3} Pa$
$61.37 \times 10^{3} = 13.6 \times 10^{3} \times 9.81 \times h$
$h = 0.4599 m$ of $Hg$
$∴$ $h = 46 cm$ of $Hg$

AP EAMCET-23.04.2019

Mechanical Properties of Fluids

143098 A man is carrying a block of certain material (density $1000 kg . m^{- 3}$ ) weighing $1 kg$ in his left hand and a bucket full of water, weighing 10 $kg$ , in his right hand. He drops the block in the bucket. How much load does he carry in his right hand now?

1

9 kg

2

10 kg

3

11 kg

4

12 kg

Explanation:

C Given,
Weighing of block in left hand of man $= 1 kg$
Weighing of bucket full of water in right hand of $man = 10 kg$
Now, After block is drops in the bucket the weighing in the right hand $=$ bucket + block
$= 10 + 1$
$= 11 kg$

AP EAMCET-25.09.2020

Mechanical Properties of Fluids

143099 The excess pressure inside a spherical soap bubble of radius $1 cm$ is balanced by a column of oil (Specific gravity $= 0.8$ ), $2 mm$ high, the surface tension of the bubble is

1

3.92 N / m

2

0.0392 N / m

3

0.392 N / m

4

0.00392 N / m

Explanation:

B Given that,
Radius of spherical soap bubble $= 1 cm = 1 \times 10^{- 2} m$ Height of column $= 2 mm = 2 \times 10^{- 3} m$
$Δ P = \frac{4 T}{R}$
$ρ gh = \frac{4 T}{R}$
$T = \frac{R \cdot ρ gh}{4}$
$= \frac{1 \times 10^{- 2} \times 2 \times 10^{- 3} \times 0.8 \times 10^{3} \times 9.8}{4}$
$= 3.92 \times 10^{- 2}$
$= 0.0392 N / m$

AP EAMCET -2010

Mechanical Properties of Fluids

143100 When a big drop of water is formed from $n$ small drops of water, the energy loss is 3E, where, $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then number of smaller drops $(n)$ is

1

\frac{n R}{r^{2}}

2

\frac{nR}{r}

3

\frac{2 R^{2}}{r}

4

\frac{4 R^{2}}{r^{2}}

Explanation:

D Given,
Number of small drops $= n$
Energy of big drop $= E$
Energy loss of $n$ small drop $(E_{L}) = 3 E$
Radius of big drop $= R$
Radius of small drop $= r$
We know that,
Total surface area of smaller drops $A_{1} = n \times 4 π r^{2}$
Surface area of bigger drop $A_{2} = 4 π R^{2}$
change in surface area $Δ A = A_{1} - A_{2} = 4 π (n r^{2} - R^{2})$
Now, Energy of bigger drop, $E = T \times A_{2} = 4 π T R^{2} \dots$ (i)
Energy loss in the formation of bigger drop, $E_{L} = T Δ A$
$∴ 3 E = T \times 4 π ({nr}^{2} - R^{2})$
$3 \times 4 π TR^{2} = 4 π T ({nr}^{2} - R^{2})$
Now, $3 R^{2} = n r^{2} - R^{2}$
$\Rightarrow n = \frac{4 R^{2}}{r^{2}}$

AP EAMCET -2014

Mechanical Properties of Fluids

143097 In a barometer, the mercury level is $76 cm$ at sea level. On a hill of height $3 km$ . If the ratio of density of $Hg$ to that of air is $10^{4}$ , the atmospheric pressure on the hill is

1

26 cm

of

Hg

2

46 cm

of

Hg

3

36 cm

of Hg

4

56 cm

of

Hg

Explanation:

B Given,
Height of mercury level $= 76 cm$
Pressure at sea level $(P_{1}) = ρ_{Hg} \times g \times 0.76$
Height of hill from sea level $= 3000 m$
Pressure at hill $(P_{2}) = ρ_{air} \times g \times 3000$
Now,
Atmospheric pressure at hill $= P_{3}$
$∴ P_{3} + P_{2} = P_{1}$
$P_{3} + ρ_{air} \times g \times 3000 = ρ_{Hg} \times g \times 0.76$
$P_{3} + 1.36 \times g \times 3000 = 13.6 \times 10^{3} \times g \times 0.76$
$P_{3} + 40.02 \times 10^{3} = 101.396 \times 10^{3}$
Now, $P_{3} = 61.37 \times 10^{3} Pa$
$61.37 \times 10^{3} = 13.6 \times 10^{3} \times 9.81 \times h$
$h = 0.4599 m$ of $Hg$
$∴$ $h = 46 cm$ of $Hg$

AP EAMCET-23.04.2019

Mechanical Properties of Fluids

143098 A man is carrying a block of certain material (density $1000 kg . m^{- 3}$ ) weighing $1 kg$ in his left hand and a bucket full of water, weighing 10 $kg$ , in his right hand. He drops the block in the bucket. How much load does he carry in his right hand now?

1

9 kg

2

10 kg

3

11 kg

4

12 kg

Explanation:

C Given,
Weighing of block in left hand of man $= 1 kg$
Weighing of bucket full of water in right hand of $man = 10 kg$
Now, After block is drops in the bucket the weighing in the right hand $=$ bucket + block
$= 10 + 1$
$= 11 kg$

AP EAMCET-25.09.2020

Mechanical Properties of Fluids

143099 The excess pressure inside a spherical soap bubble of radius $1 cm$ is balanced by a column of oil (Specific gravity $= 0.8$ ), $2 mm$ high, the surface tension of the bubble is

1

3.92 N / m

2

0.0392 N / m

3

0.392 N / m

4

0.00392 N / m

Explanation:

B Given that,
Radius of spherical soap bubble $= 1 cm = 1 \times 10^{- 2} m$ Height of column $= 2 mm = 2 \times 10^{- 3} m$
$Δ P = \frac{4 T}{R}$
$ρ gh = \frac{4 T}{R}$
$T = \frac{R \cdot ρ gh}{4}$
$= \frac{1 \times 10^{- 2} \times 2 \times 10^{- 3} \times 0.8 \times 10^{3} \times 9.8}{4}$
$= 3.92 \times 10^{- 2}$
$= 0.0392 N / m$

AP EAMCET -2010

Mechanical Properties of Fluids

143100 When a big drop of water is formed from $n$ small drops of water, the energy loss is 3E, where, $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then number of smaller drops $(n)$ is

1

\frac{n R}{r^{2}}

2

\frac{nR}{r}

3

\frac{2 R^{2}}{r}

4

\frac{4 R^{2}}{r^{2}}

Explanation:

D Given,
Number of small drops $= n$
Energy of big drop $= E$
Energy loss of $n$ small drop $(E_{L}) = 3 E$
Radius of big drop $= R$
Radius of small drop $= r$
We know that,
Total surface area of smaller drops $A_{1} = n \times 4 π r^{2}$
Surface area of bigger drop $A_{2} = 4 π R^{2}$
change in surface area $Δ A = A_{1} - A_{2} = 4 π (n r^{2} - R^{2})$
Now, Energy of bigger drop, $E = T \times A_{2} = 4 π T R^{2} \dots$ (i)
Energy loss in the formation of bigger drop, $E_{L} = T Δ A$
$∴ 3 E = T \times 4 π ({nr}^{2} - R^{2})$
$3 \times 4 π TR^{2} = 4 π T ({nr}^{2} - R^{2})$
Now, $3 R^{2} = n r^{2} - R^{2}$
$\Rightarrow n = \frac{4 R^{2}}{r^{2}}$

AP EAMCET -2014

Mechanical Properties of Fluids

143097 In a barometer, the mercury level is $76 cm$ at sea level. On a hill of height $3 km$ . If the ratio of density of $Hg$ to that of air is $10^{4}$ , the atmospheric pressure on the hill is

1

26 cm

of

Hg

2

46 cm

of

Hg

3

36 cm

of Hg

4

56 cm

of

Hg

Explanation:

B Given,
Height of mercury level $= 76 cm$
Pressure at sea level $(P_{1}) = ρ_{Hg} \times g \times 0.76$
Height of hill from sea level $= 3000 m$
Pressure at hill $(P_{2}) = ρ_{air} \times g \times 3000$
Now,
Atmospheric pressure at hill $= P_{3}$
$∴ P_{3} + P_{2} = P_{1}$
$P_{3} + ρ_{air} \times g \times 3000 = ρ_{Hg} \times g \times 0.76$
$P_{3} + 1.36 \times g \times 3000 = 13.6 \times 10^{3} \times g \times 0.76$
$P_{3} + 40.02 \times 10^{3} = 101.396 \times 10^{3}$
Now, $P_{3} = 61.37 \times 10^{3} Pa$
$61.37 \times 10^{3} = 13.6 \times 10^{3} \times 9.81 \times h$
$h = 0.4599 m$ of $Hg$
$∴$ $h = 46 cm$ of $Hg$

AP EAMCET-23.04.2019

Mechanical Properties of Fluids

143098 A man is carrying a block of certain material (density $1000 kg . m^{- 3}$ ) weighing $1 kg$ in his left hand and a bucket full of water, weighing 10 $kg$ , in his right hand. He drops the block in the bucket. How much load does he carry in his right hand now?

1

9 kg

2

10 kg

3

11 kg

4

12 kg

Explanation:

C Given,
Weighing of block in left hand of man $= 1 kg$
Weighing of bucket full of water in right hand of $man = 10 kg$
Now, After block is drops in the bucket the weighing in the right hand $=$ bucket + block
$= 10 + 1$
$= 11 kg$

AP EAMCET-25.09.2020

Mechanical Properties of Fluids

143099 The excess pressure inside a spherical soap bubble of radius $1 cm$ is balanced by a column of oil (Specific gravity $= 0.8$ ), $2 mm$ high, the surface tension of the bubble is

1

3.92 N / m

2

0.0392 N / m

3

0.392 N / m

4

0.00392 N / m

Explanation:

B Given that,
Radius of spherical soap bubble $= 1 cm = 1 \times 10^{- 2} m$ Height of column $= 2 mm = 2 \times 10^{- 3} m$
$Δ P = \frac{4 T}{R}$
$ρ gh = \frac{4 T}{R}$
$T = \frac{R \cdot ρ gh}{4}$
$= \frac{1 \times 10^{- 2} \times 2 \times 10^{- 3} \times 0.8 \times 10^{3} \times 9.8}{4}$
$= 3.92 \times 10^{- 2}$
$= 0.0392 N / m$

AP EAMCET -2010

Mechanical Properties of Fluids

143100 When a big drop of water is formed from $n$ small drops of water, the energy loss is 3E, where, $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then number of smaller drops $(n)$ is

1

\frac{n R}{r^{2}}

2

\frac{nR}{r}

3

\frac{2 R^{2}}{r}

4

\frac{4 R^{2}}{r^{2}}

Explanation:

D Given,
Number of small drops $= n$
Energy of big drop $= E$
Energy loss of $n$ small drop $(E_{L}) = 3 E$
Radius of big drop $= R$
Radius of small drop $= r$
We know that,
Total surface area of smaller drops $A_{1} = n \times 4 π r^{2}$
Surface area of bigger drop $A_{2} = 4 π R^{2}$
change in surface area $Δ A = A_{1} - A_{2} = 4 π (n r^{2} - R^{2})$
Now, Energy of bigger drop, $E = T \times A_{2} = 4 π T R^{2} \dots$ (i)
Energy loss in the formation of bigger drop, $E_{L} = T Δ A$
$∴ 3 E = T \times 4 π ({nr}^{2} - R^{2})$
$3 \times 4 π TR^{2} = 4 π T ({nr}^{2} - R^{2})$
Now, $3 R^{2} = n r^{2} - R^{2}$
$\Rightarrow n = \frac{4 R^{2}}{r^{2}}$

AP EAMCET -2014

Mechanical Properties of Fluids

143097 In a barometer, the mercury level is $76 cm$ at sea level. On a hill of height $3 km$ . If the ratio of density of $Hg$ to that of air is $10^{4}$ , the atmospheric pressure on the hill is

1

26 cm

of

Hg

2

46 cm

of

Hg

3

36 cm

of Hg

4

56 cm

of

Hg

Explanation:

B Given,
Height of mercury level $= 76 cm$
Pressure at sea level $(P_{1}) = ρ_{Hg} \times g \times 0.76$
Height of hill from sea level $= 3000 m$
Pressure at hill $(P_{2}) = ρ_{air} \times g \times 3000$
Now,
Atmospheric pressure at hill $= P_{3}$
$∴ P_{3} + P_{2} = P_{1}$
$P_{3} + ρ_{air} \times g \times 3000 = ρ_{Hg} \times g \times 0.76$
$P_{3} + 1.36 \times g \times 3000 = 13.6 \times 10^{3} \times g \times 0.76$
$P_{3} + 40.02 \times 10^{3} = 101.396 \times 10^{3}$
Now, $P_{3} = 61.37 \times 10^{3} Pa$
$61.37 \times 10^{3} = 13.6 \times 10^{3} \times 9.81 \times h$
$h = 0.4599 m$ of $Hg$
$∴$ $h = 46 cm$ of $Hg$

AP EAMCET-23.04.2019

Mechanical Properties of Fluids

143098 A man is carrying a block of certain material (density $1000 kg . m^{- 3}$ ) weighing $1 kg$ in his left hand and a bucket full of water, weighing 10 $kg$ , in his right hand. He drops the block in the bucket. How much load does he carry in his right hand now?

1

9 kg

2

10 kg

3

11 kg

4

12 kg

Explanation:

C Given,
Weighing of block in left hand of man $= 1 kg$
Weighing of bucket full of water in right hand of $man = 10 kg$
Now, After block is drops in the bucket the weighing in the right hand $=$ bucket + block
$= 10 + 1$
$= 11 kg$

AP EAMCET-25.09.2020

Mechanical Properties of Fluids

143099 The excess pressure inside a spherical soap bubble of radius $1 cm$ is balanced by a column of oil (Specific gravity $= 0.8$ ), $2 mm$ high, the surface tension of the bubble is

1

3.92 N / m

2

0.0392 N / m

3

0.392 N / m

4

0.00392 N / m

Explanation:

B Given that,
Radius of spherical soap bubble $= 1 cm = 1 \times 10^{- 2} m$ Height of column $= 2 mm = 2 \times 10^{- 3} m$
$Δ P = \frac{4 T}{R}$
$ρ gh = \frac{4 T}{R}$
$T = \frac{R \cdot ρ gh}{4}$
$= \frac{1 \times 10^{- 2} \times 2 \times 10^{- 3} \times 0.8 \times 10^{3} \times 9.8}{4}$
$= 3.92 \times 10^{- 2}$
$= 0.0392 N / m$

AP EAMCET -2010

Mechanical Properties of Fluids

143100 When a big drop of water is formed from $n$ small drops of water, the energy loss is 3E, where, $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then number of smaller drops $(n)$ is

1

\frac{n R}{r^{2}}

2

\frac{nR}{r}

3

\frac{2 R^{2}}{r}

4

\frac{4 R^{2}}{r^{2}}

Explanation:

D Given,
Number of small drops $= n$
Energy of big drop $= E$
Energy loss of $n$ small drop $(E_{L}) = 3 E$
Radius of big drop $= R$
Radius of small drop $= r$
We know that,
Total surface area of smaller drops $A_{1} = n \times 4 π r^{2}$
Surface area of bigger drop $A_{2} = 4 π R^{2}$
change in surface area $Δ A = A_{1} - A_{2} = 4 π (n r^{2} - R^{2})$
Now, Energy of bigger drop, $E = T \times A_{2} = 4 π T R^{2} \dots$ (i)
Energy loss in the formation of bigger drop, $E_{L} = T Δ A$
$∴ 3 E = T \times 4 π ({nr}^{2} - R^{2})$
$3 \times 4 π TR^{2} = 4 π T ({nr}^{2} - R^{2})$
Now, $3 R^{2} = n r^{2} - R^{2}$
$\Rightarrow n = \frac{4 R^{2}}{r^{2}}$

AP EAMCET -2014

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