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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369724 The compressibility of water is $6 \times 10^{- 10} N^{- 1} m^{2}$ . If one litre is subjected to a pressure of $4 \times 10^{7} N m^{2}$ , the decrease in its volume is

1

10 c c

2

2.4 c c

3

15 c c

4

24 c c

Explanation:

Compressibility, $k = \frac{1}{B} = - \frac{Δ V}{P V}$
Decrease in volume, $Δ V = P V k$
$= 4 \times 10^{7} \times 1 \times 6 \times 10^{- 10} = 24 \times 10^{- 3} l i t r e$
$= 24 \times 10^{- 3} \times 10^{3} c m^{3} = 24 c c .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369725 Statement A :
The strain produced by a hydraulic pressure is volumetric in nature
Statement B :
It is a ratio of change in volume $Δ V$ to the original volume $V$ .

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

The strain produced by a hydraulic pressure is called volumetric strain as pressure creates a normal force on every point and is defined as the ratio of charge in volume $Δ V$ to the original volume $V$ .
i.e., volume strain $= \frac{Δ V}{V}$
So option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369726 The compressibility of water is $5 \times 10^{- 10} m^{2} / N .$ Pressure of $15 \times 10^{6} P a$ is applied on $100 m L$ volume of water. The change in the volume of water is

1 zero

2

0.75 m l decrease

3

0.75 m l increase

4

1.50 m l increase

Explanation:

The compressibility of water is given by
$k = - \frac{Δ V}{Δ p V}$
$\Rightarrow Δ V = - k Δ p V$
$= - (5 \times 10^{- 10} \times 15 \times 10^{6} \times 100) m l$
$= - 0.75 m l$
Thus, volume of the water decreases by $0.75 m l .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369727 The average depth of indian Ocean is about $3000 m$ . The dfractional compression, $\frac{Δ V}{V}$ of water at the bottom of the ocean is (Given: Bulk modulus of the wear $= 2.2 \times 10^{9} N / m^{- 2}$ and $g = 10 m s^{- 2})$

1

0.91 %

2

0.82 %

3

1.24 %

4

1.36 %

Explanation:

The pressure exerted by a $3000 m$ column of water on the bottom layer is
$P = h ρ g = 3000 m \times 100 k g m^{- 3} \times 10 m s^{- 2}$
Fractional compression $\frac{Δ V}{V}$ is
$\frac{Δ V}{V} = \frac{P}{B} = \frac{3 \times 10^{7} N m^{- 2}}{2.2 \times 10^{9} N m^{- 2}} = 1.36 \times 10^{- 2} = 1.36 %$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369724 The compressibility of water is $6 \times 10^{- 10} N^{- 1} m^{2}$ . If one litre is subjected to a pressure of $4 \times 10^{7} N m^{2}$ , the decrease in its volume is

1

10 c c

2

2.4 c c

3

15 c c

4

24 c c

Explanation:

Compressibility, $k = \frac{1}{B} = - \frac{Δ V}{P V}$
Decrease in volume, $Δ V = P V k$
$= 4 \times 10^{7} \times 1 \times 6 \times 10^{- 10} = 24 \times 10^{- 3} l i t r e$
$= 24 \times 10^{- 3} \times 10^{3} c m^{3} = 24 c c .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369725 Statement A :
The strain produced by a hydraulic pressure is volumetric in nature
Statement B :
It is a ratio of change in volume $Δ V$ to the original volume $V$ .

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

The strain produced by a hydraulic pressure is called volumetric strain as pressure creates a normal force on every point and is defined as the ratio of charge in volume $Δ V$ to the original volume $V$ .
i.e., volume strain $= \frac{Δ V}{V}$
So option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369726 The compressibility of water is $5 \times 10^{- 10} m^{2} / N .$ Pressure of $15 \times 10^{6} P a$ is applied on $100 m L$ volume of water. The change in the volume of water is

1 zero

2

0.75 m l decrease

3

0.75 m l increase

4

1.50 m l increase

Explanation:

The compressibility of water is given by
$k = - \frac{Δ V}{Δ p V}$
$\Rightarrow Δ V = - k Δ p V$
$= - (5 \times 10^{- 10} \times 15 \times 10^{6} \times 100) m l$
$= - 0.75 m l$
Thus, volume of the water decreases by $0.75 m l .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369727 The average depth of indian Ocean is about $3000 m$ . The dfractional compression, $\frac{Δ V}{V}$ of water at the bottom of the ocean is (Given: Bulk modulus of the wear $= 2.2 \times 10^{9} N / m^{- 2}$ and $g = 10 m s^{- 2})$

1

0.91 %

2

0.82 %

3

1.24 %

4

1.36 %

Explanation:

The pressure exerted by a $3000 m$ column of water on the bottom layer is
$P = h ρ g = 3000 m \times 100 k g m^{- 3} \times 10 m s^{- 2}$
Fractional compression $\frac{Δ V}{V}$ is
$\frac{Δ V}{V} = \frac{P}{B} = \frac{3 \times 10^{7} N m^{- 2}}{2.2 \times 10^{9} N m^{- 2}} = 1.36 \times 10^{- 2} = 1.36 %$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369724 The compressibility of water is $6 \times 10^{- 10} N^{- 1} m^{2}$ . If one litre is subjected to a pressure of $4 \times 10^{7} N m^{2}$ , the decrease in its volume is

1

10 c c

2

2.4 c c

3

15 c c

4

24 c c

Explanation:

Compressibility, $k = \frac{1}{B} = - \frac{Δ V}{P V}$
Decrease in volume, $Δ V = P V k$
$= 4 \times 10^{7} \times 1 \times 6 \times 10^{- 10} = 24 \times 10^{- 3} l i t r e$
$= 24 \times 10^{- 3} \times 10^{3} c m^{3} = 24 c c .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369725 Statement A :
The strain produced by a hydraulic pressure is volumetric in nature
Statement B :
It is a ratio of change in volume $Δ V$ to the original volume $V$ .

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

The strain produced by a hydraulic pressure is called volumetric strain as pressure creates a normal force on every point and is defined as the ratio of charge in volume $Δ V$ to the original volume $V$ .
i.e., volume strain $= \frac{Δ V}{V}$
So option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369726 The compressibility of water is $5 \times 10^{- 10} m^{2} / N .$ Pressure of $15 \times 10^{6} P a$ is applied on $100 m L$ volume of water. The change in the volume of water is

1 zero

2

0.75 m l decrease

3

0.75 m l increase

4

1.50 m l increase

Explanation:

The compressibility of water is given by
$k = - \frac{Δ V}{Δ p V}$
$\Rightarrow Δ V = - k Δ p V$
$= - (5 \times 10^{- 10} \times 15 \times 10^{6} \times 100) m l$
$= - 0.75 m l$
Thus, volume of the water decreases by $0.75 m l .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369727 The average depth of indian Ocean is about $3000 m$ . The dfractional compression, $\frac{Δ V}{V}$ of water at the bottom of the ocean is (Given: Bulk modulus of the wear $= 2.2 \times 10^{9} N / m^{- 2}$ and $g = 10 m s^{- 2})$

1

0.91 %

2

0.82 %

3

1.24 %

4

1.36 %

Explanation:

The pressure exerted by a $3000 m$ column of water on the bottom layer is
$P = h ρ g = 3000 m \times 100 k g m^{- 3} \times 10 m s^{- 2}$
Fractional compression $\frac{Δ V}{V}$ is
$\frac{Δ V}{V} = \frac{P}{B} = \frac{3 \times 10^{7} N m^{- 2}}{2.2 \times 10^{9} N m^{- 2}} = 1.36 \times 10^{- 2} = 1.36 %$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369724 The compressibility of water is $6 \times 10^{- 10} N^{- 1} m^{2}$ . If one litre is subjected to a pressure of $4 \times 10^{7} N m^{2}$ , the decrease in its volume is

1

10 c c

2

2.4 c c

3

15 c c

4

24 c c

Explanation:

Compressibility, $k = \frac{1}{B} = - \frac{Δ V}{P V}$
Decrease in volume, $Δ V = P V k$
$= 4 \times 10^{7} \times 1 \times 6 \times 10^{- 10} = 24 \times 10^{- 3} l i t r e$
$= 24 \times 10^{- 3} \times 10^{3} c m^{3} = 24 c c .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369725 Statement A :
The strain produced by a hydraulic pressure is volumetric in nature
Statement B :
It is a ratio of change in volume $Δ V$ to the original volume $V$ .

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

The strain produced by a hydraulic pressure is called volumetric strain as pressure creates a normal force on every point and is defined as the ratio of charge in volume $Δ V$ to the original volume $V$ .
i.e., volume strain $= \frac{Δ V}{V}$
So option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369726 The compressibility of water is $5 \times 10^{- 10} m^{2} / N .$ Pressure of $15 \times 10^{6} P a$ is applied on $100 m L$ volume of water. The change in the volume of water is

1 zero

2

0.75 m l decrease

3

0.75 m l increase

4

1.50 m l increase

Explanation:

The compressibility of water is given by
$k = - \frac{Δ V}{Δ p V}$
$\Rightarrow Δ V = - k Δ p V$
$= - (5 \times 10^{- 10} \times 15 \times 10^{6} \times 100) m l$
$= - 0.75 m l$
Thus, volume of the water decreases by $0.75 m l .$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369727 The average depth of indian Ocean is about $3000 m$ . The dfractional compression, $\frac{Δ V}{V}$ of water at the bottom of the ocean is (Given: Bulk modulus of the wear $= 2.2 \times 10^{9} N / m^{- 2}$ and $g = 10 m s^{- 2})$

1

0.91 %

2

0.82 %

3

1.24 %

4

1.36 %

Explanation:

The pressure exerted by a $3000 m$ column of water on the bottom layer is
$P = h ρ g = 3000 m \times 100 k g m^{- 3} \times 10 m s^{- 2}$
Fractional compression $\frac{Δ V}{V}$ is
$\frac{Δ V}{V} = \frac{P}{B} = \frac{3 \times 10^{7} N m^{- 2}}{2.2 \times 10^{9} N m^{- 2}} = 1.36 \times 10^{- 2} = 1.36 %$

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