143158
Bulk modulus of water is $2 \times 10^{9} \mathrm{Nm}^{-2}$. The pressure required to increase the volume of water by $0.1 \%$ in $\mathrm{Nm}^{-2}$ is
1 $2 \times 10^{9}$
2 $2 \times 10^{0}$
3 $2 \times 10^{6}$
4 $2 \times 10^{4}$
Explanation:
C Given that, $\frac{\Delta \mathrm{V}}{\mathrm{V}} =0.1 \%=10^{-3}$ $\mathrm{~B} =2 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, Bulk modulus $=\frac{\text { Hydrostatic pressure }}{\text { Volumetric strain }}$ $\mathrm{B}=-\frac{\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}$ From equation (i) we get $2 \times 10^{9}=\frac{\Delta \mathrm{P}}{10^{-3}}$ So, $\quad \Delta \mathrm{P}=2 \times 10^{9} \times 10^{-3}$ $\Delta \mathrm{P}=2 \times 10^{6}$
EAMCET-2003
Mechanical Properties of Fluids
143159
A force of $500 \mathrm{~N}$ is executed on a hydraulic piston of cross sectional area of $100 \mathrm{~cm}^{2}$. The cross-sectional area of other piston which supports a truck of a tonne weight is [use $g=$ $9.8 \mathrm{~m} / \mathrm{s}^{2}$ ]
1 $200 \mathrm{~cm}^{2}$
2 $196 \mathrm{~cm}^{2}$
3 $1960 \mathrm{~cm}^{2}$
4 $98 \mathrm{~cm}^{2}$
Explanation:
B Given, Force $\left(\mathrm{F}_{1}\right)=500 \mathrm{~N}$ Cross - sectional area $\left(\mathrm{A}_{1}\right)=100 \mathrm{~cm}^{2}$ Cross - sectional area of other piston $\left(\mathrm{A}_{2}\right)=$ ? Weight of truck $\left(\mathrm{F}_{2}\right)=1000 \times 9.8 \mathrm{~N}$ $=980 \mathrm{~N}$ Now, from Pascal's law $\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}$ $\mathrm{A}_{2}=\frac{\mathrm{F}_{2}}{\mathrm{~F}_{1}} \times \mathrm{A}_{1}$ $\mathrm{A}_{2}=\frac{980}{500} \times 100$ $\mathrm{A}_{2}=196 \mathrm{~cm}^{2}$
TS EAMCET 30.07.2022
Mechanical Properties of Fluids
143160
If a soap bubble expands, the pressure inside the bubble
1 Remains the same
2 Is equal to the atmospheric pressure
3 Decreases
4 Increases
Explanation:
C As we know, In case of soap bubble there are two surfaces, So pressure difference, $\Delta \mathrm{P}=\frac{8 \sigma}{\mathrm{d}}$ Where, $\sigma=$ surface tension $\mathrm{d}=$ diameter of soap bubble. From equation (i) $\Delta \mathrm{P} \propto \frac{1}{\mathrm{~d}}$ As a soap bubble expands means diameter increase Therefore pressure inside the bubble decrease.
Mechanical Properties of Fluids
143161
The volume contraction of a solid copper cube, $10 \mathrm{~cm}$ on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^{6} \mathrm{~Pa}$ is (B for copper $=140 \times 10^{9} \mathrm{Nm}^{-2}$ )
143162
The density of air in the earth's atmosphere decreases with height as $\rho=\rho_{0} \mathrm{e}^{-\mathrm{kh}}$, where $\rho_{0}=$ density of air at sea level and $k$ is a constant. The atmospheric pressure at sea level is
143158
Bulk modulus of water is $2 \times 10^{9} \mathrm{Nm}^{-2}$. The pressure required to increase the volume of water by $0.1 \%$ in $\mathrm{Nm}^{-2}$ is
1 $2 \times 10^{9}$
2 $2 \times 10^{0}$
3 $2 \times 10^{6}$
4 $2 \times 10^{4}$
Explanation:
C Given that, $\frac{\Delta \mathrm{V}}{\mathrm{V}} =0.1 \%=10^{-3}$ $\mathrm{~B} =2 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, Bulk modulus $=\frac{\text { Hydrostatic pressure }}{\text { Volumetric strain }}$ $\mathrm{B}=-\frac{\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}$ From equation (i) we get $2 \times 10^{9}=\frac{\Delta \mathrm{P}}{10^{-3}}$ So, $\quad \Delta \mathrm{P}=2 \times 10^{9} \times 10^{-3}$ $\Delta \mathrm{P}=2 \times 10^{6}$
EAMCET-2003
Mechanical Properties of Fluids
143159
A force of $500 \mathrm{~N}$ is executed on a hydraulic piston of cross sectional area of $100 \mathrm{~cm}^{2}$. The cross-sectional area of other piston which supports a truck of a tonne weight is [use $g=$ $9.8 \mathrm{~m} / \mathrm{s}^{2}$ ]
1 $200 \mathrm{~cm}^{2}$
2 $196 \mathrm{~cm}^{2}$
3 $1960 \mathrm{~cm}^{2}$
4 $98 \mathrm{~cm}^{2}$
Explanation:
B Given, Force $\left(\mathrm{F}_{1}\right)=500 \mathrm{~N}$ Cross - sectional area $\left(\mathrm{A}_{1}\right)=100 \mathrm{~cm}^{2}$ Cross - sectional area of other piston $\left(\mathrm{A}_{2}\right)=$ ? Weight of truck $\left(\mathrm{F}_{2}\right)=1000 \times 9.8 \mathrm{~N}$ $=980 \mathrm{~N}$ Now, from Pascal's law $\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}$ $\mathrm{A}_{2}=\frac{\mathrm{F}_{2}}{\mathrm{~F}_{1}} \times \mathrm{A}_{1}$ $\mathrm{A}_{2}=\frac{980}{500} \times 100$ $\mathrm{A}_{2}=196 \mathrm{~cm}^{2}$
TS EAMCET 30.07.2022
Mechanical Properties of Fluids
143160
If a soap bubble expands, the pressure inside the bubble
1 Remains the same
2 Is equal to the atmospheric pressure
3 Decreases
4 Increases
Explanation:
C As we know, In case of soap bubble there are two surfaces, So pressure difference, $\Delta \mathrm{P}=\frac{8 \sigma}{\mathrm{d}}$ Where, $\sigma=$ surface tension $\mathrm{d}=$ diameter of soap bubble. From equation (i) $\Delta \mathrm{P} \propto \frac{1}{\mathrm{~d}}$ As a soap bubble expands means diameter increase Therefore pressure inside the bubble decrease.
Mechanical Properties of Fluids
143161
The volume contraction of a solid copper cube, $10 \mathrm{~cm}$ on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^{6} \mathrm{~Pa}$ is (B for copper $=140 \times 10^{9} \mathrm{Nm}^{-2}$ )
143162
The density of air in the earth's atmosphere decreases with height as $\rho=\rho_{0} \mathrm{e}^{-\mathrm{kh}}$, where $\rho_{0}=$ density of air at sea level and $k$ is a constant. The atmospheric pressure at sea level is
143158
Bulk modulus of water is $2 \times 10^{9} \mathrm{Nm}^{-2}$. The pressure required to increase the volume of water by $0.1 \%$ in $\mathrm{Nm}^{-2}$ is
1 $2 \times 10^{9}$
2 $2 \times 10^{0}$
3 $2 \times 10^{6}$
4 $2 \times 10^{4}$
Explanation:
C Given that, $\frac{\Delta \mathrm{V}}{\mathrm{V}} =0.1 \%=10^{-3}$ $\mathrm{~B} =2 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, Bulk modulus $=\frac{\text { Hydrostatic pressure }}{\text { Volumetric strain }}$ $\mathrm{B}=-\frac{\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}$ From equation (i) we get $2 \times 10^{9}=\frac{\Delta \mathrm{P}}{10^{-3}}$ So, $\quad \Delta \mathrm{P}=2 \times 10^{9} \times 10^{-3}$ $\Delta \mathrm{P}=2 \times 10^{6}$
EAMCET-2003
Mechanical Properties of Fluids
143159
A force of $500 \mathrm{~N}$ is executed on a hydraulic piston of cross sectional area of $100 \mathrm{~cm}^{2}$. The cross-sectional area of other piston which supports a truck of a tonne weight is [use $g=$ $9.8 \mathrm{~m} / \mathrm{s}^{2}$ ]
1 $200 \mathrm{~cm}^{2}$
2 $196 \mathrm{~cm}^{2}$
3 $1960 \mathrm{~cm}^{2}$
4 $98 \mathrm{~cm}^{2}$
Explanation:
B Given, Force $\left(\mathrm{F}_{1}\right)=500 \mathrm{~N}$ Cross - sectional area $\left(\mathrm{A}_{1}\right)=100 \mathrm{~cm}^{2}$ Cross - sectional area of other piston $\left(\mathrm{A}_{2}\right)=$ ? Weight of truck $\left(\mathrm{F}_{2}\right)=1000 \times 9.8 \mathrm{~N}$ $=980 \mathrm{~N}$ Now, from Pascal's law $\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}$ $\mathrm{A}_{2}=\frac{\mathrm{F}_{2}}{\mathrm{~F}_{1}} \times \mathrm{A}_{1}$ $\mathrm{A}_{2}=\frac{980}{500} \times 100$ $\mathrm{A}_{2}=196 \mathrm{~cm}^{2}$
TS EAMCET 30.07.2022
Mechanical Properties of Fluids
143160
If a soap bubble expands, the pressure inside the bubble
1 Remains the same
2 Is equal to the atmospheric pressure
3 Decreases
4 Increases
Explanation:
C As we know, In case of soap bubble there are two surfaces, So pressure difference, $\Delta \mathrm{P}=\frac{8 \sigma}{\mathrm{d}}$ Where, $\sigma=$ surface tension $\mathrm{d}=$ diameter of soap bubble. From equation (i) $\Delta \mathrm{P} \propto \frac{1}{\mathrm{~d}}$ As a soap bubble expands means diameter increase Therefore pressure inside the bubble decrease.
Mechanical Properties of Fluids
143161
The volume contraction of a solid copper cube, $10 \mathrm{~cm}$ on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^{6} \mathrm{~Pa}$ is (B for copper $=140 \times 10^{9} \mathrm{Nm}^{-2}$ )
143162
The density of air in the earth's atmosphere decreases with height as $\rho=\rho_{0} \mathrm{e}^{-\mathrm{kh}}$, where $\rho_{0}=$ density of air at sea level and $k$ is a constant. The atmospheric pressure at sea level is
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Mechanical Properties of Fluids
143158
Bulk modulus of water is $2 \times 10^{9} \mathrm{Nm}^{-2}$. The pressure required to increase the volume of water by $0.1 \%$ in $\mathrm{Nm}^{-2}$ is
1 $2 \times 10^{9}$
2 $2 \times 10^{0}$
3 $2 \times 10^{6}$
4 $2 \times 10^{4}$
Explanation:
C Given that, $\frac{\Delta \mathrm{V}}{\mathrm{V}} =0.1 \%=10^{-3}$ $\mathrm{~B} =2 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, Bulk modulus $=\frac{\text { Hydrostatic pressure }}{\text { Volumetric strain }}$ $\mathrm{B}=-\frac{\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}$ From equation (i) we get $2 \times 10^{9}=\frac{\Delta \mathrm{P}}{10^{-3}}$ So, $\quad \Delta \mathrm{P}=2 \times 10^{9} \times 10^{-3}$ $\Delta \mathrm{P}=2 \times 10^{6}$
EAMCET-2003
Mechanical Properties of Fluids
143159
A force of $500 \mathrm{~N}$ is executed on a hydraulic piston of cross sectional area of $100 \mathrm{~cm}^{2}$. The cross-sectional area of other piston which supports a truck of a tonne weight is [use $g=$ $9.8 \mathrm{~m} / \mathrm{s}^{2}$ ]
1 $200 \mathrm{~cm}^{2}$
2 $196 \mathrm{~cm}^{2}$
3 $1960 \mathrm{~cm}^{2}$
4 $98 \mathrm{~cm}^{2}$
Explanation:
B Given, Force $\left(\mathrm{F}_{1}\right)=500 \mathrm{~N}$ Cross - sectional area $\left(\mathrm{A}_{1}\right)=100 \mathrm{~cm}^{2}$ Cross - sectional area of other piston $\left(\mathrm{A}_{2}\right)=$ ? Weight of truck $\left(\mathrm{F}_{2}\right)=1000 \times 9.8 \mathrm{~N}$ $=980 \mathrm{~N}$ Now, from Pascal's law $\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}$ $\mathrm{A}_{2}=\frac{\mathrm{F}_{2}}{\mathrm{~F}_{1}} \times \mathrm{A}_{1}$ $\mathrm{A}_{2}=\frac{980}{500} \times 100$ $\mathrm{A}_{2}=196 \mathrm{~cm}^{2}$
TS EAMCET 30.07.2022
Mechanical Properties of Fluids
143160
If a soap bubble expands, the pressure inside the bubble
1 Remains the same
2 Is equal to the atmospheric pressure
3 Decreases
4 Increases
Explanation:
C As we know, In case of soap bubble there are two surfaces, So pressure difference, $\Delta \mathrm{P}=\frac{8 \sigma}{\mathrm{d}}$ Where, $\sigma=$ surface tension $\mathrm{d}=$ diameter of soap bubble. From equation (i) $\Delta \mathrm{P} \propto \frac{1}{\mathrm{~d}}$ As a soap bubble expands means diameter increase Therefore pressure inside the bubble decrease.
Mechanical Properties of Fluids
143161
The volume contraction of a solid copper cube, $10 \mathrm{~cm}$ on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^{6} \mathrm{~Pa}$ is (B for copper $=140 \times 10^{9} \mathrm{Nm}^{-2}$ )
143162
The density of air in the earth's atmosphere decreases with height as $\rho=\rho_{0} \mathrm{e}^{-\mathrm{kh}}$, where $\rho_{0}=$ density of air at sea level and $k$ is a constant. The atmospheric pressure at sea level is
143158
Bulk modulus of water is $2 \times 10^{9} \mathrm{Nm}^{-2}$. The pressure required to increase the volume of water by $0.1 \%$ in $\mathrm{Nm}^{-2}$ is
1 $2 \times 10^{9}$
2 $2 \times 10^{0}$
3 $2 \times 10^{6}$
4 $2 \times 10^{4}$
Explanation:
C Given that, $\frac{\Delta \mathrm{V}}{\mathrm{V}} =0.1 \%=10^{-3}$ $\mathrm{~B} =2 \times 10^{9} \mathrm{Nm}^{-2}$ We know that, Bulk modulus $=\frac{\text { Hydrostatic pressure }}{\text { Volumetric strain }}$ $\mathrm{B}=-\frac{\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}$ From equation (i) we get $2 \times 10^{9}=\frac{\Delta \mathrm{P}}{10^{-3}}$ So, $\quad \Delta \mathrm{P}=2 \times 10^{9} \times 10^{-3}$ $\Delta \mathrm{P}=2 \times 10^{6}$
EAMCET-2003
Mechanical Properties of Fluids
143159
A force of $500 \mathrm{~N}$ is executed on a hydraulic piston of cross sectional area of $100 \mathrm{~cm}^{2}$. The cross-sectional area of other piston which supports a truck of a tonne weight is [use $g=$ $9.8 \mathrm{~m} / \mathrm{s}^{2}$ ]
1 $200 \mathrm{~cm}^{2}$
2 $196 \mathrm{~cm}^{2}$
3 $1960 \mathrm{~cm}^{2}$
4 $98 \mathrm{~cm}^{2}$
Explanation:
B Given, Force $\left(\mathrm{F}_{1}\right)=500 \mathrm{~N}$ Cross - sectional area $\left(\mathrm{A}_{1}\right)=100 \mathrm{~cm}^{2}$ Cross - sectional area of other piston $\left(\mathrm{A}_{2}\right)=$ ? Weight of truck $\left(\mathrm{F}_{2}\right)=1000 \times 9.8 \mathrm{~N}$ $=980 \mathrm{~N}$ Now, from Pascal's law $\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}$ $\mathrm{A}_{2}=\frac{\mathrm{F}_{2}}{\mathrm{~F}_{1}} \times \mathrm{A}_{1}$ $\mathrm{A}_{2}=\frac{980}{500} \times 100$ $\mathrm{A}_{2}=196 \mathrm{~cm}^{2}$
TS EAMCET 30.07.2022
Mechanical Properties of Fluids
143160
If a soap bubble expands, the pressure inside the bubble
1 Remains the same
2 Is equal to the atmospheric pressure
3 Decreases
4 Increases
Explanation:
C As we know, In case of soap bubble there are two surfaces, So pressure difference, $\Delta \mathrm{P}=\frac{8 \sigma}{\mathrm{d}}$ Where, $\sigma=$ surface tension $\mathrm{d}=$ diameter of soap bubble. From equation (i) $\Delta \mathrm{P} \propto \frac{1}{\mathrm{~d}}$ As a soap bubble expands means diameter increase Therefore pressure inside the bubble decrease.
Mechanical Properties of Fluids
143161
The volume contraction of a solid copper cube, $10 \mathrm{~cm}$ on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^{6} \mathrm{~Pa}$ is (B for copper $=140 \times 10^{9} \mathrm{Nm}^{-2}$ )
143162
The density of air in the earth's atmosphere decreases with height as $\rho=\rho_{0} \mathrm{e}^{-\mathrm{kh}}$, where $\rho_{0}=$ density of air at sea level and $k$ is a constant. The atmospheric pressure at sea level is
