142710
A piece of ice, with a stone embedded inside it, is floating in water contained in a vessel. When the ice melts completely, the level of water in the vessel
1 remains unchanged
2 rises
3 falls
4 falls in the beginning and rises to the same level later
Explanation:
C Let's the mass and density of stone be $m$ and $\rho_{\mathrm{s}}$ and mass of ice cube is $\mathrm{M}$. If $\rho_{\mathrm{s}}>\rho_{\mathrm{w}}$, then, The volume of solid ice is more than the same amount of water. When the ice will melt then total surface of level of water will fall.
J and K CET- 2009
Mechanical Properties of Fluids
142711
When the temperature increases, the viscosity of
1 gases decreases and liquid increases
2 gases increases and liquids decreases
3 gases and liquids increases
4 gases and liquids decreases
Explanation:
B Viscosity of liquids depends on cohesion force. As we increases the temperature, cohesion force is decreases, hence viscosity of liquids also decreases. In gases, viscosity depends on molecular momentum collision. As we increases temperature kinetic energy of molecules of gases will increased and molecules moved very fast. Hence, viscosity of gasses increases.
UPCPMT-2010
Mechanical Properties of Fluids
142715
An object weighs $m_{1}$ in a liquid of density $d_{1}$ and that in liquid of density $d_{2}$ is $m_{2}$. The density $d$ of the object is
D Given that, an object weighting $\mathrm{m}_{1}$ in a liquid of density $\mathrm{d}_{1}$ and $\mathrm{m}_{2}$ in a liquid of density $\mathrm{d}_{2}$. So, when the density of the object is $d$ then we get- $\mathrm{V}\left(\mathrm{d}-\mathrm{d}_{1}\right)=\mathrm{m}_{1} \text { and } \mathrm{V}\left(\mathrm{d}-\mathrm{d}_{2}\right)=\mathrm{m}_{2}$ $\frac{\mathrm{d}-\mathrm{d}_{1}}{\mathrm{~d}-\mathrm{d}_{2}}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ So, $\quad \mathrm{d}=\frac{\mathrm{m}_{1} \mathrm{~d}_{2}-\mathrm{m}_{2} \mathrm{~d}_{1}}{\mathrm{~m}_{1}-\mathrm{m}_{2}}$
WB JEE 2011
Mechanical Properties of Fluids
142721
A column of water $60 \mathrm{~cm}$ high supports and a $32 \mathrm{~cm}$ column of an unknown liquid. What is density of the liquid?
142710
A piece of ice, with a stone embedded inside it, is floating in water contained in a vessel. When the ice melts completely, the level of water in the vessel
1 remains unchanged
2 rises
3 falls
4 falls in the beginning and rises to the same level later
Explanation:
C Let's the mass and density of stone be $m$ and $\rho_{\mathrm{s}}$ and mass of ice cube is $\mathrm{M}$. If $\rho_{\mathrm{s}}>\rho_{\mathrm{w}}$, then, The volume of solid ice is more than the same amount of water. When the ice will melt then total surface of level of water will fall.
J and K CET- 2009
Mechanical Properties of Fluids
142711
When the temperature increases, the viscosity of
1 gases decreases and liquid increases
2 gases increases and liquids decreases
3 gases and liquids increases
4 gases and liquids decreases
Explanation:
B Viscosity of liquids depends on cohesion force. As we increases the temperature, cohesion force is decreases, hence viscosity of liquids also decreases. In gases, viscosity depends on molecular momentum collision. As we increases temperature kinetic energy of molecules of gases will increased and molecules moved very fast. Hence, viscosity of gasses increases.
UPCPMT-2010
Mechanical Properties of Fluids
142715
An object weighs $m_{1}$ in a liquid of density $d_{1}$ and that in liquid of density $d_{2}$ is $m_{2}$. The density $d$ of the object is
D Given that, an object weighting $\mathrm{m}_{1}$ in a liquid of density $\mathrm{d}_{1}$ and $\mathrm{m}_{2}$ in a liquid of density $\mathrm{d}_{2}$. So, when the density of the object is $d$ then we get- $\mathrm{V}\left(\mathrm{d}-\mathrm{d}_{1}\right)=\mathrm{m}_{1} \text { and } \mathrm{V}\left(\mathrm{d}-\mathrm{d}_{2}\right)=\mathrm{m}_{2}$ $\frac{\mathrm{d}-\mathrm{d}_{1}}{\mathrm{~d}-\mathrm{d}_{2}}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ So, $\quad \mathrm{d}=\frac{\mathrm{m}_{1} \mathrm{~d}_{2}-\mathrm{m}_{2} \mathrm{~d}_{1}}{\mathrm{~m}_{1}-\mathrm{m}_{2}}$
WB JEE 2011
Mechanical Properties of Fluids
142721
A column of water $60 \mathrm{~cm}$ high supports and a $32 \mathrm{~cm}$ column of an unknown liquid. What is density of the liquid?
142710
A piece of ice, with a stone embedded inside it, is floating in water contained in a vessel. When the ice melts completely, the level of water in the vessel
1 remains unchanged
2 rises
3 falls
4 falls in the beginning and rises to the same level later
Explanation:
C Let's the mass and density of stone be $m$ and $\rho_{\mathrm{s}}$ and mass of ice cube is $\mathrm{M}$. If $\rho_{\mathrm{s}}>\rho_{\mathrm{w}}$, then, The volume of solid ice is more than the same amount of water. When the ice will melt then total surface of level of water will fall.
J and K CET- 2009
Mechanical Properties of Fluids
142711
When the temperature increases, the viscosity of
1 gases decreases and liquid increases
2 gases increases and liquids decreases
3 gases and liquids increases
4 gases and liquids decreases
Explanation:
B Viscosity of liquids depends on cohesion force. As we increases the temperature, cohesion force is decreases, hence viscosity of liquids also decreases. In gases, viscosity depends on molecular momentum collision. As we increases temperature kinetic energy of molecules of gases will increased and molecules moved very fast. Hence, viscosity of gasses increases.
UPCPMT-2010
Mechanical Properties of Fluids
142715
An object weighs $m_{1}$ in a liquid of density $d_{1}$ and that in liquid of density $d_{2}$ is $m_{2}$. The density $d$ of the object is
D Given that, an object weighting $\mathrm{m}_{1}$ in a liquid of density $\mathrm{d}_{1}$ and $\mathrm{m}_{2}$ in a liquid of density $\mathrm{d}_{2}$. So, when the density of the object is $d$ then we get- $\mathrm{V}\left(\mathrm{d}-\mathrm{d}_{1}\right)=\mathrm{m}_{1} \text { and } \mathrm{V}\left(\mathrm{d}-\mathrm{d}_{2}\right)=\mathrm{m}_{2}$ $\frac{\mathrm{d}-\mathrm{d}_{1}}{\mathrm{~d}-\mathrm{d}_{2}}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ So, $\quad \mathrm{d}=\frac{\mathrm{m}_{1} \mathrm{~d}_{2}-\mathrm{m}_{2} \mathrm{~d}_{1}}{\mathrm{~m}_{1}-\mathrm{m}_{2}}$
WB JEE 2011
Mechanical Properties of Fluids
142721
A column of water $60 \mathrm{~cm}$ high supports and a $32 \mathrm{~cm}$ column of an unknown liquid. What is density of the liquid?
142710
A piece of ice, with a stone embedded inside it, is floating in water contained in a vessel. When the ice melts completely, the level of water in the vessel
1 remains unchanged
2 rises
3 falls
4 falls in the beginning and rises to the same level later
Explanation:
C Let's the mass and density of stone be $m$ and $\rho_{\mathrm{s}}$ and mass of ice cube is $\mathrm{M}$. If $\rho_{\mathrm{s}}>\rho_{\mathrm{w}}$, then, The volume of solid ice is more than the same amount of water. When the ice will melt then total surface of level of water will fall.
J and K CET- 2009
Mechanical Properties of Fluids
142711
When the temperature increases, the viscosity of
1 gases decreases and liquid increases
2 gases increases and liquids decreases
3 gases and liquids increases
4 gases and liquids decreases
Explanation:
B Viscosity of liquids depends on cohesion force. As we increases the temperature, cohesion force is decreases, hence viscosity of liquids also decreases. In gases, viscosity depends on molecular momentum collision. As we increases temperature kinetic energy of molecules of gases will increased and molecules moved very fast. Hence, viscosity of gasses increases.
UPCPMT-2010
Mechanical Properties of Fluids
142715
An object weighs $m_{1}$ in a liquid of density $d_{1}$ and that in liquid of density $d_{2}$ is $m_{2}$. The density $d$ of the object is
D Given that, an object weighting $\mathrm{m}_{1}$ in a liquid of density $\mathrm{d}_{1}$ and $\mathrm{m}_{2}$ in a liquid of density $\mathrm{d}_{2}$. So, when the density of the object is $d$ then we get- $\mathrm{V}\left(\mathrm{d}-\mathrm{d}_{1}\right)=\mathrm{m}_{1} \text { and } \mathrm{V}\left(\mathrm{d}-\mathrm{d}_{2}\right)=\mathrm{m}_{2}$ $\frac{\mathrm{d}-\mathrm{d}_{1}}{\mathrm{~d}-\mathrm{d}_{2}}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ So, $\quad \mathrm{d}=\frac{\mathrm{m}_{1} \mathrm{~d}_{2}-\mathrm{m}_{2} \mathrm{~d}_{1}}{\mathrm{~m}_{1}-\mathrm{m}_{2}}$
WB JEE 2011
Mechanical Properties of Fluids
142721
A column of water $60 \mathrm{~cm}$ high supports and a $32 \mathrm{~cm}$ column of an unknown liquid. What is density of the liquid?
