143137
A body floats in water with one fourth of its volume above the surface of water. If placed in oil if floats with one third of its volume above the surface of oil. The density of oil is-
1 $\frac{3}{4}$
2 $\frac{4}{9}$
3 $\frac{2}{3}$
4 $\frac{9}{8}$
Explanation:
D Let the volume of the body $\mathrm{v}$ and density $\rho$ Then for the case of water, $\mathrm{v} \times \rho \times \mathrm{g}=\frac{3}{4} \times \mathrm{v} \times \rho_{\text {water }} \times \mathrm{g}$ $\left\{\because \mathrm{v}^{\prime}=\left(1-\frac{1}{4}\right) \mathrm{v}=\frac{3}{4} \mathrm{v}\right\}$ For oil, $v \rho g=v^{\prime \prime} \rho_{\text {oil }}$ $v \times \rho \times g=\frac{2}{3} \times v \times \rho_{\text {oil }} \times g$ $\left\{\because v "=\left(1-\frac{1}{3}\right) v=\frac{2}{3} v\right\}$ $\frac{\rho_{\text {oil }}}{\rho_{\text {water }}}=\frac{9}{8}$
BCECE-2013
Mechanical Properties of Fluids
143141
A force of $50 \mathrm{~N}$ is applied to the smaller piston of a hydraulic press. If the diameters of the piston are in the ratio $5: 1$, the force on the larger piston is
1 $250 \mathrm{~N}$
2 $1000 \mathrm{~N}$
3 $1250 \mathrm{~N}$
4 $100 \mathrm{~N}$
Explanation:
C Given, Force on smaller piston $=50 \mathrm{~N}$ Force on larger piston $=$ ? Diameter of larger piston $=5$ Diameter of smaller piston $=1$ $\frac{\text { Force on larger piston }}{\text { Area of larger piston }}=\frac{\text { Force on smaller piston }}{\text { Area of smaller piston }}$ Force on larger position $=\frac{50}{(1)^{2}} \times(5)^{2}$ $=1250 \mathrm{~N}$
MHT-CET 2020
Mechanical Properties of Fluids
143144
From the adjacent figure, the correct observation is:
1 the pressure on the bottom of tank (a) is greater than at the bottom of (b)
2 the pressure on the bottom of the tank (a) is smaller than at the bottom of (b)
3 the pressure depend on the shape of the container
4 the pressure on the bottom of (a) and (b) is the same
Explanation:
D Pressure, $\mathrm{P}=\rho$ gh Pressure depends on density of liquid ( $\rho$ ) gravitational acceleration $(\mathrm{g})$ and height of liquid $(\mathrm{h})$. But pressure does not depends on the shape of the container. So, pressure on the bottom of (a) and (b) is same.
Karnataka CET-2005
Mechanical Properties of Fluids
143145
A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ is punctured with a pin, the thread
1 becomes concave towards $\mathrm{A}$
2 becomes convex towards $\mathrm{A}$
3 either (a) or (b) depending on the size of A with respect to $B$
4 remains in the initial position
Explanation:
A A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion ' $A$ ' is punctured with a pin the thread becomes concave towards ' $\mathrm{A}$ '.
143137
A body floats in water with one fourth of its volume above the surface of water. If placed in oil if floats with one third of its volume above the surface of oil. The density of oil is-
1 $\frac{3}{4}$
2 $\frac{4}{9}$
3 $\frac{2}{3}$
4 $\frac{9}{8}$
Explanation:
D Let the volume of the body $\mathrm{v}$ and density $\rho$ Then for the case of water, $\mathrm{v} \times \rho \times \mathrm{g}=\frac{3}{4} \times \mathrm{v} \times \rho_{\text {water }} \times \mathrm{g}$ $\left\{\because \mathrm{v}^{\prime}=\left(1-\frac{1}{4}\right) \mathrm{v}=\frac{3}{4} \mathrm{v}\right\}$ For oil, $v \rho g=v^{\prime \prime} \rho_{\text {oil }}$ $v \times \rho \times g=\frac{2}{3} \times v \times \rho_{\text {oil }} \times g$ $\left\{\because v "=\left(1-\frac{1}{3}\right) v=\frac{2}{3} v\right\}$ $\frac{\rho_{\text {oil }}}{\rho_{\text {water }}}=\frac{9}{8}$
BCECE-2013
Mechanical Properties of Fluids
143141
A force of $50 \mathrm{~N}$ is applied to the smaller piston of a hydraulic press. If the diameters of the piston are in the ratio $5: 1$, the force on the larger piston is
1 $250 \mathrm{~N}$
2 $1000 \mathrm{~N}$
3 $1250 \mathrm{~N}$
4 $100 \mathrm{~N}$
Explanation:
C Given, Force on smaller piston $=50 \mathrm{~N}$ Force on larger piston $=$ ? Diameter of larger piston $=5$ Diameter of smaller piston $=1$ $\frac{\text { Force on larger piston }}{\text { Area of larger piston }}=\frac{\text { Force on smaller piston }}{\text { Area of smaller piston }}$ Force on larger position $=\frac{50}{(1)^{2}} \times(5)^{2}$ $=1250 \mathrm{~N}$
MHT-CET 2020
Mechanical Properties of Fluids
143144
From the adjacent figure, the correct observation is:
1 the pressure on the bottom of tank (a) is greater than at the bottom of (b)
2 the pressure on the bottom of the tank (a) is smaller than at the bottom of (b)
3 the pressure depend on the shape of the container
4 the pressure on the bottom of (a) and (b) is the same
Explanation:
D Pressure, $\mathrm{P}=\rho$ gh Pressure depends on density of liquid ( $\rho$ ) gravitational acceleration $(\mathrm{g})$ and height of liquid $(\mathrm{h})$. But pressure does not depends on the shape of the container. So, pressure on the bottom of (a) and (b) is same.
Karnataka CET-2005
Mechanical Properties of Fluids
143145
A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ is punctured with a pin, the thread
1 becomes concave towards $\mathrm{A}$
2 becomes convex towards $\mathrm{A}$
3 either (a) or (b) depending on the size of A with respect to $B$
4 remains in the initial position
Explanation:
A A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion ' $A$ ' is punctured with a pin the thread becomes concave towards ' $\mathrm{A}$ '.
143137
A body floats in water with one fourth of its volume above the surface of water. If placed in oil if floats with one third of its volume above the surface of oil. The density of oil is-
1 $\frac{3}{4}$
2 $\frac{4}{9}$
3 $\frac{2}{3}$
4 $\frac{9}{8}$
Explanation:
D Let the volume of the body $\mathrm{v}$ and density $\rho$ Then for the case of water, $\mathrm{v} \times \rho \times \mathrm{g}=\frac{3}{4} \times \mathrm{v} \times \rho_{\text {water }} \times \mathrm{g}$ $\left\{\because \mathrm{v}^{\prime}=\left(1-\frac{1}{4}\right) \mathrm{v}=\frac{3}{4} \mathrm{v}\right\}$ For oil, $v \rho g=v^{\prime \prime} \rho_{\text {oil }}$ $v \times \rho \times g=\frac{2}{3} \times v \times \rho_{\text {oil }} \times g$ $\left\{\because v "=\left(1-\frac{1}{3}\right) v=\frac{2}{3} v\right\}$ $\frac{\rho_{\text {oil }}}{\rho_{\text {water }}}=\frac{9}{8}$
BCECE-2013
Mechanical Properties of Fluids
143141
A force of $50 \mathrm{~N}$ is applied to the smaller piston of a hydraulic press. If the diameters of the piston are in the ratio $5: 1$, the force on the larger piston is
1 $250 \mathrm{~N}$
2 $1000 \mathrm{~N}$
3 $1250 \mathrm{~N}$
4 $100 \mathrm{~N}$
Explanation:
C Given, Force on smaller piston $=50 \mathrm{~N}$ Force on larger piston $=$ ? Diameter of larger piston $=5$ Diameter of smaller piston $=1$ $\frac{\text { Force on larger piston }}{\text { Area of larger piston }}=\frac{\text { Force on smaller piston }}{\text { Area of smaller piston }}$ Force on larger position $=\frac{50}{(1)^{2}} \times(5)^{2}$ $=1250 \mathrm{~N}$
MHT-CET 2020
Mechanical Properties of Fluids
143144
From the adjacent figure, the correct observation is:
1 the pressure on the bottom of tank (a) is greater than at the bottom of (b)
2 the pressure on the bottom of the tank (a) is smaller than at the bottom of (b)
3 the pressure depend on the shape of the container
4 the pressure on the bottom of (a) and (b) is the same
Explanation:
D Pressure, $\mathrm{P}=\rho$ gh Pressure depends on density of liquid ( $\rho$ ) gravitational acceleration $(\mathrm{g})$ and height of liquid $(\mathrm{h})$. But pressure does not depends on the shape of the container. So, pressure on the bottom of (a) and (b) is same.
Karnataka CET-2005
Mechanical Properties of Fluids
143145
A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ is punctured with a pin, the thread
1 becomes concave towards $\mathrm{A}$
2 becomes convex towards $\mathrm{A}$
3 either (a) or (b) depending on the size of A with respect to $B$
4 remains in the initial position
Explanation:
A A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion ' $A$ ' is punctured with a pin the thread becomes concave towards ' $\mathrm{A}$ '.
143137
A body floats in water with one fourth of its volume above the surface of water. If placed in oil if floats with one third of its volume above the surface of oil. The density of oil is-
1 $\frac{3}{4}$
2 $\frac{4}{9}$
3 $\frac{2}{3}$
4 $\frac{9}{8}$
Explanation:
D Let the volume of the body $\mathrm{v}$ and density $\rho$ Then for the case of water, $\mathrm{v} \times \rho \times \mathrm{g}=\frac{3}{4} \times \mathrm{v} \times \rho_{\text {water }} \times \mathrm{g}$ $\left\{\because \mathrm{v}^{\prime}=\left(1-\frac{1}{4}\right) \mathrm{v}=\frac{3}{4} \mathrm{v}\right\}$ For oil, $v \rho g=v^{\prime \prime} \rho_{\text {oil }}$ $v \times \rho \times g=\frac{2}{3} \times v \times \rho_{\text {oil }} \times g$ $\left\{\because v "=\left(1-\frac{1}{3}\right) v=\frac{2}{3} v\right\}$ $\frac{\rho_{\text {oil }}}{\rho_{\text {water }}}=\frac{9}{8}$
BCECE-2013
Mechanical Properties of Fluids
143141
A force of $50 \mathrm{~N}$ is applied to the smaller piston of a hydraulic press. If the diameters of the piston are in the ratio $5: 1$, the force on the larger piston is
1 $250 \mathrm{~N}$
2 $1000 \mathrm{~N}$
3 $1250 \mathrm{~N}$
4 $100 \mathrm{~N}$
Explanation:
C Given, Force on smaller piston $=50 \mathrm{~N}$ Force on larger piston $=$ ? Diameter of larger piston $=5$ Diameter of smaller piston $=1$ $\frac{\text { Force on larger piston }}{\text { Area of larger piston }}=\frac{\text { Force on smaller piston }}{\text { Area of smaller piston }}$ Force on larger position $=\frac{50}{(1)^{2}} \times(5)^{2}$ $=1250 \mathrm{~N}$
MHT-CET 2020
Mechanical Properties of Fluids
143144
From the adjacent figure, the correct observation is:
1 the pressure on the bottom of tank (a) is greater than at the bottom of (b)
2 the pressure on the bottom of the tank (a) is smaller than at the bottom of (b)
3 the pressure depend on the shape of the container
4 the pressure on the bottom of (a) and (b) is the same
Explanation:
D Pressure, $\mathrm{P}=\rho$ gh Pressure depends on density of liquid ( $\rho$ ) gravitational acceleration $(\mathrm{g})$ and height of liquid $(\mathrm{h})$. But pressure does not depends on the shape of the container. So, pressure on the bottom of (a) and (b) is same.
Karnataka CET-2005
Mechanical Properties of Fluids
143145
A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ is punctured with a pin, the thread
1 becomes concave towards $\mathrm{A}$
2 becomes convex towards $\mathrm{A}$
3 either (a) or (b) depending on the size of A with respect to $B$
4 remains in the initial position
Explanation:
A A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion ' $A$ ' is punctured with a pin the thread becomes concave towards ' $\mathrm{A}$ '.
