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CBSE Questions for Class 8 Physics Force And Pressure Quiz 11 - MCQExams.com
CBSE
Class 8 Physics
Force And Pressure
Quiz 11
Tick the most appropriate answer.
If $$F$$ is a force acting normally on a surface of area $$A$$, then the pressure $$P$$ is given by:
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$$\;F\times A$$
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$$\;F+A$$
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$$\;F/A$$
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None of these.
Explanation
We know that pressure is defined as force acting normally per unit area of a surface.
It is given by
$$Pressure=\dfrac{Force}{Area}=\dfrac{F}{A}$$
So option C is correct.
Which of the following activities uses the most force?
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Explanation
Friction opposes the relative motion between two surfaces in contact. It acts on both surfaces. Friction depends on the nature of surfaces in contact. For a given pair of surfaces, friction depends upon the state of smoothness of those surfaces. When one body rolls over another body, rolling friction comes into play. Rolling friction is smaller than sliding friction. Friction is reduced by using ball bearings.
The force will be maximum for the rough surface.
A tall vertical cylinder filled with water is kept on a horizontal table top. Two small holes A and B are made on the wall of the cylinder, A near the middle and B just below the free surface of water.
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Water jet through A will have larger range
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Water jet through B will have larger range
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Water jets through A and B will have equal range
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There will be no water jet through B
Explanation
The water jet through A will have larger horizontal range because water pressure is more at that point. While B is just below the surface so water will come out of it without any significant pressure or jet. It will just leak through it.
At sea-level, atmospheric pressure is
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$$\;76\;cm\;of\;Hg$$ column
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$$\;0.76\;cm\;of\;Hg$$ column
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$$\;76\;m\;of\;Hg$$ column
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$$\;76\;cm\;of\;water$$ column
Explanation
At sea -level , atmospheric pressure is equal to $$76cm$$ of $$Hg$$(MERCURY) column.That is , at sea -level , atmospheric pressure is equal to pressure due to weight of $$76cm$$ height of $$Hg$$ column per unit area.
The pressure exerted by a force of $$50 \ N$$, normal to the nail of area $$1 \ mm^2$$ is
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$$5 \times 10^7$$ Pa
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$$5 \times 10^6$$ Pa
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$$5 \times 10^5$$ Pa
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$$5 \times 10^4$$ Pa
Explanation
Given:
Force $$F=50 \ N$$
Area $$A= 1 \ mm^2=(10^{-3} \ m) \times (10^{-3} \ m)=10^{-6} \ m^2 $$
We know that,
Pressure $$P=\dfrac{F}{A}=\dfrac{50}{10^{-6}}=5 \times 10^7 \ N/m^2$$
Also, $$1 \ N/m^2= 1 \ Pa$$
So pressure $$P=5 \times 10^7 \ Pa$$
Answer: $$(A)$$
A perpendicular force is applied to a certain area and produces a pressure $$P$$. If the same force is applied to a twice bigger area, the new pressure on the surface is:
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$$\dfrac{P}{2}$$
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$$\dfrac{P}{3}$$
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$$\dfrac{P}{4}$$
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none
Explanation
We know that pressure is force per unit area
$$\Rightarrow P=\frac{F}{A}$$
Given that for the former one , $$P=\frac{F}{A}$$
For the later one , the area is doubled and force is kept constant. So the pressure will become half
Therefore the pressure for new system is $$\frac{P}{2}$$
So the correct option is $$A$$
Two forces $$F_{1}$$ and $$F_{2}$$ are acting on a body. $$F_{1}$$ is a contact force. $$F_{2}$$ is a non - contact force. Then, the body (if at all moving) always moves in direction of
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$$F_{1}$$
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Force whose magnitude is greater
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$$F_{2}$$
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None of the above
Which of the following statement is incorrect:
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At high altitude air pressure is relatively low.
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At high altitude density of air is relatively high
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Density of a gas is directly proportional to its pressure at constant temperature.
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Air is compressible at constant temperature
The U-tube acts as a water siphon. The bend in the tube is lm above the water surface. The tube outlet is 7m below the water surface. The water issues from the bottom of the siphon as a free jet at atmospheric pressure. Determine the speed of the free jet and the minimum absolute pressure of the water in the bend. Given atm pressure $$=1.01\times 10^5 \,N / m^2.g =9.8m /s^2, p_w =10^3\, kg/g/m^3) $$
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$$10 \, m/s, 2\times 10^4 \, N/m^2$$
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$$11.7 \, m/s, 2.27\times 10^4 \, N/m^2$$
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$$5 \, m/s, 4\times 10^4 \, N/m^2$$
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$$12 \, m/s, 3\times 10^4 \, N/m^2$$
It is recommended that the air pressure in motor car tyres be reduced for a motion of the motor cars on sand because
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inflated tyres cause high pressure on the sand
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deflated tyres cause high pressure on the sand
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inflated tyres cause low pressure on the sand
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deflated tyres cause low pressure on the sand
The ratio of forces exerted by the larger piston to the smaller piston is
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1 : 144
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1 : 72
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144 : 1
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72 : 1
The folllowing forces act on an object $$13.5\,N$$ towards West, $$21.2\,N$$ towards East, $$33.0\,N$$ towards East, and $$25.3\,N$$ towards West. Calculate the net force acting on the object.
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$$15.4\,N$$ towards East
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$$15.4\,N$$ towards West
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$$23.6\,N$$ towards West
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$$23.6\,N$$ towards East
Sharp nails are easy to fix than a blunt nail because it has
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Less area
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Less volume
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More volume
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None of these
Explanation
As we know, $$Pressure = \dfrac {Force}{Area} \left (P\propto \dfrac {1}{A}\right )$$
Pressure is inversely proportional to area.
Thus, sharp nails has less area so they will produce more amount of pressure than the blunt nails with more area for same applied force.
Hence, sharp nails are easy to fix.
$$\therefore$$ Correct option is (A) Less area.
A uniform rod of length $$2L$$, area of cross section $$A$$ and Young's modulus $$Y$$ is lying at rest under the action of three forces as shown. The elongation of the rod will be
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$$\cfrac{FL}{AY}$$
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$$\cfrac{4FL}{AY}$$
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$$\cfrac{6FL}{AY}$$
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$$\cfrac{5FL}{AY}$$
The diagram shows students playing a tug-of-war game. If each student is pulling with the same force towards which dot will the flag move?
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$$dot\ A$$
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$$dot\ B$$
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$$dot\ C$$
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$$dot\ D$$
Explanation
Let's consider the forces applied on the middle flag $$O$$.
The student at $$P$$ and at $$Q$$ apply a force of $$F$$ each.
There are $$2$$ students at $$R$$ and at $$S$$ each. So they apply a force of $$2F$$ each.
Let us consider the forces along $$PR$$
The net force will be $$2F-F=F$$ towards $$R.$$
Let us consider the forces along $$QS$$
The net force will be $$2F-F=F$$ towards $$S.$$
The resultant of these two forces will be along the angle bisector of $$OR$$ and $$OS$$ towards point $$C.$$
The magnitude of the resultant will be $$F_{net}=\sqrt{F^2+F^2}=F\sqrt2$$
So option (C) is the answer.
Three vessels of different shapes are filled to the same level with water as in Figure $$OQ14.7$$. The area of the base is the same for all three vessels. Which of the following
statements are valid? (Choose all correct statements.)
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The pressure at the top surface of vessel $$A$$ is greatest because it has the largest surface area.
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The pressure at the bottom of vessel $$A$$ is greatest because it contains the most water
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The pressure at the bottom of each vessel is the same.
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The force on the bottom of each vessel is not the same.
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At a given depth below the surface the surface of each vessel, the pressure on the side of vessel $$A$$ is greatest because of its slope.
Explanation
The vessels (ii) is narrowest and vessel (i) is widest at the top. Therefore when the vessel are filled by same quantity of liquid, the height column of the liquid in vessels (ii) will be maximum and in vessels (i) is minimum. Thus the pressure due to height column of liquid at the base in maximum in vessels (ii) and minimum in vessels (i)
As the base area of all the vessels is same, hence the force at the base of the vessels (ii) will be Maximum and that at the base of the vessels (i) will be minimum
The pressure at the bottom of a tank containing a liquid does not depend on
[CBSE PMT 1994] [Kerala (Engg.)2002]
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Acceleration due to gravity
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Height of the liquid column
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Area of the bottom surface
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Nature of the liquid
Explanation
$$P=hpg$$ i.e.
Also it
depend
on the
bottom
surface area ,where the
bottom
surface area is smaller than the top it will have higher
pressure
than wide
bottom
. This shows that
pressure
of a given place in inversely proportional to the area,
pressure
in the
bottom
of the
tank does not depend
factors such as the material of the
tank
.
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