Question 1

Question 7The pressure on a surface is reduced by :using high thrustincreasing the area of surfacedecreasing the area of surfacenone of these

Accepted Answer

increasing the area of surfaceReason — Larger the area on which a given thrust acts, lesser is the pressure exerted by it, as pressure is inversely proportional to the area of surface.Pressure=ThrustAreaorP=FA\text{Pressure} = \dfrac{\text{Thrust}}{\text{Area}} \[0.5em] \text{or} \[0.5em] \text P = \dfrac{\text{F}}{\text{A}}Pressure=AreaThrust​orP=AF​

Question 2

Question 16A rectangular cement block has length, breadth and height of 60 cm, 30 cm and 15 cm respectively. Identify the correct position of the block and pressure exerted by it on the ground.Minimum pressure is exerted when breadth and height form the base.Maximum pressure is exerted when breadth and height form the base.Maximum pressure is exerted when length and breadth form the base.Minimum pressure is exerted when length and height form the base.

Accepted Answer

Maximum pressure is exerted when breadth and height form the baseReason — Pressure is given by the formula : Pressure=ThrustArea	ext {Pressure} = \dfrac {	ext {Thrust}}{	ext {Area}}Pressure=AreaThrust​For the same weight (thrust) of the block, the pressure is inversely proportional to the contact area. So, larger base area → smaller pressure, and smaller base area → greater pressure.As, minimum area → side with minimum dimensions → 30 cm breadth and 15 cm height.Hence, maximum pressure is exerted when breadth and height form the base because it has smallest contact area.

Question 3

Question 1A hammer exerts a force of 1.5 N on each of the two nails A and B. The area of cross section of tip of nail A is 2 mm2 while that of nail B is 6 mm2 . Calculate pressure on each nail in pascal.

Accepted Answer

As we know,Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​Given,F = 1.5 NA = 2 mm2Converting 2 mm2 in metre2As 1 mm = 11000\dfrac{1}{1000}10001​ mSo, 1 mm2 = (1) mm x (1) mm= (11000)m×(11000)m(\dfrac{1}{1000}) m \times (\dfrac{1}{1000}) m(10001​)m×(10001​)m= 1 x 10 -6 m2Hence,2 mm2 = 2 x 10-6 m2AA = 2 x 10 -6 m2AB = 6 x 10 -6 m2Substituting the values in the formula above, we get,Pressure on AP=1.52×10−6P=0.75×106⇒P=7.5×105 Pa\text P = \dfrac{1.5}{2 \times 10^{-6}} \[0.5em] \text P = 0.75 \times 10^{6} \[0.5em] \Rightarrow \text P = 7.5 \times 10^{5} \text { Pa} \[0.5em]P=2×10−61.5​P=0.75×106⇒P=7.5×105 PaHence, Pressure on A = 7.5 x 105 PaPressure on BP=1.56×10−6P=0.25×106⇒P=2.5×105 Pa\text P = \dfrac{1.5}{6 \times 10^{-6}} \[0.5em] \text P = 0.25 \times 10^{6} \[0.5em] \Rightarrow \text P = 2.5 \times 10^{5} \text { Pa} \[0.5em]P=6×10−61.5​P=0.25×106⇒P=2.5×105 PaHence, Pressure on B = 2.5 x 105 Pa

Question 4

Question 3A vessel contains water up to a height of 1.5 m. Taking the density of water 10 3 kg m -3, acceleration due to gravity 9.8 m s -2 and area of base vessel 100 cm 2, calculate (a) the pressure and (b) the thrust, at the base of vessel.

Accepted Answer

(a) As we know,Pressure due to water column of height h = h ρ gGiven,h = 1.5 mρ = 1000 kg m -3g = 9.8 m s 2Substituting the values in the formula above, we get,P=1.5×1000×9.8P=14.7×103 N m−2\text P = 1.5 \times 1000 \times 9.8 \[0.5em] \text P = 14.7 \times 10^{3}\ \text {N m}^{-2} \[0.5em]P=1.5×1000×9.8P=14.7×103 N m−2Hence, pressure = 1.47 x 104 N m-2(b) As we know,Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​Given,A = 100 cm 2Converting cm 2 into m 2100 cm = 1 mSo, 100 cm x 100 cm = 1 m 2Hence, 100 cm 2 = 1×10010000\dfrac{1 \times 100}{10000}100001×100​Therefore, A = 10-2 m 2Substituting the values in the formula above, we get,1.47×104=Thrust (F)10−2⇒Thrust (F)=1.47×102⇒Thrust (F)=147 N1.47 \times 10^{4} = \dfrac{\text {Thrust (F)}}{10^{-2}} \[0.5em] \Rightarrow \text {Thrust (F)} = 1.47 \times 10^{2} \[0.5em] \Rightarrow \text {Thrust (F)} = 147\ \text N \[0.5em]1.47×104=10−2Thrust (F)​⇒Thrust (F)=1.47×102⇒Thrust (F)=147 NHence, Thrust = 147 N

Question 5

Question 4The area of base of a cylindrical vessel is 300 cm 2. Water (density = 1000 kg m -3) is poured into it up to a depth of 6 cm. Calculate (a) the pressure and (b) the thrust of water on the base. (g = 10 m s-2).

Accepted Answer

(a) As we know,Pressure due to water column of height h = hρgGiven,ρ = 1000 kg m -3g = 10 m s-2h = 6 cmConverting cm to m100 cm = 1 mSo, 6 cm = 1100\dfrac{1}{100}1001​ x 6 = 0.06 mHence, h = 0.06 mSubstituting the values in the formula above we get,P=0.06×1000×10P=600 Pa\text P = 0.06 \times 1000 \times 10 \[0.5em] \text P = 600 \text { Pa} \[0.5em]P=0.06×1000×10P=600 PaHence, Pressure = 600 Pa(b) As we know,Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​A = 300 cm 2Converting cm 2 into m 2100 cm = 1 mSo, 100 cm x 100 cm = 1 m 2Hence, 300 cm 2 = 1×30010000\dfrac{1 \times 300}{10000}100001×300​Therefore, A = 3 x 10-2 m 2Substituting the values in the formula we get,600=Thrust (F)3×10−2⇒Thrust (F)=600×3×10−2⇒Thrust (F)=18 N600 = \dfrac{\text {Thrust (F)}}{3 \times 10^{-2}} \[0.5em] \Rightarrow \text {Thrust (F)} = 600 \times 3 \times 10^{-2}\[0.5em] \Rightarrow \text {Thrust (F)} = 18\ \text N \[0.5em]600=3×10−2Thrust (F)​⇒Thrust (F)=600×3×10−2⇒Thrust (F)=18 NHence, thrust = 18 N

Question 6

Question 5(a) Calculate the height of a water column which will exert on it's base the same pressure as the 70 cm column of mercury. Density of mercury is 13.6 g cm-3.(b) Will the height of the water column in part (a) change if the cross section of the water column is made wider ?

Accepted Answer

(a) As we know,Pressure due to water column of height h = hρgandPascal's law, states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.Hence,Pressure due to water column = Pressure due to mercury columnHence,hw ρw g = hm ρm gGiven,hm = 70 cmρm = 13.6 g cm-3ρw = 1 g cm-3From the above formula, we get,hw = hmρmρw\dfrac{\text h_\text m \text ρ_\text m}{\text ρ_\text w}ρw​hm​ρm​​Substituting the values, we get,hw=(13.61)×70⇒hw=952 cm⇒hw=9.52 m\text h_\text w = (\dfrac{13.6}{1}) \times 70 \[0.5em] \Rightarrow \text h_\text w = 952\ \text{cm} \[0.5em] \Rightarrow \text h_\text w = 9.52\ \text{m} \[0.5em]hw​=(113.6​)×70⇒hw​=952 cm⇒hw​=9.52 mHence, height of a water column = 9.52 m(b) No, if the cross section of the water column is made wider, the height of the water column will be unaffected.

Question 7

Question 6The pressure of water on the ground floor is 40,000 Pa and on the first floor is 10,000 Pa. Find the height of the first floor.(Take : density of water = 1000 kg m-3, g = 10 m s-2)

Accepted Answer

As we know,Pressure due to water column of height h = h ρ gGiven,density of water = 1000 kg m -3g = 10 m s-2Pw on ground floor = 40,000 PaPw on first floor Pwf = 10,000 PaIn order to know the height of the first floor, let us calculate the difference in pressureP = Pwg - Pwf = 40,000 – 10,000 = 30,000 PaSubstituting the values in the formula,P=h ρ g30,000=1000×10×hh=3 m\text P = \text {h ρ g} \[0.5em] 30,000 = 1000 \times 10 \times \text h \[0.5em] \text h = 3\ \text m \[0.5em]P=h ρ g30,000=1000×10×hh=3 mHence, the height of the first floor = 3 m.

Question 8

Question 8In a hydraulic machine, a force of 2 N is applied on the piston of area of cross section 10 cm 2. What force is obtained on it's piston of area of cross section 100 cm 2?

Accepted Answer

As we know,Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​Given,A = 10 cm 2Converting cm 2 into m 2100 cm = 1 mSo, 100 cm x 100 cm = 1 m 2Hence, 10 cm 2 = 1×1010000\dfrac{1 \times 10}{10000}100001×10​Therefore, A = 10-3 m 2F = 2 N at area of cross section 10 cm 2Substituting the values in the formula we get,P = 210−3\dfrac{2}{10^{-3}}10−32​     [Equation 1]Now when, A = 100 cm 2Converting cm 2 into m 2100 cm = 1 mSo, 100 cm x 100 cm = 1 m 2Hence, 100 cm 2 = 1×10010000\dfrac{1 \times 100}{10000}100001×100​Therefore, A = 10-2 m 2Substituting the value in the formula above we get,Therefore, P = F10−2\dfrac{\text F}{10^{-2}}10−2F​     [Equation 2]Equating 1 and 2 we get,210−3=F10−2⇒F=2×10−210−3⇒F=20 N\dfrac{2}{10^{-3}} = \dfrac{\text F}{10^{-2}} \[0.5em] \Rightarrow \text F = \dfrac{2 \times 10^{-2} }{10^{-3}} \[0.5em] \Rightarrow \text F = 20\ \text N10−32​=10−2F​⇒F=10−32×10−2​⇒F=20 NHence, force at 100 cm 2 = 20 N

Question 9

Question 9What should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of it's brake shoe by exerting a force of 0.5 N on the pedal ?

Accepted Answer

As we know,Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​Let,area of cross section of the master cylinder = A1area of cross section of the wheel cylinder = A2force applied on pedal be F1 = 0.5 Nforce applied on brake shoe F2 = 15 NBy the principle of hydraulic brakes which works on Pascal's lawPressure on narrow piston = pressure on broader pistonHence,F1A1=F2A20.5A1=15A2A1A2=0.515⇒A1A2=5150⇒A1A2=130\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \[0.5em] \dfrac{0.5}{\text A_{1}} = \dfrac{15}{\text A_{2}} \[0.5em] \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{0.5}{15} \[0.5em] \Rightarrow \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{5}{150} \[0.5em] \Rightarrow \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{1}{30} \[0.5em]A1​F1​​=A2​F2​​A1​0.5​=A2​15​A2​A1​​=150.5​⇒A2​A1​​=1505​⇒A2​A1​​=301​Hence, the ratio of area of cross section = 1 : 30

Question 10

Question 10The areas of pistons in a hydraulic machine are 5 cm 2 and 625 cm 2. What force on the smaller piston will support a load of 1250 N on the larger piston ? State any assumption which you make in your calculation.

Accepted Answer

As we know,Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​and by the principle of hydraulic machinePressure on narrow piston = pressure on broader pistonHence,F1A1=F2A2\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \[0.5em]A1​F1​​=A2​F2​​Given,Area of narrow piston (A1) = 5 cm 2Area of wider piston (A2) = 625 cm 2force (F2) = 1250 NSubstituting the values in the formula above we get,F15=1250625F1=1250×5625⇒F1=10 N\dfrac{\text F_{1}}{5} = \dfrac{1250}{625} \[0.5em] \text F_{1} = \dfrac{1250 \times 5 }{625} \[0.5em] \Rightarrow \text F_{1} = 10\ \text N \[0.5em]5F1​​=6251250​F1​=6251250×5​⇒F1​=10 NHence, force acting on the smaller piston = 10 NAssumption — There is no friction and no leakage of liquid.

Question 11

Question 11(a) The diameter of neck and bottom of a bottle are 2 cm and 10 cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2 kgf, what force is exerted on the bottom of the bottle ?(b) Name the law/principle you have used to find the force in part (a)

Accepted Answer

(a) As we know,Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​and by the principle of hydraulic machinePressure on neck = pressure on bottom of bottleHence,F1A1=F2A2\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \[0.5em]A1​F1​​=A2​F2​​Given,Diameter of neck (d1) = 2 cmHence, A1 = 𝜋 (22)2(\dfrac{2}{2})^{2}(22​)2 = 𝜋Diameter of bottom of bottle (d2) = 10 cmHence, A2 = 𝜋 (102)2(\dfrac{10}{2})^{2}(210​)2 = 25 𝜋Force applied on the cork in the neck (F1) = 1.2 kgfSubstituting the values in the formula above we get,1.2π=F225πF2=1.2×25F2=30 kgf\dfrac{1.2}{\text π} = \dfrac{\text F_{2}}{25\text π} \[0.5em] \text F_{2} = 1.2 \times 25 \[0.5em] \text F_{2} = 30 \text{ kgf} \[0.5em]π1.2​=25πF2​​F2​=1.2×25F2​=30 kgfHence, force exerted at the bottom of the neck = 30 kgf.(b) The Pascal's law is applied to solve the part (a) which states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.Hence,F1A1=F2A2\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \[0.5em]A1​F1​​=A2​F2​​

Question 12

Question 12A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.

Accepted Answer

(a) As we know, by the principle of hydraulic machinePressure on the smaller piston = pressure on the larger pistonHence,F1A1=F2A2\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \[0.5em]A1​F1​​=A2​F2​​Given,Diameter of the smaller piston (d1) = 5 cmHence, A1 = 𝜋 (52)2(\dfrac{5}{2})^{2}(25​)2 = 6.25 𝜋Diameter of the larger piston (d2) = 25 cmHence, A2 = 𝜋 (252)2(\dfrac{25}{2})^{2}(225​)2 = 156.25 𝜋Force applied on the smaller piston (F1) = 50 kgfSubstituting the values in the formula above we get,506.25π=F2156.25πF2=50×156.256.25=1250F2=1250 kgf\dfrac{50}{6.25\text π} = \dfrac{\text F_{2}}{156.25\text π} \[1 em] \text F_{2} = \dfrac{50 \times 156.25}{6.25} = 1250 \[0.5em] \text F_{2} = 1250 \text { kgf} \[0.5em]6.25π50​=156.25πF2​​F2​=6.2550×156.25​=1250F2​=1250 kgfHence, force exerted on the larger piston = 1250 kgf.

Question 13

Question 13Two cylindrical vessels fitted with pistons A and B of area of cross section 8 cm2 and 320 cm2 respectively, are joined at their bottom by a tube and they are completely filled with water. When a mass of 4 kg is placed on piston A, find : (i) the pressure on piston A, (ii) the pressure on piston B, and (iii) the thrust on piston B.

Accepted Answer

(i) As we know,Pressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​Given,Force on the narrow piston A = 4 kgAA = 8 cm 2AB = 320 cm 2Substituting the values in the formula above, we get,PA=48⇒PA=0.5 kg cm−2\text P_\text A = \dfrac{4}{8} \[0.5em] \Rightarrow \text P_\text A = 0.5 \text { kg cm}^{-2} \[0.5em]PA​=84​⇒PA​=0.5 kg cm−2Hence, PA = 0.5 kg cm-2(ii) As we know, by the principle of hydraulic machinePressure on piston A = pressure on piston BHence, PB = 0.5 kg cm-2(iii) Thrust on piston B is acting in the upward direction, which is given byPressure (P) = Thrust (F)Area (A)\dfrac{\text {Thrust (F)}}{\text {Area (A)}}Area (A)Thrust (F)​Substituting the values, we get,0.5=thrust320⇒thrust=0.5×320⇒thrust=160kgf0.5 = \dfrac{\text{thrust}}{320} \[0.5em] \Rightarrow \text{thrust} = 0.5 \times 320 \[0.5em] \Rightarrow \text{thrust} = 160 \text{kgf} \[0.5em]0.5=320thrust​⇒thrust=0.5×320⇒thrust=160kgfHence, thrust on piston B = 160 kgf

Question 14

Question 14What force is applied on a piston of area of cross section 2 cm2 to obtain a force 150 N on the piston of area of cross section 12 cm2 in a hydraulic machine ?

Accepted Answer

As we know, by the principle of hydraulic machinePressure on piston A = pressure on piston BHence,F1A1=F2A2\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \[0.5em]A1​F1​​=A2​F2​​Given,A1 = 2 cm 2Converting cm 2 into m 2100 cm = 1 mSo, 100 cm x 100 cm = 1 m 2Hence, 2 cm 2 = 1×210000\dfrac{1 \times 2}{10000}100001×2​Therefore, A1 = 2 x 10-4 m 2A2 = 12 cm 2Hence, 12 cm 2 = 1×1210000\dfrac{1 \times 12}{10000}100001×12​Therefore, A2 = 12 x 10-4 m 2F2 = 150 NSubstituting the values in the formula above we get,F12×10−4=15012×10−4⇒F1=150×2×10−412×10−4⇒F1=25 N\dfrac{\text F_1}{2 \times 10^{-4}} = \dfrac{150}{12 \times 10^{-4}} \[1 em] \Rightarrow \text F_1 = \dfrac{150 \times 2 \times 10^{-4}}{12 \times 10^{-4}} \[0.5em] \Rightarrow \text F_1 = 25\ \text N \[0.5em]2×10−4F1​​=12×10−4150​⇒F1​=12×10−4150×2×10−4​⇒F1​=25 NHence, force applied = 25 N

Question 15

Question 15A 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. State and explain your observation.

Accepted Answer

We observe that the throw of the liquid from hole A is more than the B i.e., the liquid reaches to a greater distance on the horizontal surface from hole A than hole B. This shows that liquid pressure at a point increases with the increase of depth from it's free surface.

Question 16

Question 2At a given place, a mercury barometer records a pressure of 0.70 m of Hg. What would be the height of water column if mercury in barometer is replaced by water ? Take density of mercury to be 13.6 x 103 kg m-3

Accepted Answer

We know that,Pressure = h ρ gGiven,ρ = 13.6 x 103 kg m-3g = 9.8 m s -2h = 0.70 mSubstituting the values in the formula above we get,P=(0.70)×(13.6×103)×(9.8)⇒P=93.3×103 Pa\text P = (0.70) \times (13.6 \times 10^{3}) \times (9.8) \[0.5em] \Rightarrow \text P = 93.3 \times 10^{3} \text { Pa} \[0.5em]P=(0.70)×(13.6×103)×(9.8)⇒P=93.3×103 PaHence, P = 93.3 x 103PaNow, if we use a water barometer,P = 93.3 x 103 Paρ = 1 x 103 kg m-3g = 9.8 m s -2Substituting the values in the formula above we get,93.3×103=(h)×(1×103)×(9.8)h=93.3×1031×103×9.8h=9.52 m93.3 \times 10^{3} = (\text h) \times (1 \times 10^{3}) \times (9.8) \[0.5em] \text h = \dfrac{93.3 \times 10^{3}}{1 \times 10^{3} \times 9.8} \[0.5em] \text h = 9.52 \text { m} \[0.5em]93.3×103=(h)×(1×103)×(9.8)h=1×103×9.893.3×103​h=9.52 mHence, height of water colum = 9.52 m

Question 17

Question 15State two factors which affect the atmospheric pressure as we go up.

Accepted Answer

The atmospheric pressure decreases as we go up because —Decrease in height of air column which causes a linear decrease in the atmospheric pressure.Decrease in density of air which causes a non-linear decrease in atmospheric pressure.

Question 18

Question 16Why does a fountain pen leak at a high altitude?

Accepted Answer

At high altitudes, a fountain pen leaks because normally a fountain pen filled with ink contains some air also which is at a pressure equal to the atmospheric pressure on the earths' surface. When the pen is taken at an altitude, the atmospheric pressure at this altitude is low so the excess pressure due to air inside the rubber tube forces the ink to leak out.

Question 19

Question 17Why does our nose start bleeding on high mountains?

Accepted Answer

Our blood pressure is slightly more than atmospheric pressure. However, at high altitude, the atmospheric pressure becomes less because the height of air column above that altitude is less than at the earth's surface. As a result, at high altitudes the blood pressure becomes much more than the atmospheric pressure and nose bleeding may occur due to excess blood pressure.

Pressure in Fluids and Atmospheric Pressure

Exercise 4A Multiple Choice Type

Exercise 4A Numericals

Exercise 4A Short Answer Type

Exercise 4B Numericals

Exercise 4B Short Answer Type