{"id":59597,"date":"2022-01-28T10:33:51","date_gmt":"2022-01-28T05:03:51","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=59597"},"modified":"2024-10-07T16:36:36","modified_gmt":"2024-10-07T11:06:36","slug":"ncert-exemplar-for-class-11-physics-chapter-10-mechanical-properties-of-fluids","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-exemplar-solutions\/class-11\/physics\/chapter-10\/mechanical-properties-of-fluids\/","title":{"rendered":"NCERT Exemplar for Class 11 Physics Chapter 10 &#8211; Mechanical Properties of Fluids"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" style=\"display: none;\"><label for=\"item\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-exemplar-solutions\/class-11\/physics\/chapter-10\/mechanical-properties-of-fluids\/#FAQ_on_NCERT_Exemplar_Solutions_for_Class_11_Physics_Chapter_10\" title=\"FAQ on NCERT Exemplar Solutions for Class 11 Physics Chapter 10\">FAQ on NCERT Exemplar Solutions for Class 11 Physics Chapter 10<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-exemplar-solutions\/class-11\/physics\/chapter-10\/mechanical-properties-of-fluids\/#Q_How_to_solve_the_questions_present_in_Chapter_10_of_NCERT_Exemplar_Solutions_for_Class_11_Physics_faster\" title=\"Q. How to solve the questions present in Chapter 10 of NCERT Exemplar Solutions for Class 11 Physics faster?\">Q. How to solve the questions present in Chapter 10 of NCERT Exemplar Solutions for Class 11 Physics faster?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-exemplar-solutions\/class-11\/physics\/chapter-10\/mechanical-properties-of-fluids\/#Q_What_are_the_key_features_of_NCERT_Exemplar_Solutions_for_Class_11_Physics_Chapter_10_Mechanical_Properties_of_Fluids\" title=\"Q. What are the key features of NCERT Exemplar Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids?\">Q. What are the key features of NCERT Exemplar Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids?<\/a><\/li><\/ul><\/nav><\/div>\n<p><span style=\"font-size: 14pt;\"><strong>Class 11 Physics Chapter 10 &#8211; Mechanical Properties of Fluids<\/strong><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong><span style=\"color: #0000ff;\">Question.1<\/span><\/strong>: Explain why<\/span><\/p>\n<p><span style=\"color: #000000;\">(a) The blood pressure in humans is greater at the feet than at the brain<\/span><br \/>\n<span style=\"color: #000000;\">(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km<\/span><br \/>\n<span style=\"color: #000000;\">(c) Hydrostatic pressure is a scalar quantity even though the pressure is force divided by area<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Ans:<\/strong><\/span><\/p>\n<p><strong>(i).<\/strong> The blood column to the feet is at a greater height than the head, thus the blood pressure in the feet is greater than that in the brain.<\/p>\n<p><strong>(ii).<\/strong> The density of the atmosphere does not decrease linearly with the increase in altitude, in fact, most of the air molecules are close to the surface. Thus, there is this nonlinear variation of atmospheric pressure.<\/p>\n<p><strong>(iii).<\/strong> In hydrostatic pressure the force is transmitted equally in all direction in the liquid, thus there is no fixed direction of pressure making it a scalar quantity.<\/p>\n<p><span style=\"color: #000000;\"><span style=\"color: #0000ff;\"><strong>Question 2:<\/strong> <\/span> Explain why<br \/>\n(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.<br \/>\n(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)<br \/>\n(c) Surface tension of a liquid is independent of the area of the surface<br \/>\n(d) Water with detergent dissolved in it should have small angles of contact.<br \/>\n(e) A drop of liquid under no external forces is always spherical in shape<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Ans:<\/strong><\/span><\/p>\n<p><strong>(a)<\/strong> Water molecules have weak intermolecular forces and a strong force of attraction towards solids. Thus, they spill out. Whereas mercury molecules have a stronger intermolecular force of attraction and a weak attraction force towards solids, thus they form droplets.<\/p>\n<p><strong>(b)<\/strong> The angle of contact is the angle between the line tangent to the liquid surface at the point of contact and the surface of the liquid. It is donated by \u03b8 in the following diagram:<\/p>\n<p>In the diagram <strong>S<\/strong><strong><sub>sl<\/sub><\/strong><strong>, S<\/strong><strong><sub>la<\/sub><\/strong><strong> and   S<sub>sa<\/sub><\/strong>, are the respective interfacial tensions between the liquid-solid, liquid-air, and solid-air interfaces. At the line of contact, the surface forces between the three media are in equilibrium, i.e.,<\/p>\n<p><span class=\"base\"><span class=\"mord mathdefault\">c<\/span><span class=\"mord mathdefault\">o<\/span><span class=\"mord mathdefault\">s<\/span><span class=\"mord\">\u0398<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size2\">(<\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">S<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">l<\/span><span class=\"mord mathdefault mtight\">a<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">S<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">a<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"mbin mtight\">\u2212<\/span><span class=\"mord mathdefault mtight\">S<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">l<\/span><span class=\"mord mathdefault mtight\">a<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size2\">)<\/span><\/span><\/span><\/span><\/p>\n<p>Thus, for mercury, the angle of contact \u03b8, is obtuse because <strong>S<sub>sa<\/sub> &lt; S<sub>la<\/sub><\/strong>. And for water, the angle is acute because <strong>S<sub>sl<\/sub> &lt; S<sub>la<\/sub><\/strong><\/p>\n<p><strong>(c)<\/strong> A liquid always tends to acquire minimum surface area because of the presence of surface tension. And as a sphere always has the smallest surface area for a given volume, a liquid drop will always take the shape of a sphere under zero external forces.<\/p>\n<p><strong>(d)<\/strong> Surface tension is independent of the area of the liquid surface because it is a force depending upon the unit length of the interface between the liquid and the other surface, not the area of the liquid.<\/p>\n<p><strong>(e)<\/strong> Clothes have narrow pores that behave like capillaries, now we know that the rise of liquid in a capillary tube is directly proportional to <strong>cos<\/strong><strong> \u03b8<\/strong>. So a soap decreases the value of <strong>\u03b8<\/strong> in order to increase the value of <strong>cos<\/strong><strong> \u03b8<\/strong>, allowing the faster rise of water through the pores of the clothes.<\/p>\n<p><strong>Q.3: Fill in the blanks using the word(s) from the list appended with each statement:<\/strong><\/p>\n<p><strong>(a) The surface tension of liquids generally \u2026 with temperatures (increases\/decreases)<br \/>\n(b) The viscosity of gases \u2026 with temperature, whereas the viscosity of liquids \u2026 with temperature (increases\/decreases)<br \/>\n(c) For solids with an elastic modulus of rigidity, the shearing force is proportional to \u2026 , while for fluids, it is proportional to \u2026 (shear strain\/rate of shear strain)<br \/>\n(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass \/ Bernoulli\u2019s principle)<br \/>\n(e) For the model of a plane in a wind tunnel, turbulence occurs at a \u2026 speed for turbulence for an actual plane (greater \/ smaller)<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p>(a) Decreases<\/p>\n<p>(b) increases; decreases<\/p>\n<p>(c) shear strain; rate of shear strain<\/p>\n<p>(d) conservation of mass; Bernoulli\u2019s principle<\/p>\n<p>(e) greater.<\/p>\n<p><strong>Q.4: Explain why?<\/strong><\/p>\n<p><strong>(a) To keep a piece of paper horizontal, you should blow over, not under, it<br \/>\n(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers<br \/>\n(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection<br \/>\n(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel<br \/>\n(e) A spinning cricket ball in air does not follow a parabolic trajectory<\/strong><\/p>\n<p><strong> <\/strong><strong>Ans: <\/strong><\/p>\n<p><strong>(a)<\/strong> If we blow over a piece of paper, velocity of air above the paper becomes more than that below it. As K.E. of air above the paper increases, so in accordance with Bernoulli\u2019s theorem its pressure energy and hence its pressure decreases.<\/p>\n<p>Due to greater value of pressure below the piece of paper = atmospheric  pressure, it remains horizontal and does not fall.<\/p>\n<p><strong>(b)<\/strong> As per the equation of continuity area \u00d7 velocity = constant. When we try to close a water tap with our fingers, the area of cross-section of the outlet of water jet is reduced considerably as the openings between our fingers provide constriction  (regions of smaller area)<br \/>\nThus, velocity of water increases greatly and fast jets of water come through the openings between our fingers.<\/p>\n<p><strong>(c)<\/strong> The size of the needle controls the velocity of flow and the thumb pressure controls pressure.  According to the Bernoulli\u2019s theorem  P +<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 1 }{ 2 } pv^{ 2 }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord mathdefault\">p<\/span><span class=\"mord\"><span class=\"mord mathdefault\">v<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = Constant<\/p>\n<p>In this equation, the pressure P occurs with a single power whereas the velocity occurs with a square power. Therefore, the velocity has more effect compared to the pressure. It is for this reason that needle of the syringe controls flow rate better than the thumb pressure exerted by the doctor.<\/p>\n<p><strong>(d)<\/strong> This is because of principle of conservation of momentum. While the flowing fluid carries forward momentum, the vessel gets a backward momentum.<br \/>\n<strong>(e)<\/strong> A spinning cricket ball would have followed a parabolic trajectory has there been no air. But because of air the Magnus effect takes place. Due to the Magnus effect the spinning cricket ball deviates from its parabolic trajectory.<\/p>\n<p><strong>Question.5: A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?<\/strong><\/p>\n<p><strong>Ans: <\/strong><strong>Given: <\/strong>Radius of the heel,<\/p>\n<p><strong>r =<\/strong><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">d<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> <strong>= 0.005 m<\/strong><\/p>\n<p>Mass of the lady, <strong>m = 50 kg<\/strong><\/p>\n<p>Area of the heel, <strong>A = <\/strong>\u03c0r<sup>2<\/sup> = \u03c0 (0.005)<sup>2<\/sup><strong>= 7.85 \u00d7 10<sup>-5 <\/sup>m<sup>2<\/sup><\/strong><\/p>\n<p>Force on the floor due to the heel: <strong>F =<\/strong> mg = 50 \u00d7 9.8 <strong>= 490 N<\/strong><\/p>\n<p>Pressure exerted by the heel on the floor:<\/p>\n<p>P = <span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">A<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">F<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>  =<span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"> 7.85<span class=\"mbin mtight\">\u00d7<\/span> 10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22125<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">490<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/p>\n<p><strong>P = <\/strong><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">6<\/span><span class=\"mord\">.<\/span><span class=\"mord\">2<\/span><span class=\"mord\">4<\/span><span class=\"mbin\">\u00d7<\/span><\/span><span class=\"base\"><span class=\"mord\">1<\/span><span class=\"mord\">0<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">6<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> Nm <sup>-2<\/sup><\/p>\n<p><strong><sup> <\/sup><\/strong><strong>Q-6: Toricelli\u2019s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m<sup>\u20133<\/sup>. Determine the height of the wine column for normal atmospheric pressure<\/strong><\/p>\n<p><strong>Ans: <\/strong>We know:<br \/>\nDensity of mercury, <strong>\u03c1<sub>1<\/sub> = 13.6 \u00d7 10<sup>3<\/sup> kg\/m<sup>3<\/sup><\/strong><br \/>\nHeight of the mercury column, h<sub>1<\/sub> = 0.76 m<br \/>\nDensity of French wine, <strong>\u03c1<sub>2<\/sub> = 984 kg\/m<sup>3<\/sup><\/strong><\/p>\n<p>Let the height of the French wine column<strong> = h<sub>2<\/sub><br \/>\n<\/strong>Acceleration due to gravity, <strong>g = 9.8 m\/s<sup>2<\/sup>.<\/strong><\/p>\n<p>We know that:<br \/>\nPressure in the mercury column = Pressure in the wine column<br \/>\n<strong>\u03c1<sub>1<\/sub>h<sub>1 <\/sub>g =\u03c1<sub>2<\/sub>h<sub>2 <\/sub>g<\/strong><\/p>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel\">\u21d2<\/span><\/span><\/span>  h<sub>2<\/sub> =<span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">2<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">1<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"mord mathdefault mtight\">h<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">1\/g<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\Rightarrow<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel\">\u21d2<\/span><\/span><\/span>  h<sub>2<\/sub> =<span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">984<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">13.6<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">3<\/span><\/span><\/span><span class=\"mbin mtight\">\u00d7<\/span>0.76<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><strong>= 10.5 m<\/strong><\/p>\n<p><strong>Q-7: A vertical off-shore structure is built to withstand a maximum stress of 10<sup>9<\/sup> Pa. Is the structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents<\/strong><\/p>\n<p><strong>Ans: <\/strong><strong>Given: <\/strong>The maximum stress the structure can handle, <strong>P = 10<sup>9<\/sup> Pa<\/strong><br \/>\nDepth of the sea, <strong>d =<\/strong> 3 km <strong>= 3 \u00d7 10<sup>3<\/sup> m<br \/>\n<\/strong>Density of water, <strong>\u03c1 =<\/strong> <strong>10<sup>3<\/sup> kg\/m<sup>3<\/sup><\/strong><br \/>\nAcceleration due to gravity, <strong>g = 9.8 m\/s<sup>2<\/sup><\/strong><\/p>\n<p>We know:<br \/>\nThe pressure exerted by the seawater at depth,<\/p>\n<p><strong>d = \u03c1dg<\/strong> = 10<sup>3<\/sup> x 3 \u00d7 10<sup>3<\/sup> \u00d7 9.8 <strong>= 2.94 \u00d7 10<sup>7<\/sup> Pa<\/strong><\/p>\n<p>As the sea exerts a pressure lesser than the maximum stress the structure can handle, the structure can survive on the oil well in the sea.<\/p>\n<p><strong>Q-8: A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm<sup>2<\/sup>. What maximum pressure would the smaller piston have to bear ?<\/strong><\/p>\n<p><strong>Ans: <\/strong><strong>Given: <\/strong>Maximum mass that can be lifted, <strong>m = 3000 kg<\/strong><br \/>\nArea of cross-section of the load-carrying piston, <strong>A =<\/strong> 425 cm<sup>2<\/sup><strong>= 425 \u00d7 10<sup>-4<\/sup> m<sup>2<\/sup><\/strong><\/p>\n<p>The maximum force exerted by the load,<br \/>\nF = <strong>mg =<\/strong> 3000 x 9.8 <strong>=  <sup> <\/sup>29400 N<\/strong><br \/>\nThe maximum pressure  on the load carrying piston ,<\/p>\n<p><strong>P = F \/ A<\/strong><\/p>\n<p><strong>P = <\/strong><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"> 425<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22124<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">29400<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> <strong>= 6.917 x 10<sup>5<\/sup> Pa<\/strong><\/p>\n<p>In a liquid, the pressure is transmitted equally in all directions. Therefore, the maximum pressure on the smaller is 6.917 \u00d7 10<sup>5<\/sup> Pa<\/p>\n<p><strong>Q-9: A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?<\/strong><\/p>\n<p><strong>Ans: <\/strong><strong>Given:<\/strong><\/p>\n<p>Height of the spirit column, <strong>h<sub>1<\/sub> = <\/strong>12.5 cm <strong>= 0.125 m<\/strong><\/p>\n<p>Height of the water column, <strong>h<sub>2<\/sub> =<\/strong> 10 cm <strong>= 0.1 m<\/strong><\/p>\n<p>Let, A and B be the points of contact between spirit and mercury and water and mercury, respectively.<br \/>\n<strong>P<sub>0<\/sub> = Atmospheric pressure<br \/>\n\u03c1<sub>1<\/sub> = Density of spirit<br \/>\n\u03c1<sub>2<\/sub> = Density of water<\/strong><\/p>\n<p>Pressure a <strong>point A<\/strong> = P<sub>0 <\/sub>+ \u03c1<sub>1<\/sub>h<sub> 1<\/sub>g<\/p>\n<p>Pressure at <strong>point B<\/strong> = P<sub>0 <\/sub>+ \u03c1<sub>2<\/sub>h<sub> 2<\/sub>g<\/p>\n<p>We know pressure at B and D is the same so;<\/p>\n<p><strong>P<sub>0 <\/sub>+ \u03c1<sub>1<\/sub>h<sub>1<\/sub>g = P<sub>0 <\/sub>+ \u03c1<sub>2<\/sub>h<sub> 2<\/sub>g<\/strong><\/p>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">2<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">1<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">h<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">1<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">h<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">2<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">12.5<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">10<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 0.8<\/p>\n<p>Therefore the specific gravity of water is 0.8.<\/p>\n<p><span style=\"color: #0000ff;\"><strong>Q-10: In the previous problem, if 15.0 cm of water and spirit each is further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)<\/strong><\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Ans: <\/strong><\/span><\/p>\n<p><strong>Given: <\/strong>h<sub>1<\/sub> = 10.0 cm<br \/>\n\u03c1<sub>1<\/sub> = 1 g cm<sup>-3<br \/>\n<\/sup>For spirit column in other arm of U-tube,<br \/>\nh<sub>2<\/sub> = 12.5 cm<br \/>\np<sub>2<\/sub> = ?<\/p>\n<p>Let h be the difference between the levels of mercury in the two arms.<br \/>\nPressure exerted by height h, of the mercury column:<br \/>\n= h\u03c1g<br \/>\n= h \u00d7 13.6g \u2026 (i)<br \/>\nDifference between the pressures exerted by water and spirit:<br \/>\n= \u03c1<sub>1<\/sub>h<sub>1<\/sub>g \u2013 \u03c1<sub>2<\/sub>h<sub>2<\/sub>g<br \/>\n= g(25 \u00d7 1 \u2013 27.5 \u00d7 0.8)<br \/>\n= 3g \u2026 (ii)<br \/>\nEquating equations (i) and (ii), we get:<br \/>\n13.6 hg = 3g<br \/>\nh = 0.220588 \u2248 0.221 cm<\/p>\n<p><strong>Q-11: Can Bernoulli\u2019s equation be used to describe the flow of water through a rapid in a river? Explain.<\/strong><\/p>\n<p><strong>Ans: <\/strong>Bernoulli\u2019s equation cannot be applied to the water flowing in a river because it is applicable only for <strong>ideal liquids in a streamlined flow<\/strong> and the water in a stream is turbulent.<\/p>\n<p><strong>Q-12: Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli\u2019s equation? Explain<\/strong><\/p>\n<p><strong>Ans: <\/strong>No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli\u2019s equation, provided the atmospheric pressure at the two points where Bernoulli\u2019s equation applied to the system are significantly different.<\/p>\n<p><strong> <\/strong><strong>Q-13: Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 \u00d7 10<sup>\u20133<\/sup> kg s<sup>\u20131<\/sup>, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 \u00d7 10<sup>3<\/sup> kg m<sup>\u20133<\/sup> and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].<\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\n<strong>Length<\/strong> of the horizontal tube, <strong>l = 1.5 m<\/strong><\/p>\n<p><strong>Radius<\/strong> of the tube, r = 1 cm <strong>= 0.01 m<\/strong><\/p>\n<p><strong>Diameter<\/strong> of the tube, d = 2r <strong>= 0.02 m<\/strong><\/p>\n<p>Glycerine is flowing at the rate of 4<strong>.0 \u00d7 10<sup>-3<\/sup> kg\/s<\/strong><\/p>\n<p><strong>M = 4.0 \u00d7 10<sup>-3<\/sup> kg\/s<\/strong><\/p>\n<p><strong>Density of glycerine, \u03c1 = 1.3 \u00d7 10<sup>3<\/sup> kg m<sup>-3<\/sup><\/strong><\/p>\n<p><strong>Viscosity of glycerine, \u03b7 = 0.83 Pa s<\/strong><\/p>\n<p>We know, volume of glycerine flowing per sec:<\/p>\n<p>V =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ M }{ density }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">d<\/span><span class=\"mord mathdefault mtight\">e<\/span><span class=\"mord mathdefault mtight\">n<\/span><span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">i<\/span><span class=\"mord mathdefault mtight\">t<\/span><span class=\"mord mathdefault mtight\">y <\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">M<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> =<span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1.3<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">3<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">4<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 3.08 x 10 <sup>-6<\/sup> m<sup>3<\/sup> \/s<\/p>\n<p><strong>Using Poiseville\u2019s formula, we get:<\/strong><\/p>\n<p>V = <span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">8<span class=\"mord mathdefault mtight\">\u03b7<\/span><span class=\"mord mathdefault mtight\">l<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c0<\/span><span class=\"mord mathdefault mtight\">p<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u2018<\/span><\/span><\/span><span class=\"mord mathdefault mtight\">r<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">4<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">p^{\u2018} = \\frac{ V \\;8\\; \\eta \\;l}{ \\pi r^{ 4 }}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mord mathdefault\">p <\/span><\/span><span class=\"mrel\">= <\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c0<\/span><span class=\"mord mathdefault mtight\">r<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">4<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">V<\/span>8<span class=\"mord mathdefault mtight\">\u03b7<\/span><span class=\"mord mathdefault mtight\">l<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p>Where <strong>p\u2019<\/strong> is the pressure difference between the two ends of the pipe.<\/p>\n<p>p = <span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c0<\/span><span class=\"mbin mtight\">\u00d7<\/span><span class=\"mopen mtight\">(<\/span>0.01<span class=\"mclose mtight\">)<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">4<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3.08<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22126<\/span><\/span><\/span><span class=\"mbin mtight\">\u00d7<\/span>8<span class=\"mbin mtight\">\u00d7<\/span>0.83<span class=\"mbin mtight\">\u00d7<\/span>2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 9.8 x 10<sup>2 <\/sup>Pa<\/p>\n<p>And, we know:<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">R = \\frac{ 4 \\;\\rho \\;V }{ \\pi \\;d\\; \\eta }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathdefault\">R<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c0<\/span><span class=\"mord mathdefault mtight\">d<\/span><span class=\"mord mathdefault mtight\">\u03b7<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">4<span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"mord mathdefault mtight\">V<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>, <strong>[Where R = Reynolds\u2019s number]<\/strong><\/p>\n<p>R = <span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c0<\/span><span class=\"mbin mtight\">\u00d7<\/span>0.83<span class=\"mbin mtight\">\u00d7<\/span>0.02<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">4<span class=\"mbin mtight\">\u00d7<\/span>1.3<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">3<\/span><\/span><\/span><span class=\"mbin mtight\">\u00d7<\/span>3.08<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22126<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 0.3<\/p>\n<p><strong>Since the Reynolds\u2019s number is 0.3   which is way smaller than 2000 , the flow of glycerine in the pipe is laminar.<\/strong><\/p>\n<p><strong>Q-14:  In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s<sup>\u20131<\/sup> and 63 m s<sup>-1<\/sup> respectively. What is the lift on the wing if its area is 2.5 m<sup>2<\/sup> ? Take the density of air to be 1.3 kg m<sup>\u20133<\/sup><\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\nSpeed of wind on the upper side of the wing, <strong>V<sub>1<\/sub> = 70 m\/s<\/strong><br \/>\nSpeed of wind on the lower side of the wing, <strong>V<sub>2<\/sub> = 63 m\/s<\/strong><br \/>\nArea of the wing, <strong>A = 2.5 m<sup>2<\/sup><\/strong><br \/>\nDensity of air, <strong>\u03c1 = 1.3 kg m <sup>-3<\/sup><\/strong><\/p>\n<p><strong>Using Bernoulli\u2019s theorem, we get :<br \/>\n<\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">P_{ 1 } + \\frac{ 1 }{ 2 }\\rho V_{ 1 }^{ 2 } = P_{ 2 } + \\frac{ 1 }{ 2 }\\rho V_{ 2 }^{ 2 }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mord mathdefault\">P<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mbin\">+<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"mord\"><span class=\"mord mathdefault\">V<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mord mathdefault\">P<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mbin\">+<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"mord\"><span class=\"mord mathdefault\">V<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><span class=\"base\"><span class=\"mord\"><span class=\"mord mathdefault\">P<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord\">\u2013<\/span><span class=\"mord\"><span class=\"mord mathdefault\">P<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size1\">(<\/span><\/span><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"mord\"><span class=\"mord mathdefault\">V<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord\">\u2013<\/span><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"mord\"><span class=\"mord mathdefault\">V<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size1\">)<\/span><\/span><\/span><\/span><\/p>\n<p><strong>Where, P<sub>1<\/sub> = Pressure on the upper side of the wing<br \/>\nP<sub>2<\/sub> = Pressure on the lower side of the wing<\/strong><\/p>\n<p><strong>Now the lift on the wing = ( P<sub>2<\/sub> \u2013 P<sub>1<\/sub> ) x Area<\/strong><br \/>\n= <span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size1\">(<\/span><\/span><span class=\"mord\"><span class=\"mord mathdefault\">V<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord\">\u2013<\/span><span class=\"mord\"><span class=\"mord mathdefault\">V<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size1\">)<\/span><\/span><\/span><span class=\"mbin\">\u00d7<\/span><\/span><span class=\"base\"><span class=\"mord mathdefault\">A<\/span><\/span><\/p>\n<p>= <span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mbin\">\u00d7<\/span><\/span><span class=\"base\"><span class=\"mord\">1<\/span><span class=\"mord\">.<\/span><span class=\"mord\">3<\/span><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size2\">(<\/span><\/span><span class=\"mopen delimcenter\">(<\/span><span class=\"mord\">7<\/span><span class=\"mord\">0<\/span><span class=\"mclose delimcenter\">)<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mord\">\u2013<\/span><span class=\"mopen delimcenter\">(<\/span><span class=\"mord\">6<\/span><span class=\"mord\">3<\/span><span class=\"mclose delimcenter\">)<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size2\">)<\/span><\/span><\/span><span class=\"mbin\">\u00d7<\/span><\/span><span class=\"base\"><span class=\"mord\">2<\/span><span class=\"mord\">.<\/span><span class=\"mord\">5<\/span><\/span><\/p>\n<p>= 1.51 x 10<sup>3<\/sup> N<\/p>\n<p>Therefore the lift experienced by the wings of the air craft is 1.51 x 10<sup>3<\/sup> N.<\/p>\n<p><span style=\"color: #0000ff;\"><strong>Question-15: <\/strong><\/span><\/p>\n<p><strong>Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?<\/strong><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Ans:<\/strong><\/span><\/p>\n<p>Figure (a) is incorrect. The reason is that at the kink, the velocity of flow of liquid is large and hence using the Bernoulli\u2019s theorem the pressure is less. As a result, the water should not rise higher in the tube where there is a kink (i.e., where the area of cross-section is small).<\/p>\n<p><strong>Q-16: The cylindrical tube of a spray pump has a cross-section of 8.0 cm<sup>2<\/sup> one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min<sup>\u20131<\/sup>, what is the speed of ejection of the liquid through the holes?<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p><strong>Given:<br \/>\n<\/strong><strong>Number of holes, n = 40<br \/>\n<\/strong>Cross-sectional area  of the spray pump, A<sub>1 <\/sub>= 8 cm <sup>-2<\/sup><strong>= 8 \u00d7 10<sup>-4<\/sup> m<sup>-2<\/sup><\/strong><br \/>\nRadius of each hole, <strong>r = 0.5 \u00d7 10<sup>-3<\/sup> m<\/strong><br \/>\nCross-sectional area of each hole, a = \u03c0r<sup>2<\/sup> = <strong>\u03c0 (0.5 \u00d7 10<sup>-3<\/sup> )<sup>2<\/sup> m<sup>2<\/sup><\/strong><br \/>\nTotal area of 40 holes, A<sub>2<\/sub>= n \u00d7 a =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">40 \\times \\pi \\left ( 0.5 \\times 10^{ -3 } \\right )^{ 2 }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">4<\/span><span class=\"mord\">0<\/span><span class=\"mbin\">\u00d7<\/span><\/span><span class=\"base\"><span class=\"mord mathdefault\">\u03c0<\/span><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size1\">(<\/span><\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><span class=\"mord\">5<\/span><span class=\"mbin\">\u00d7<\/span><span class=\"mord\">1<\/span><span class=\"mord\">0<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size1\">)<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> m<sup>2<\/sup> = 3.14 x 10<sup>-5<\/sup> m<sup>2<\/sup><br \/>\nSpeed of flow of water inside the tube, V<sub>1<\/sub> = 1.5 m\/min <strong>= 0.025 m\/s<\/strong><br \/>\nLet, the water ejected through the holes at a speed <strong>= V<sub>2<\/sub><\/strong><\/p>\n<p><strong>Using the law of continuity:<\/strong><br \/>\n<strong>A<sub>1<\/sub>V<sub>1<\/sub> = A<sub>2 <\/sub>V<sub>2<\/sub><\/strong><\/p>\n<p>V<sub>2<\/sub> =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ A_{ 1 } V_{ 1 } }{ A_{ 2 } }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">A<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">2<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">A<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">1<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"mord mathdefault mtight\">V<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">1<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>  =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 8 \\times 10^{ -4 }\\times 0.025 }{ 3.14 \\times 10^{ -5 } }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3.14<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22125<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">8<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22124<\/span><\/span><\/span><span class=\"mbin mtight\">\u00d7<\/span>0.025<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><strong>Therefore, V<sub>2 <\/sub>= 0.636 m\/s<\/strong><\/p>\n<p><strong>Q-17: A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 \u00d7 10<sup>\u20132<\/sup> N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ?<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\nThe maximum weight the film can support, <strong>W = 1.5 \u00d7 10<sup>-2<\/sup> N<\/strong><br \/>\nLength of the slider, l = 30 cm <strong>= 0.3 m<\/strong><\/p>\n<p>Total length of liquid film, l = 2 x 30 cm = 60 cm = 0.6 m because the liquid film has two surfaces.<br \/>\nSurface tension, T = F\/l =1.5 x 10<sup>-2<\/sup> N\/0.6m =2.5 x 10<sup>-2<\/sup> Nm<sup>-1<\/sup><\/p>\n<p>&nbsp;<\/p>\n<p><strong>Q-18: Figure (a) shows a thin liquid film supporting a small weight = 4.5 \u00d7 10<sup>\u20132<\/sup> N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.<\/strong><\/p>\n<p><strong>Case (a):<\/strong><\/p>\n<p>(a) Given, the length of the film supporting the weight = 40 cm = 0.4 m.<\/p>\n<p>Total weight supported (or force) = 4.5 x 10<sup>-2 <\/sup>N.<br \/>\nFilm has two free surfaces, Surface tension, S =4.5 x 10<sup>-2<\/sup>\/2 x 0.4 =5.625 x 10<sup>-2 <\/sup>Nm<sup>-1<\/sup><br \/>\nSince the liquid is same for all the given cases (a), (b) and (c), and temperature is also the same, therefore surface tension for cases (b) and (c) will also be the same = 5.625 x 10<sup>-2<\/sup>.<\/p>\n<p>In Fig. (b) and (c), the length of the film supporting the weight is also the same as that of (a), hence the total weight supported in each case is 4.5 x 10<sup>-2<\/sup> N.<\/p>\n<p><strong>Q-19: What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 \u00b0C) is 4.65 \u00d7 10<sup>\u20131<\/sup> N m<sup>\u20131<\/sup>. The atmospheric pressure is 1.01 \u00d7 10<sup>5<\/sup> Pa. Also give the excess pressure inside the drop.<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p><strong>Given:<br \/>\nSurface tension of mercury, S = 4.65 \u00d7 10<sup>-1<\/sup> N m<sup>-1<\/sup><br \/>\nRadius of the mercury drop, r = 3.00 mm = 3 \u00d7 10<sup>-3<\/sup> m<br \/>\nAtmospheric pressure, P<sub>0<\/sub> = 1.01 \u00d7 10<sup>5<\/sup> Pa<br \/>\n<\/strong><br \/>\nWe know:<\/p>\n<p>Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure<br \/>\n= 2S\/r + P<sub>0 <\/sub>=<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\left (\\frac{ 2 \\times 4.65 \\times 10^{ -1 } }{ 3 \\times 10^{ -3 } } \\right ) + 1.01 \\times 10^{ 5 }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size2\">(<\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mbin mtight\">\u00d7<\/span>4.65<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22121<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size2\">)<\/span><\/span><\/span><span class=\"mbin\">+<\/span><\/span><span class=\"base\"><span class=\"mord\">1<\/span><span class=\"mord\">.<\/span><span class=\"mord\">0<\/span><span class=\"mord\">1<\/span><span class=\"mbin\">\u00d7<\/span><\/span><span class=\"base\"><span class=\"mord\">1<\/span><span class=\"mord\">0<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">5<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 1.0131 x 10<sup>5<\/sup> Pa<\/p>\n<p>= 1.01 x 10<sup>5<\/sup> Pa<\/p>\n<p>Excess pressure =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 2S }{ r }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">r<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">S<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><strong>= <\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\left (\\frac{ 2 \\times 4.65 \\times 10^{ -1 } }{ 3 \\times 10^{ -3 } } \\right )<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size2\">(<\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mbin mtight\">\u00d7<\/span>4.65<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22121<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size2\">)<\/span><\/span><\/span><\/span><\/span> = 310<sup> <\/sup>Pa<\/p>\n<p><strong>Q-20: What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 \u00b0C) is 2.50 \u00d7 10<sup>\u20132<\/sup> N m<sup>\u20131<\/sup> ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 \u00d7 10<sup>5<\/sup> Pa)<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\nSurface tension of the soap solution, <strong>S = 2.50 \u00d7 10<sup>-2<\/sup> N\/m<\/strong><br \/>\n<strong> r = 5.00 mm = 5 \u00d7 10<sup>-3<\/sup> m<\/strong><br \/>\nDensity of the soap solution, <strong>\u03c1 = 1.2 \u00d7 10<sup>3<\/sup> kg\/m<sup>3<\/sup><\/strong><sup><br \/>\n<\/sup>Relative density of the soap solution <strong>= 1.20<\/strong><br \/>\nAir bubble is at a depth, h = 30 cm <strong>= 0.3 m<\/strong><br \/>\nRadius of the air bubble, r = 4 mm <strong>= 5\u00d7 10<sup>-3<\/sup> m<\/strong><br \/>\nAcceleration due to gravity, <strong>g = 9.8 m\/s<sup>2<\/sup><\/strong><sup><br \/>\n<\/sup>1 atmospheric pressure = <strong>1.01 \u00d7 10<sup>5<\/sup> Pa<\/strong><sup><br \/>\n<\/sup>We know;<br \/>\nP =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 4S } { r }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">r<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">4<span class=\"mord mathdefault mtight\">S<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p>=<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\left (\\frac{ 4 \\times 2.5 \\times 10^{ -2 } }{ 5 \\times 10^{ -3 } } \\right )<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size2\">(<\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">5<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">4<span class=\"mbin mtight\">\u00d7<\/span>2.5<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22122<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size2\">)<\/span><\/span><\/span><\/span><\/span> <strong>= 20 Pa<\/strong><\/p>\n<p><strong>Thus the excess pressure inside the soap bubble is 20 Pa.<br \/>\n<\/strong>Now for the excess pressure inside the air bubble, P\u2019 =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 2S }{ r }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">r<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">S<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><br \/>\nP\u2019 =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\left (\\frac{ 2 \\times 2.5 \\times 10^{ -2 } }{ 5 \\times 10^{ -3 } } \\right )<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size2\">(<\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">5<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mbin mtight\">\u00d7<\/span>2.5<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22122<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size2\">)<\/span><\/span><\/span><\/span><\/span> = 10 Pa<\/p>\n<p><strong>Thus, the excess pressure inside the air bubble is 10 Pa<\/strong><\/p>\n<p>At the depth of 0.4 m, the total pressure inside the air bubble = Atmospheric pressure + h\u03c1g + P\u2019<br \/>\n= 1.01 x 10<sup>5<\/sup> + 0.4 x 1.2 x 10<sup>3<\/sup> x 9.8 + 10<br \/>\n<strong>= 1.06 x 10<sup>5<\/sup> Pa.<\/strong><\/p>\n<p><strong>Q-21: A tank with a square base of area 1.0 m<sup>2<\/sup> is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm<sup>2<\/sup>. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close. <\/strong><\/p>\n<p><strong>Ans:<\/strong><br \/>\n<strong>Given:<\/strong><br \/>\nArea of the hinged door, a = 20 cm<sup>2<\/sup><strong>= 20 \u00d7 10<sup>-4<\/sup> m<\/strong><br \/>\nBase area of the given tank, <strong>A = 2 m<sup>2<\/sup><\/strong><br \/>\nDensity of water, <strong>\u03c1<sub>1<\/sub> = 10<sup>3<\/sup> kg\/m<sup>3<\/sup><\/strong><br \/>\nDensity of acid, <strong>\u03c1<sub>2<\/sub> = 1.7 \u00d7 10<sup>3<\/sup> kg\/m<sup>3<\/sup><\/strong><br \/>\nHeight of the water column, <strong>h<sub>1<\/sub> = 4 m<\/strong><br \/>\nHeight of the acid column, <strong>h<sub>2<\/sub> = 4 m<\/strong><br \/>\nAcceleration due to gravity, <strong>g = 9.8 ms<sup>-2<\/sup><\/strong><br \/>\nPressure exerted by water , <strong>P<sub>1\u00ad<\/sub> = h<sub>1<\/sub>\u03c1<sub>1<\/sub>g  =<\/strong> 4 x 10<sup>3<\/sup> x 9.8  <strong>= 3.92 x 10<sup>4 <\/sup>Pa<\/strong><br \/>\nthe pressure exerted by acid , <strong>P<sub>2 <\/sub>= h<sub>2<\/sub>\u03c1<sub>2<\/sub>g  =<\/strong> 4 x 1.7 x 10<sup>3<\/sup> x 9.8 <strong>= 6.664 x 10<sup>4<\/sup> Pa<\/strong><br \/>\nPressure difference between the above two :<br \/>\n<strong>\u0394<\/strong><strong>P = <\/strong><strong>P<sub>2<\/sub> \u2013 P<sub>1<\/sub><\/strong><br \/>\n= (6.664 \u2013 3.92) X 10<sup>4<\/sup><strong>= 2.744 x 10<sup>4<\/sup>Pa<\/strong><br \/>\n<strong>Thus, the force on the door = <\/strong><strong>\u0394<\/strong><strong>P x a<\/strong><br \/>\n=2.744 x 10<sup>4<\/sup> x 20 \u00d7 10<sup>-4<\/sup><strong>= 54.88 N<\/strong><br \/>\n<strong>Hence the force required to keep the door closed is 54.88N<\/strong><\/p>\n<p><strong>Q-22: A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.<br \/>\n(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.<br \/>\n(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) is poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).<\/strong><\/p>\n<p><strong>(a) For diagram (a):<\/strong><br \/>\nGiven,  Atmospheric pressure, P<sub>0<\/sub> = 76 cm of Hg<br \/>\nThe difference between the levels of mercury in the two arms is gauge pressure.<br \/>\nThus, gauge pressure is 20 cm of Hg.<br \/>\nWe know, Absolute pressure = Atmospheric pressure + Gauge pressure<br \/>\n<strong>= 76 + 20 = 96 cm of Hg<\/strong><\/p>\n<p><strong>For diagram (b):<\/strong><br \/>\nDifference between the levels of mercury in the two arms = \u201318 cm<br \/>\nHence, gauge pressure is \u201318 cm of Hg.<br \/>\nAnd, Absolute pressure = Atmospheric pressure + Gauge pressure<br \/>\n<strong>= 76 cm \u2013 18 cm = 58 cm<\/strong><br \/>\n<strong>(b)<\/strong> It is given that 13.6 cm of water is poured into the right arm of figure (b).<br \/>\nWe know that relative density of mercury = 13.6<br \/>\n=&gt; A 13.6 cm column of water is equivalent to 1 cm of mercury.<br \/>\nLet, h be the difference in the mercury levels of the two arms.<br \/>\nNow, pressure in the right arm P<sub>R<\/sub> = Atmospheric pressure + 1 cm of Hg<\/p>\n<p><strong>= 76 + 1 = 77 cm of Hg . . . . . . (a)<\/strong><\/p>\n<p>The mercury column rises in the left arm, thus the pressure in the left limb, <strong>P<sub>L<\/sub> = 58 + h . . . . . . (b)<\/strong><br \/>\n<strong>Equating equations (a) and (b) we get :<\/strong><\/p>\n<p><strong>77 = 58 + h<\/strong><\/p>\n<p><strong>Therefore the difference in the mercury levels of the two arms, h = 19 cm<\/strong><\/p>\n<p><strong>Q-23: Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases ? If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p>As the base area is the same the pressure and thus the force acting on the two vessels will also be the same. However, force is also exerted on the walls of the vessel, which have a nonvertical component when the walls are not perpendicular to the base. The net non-vertical component on the sides of the vessel is lesser for the second vessel than the first.  Therefore, <strong>the vessels have different weights despite having the same force on the base.<\/strong><\/p>\n<p><strong>Q-24: During a blood transfusion, the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [Use the density of whole blood from Table 10.1]<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\nDensity of whole blood, <strong>\u03c1 = 1.06 \u00d7 10<sup>3<\/sup> kg m<sup>-3<\/sup><\/strong><br \/>\nGauge pressure, <strong>P = 2000 Pa<br \/>\n<\/strong>Acceleration due to gravity, <strong>g = 9.8 m\/s<sup>2<\/sup><\/strong><br \/>\n<strong>=P\/\u03c1g =200\/(1.06 x 10<sup>3<\/sup> x 9.8) =0.1925 m<br \/>\n<\/strong>The blood may just enter the vein if the height at which the blood container be kept must be slightly greater than 0.1925 m i.e\u201e 0.2 m<strong>.<\/strong><\/p>\n<p><strong>Q-25: In deriving Bernoulli\u2019s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter 2 \u00d7 10<sup>\u20133<\/sup> m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p>(a) If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces, due to which the pressure drop becomes large.<br \/>\n(b) The dissipative forces become more important with increasing flow velocity, because of turbulence.<\/p>\n<p><strong>Q-26: (a) What is the largest average velocity of blood flow in an artery of radius 2\u00d710<sup>\u20133<\/sup>m if the flow must remain lanimar? (b) What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 \u00d7 10<sup>\u20133<\/sup> Pa s).<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\nRadius of the vein, <strong>r = 2 \u00d7 10<sup>-3<\/sup> m<\/strong><br \/>\nDiameter of the vein, d = 2 \u00d7 1 \u00d7 10<sup>-3<\/sup> m <strong>= 2 \u00d7 10<sup>-3<\/sup> m<\/strong><br \/>\nViscosity of blood ,<strong>\u03b7 <\/strong><strong>= 2.08 x 10<sup>-3<\/sup> m<br \/>\n<\/strong>Density of blood, <strong>\u03c1 = 1.06 \u00d7 10<sup>3<\/sup> kg\/m<sup>3<\/sup><\/strong><\/p>\n<p><strong>(a) We know, Reynolds\u2019 number for laminar flow, N<sub>R<\/sub> = 2000<\/strong><br \/>\nTherefore, greatest  average velocity of blood is:<br \/>\nV<sub>AVG<\/sub>=<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ N_{ R }\\eta }{ \\rho d }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"mord mathdefault mtight\">d<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">N<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">R<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"mord mathdefault mtight\">\u03b7<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p>=<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 2000 \\times 2.084 \\times 10^{-3} }{ 1.06 \\times 10^{ 3 } \\times 4 \\times 10^{ -3 }}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1.06<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">3<\/span><\/span><\/span><span class=\"mbin mtight\">\u00d7<\/span>4<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2000<span class=\"mbin mtight\">\u00d7<\/span>2.084<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><strong>= 0.983m\/s<\/strong><br \/>\n(b) And, flow rate <strong>R = V<sub>AVG<\/sub> \u03c0 r<sup>2<\/sup><\/strong> = 0.983 x 3.14 x ( 10<sup>-3<\/sup>)<sup>2<\/sup><strong>= 1.235 x 10<sup>-6<\/sup> m<sup>3<\/sup>\/s<\/strong><\/p>\n<p><strong>Q.27: A plane is in level flight at constant speed and each of its wings has an area of 25 m<sup>2<\/sup>. If the speed of the air is 180 km\/h over the lower wing and 234 km\/h over the upper wing surface, determine the plane\u2019s mass. (Take air density to be 1 kg\/m<sup>3<\/sup>), g = 9.8 m\/s<sup>2<\/sup><\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p>Area of the wings of the plane, A=2\u00d725=50 m<sup>2<\/sup><br \/>\nSpeed of air over the lower wing, V<sub>1<\/sub><br \/>\n\u200b=180km\/h= 180 x (5\/18) = 50 m\/s<\/p>\n<p>Speed of air over the upper wing, V<sub>2<\/sub><br \/>\n=234km\/h= 234 x (5\/18) = 65 m\/s<\/p>\n<p>Density of air, =1kg\/m<sup>3<\/sup><br \/>\nPressure of air over the lower wing =P<sub>1<\/sub><br \/>\n\u200bPressure of air over the upper wing =P<sub>2<\/sub><br \/>\n\u200bPressure difference,\u0394P = P<sub>1<\/sub>\u200b\u2212P<sub>2<\/sub> = (1\/2) \u03c1 (V<sub>2<\/sub><sup>2<\/sup> \u2013 V<sub>1<\/sub><sup>2<\/sup>)<\/p>\n<p>= (1\/2) x 1 x (65<sup>2<\/sup> \u2013 50<sup>2<\/sup>) = 862.5 Pa<\/p>\n<p>\u200bThe net upward force F=\u0394P x A<br \/>\nThe upward forces balances the weight of the plane<\/p>\n<p>mg = \u0394P x A<\/p>\n<p>m = (\u0394P x A)\/g<\/p>\n<p>= (862.5 x 50)\/9.8\u200b<br \/>\n=4400kg<\/p>\n<p>The mass of the plane is 4400kg<\/p>\n<p><strong><br \/>\nQ-28: In Millikan\u2019s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 \u00d7 10<sup>\u20135<\/sup> m and density 1.2 \u00d7 10<sup>3<\/sup> kg m<sup>\u20133<\/sup>. Take the viscosity of air at the temperature of the experiment to be 1.8 \u00d7 10<sup>\u20135<\/sup> Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.<\/strong><\/p>\n<p><strong>Ans:<\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\nAcceleration due to gravity, <strong>g = 9.8 m\/s<sup>2<\/sup><\/strong><br \/>\nRadius of the uncharged drop, <strong>r = 2.0 \u00d7 10<sup>-5<\/sup> m<\/strong><br \/>\nDensity of the uncharged drop, <strong>\u03c1 = 1.2 \u00d7 10<sup>3<\/sup> kg m<sup>-3<\/sup><\/strong><sup><br \/>\n<\/sup>Viscosity of air, <strong>\u03b7<\/strong><strong> =  1.8 \u00d7 10<sup>-5<\/sup> Pa s<\/strong><br \/>\nWe consider the density of air to be zero in order to neglect the buoyancy of air.<br \/>\nTherefore terminal velocity (v) is :<\/p>\n<p>v =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 2 r^{ 2 } g \\rho }{ 9 \\eta }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">9<span class=\"mord mathdefault mtight\">\u03b7<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">r<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">2<\/span><\/span><\/span><span class=\"mord mathdefault mtight\">g<\/span><span class=\"mord mathdefault mtight\">\u03c1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p>=<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 2 \\left ( 2.0 \\times 10^{ -5 } \\right )^{ 2 } \\times 9.8 \\times 1.2 \\times 10^{ 3 } }{ 9 \\times 1.8 \\times 10^{ -5 } }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">9<span class=\"mbin mtight\">\u00d7<\/span>1.8<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22125<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"minner mtight\"><span class=\"mopen sizing reset-size3 size6 mtight delimcenter\"><span class=\"mtight\">(<\/span><\/span>2.0<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22125<\/span><\/span><\/span><span class=\"mclose sizing reset-size3 size6 mtight delimcenter\"><span class=\"mtight\">)<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">2<\/span><\/span><\/span><\/span><span class=\"mbin mtight\">\u00d7<\/span>9.8<span class=\"mbin mtight\">\u00d7<\/span>1.2<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> <strong>= 5.8cm<sup>-1<\/sup><\/strong><br \/>\nAnd the viscous force on the drop is :<br \/>\nF = 6\u03c0\u03b7rv<br \/>\n= 6 x 3.14 x 1.8 \u00d7 10<sup>-5<\/sup>x2 x 10<sup>-5<\/sup> x  5.8 10<sup>-2<\/sup><br \/>\n<strong>= 3.91 x 10<sup>-10<\/sup> N<\/strong><\/p>\n<p><strong>Q-29:  Mercury has an angle of contact equal to 140\u00b0 with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m<sup>\u20131<\/sup>. Density of mercury = 13.6 \u00d7 10<sup>3<\/sup> kg m<sup>\u20133<\/sup><\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\nDensity of mercury, <strong>\u03c1 =13.6 \u00d7 10<sup>3<\/sup> kg\/m<sup>3<\/sup><\/strong><br \/>\nAngle of contact between mercury and soda lime glass, <strong>\u03b8 = 140\u00b0<\/strong><br \/>\nSurface tension of mercury at that temperature , <strong>s = 0.465 N m<sup>-3<\/sup><\/strong><br \/>\nRadius of the narrow tube, r = 2\/2  = 1 mm <strong>= 1 \u00d7 10<sup>-3<\/sup> m<\/strong><br \/>\n<strong>Let, the dip in the depth of mercury = h<br \/>\n<\/strong>Acceleration due to gravity, <strong>g = 9.8 m\/s<sup>2<\/sup><\/strong><br \/>\nWe know, surface tension S =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ h g \\rho r }{ 2 cos\\Theta }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">c<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">s<\/span>\u0398<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">h<\/span><span class=\"mord mathdefault mtight\">g<\/span><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"mord mathdefault mtight\">r<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\Rightarrow h = \\frac{ 2 S cos\\Theta }{ g \\rho r }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel\">\u21d2<\/span><\/span><span class=\"base\"><span class=\"mord mathdefault\">h<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">g<\/span><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"mord mathdefault mtight\">r<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">S<\/span><span class=\"mord mathdefault mtight\">c<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">s<\/span>\u0398<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\Rightarrow h = \\frac{ 2 \\times 0.465 \\times cos140^{\\circ} }{ 13.6 \\times 9.8 }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel\">\u21d2<\/span><\/span><span class=\"base\"><span class=\"mord mathdefault\">h<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">13.6<span class=\"mbin mtight\">\u00d7<\/span>9.8<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mbin mtight\">\u00d7<\/span>0.465<span class=\"mbin mtight\">\u00d7<\/span><span class=\"mord mathdefault mtight\">c<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">s<\/span>140<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> = -5.34 mm<\/p>\n<p><strong>The negative sign indicates the falling level of mercury. Thus, the mercury dips by 5.34 mm.<\/strong><\/p>\n<p><strong><sup> <\/sup><\/strong><strong>Q-30:  Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 \u00d7 10<sup>\u20132<\/sup> N m<sup>\u20131<\/sup>. Take the angle of contact to be zero and density of water to be 1.0 \u00d7 10<sup>3<\/sup> kg m<sup>\u20133<\/sup> (g = 9.8 m s<sup>\u20132<\/sup>).<\/strong><br \/>\n<strong>Ans:<\/strong><\/p>\n<p><strong>Given:<\/strong><br \/>\nDiameter of the first bore, d<sub>1<\/sub> = 3.0 mm <strong>= 3 \u00d7 10<sup>-3<\/sup> m<br \/>\n<\/strong>Radius of the first bore, r<sub>1 <\/sub>= 3\/2 <strong>= 1.5 x 10<sup>-3<\/sup> m.<\/strong><br \/>\nDiameter of the second bore, <strong>d<sub>2<\/sub> =6mm<\/strong><br \/>\nRadius of the second bore, r<sub>2 <\/sub>= 6\/2 <strong>= 3 x 10<sup>-3<\/sup> mm<\/strong><br \/>\nSurface tension of water, <strong>s = 7.3 \u00d7 10<sup>-2<\/sup> N \/m<\/strong><br \/>\nAngle of contact between the bore surface and water<strong>, \u03b8= 0<\/strong><br \/>\nDensity of water, <strong>\u03c1 =1.0 \u00d7 10<sup>3<\/sup> kg\/m<sup>-3<\/sup><\/strong><br \/>\nAcceleration due to gravity,<strong> g = 9.8 m\/s<sup>2<\/sup><\/strong><br \/>\n<strong>Let, h<sub>1<\/sub> and h<sub>2<\/sub> be the heights to which water rises in the first and second tubes respectively.<\/strong><br \/>\nThus, the difference in the height:<br \/>\n<strong>h<sub>1<\/sub> \u2013 h<sub>2<\/sub> =<\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{2sCos\\Theta }{r_{1}\\rho g} -\\frac{2sCos\\Theta }{r_{2}\\rho g}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">r<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">1<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"mord mathdefault mtight\">g<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">C<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">s<\/span>\u0398<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mbin\">\u2212<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">r<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">2<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"mord mathdefault mtight\">g<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">C<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">s<\/span>\u0398<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><strong>Since, h =<\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{2sCos\\Theta }{r\\rho g}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">r<\/span><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"mord mathdefault mtight\">g<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">C<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">s<\/span>\u0398<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><strong>h<sub>1<\/sub> \u2013 h<sub>2<\/sub> =<\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{2sCos\\Theta }{\\rho g}[\\frac{1}{r_{1}} -\\frac{1}{r_{2}}]<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"mord mathdefault mtight\">g<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">C<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">s<\/span>\u0398<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mopen\">[<\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">r<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">1<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mbin\">\u2212<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">r<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">2<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose\">]<\/span><\/span><\/span><br \/>\n=<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{2\\times 7.3\\times 10^{-2}\\times 1 }{10^{3}\\times 9.8}\\left [ \\frac{1}{1.5\\times 10^{-3}} -\\frac{1}{3 \\times 10^{-3}} \\right ]<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">3<\/span><\/span><\/span><span class=\"mbin mtight\">\u00d7<\/span>9.8<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<span class=\"mbin mtight\">\u00d7<\/span>7.3<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22122<\/span><\/span><\/span><span class=\"mbin mtight\">\u00d7<\/span>1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size1\">[<\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1.5<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mbin\">\u2212<\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<span class=\"mbin mtight\">\u00d7<\/span>10<span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"sizing reset-size3 size1 mtight\">\u22123<\/span><\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size1\">]<\/span><\/span><\/span><\/span><\/span> = 4.97 mm<br \/>\n<strong>Therefore, the difference in the water levels of the two arms =2.482 mm.<\/strong><\/p>\n<p><strong>Q-31: (a) According to the law of atmospheres density of air decreases with increase in height y as \u03c1=<\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\rho _{0}e^{\\frac{-y}{y_{0}}}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">0<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mord mathdefault\">e<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">y<\/span>0<span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"sizing reset-size3 size1 mtight\">\u2212<span class=\"mord mathdefault mtight\">y<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>. Where \u03c1<sub>0<\/sub> =1.25 kg m<sup>-3<\/sup> is the density of air at sea level and y<sub>0 <\/sub>is a constant. Derive this equation\/law considering that the atmosphere and acceleration due to gravity remain constant.<\/p>\n<p><strong><br \/>\n(b) A zeppelin of volume 1500 m<sup>3<\/sup> is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y<sub>0<\/sub> = 8000 m and \u03c1H<sub>e<\/sub> =0.18 kg m<sup>-3<\/sup>]<\/strong>.<br \/>\n<strong>Ans:<\/strong><\/p>\n<p><strong>(a). <\/strong>We know that rate of decrease of density \u03c1 of air is directly proportional  to the height y.<br \/>\ni.e.,<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ d \\rho }{ d y } = \\;- \\frac{ \\rho }{ y_{ 0 }}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">d<\/span><span class=\"mord mathdefault mtight\">y<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">d<\/span><span class=\"mord mathdefault mtight\">\u03c1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\">\u2212<\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">0<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> <strong>. . . . . . . . . . . . . . ( 1 )<sub><br \/>\n<\/sub><\/strong>Where y is the constant of proportionality and the \u2013ve sign indicates the decrease in density with increase in height.<\/p>\n<p><strong>Integrating equation (1) , we get :<\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\int_{\\rho _{0}}^{\\rho }\\frac{d}{\\rho } =\\; \u2013 \\int_{0}^{y } \\frac{1}{y_{0}}dy\\\\<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mop\"><span class=\"mop op-symbol small-op\">\u222b<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"sizing reset-size3 size1 mtight\">0<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">d<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\">\u2013<\/span><span class=\"mop\"><span class=\"mop op-symbol small-op\">\u222b<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">0<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">0<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord mathdefault\">d<\/span><span class=\"mord mathdefault\">y<\/span><\/span><\/span><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">[log \\rho ]_{\\rho _{0}}^{\\rho } = -[\\frac{y}{y_{0}}]_{0}^{y}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mopen\">[<\/span><span class=\"mord mathdefault\">l<\/span><span class=\"mord mathdefault\">o<\/span><span class=\"mord mathdefault\">g<\/span><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"mclose\">]<span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"sizing reset-size3 size1 mtight\">0<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\">\u2212<\/span><span class=\"mopen\">[<\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">0<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose\">]<span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">0<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><br \/>\nWhere \u03c1<sub>0<\/sub> = density of air at sea level ie y =0<br \/>\nOr, log<sub>e<\/sub>(\u03c1<sub>0<\/sub> \/\u03c1 ) = -y\/y<sub>0<br \/>\n<\/sub>Therefore,<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\rho =\\rho _{0}\\;e^{\\frac{y}{y_{0}}}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">0<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mord mathdefault\">e<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">y<\/span>0<span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">y<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><strong>(b).  Given:<br \/>\nVolume of zeppelin  = 1500 m<sup>3<br \/>\n<\/sup>Mass of payload, m = 400 kg<br \/>\ny<sub>0<\/sub> = 8000 m<br \/>\n\u03c1<sub>0<\/sub> =1.25 kg m<sup>-3<\/sup><br \/>\ndensity of helium, \u03c1 H<sub>e<\/sub> =0.18 kg m<sup>-3<\/sup><\/strong><br \/>\n<strong>Density \u03c1 =<\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ Mass }{ Volume }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">V<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">l<\/span><span class=\"mord mathdefault mtight\">u<\/span><span class=\"mord mathdefault mtight\">m<\/span><span class=\"mord mathdefault mtight\">e <\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">M<\/span><span class=\"mord mathdefault mtight\">a<\/span><span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">s<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span> =<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ Mass \\;of \\;payload + Mass \\;of \\;helium }{ Volume }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">V<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">l<\/span><span class=\"mord mathdefault mtight\">u<\/span><span class=\"mord mathdefault mtight\">m<\/span><span class=\"mord mathdefault mtight\">e <\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">M<\/span><span class=\"mord mathdefault mtight\">a<\/span><span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">s <\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">f <\/span><span class=\"mord mathdefault mtight\">p<\/span><span class=\"mord mathdefault mtight\">a<\/span><span class=\"mord mathdefault mtight\">y<\/span><span class=\"mord mathdefault mtight\">l<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">a<\/span><span class=\"mord mathdefault mtight\">d <\/span><span class=\"mbin mtight\">+ <\/span><span class=\"mord mathdefault mtight\">M<\/span><span class=\"mord mathdefault mtight\">a<\/span><span class=\"mord mathdefault mtight\">s<\/span><span class=\"mord mathdefault mtight\">s <\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">f <\/span><span class=\"mord mathdefault mtight\">h<\/span><span class=\"mord mathdefault mtight\">e<\/span><span class=\"mord mathdefault mtight\">l<\/span><span class=\"mord mathdefault mtight\">i<\/span><span class=\"mord mathdefault mtight\">u<\/span><span class=\"mord mathdefault mtight\">m<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p>=<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\frac{ 400 + 1500 \\times  0.18 }{ 1500 }<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1500<\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">400<span class=\"mbin mtight\">+<\/span>1500<span class=\"mbin mtight\">\u00d7<\/span> 0.18<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>  <strong>[ Mass = volume <\/strong><strong>\u00d7<\/strong><strong> density]       <\/strong><br \/>\n<strong>=  0.45 kg m<sup>-3<\/sup><\/strong><br \/>\n<strong>Using equation (1), we will get:<\/strong><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\rho =\\rho _{0}e^{\\frac{y}{y_{0}}}\\\\<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mord mathdefault\">\u03c1<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">0<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mord mathdefault\">e<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">y<\/span>0<span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">y<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\\\\\Rightarrow log_{e}(\\frac{\\rho _{0}}{\\rho })= \\frac{y_{0}}{y}\\\\<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel\">\u21d2<\/span><\/span><span class=\"base\"><span class=\"mord mathdefault\">l<\/span><span class=\"mord mathdefault\">o<\/span><span class=\"mord\"><span class=\"mord mathdefault\">g<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">e<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mopen\">(<\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">0<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose\">)<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">0<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><br \/>\nOr,<\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\\\y= \\frac{y_{0}}{log_{e}(\\frac{\\rho _{0}}{\\rho })}\\\\<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathdefault\">y<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">l<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">g<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">e<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"mopen mtight\">(<\/span><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><\/span><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">\u03c1<\/span><span class=\"msupsub\">0<span class=\"vlist-s\">\u200b<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><span class=\"mclose mtight\">)<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">y<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\">0<\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><annotation encoding=\"application\/x-tex\">\\\\\\Rightarrow y= \\frac{8000}{log_{e}(\\frac{1.25}{0.45 })}<\/annotation><\/semantics><\/math>\n<p><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel\">\u21d2<\/span><\/span><span class=\"base\"><span class=\"mord mathdefault\">y<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">l<\/span><span class=\"mord mathdefault mtight\">o<\/span><span class=\"mord mathdefault mtight\">g<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathdefault mtight\">e<\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"mopen mtight\">(<\/span><span class=\"sizing reset-size3 size1 mtight\">0.45<\/span><span class=\"sizing reset-size3 size1 mtight\">1.25<\/span><span class=\"vlist-s\">\u200b<\/span><span class=\"mclose mtight\">)<\/span><\/span><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">8000<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><br \/>\n<strong>Therefore,  y <\/strong><strong>\u2248 8km<\/strong><\/p>\n<h4><span class=\"ez-toc-section\" id=\"FAQ_on_NCERT_Exemplar_Solutions_for_Class_11_Physics_Chapter_10\"><\/span><span style=\"font-size: 12pt;\"><b>FAQ on NCERT Exemplar Solutions for Class 11 Physics Chapter 10<\/b><\/span><span class=\"ez-toc-section-end\"><\/span><\/h4>\n<h3><span class=\"ez-toc-section\" id=\"Q_How_to_solve_the_questions_present_in_Chapter_10_of_NCERT_Exemplar_Solutions_for_Class_11_Physics_faster\"><\/span><span style=\"font-weight: 400; color: #0000ff; font-size: 12pt;\">Q. How to solve the questions present in Chapter 10 of NCERT Exemplar Solutions for Class 11 Physics faster?<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400; font-size: 12pt;\">Students in Class 11 should answer the questions in the NCERT textbook to gain an understanding of the key concepts covered. Students can use the NCERT Exemplar Solutions for Class 11 if they have any questions while solving the tasks. From infinity, physics is learned, and each and every problem is answered precisely. The solutions include detailed explanations for each step to ensure that students understand how to solve problems without trouble.<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"Q_What_are_the_key_features_of_NCERT_Exemplar_Solutions_for_Class_11_Physics_Chapter_10_Mechanical_Properties_of_Fluids\"><\/span><span style=\"font-weight: 400; color: #0000ff; font-size: 12pt;\">Q. What are the key features of NCERT Exemplar Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids?<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400; font-size: 12pt;\">Following are the key features of NCERT Exemplar Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids:<\/span><\/p>\n<ol>\n<li><span style=\"font-weight: 400; font-size: 12pt;\"> The solutions have been prepared by highly qualified physicists.<\/span><\/li>\n<li><span style=\"font-weight: 400; font-size: 12pt;\"> Based on the most recent CBSE Syllabus, the solutions are accurate and error-free.<\/span><\/li>\n<li><span style=\"font-weight: 400; font-size: 12pt;\"> Every significant topic is taught in clear English to ensure that students do well on their second-term exams.<\/span><\/li>\n<li><span style=\"font-weight: 400; font-size: 12pt;\"> Using Infinity Learn&#8217;s NCERT Exemplar Solutions, students gain a full comprehension of key ideas.<\/span><\/li>\n<\/ol>\n<p><span style=\"color: #0000ff; font-size: 12pt;\">Q. Why Opt for Infinity Learn?<\/span><\/p>\n<p><span style=\"font-weight: 400; font-size: 12pt;\">In this chapter Infinity Learn provides videos, notes, NCERT text book solutions, other practice book solutions and assignments that help you learn the concepts and to memorize the concepts for entrance exams and board exams. Most important, Infinity learn provide instant doubt support by subject experts.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Class 11 Physics Chapter 10 &#8211; Mechanical Properties of Fluids Question.1: Explain why (a) The blood pressure in humans is [&hellip;]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"NCERT Exemplar for Class 11 Physics Chapter 10","_yoast_wpseo_title":"NCERT Exemplar Solutions Class 11 Physics Chapter 10 Solved Examples","_yoast_wpseo_metadesc":"Get the most important questions of NCERT Exemplar solutions for Class 11 Physics Chapter 10 - Mechanical Properties of fluid.","custom_permalink":"study-materials\/ncert-exemplar-solutions\/class-11\/physics\/chapter-10\/mechanical-properties-of-fluids\/"},"categories":[107,105,114,21],"tags":[8052],"table_tags":[],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v17.9 - 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