ANSWER KEY. 3. B 15. A 27. 3m 39. B 51. D 63. C 40. C 52. A 64. D

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Vincent Butler
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1 NSWE KEY CUMUIVE ES (5-7-8)_Solutions.. N/m. C C D. m 4. D 6. C 8. D D m D 6. C W 9 4. C D 5. C 7. C m 4. D 5. C 65. D 6. D 8. 4%. 4. a 4k M D N/m kgm m kpa D. C Pa 45. C C 4..9 Pa kpa 47. C 59.. D 4. D D SOUIONS. (..4) P avg N / m hen, ma avg. N / m m. ( ) d Consider an element of length d at a distance from end. So, df dm Now, So, dm m d m d df Maimum force will be at a cross section about which it is rotating F ma F m d ma m Fma m Maimum stress will be. () he density of fluid, S. G kg/ m 85 kg/ m Pressure at depth h from free surface gh Pressure at inside tank, P gh atm P 96 kpa.55 kpa P.586 kpa GEMENOEDUSEVICES PV. D., -, SEE-7, SMII NG, HII 49 CONC

2 CUMUIVE ES (5-7-8)_Solutions 4. () For free falling body relative acceleration due to gravity is zero P gh if g then P (but it is only hydrostatic pressure) and these will be atm. Pressure throughout the liquid. 5. (C) bsolute pressure = tm pressure + gauge pressure 5. bar 6. bar 6. (D) Since there is no flow of water in pipe. So there no raise or fall of meniscus in tube. Mercury will balance. 7. (D) ension in string and weight of body is equation to the upward buoyancy force. Mg uoyancy force S Hg hg g h H S 8. (.9 m from bottom surface.) pply the hydrostatic formula from point to point. P gh gh g h z P P 5 Pa, P air gasoline glycerin z z Solving for z, z.9 m 9. (D) 4 = and v = -y he equation of stream line in two Dimensional flow is dy v d dy d u u v On integration d dy y. () n ny nc where c constant ny nc or ny c For stream line passing through,), c = he required stream line equation is y = From the definition stream function u y and v, as =y v y, u y u, v y For irrotational flow v u, u, v y y, flow is irrotational for irrotational flow vorticity is zero (). () valid potential function must satisfy the laplace equation (i) 5y and y 5 and y Hence y 5y is a valid potential function (ii) 4 5y 8 and y y 8 and y Hence y Hence 4 5y 5y is not a valid potential function.. (D) For irritation flow following condition must be satisfied. (C) 4. (D) v u z y (i) u y; v y v u y Hence flow is irrotational (ii) u y, v v u z y y Hence flow is not irrotational. GEMENOEDUSEVICES PV. D., -, SEE-7, SMII NG, HII 49 CONC

3 5. () 6. () V V d V d V V 4 6 m / s V 6 Velocity head at sec = m g (C) head total P V at section = z g g head 8. (4) Given: total P V at section = z g g Difference between the two sections = 5. m ma min.5 ma min ma Now, C f.5.5 4% (.74) Case : k k m Now, Case : min ma min ma min k k k mg m. () Given: P 5 kw, N 8 rpm, F.48 kn, 4 So, N 8 P kNm 8 F D lso, CUMUIVE ES (5-7-8)_Solutions.6 D.997 m mm F.48 D D Now, 66 D.6 mm lso, by equal module of gear and, D D D.6 D (C) Given: C 4 N / m / s, C 4 N / m / s. (C) C 4 So,.95 C 4 c Now, logarithmic decrement C c C Fluid Film.4m From Fourier s law mof heat conduction d.4 4 Q K d.4 Q 5 W / m. () U b h k h U 5 U (D) U Q th GEMENOEDUSEVICES PV. D., -, SEE-7, SMII NG, HII 49 CONC

4 4 CUMUIVE ES (5-7-8)_Solutions 5. () W 6. (C) yma m 6 t 6MPa () V 4 kn 8. (D) M, So, 7 kn 4 4 nd kn ending moment at a distance ' ' from end will be M Now, the MD changes sign in section, so the point of contrafleure is where the M is zero. So, m w w V w M w So, w w w 6 ending moment at distance from left end, M w w w w dm For maimum M, d w w So, 6 6 w 6 6 w So, from equation (i), M ma w w 6 6 w w w w Now, w W w W Substituting in (ii), W So, M ma 9 9. Marks to ll (.5.56) Point of contrafleure is the position where bending moment changes sign. he point itself have zero bending moment 5.5kN m m m X aking moment about kn X kn ending moment at a section XX at a distance from end. M 4 M 4 8 aking M, 8 4 on solving.559 mas other value of is not admissible.. (-) P.8kPa M Z GEMENOEDUSEVICES PV. D., -, SEE-7, SMII NG, HII 49 CONC

5 So,.8kPa t ma. (5 54) Given, h W W N, d. m, h 5 m, E GN / m, l m bar From the relation, W he W N / m 5.6 MN / m l 9. (9.47 kpa) Equating the pressure on both the limbs at horizontal plane P.8..8 g P. g.8 g P m P P P P P Pa P P 9.47 kpa. (478.8 Pa) ake reference line passing through point Pressure will same on reference line, Pressure at, gh P P Pa pa nd pressure at, 8 6 g Pa Pa Difference in pressure, P P Pa CUMUIVE ES (5-7-8)_Solutions 5 4. (.9 Kpa) eassure at, P P P P 9.45 Pa P P Pa C P P 4Pa C D P P Pa E D.9 kpa gauge 5. ( kPa ) ccording to the condition for static equilibrium, it y is measured downwards from free surface. P g or dp gdy y Hence, dp.y g dy Integration of above equation gives, y P y. g c Where c is constant of integration. Since gauge pressure p at the free surface (where y = ) is zero, c. y P y. g. 8 t y 8m P P Pa 6. () o check for steady flow use continuity equation. (i) u, v u v y y, So the flow is steady (ii) u v t, t Statistics the continuity y equation so the flow is steady (iii) u v t, t y y his does not satisfy the requirement for steady flow. o check for irrotational flow u v y (i) u v, Flow is irrotational y (ii) u v, Flow is not irrotational y (iii) u v yt Flow is not irrotational y GEMENOEDUSEVICES PV. D., -, SEE-7, SMII NG, HII 49 CONC

6 6 CUMUIVE ES (5-7-8)_Solutions 7. (C) y y u y y y y f (i) v y y (ii) nd from (i) y f ' hus f ' and hence y he required stream function is y y y y t point (,) t point (,) 4 8.5unit Flow rate between the stream line passing through (,) and (,) unit 8. Marks to all (6.55 unit) 5 y ; y y ut velocity component u and v are given by definition as u v y y y cceleration can be calculated by uu vu a y a y a y uy vv y a y y y t (,) a a y y cceleration a a a 6.55 unit 9. () We know that where - potential function y nd y - stream function. y y y...() y y Integrate equation () w.r.t y we get...() y d y dy or y c C-constant It can be function of let the of constant K then y y K...() Differentiating the above equation w.r.t. we get k ut from equation () =- hen k Integrating this equation K d substitute this value of K in () we get y y 4. (C) We know that where - potential function y nd y y y y y y - stream function....()...() Integrating equation () we get d d c c...() Where c is constant It can be function of y only c differentiating equation () w.r.t. we get y y From equation () c y y y y GEMENOEDUSEVICES PV. D., -, SEE-7, SMII NG, HII 49 CONC

7 Integrating this equation we get y c ydy y Substituting this value of c in equation y 4. (D) In the top plane, pressure at any radial distance r is r P and thrust on top plane 4 F r dr 4 F P. rdr nd,.94 rad /sec 6 putting the values F 55N CUMUIVE ES (5-7-8)_Solutions 7 4. (..) et the energy at So, E E E E.5 E E.5. E.7 C E E E.5 D E E E. E E E E.5 F EG E E.6 E E E H So, E Ema Emin EE E E. E.5.7cm Nm 8 lso, E I c 6.8 I.75kgm a 4k M O O a 44. () S qro h S 7 C Displacement, ' a sin a (For small angle, sin ) So, ' a he total energy of the system, U M I ky M Where, y ' a and I M So, U M k a U M k a 4 du Now, dt M k a 4k a M 4k a a 4k n M M 45. (C) For sphere gm 8 4m 46. () Heat lost by convectioin = Heat conducted by conduction h C K K K C K K K K.5 W / mk 47. (C) From the Fourier s law of conduction GEMENOEDUSEVICES PV. D., -, SEE-7, SMII NG, HII 49 CONC

8 8 CUMUIVE ES (5-7-8)_Solutions 48. () 49. () d Q K d K d W W 5. () q f h q f h ma W ir Surface heat flu qo h W f q qo h qo q K I 5. (D) K Fe C I mm K l Q l KFe Q Fe K K l Fe C Fe mm C C 7 5. () hermal resistance. 4 K / W K 5 5. (C) K Q C 54. () From the fourier law of heat conduction K 55. () t p t c 7 K 6 K K cold Hence MD = K hot K t p 4 K 4 K 56. () 57. () 58. () 59. () 6. (D) dart is a small type of spear while a gun is a small type of canon 6. (D) let N be the number of days Nikhil takes to complete the job working alone and be the number of days nkit takes to finish the job working alone. hus we have, /N = /4. Now, if Nikhil worked twice as efficiently, he will take N/ days to complete the job alone and if nkit works / rd as efficiently, he will take days to finish the same job alone, hus we can say that /(N/) + / = /8 Solving we get /N = /8 /7 => 5/N = /7 = > N = 4 days lternate Method: CM of 4 and 8 is 7. So let us assume the total work to be 7 units If rate of working for Nikhil is n units /day and rate of working for nkit is a units/day, we can say that n + a = 7/4 = (otal rate of working = otal work/otal time taken). lso, n + a/ = 7/8 = 4. Solving both equations, we get n = 9/5 units/day. hus the time taken by Nikhil to finish the job alone = 7/(9/5) = 4 days, 6. (D) Sum of n terms = n + n. Sum of (n-) terms = (n-) + (n-) = n + n Now, we know, n th term = Sum of n terms Sum of (n-) terms, herefore, n th in this case = n + n (n + n ) = n +. herefore, 5 th terms is5 + = 6. (C) 64. (D) he net movement of the monkey = steps in 4 seconds ie step for every second In our bid to solve question quickly, we may tend to directly multiply steps by seconds and arrive at 4 seconds. ut consider this after the monkey reaches the 8 th step, he has to climb another steps to reach which he will do in the net second (the fact that he slips another step after reaching the st is of no concern to us). herefore total time taken by the monkey is 8 + = 9 seconds. 65. (D) Since we do not know that total number of students graduating from all the IIs put together, we cannot find the percentage of students who did not get placed and hence the data is insufficient t c GEMENOEDUSEVICES PV. D., -, SEE-7, SMII NG, HII 49 CONC

CHAPTER 2 Fluid Statics

CHAPTER 2 Fluid Statics Chapter / Fluid Statics CHPTER Fluid Statics FE-type Eam Review Problems: Problems - to -9. (C). (D). (C).4 ().5 () The pressure can be calculated using: p = γ h were h is the height of mercury. p= γ h=