KVPY SA Stream Solution 2008

(1)

PART-I (1 Mark)

MATHEMATICS

1.

1. Obvious (A) is greatestObvious (A) is greatest

2. 2. _{S =}_{S =} ... 10 10 n n ... ... 10 10 3 3 10 10 2 2 10 10 1 1 n n 3 3 2 2





10 10 S S = = 22 10 10 1 1 + + 33 10 10 2 2 + + ...



Subtracting, Subtracting, 10 10 S S 9 9 = = 10 10 1 1 + + ₂₂ 10 10 1 1 + + ₃₃ 10 10 1 1 + + ...



10 10 S S 9 9 = = 10 10 1 1 – – 1 1 10 10 1 1 10 10 S S 9 9 = = 9 9 1 1 S = S = 81 81 10 10 3. 3. ₍₁₀₂₄₎₍₁₀₂₄₎10241024_{= (16)}_{= (16)}16n16n (2 (21010₎₎10241024₌_{= (2}₍₂44₎₎16n16n 10 × 1024 = 4 × 16n 10 × 1024 = 4 × 16n n = n = 16 16 4 4 1024 1024 10 10



n = 160 n = 160 4. 4. _x_x22_{+ 6x + 8}_{+ 6x + 8} x x



R R x x22_{– 2x – 8}_{– 2x – 8}



₀₀ x x22_–_–₂_2x_x_–_–₈₈₌₌ _x_x22_{+ 2x – 4x – 8}_{+ 2x – 4x – 8} x(x + 2) – 4(x + 2) x(x + 2) – 4(x + 2)



0 0

ANSWER KEY

HINTS & SOLUTIONS (YEAR-2008)

Q

Quueess. . 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 110 0 111 1 112 2 113 3 114 4 1155 Ans.

Ans. _A_A _B_B _D_D _A_A _C_C _C_C _C_C _A_A _C_C _D_D _B_B _B_B _D_D _B_B _A_A

Q Quueess. . 116 6 117 7 118 8 119 9 220 0 221 1 222 2 223 3 224 4 225 5 226 6 227 7 228 8 229 9 3300 Ans. Ans. _D_D _D_D _A_A _A_A _B_B _C_C _A_A _D_D _A_A _B_B _C_C _B_B _D_D _C_C _A_A Q Quueess. . 331 1 332 2 333 3 334 4 335 5 336 6 337 7 338 8 339 9 440 0 441 1 442 2 443 3 444 4 4455 Ans. Ans. _B_B _A_A _A_A _C_C _D_D _A_A _C_C _A_A _D_D _C_C _A_A _B_B _B_B _C_C _D_D Q Quueess. . 446 6 447 7 448 8 449 9 550 0 551 1 552 2 553 3 554 4 555 5 556 6 557 7 558 8 559 9 6600 Ans.

Ans. _B_B _B_B _B_B _B_B _A_A _C_C _A_A&_&C_C _C_C _A_A _D_D _C_C _A_A _D_D _B_B _D_D

Q Quueess. . 661 1 662 2 663 3 664 4 665 5 666 6 667 7 668 8 669 9 770 0 771 1 772 2 773 3 774 4 7755 Ans. Ans. _B_B _A_A _D_D _A_A _C_C _D_D _A_A _B_B _A_A _A_A _C_C _B_B _D_D _B_B _A_A Q Quueess. . 776 6 777 7 778 8 779 9 8800 Ans. Ans. _B_B _C_C _C_C _A_A _D_D

(2)

–4 –4 –3 –3 –2 –2 x x + 622+ 6x + x + 88 0 0 ₂₂ ₄₄ x x



[–2, 4] [–2, 4]

clearly min value of expression is 0 clearly min value of expression is 0 at x = – 2

at x = – 2

5.

5. _{Check by option}_{Check by option}

P

P₁₂₁₂ = {24, 36, 60, 84, ....} = {24, 36, 60, 84, ....} P

P₂₀₂₀ = {40, 60, 100, ...} = {40, 60, 100, ...} P

P₁₂₁₂



P P₂₀₂₀ has common element has common element

6.

6. _{All even values}_{All even values of a i.e. 50 and}_{of a i.e. 50 and 1, 9, 25, 49, 81, total}_{1, 9, 25, 49, 81, total 55}₅₅ 7.

7. _{If any statement is true then remaining 2 are false.}_{If any statement is true then remaining 2 are false.}

8. 8. C C F F A A EE BB D D 4 4 22 4 4 22 4 4 22 P P Angle bise

Angle bisector ctor



Incircle is formed whose radius =Incircle is formed whose radius = ₄₄ ₂₂ PE = r =

PE = r = ₄₄ ₂₂ PF = r =

PF = r = ₄₄ ₂₂ also PF = AEalso PF = AE





APE, (AP) APE, (AP)22_{= (AE)}_{= (AE)}22_{+ (PE)}_{+ (PE)}22

= ( = (₄₄ ₂₂ ))22_{+ (}_{+ (} 2 2 4 4 ))22 = 64 = 64



AP = 8 AP = 8 9.

9. _{Area of}_{Area of rhombus =}_{rhombus =}

2 2 1 1 d d₁₁dd₂₂ A A D D _C_C B B FF E E y y y y y y y y 22– h– h22 y y 22– h– h22 h h Let one diagonal = x

Let one diagonal = x = = 2 2 1 1 ×(x)(2x) = x ×(x)(2x) = x22 A = x A = x22

Let side of rhombus = y & height = h Let side of rhombus = y & height = h



BFC side BF =BFC side BF = yy22



hh22 In

In



AFC, (y + AFC, (y + yy22



hh22 ))22_{+ h}_{+ h}22_{= (AC)}_{= (AC)}22_{= 4x}_{= 4x}22



DEB (y –DEB (y – _y_y22

_

_h_h22 ₎₎22_{+ h}_{+ h}22_{= (BD)}_{= (BD)}22_{= x}_{= x}22 Addi Addingng 4y 4y22_{= 5x}_{= 5x}22 y = y = 4 4 x x 5 5 22 = = 2 2 A A 5 5

(3)

10. 10. A(2a) A(2a) B B CC ((33bb)) P P D D E E (a) (a) (2b) (2b)

Let B is origin and

Let B is origin and the position vector of A and C arethe position vector of A and C are 22aa and and b33b

 

Then P.V. of E =

Then P.V. of E = aa and P.V. of D = and P.V. of D = 22bb

 

Now, let P divides AD in



: 1 ratio : 1 ratio a

anndd P P ddiivviiddees s EEC C iinn



: 1 : 1



1 1 a a 2 2 b b 2 2









   = = 1 1 a a b b 3 3









   b b 2 2  



+ + 22bb  



+ + 22aa



+ + 22aa = = 33bb  



+ + aa



+ + 33bb  



++ aa a a(2(2



+ 2 – + 2 –



– 1) = – 1) = bb   (3 (3



+ 3 + 3



– 2 – 2



– 2 – 2



)) But

But aa and and bb

 

are not

are not collinear.collinear... 2 2



– –



+ 1 = 0 and + 1 = 0 and



+ 3 + 3



– 2 – 2



= 0 = 0 We get We get



= 1 = 1 Now, P.V. of P is = Now, P.V. of P is = 2 2 b b 3 3 a a    



Now,

Now, _ar_arar ar

_



_ABC_ABCPEDPED = =

b b 3 3 a a 2 2 2 2 1 1 2 2 b b 3 3 a a b b 2 2 2 2 b b 3 3 a a a a 2 2 1 1                















_

















_



= =



 

 

 

b b a a 6 6 a a b b b b 3 3 a a 4 4 1 1            









= = 12 12 1 1 1 11.1. C C D D A A BB y y _P_P MM O O L L N N L Leett AAB B = = aa, , BBC C = = bb a anndd PPL L = = hh₁₁ tthheenn PPN N = = b b – – hh₁₁ a anndd OOP P = = hh₂₂, then PM = a – h, then PM = a – h₂₂



ar(ar(



PAB +PAB +



PCD) =PCD) = 2 2 1 1 a(h a(h₁₁ + b – h + b – h₁₁)) = = 2 2 ab ab and area (

and area (



PBC +PBC +



PAD) =PAD) = 2 2 1 1 × b × (h × b × (h₂₂ + a – h + a – h₂₂) =) = 2 2 ab ab From this only option B is correct

(4)

12. 12. _{Let x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6}_{Let x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6} & x + 7, x + 8, x + 9, x + 10, x + 11 & x + 7, x + 8, x + 9, x + 10, x + 11 7x + 21 = 5x + 45 7x + 21 = 5x + 45 2x = 24 2x = 24 x = 12 x = 12 largest = x + 11 = 23 largest = x + 11 = 23 13.

13. _{Let x minite will be taken. I}_{Let x minite will be taken. I n one minute A can fill the}_{n one minute A can fill the}

60 60 1 1

part of tanker and in one minute B can fill the part of tanker and in one minute B can fill the

40 40 1 1 part. part.

Both can fill in t Both can fill in t

60 60 tt + + 40 40 tt = 1 = 1 t = t = 100 100 40 40 60 60



t = 24 t = 24

both can fill in one minute both can fill in one minute

24 24 1 1 part of tanker. part of tanker... 1 = 1 =













2 2 x x 40 40 1 1 + +













24 24 1 1 2 2 x x 1 = 1 = 80 80 x x + + 48 48 x x x = x = 128 128 48 48 80 80



= 3 = 300 14. 14. _{Given a < b < c}_{Given a < b < c}



6 three digit number 6 three digit number are possible with distinct a, b, c.are possible with distinct a, b, c.

4 4 5 5 5 5 1 1 a a b b c c b b a a c c a a c c b b c c a a b b b b c c a a c c b b a a     

= 100[2(a + b + c)] + 10[2(a + b + c)] + [2(a + b + c)] = 1554 = 100[2(a + b + c)] + 10[2(a + b + c)] + [2(a + b + c)] = 1554 111[2(a + b + c)] = 1554 111[2(a + b + c)] = 1554



a + b + c = 7a + b + c = 7 Given a < b < c Given a < b < c



1, 2, 1, 2, 4 only satisly above two condition.4 only satisly above two condition. Hence, c = 4.

Hence, c = 4.

15.

15. _{Since factor of 128 are =}_{Since factor of 128 are = 1, 2, 4, 8 ,}_{1, 2, 4, 8 , 16,}_{16, 32, 64, 128}_{32, 64, 128}

Hence it will be incr

(5)

PHYSICS

16.

16. _{speed will not decrease, so answer is (D)}_{speed will not decrease, so answer is (D)}

17.

17. _{For electron, t}_{For electron, t}

1 1 = = e e a a s s 2 2 For protion, t For protion, t₂₂ = = p p a a s s 2 2 or or _eE_eE m m m m eE eE a a a a tt tt pp e e p p e e 1 1 2 2

_

_

= = e e p p m m m m 18.

18. _{Focal length, f = 6 cm}_{Focal length, f = 6 cm}

u = 1.5m = 150 cm u = 1.5m = 150 cm v = v = ?? u u 1 1 v v 1 1 f f 1 1





150 150 1 1 v v 1 1 6 6 1 1





150 150 1 1 25 25 150 150 1 1 6 6 1 1 v v 1 1









v = v = 66..2525 4 4 25 25 12 12 75 75 24 24 150 150



change in distance = 6.25 – 6 = 0.25 cm change in distance = 6.25 – 6 = 0.25 cm = 0.25cm = 2.5 mm decreased = 0.25cm = 2.5 mm decreased 19.

19. _{Initial momentum, P}_{Initial momentum, P}

1

1 = mvcos30 = mvcos30

and final momentum, P

and final momentum, P₂₂ = mvcos30 = mvcos30 change in momentum change in momentum



P = – 2mv cos30P = – 2mv cos30



P = –P = – ₃_{3 mv}mv Force on wall-1 Force on wall-1 F F₁₁ = = tt mv mv 2 2



Force on wall-2 Force on wall-2 F F₂₂ = = tt mv mv 3 3



, so F, so F11 > F > F22 22. 22. _A_A 1 1uu11 = A = A22uu22 16 16 1 1 A A A A u u u u 1 1 2 2 2 2 1 1

_

23.

23. resultant force at centre is zero. On removing the charge from the position 6, the resultant force at centreresultant force at centre is zero. On removing the charge from the position 6, the resultant force at centre will be will be 22 r r kq kq downward. downward.

(6)

25. 25. L L w w d d d d V V v v _ 840 840 d d 2 2 1 1 _w_w





d d_w_w = 420 kg/m = 420 kg/m33 R.D. = R.D. = 00..4242 10 10 420 420 3 3



26. 26. _R_R S

S = nR for maximum resistance = nR for maximum resistance

R

R_P_P = R/n = R/n for minimum resistancefor minimum resistance or or P P S S R R R R = n = n22 27.

27. _{For block 2kg}_{For block 2kg}

T T – – 22g g = = 22aa ...((ii)) For 6 kg For 6 kg 6 6g g – – T T = = 66aa ...((iiii)) aa 6g 6g 2g 2g a a T T T T

From (i) and (ii) T = 30 N From (i) and (ii) T = 30 N

28.

28. _v_v_e_e



₂₂_gR_gR with height g will change, so answer is with height g will change, so answer is (D)(D)

29. 29. _{R.H. =}_{R.H. =} pressure pressure vapour vapour saturated saturated pressure pressure partial partial 100 100 90 90 = = 55 10 10 0169 0169 .. 0 0 pressure pressure partial partial



partial pressue = 0.0152 × 10 partial pressue = 0.0152 × 1055_Pa_Pa

CHEMISTRY

31. 31. _N_Na_aO_OH_H _H_HC_Cll N N₁₁VV₁₁ == NN₂₂VV₂₂ 0 0..5 5 × × V V == 2 2 × × 1100 V = 40 mL V = 40 mL 35. 35. _{Ethanol (C}_{Ethanol (C} 2

2HH55OH) and dimethyl ether (CHOH) and dimethyl ether (CH33 –O – CH –O – CH33) have same molecular formula but different functional) have same molecular formula but different functional

groups, so they are isomers. groups, so they are isomers.

36.

36. _{For the elements belonging to one period, increase in atomic number results in decrease in atomic}_{For the elements belonging to one period, increase in atomic number results in decrease in atomic}

radius. So Li has the largest atomic

radius. So Li has the largest atomic radius.radius.

37. 37. _2H_2H 2 2OO22





2H 2H22O + O + OO22 39. 39. _S_S ₊₊ _O_O 2 2





SOSO22 1

1 mmoollee 1 1 momollee 1 1 mmoollee

2 2 1 1 mole mole 2 2 1 1 mole mole 2 2 1 1 mole mole 3.01 × 10 3.01 × 102323 ₀_0..5₅_m_mo_olle_e _?_?



3.01 × 10 3.01 × 102323_{molecules of SO}_{molecules of SO} 2

(7)

40.

40. _{Zn and Pb are placed}_{Zn and Pb are placed above hydrogen in the metal activity series, so}_{above hydrogen in the metal activity series, so they}_{they will produce hydrogen gas}_{will produce hydrogen gas with}_with

dilute acids. dilute acids.

41.

41. _{Milk of magnesia is basic, water is neutral and lemon juice is acidic in nature.}_{Milk of magnesia is basic, water is neutral and lemon juice is acidic in nature.} 42.

42. _{As pressure is increased, solubility}_{As pressure is increased, solubility of gas in liquid increases.}_{of gas in liquid increases.}

45. 45. CH CH 33– C– CH –H –CH CH 22– C– CH H 22– C– CHH33 OH OH CH CH 33– C– CH = H = CH CH – C– CH H 22– C– CHH33+ C+ CH H 22= C= CH – H – CH CH 22– C– CH H 22– C– CHH33 2-Pentene (Major) 2-Pentene (Major) Conc Conc. H. H22SOSO44 –H –H22OO 1-Pentene (minor) 1-Pentene (minor) 2-Pentanol 2-Pentanol

PART-II (2 Mark)

MATHEMATICS

61. 61.

 

_x_x₁₁



₂₂ _x_x₁₁



₁₁



₊₊

 

_x_x₂₂



₄₄ _x_x₂₂



₄₄



₊₊

 

_x_x₃₃



₆₆ _x_x₃₃



₉₉



₊₊

 

_x_x₄₄



₈₈ _x_x₄₄



₁₆₁₆



₊₊

 

_x_x₅₅



₈₈ _x_x₅₅



₂₅₂₅



_{= 0}_{= 0}



 

_x_x₁₁

_

₁₁

_

₁₁



22 + +

 

xx₂₂



44



22



22 + +

 

xx₃₃



99



33



22 + +

 

xx₄₄



1616



44



22 + +

 

xx₅₅



2525



55



22 Now, Now, xx₁₁



11 – 1 = 0, – 1 = 0, xx₂₂



44 – 2 = 0, – 2 = 0, xx₃₃



99 – 3 = 0, – 3 = 0, xx₄₄



1616 – 4, – 4, xx₅₅



2525 – 5 = 0 – 5 = 0 x x₁₁ = 2, x = 2, x₂₂ - 8, x - 8, x₃₃ = 18, x = 18, x₄₄ = 32, x = 32, x₅₅ = 50 = 50



2 2 x x x x x x x x x x₁₁



₂₂



₃₃



₄₄



₅₅ = 55 = 55 62. 62. _{(1 + 2x + 3x}_{(1 + 2x + 3x}22_{+ .... 21x}_{+ .... 21x}2020_{)) (2}_(21x_1x2020_{+ 20x}_{+ 20x}1919_{+ 19x}_{+ 19x}1818_{+ ... 2x + 1)}_{+ ... 2x + 1)} coeff. of x

coeff. of x3030_{is 11.21 + 12.20 + ... + 21.11}_{is 11.21 + 12.20 + ... + 21.11}

= 2[11.21 + 12.20 + 13.19 + 14.18 + 15.17] + 16.16 = 2[11.21 + 12.20 + 13.19 + 14.18 + 15.17] + 16.16 = 2[231 + 240 + 247 + 252 + 255] + 256 = 2[231 + 240 + 247 + 252 + 255] + 256 = 2[1225] + 256 = 2[1225] + 256 = 2450 + 256 = 2450 + 256 = 2706 = 2706 63.

63. _{If we follow the pattern accordin}_{If we follow the pattern according the rule that angle made by incident ray with normal is equal to the angle}_{g the rule that angle made by incident ray with normal is equal to the angle}

made by reflected ray with normal then we find that ball will not go in any hole. made by reflected ray with normal then we find that ball will not go in any hole.

64.

64. _{C can be collin}_{C can be collin ear with A & B.}_{ear with A & B.} 65.

65. _{Let initial prize is P}_{Let initial prize is P}

after X% increment after X% increment





_



PP



₁₀₀₁₀₀PxPx



_



after decrement y% after decrement y%

























_













_

100 100 y y 100 100 Px Px P P – – 100 100 Px Px P P ₌_{= P}_P 1 + 1 + 100 100 x x – – 100 100 y y – – ₍₍₁₀₀₁₀₀₎₎22 xy xy = 1 = 1 100 100 y y – – x x = = ₍₍₁₀₀₁₀₀₎₎22 xy xy y y 1 1 – – x x 1 1 = = 100 100 1 1

(8)

PHYSICS

66.

66. _{mass of water, m}_{mass of water, m}

1

1= 0.4 kg= 0.4 kg

temperature of

temperature of waterwater,,



₁₁0ºC0ºC mass of ice, m mass of ice, m₂₂= 0.1 kg= 0.1 kg tempeature of ice, tempeature of ice,



₂₂ = – 15ºC = – 15ºC mass of steam = m kg mass of steam = m kg final temperature of mixture

final temperature of mixture



= 40ºC = 40ºC specific hea

specific heat of t of ice, ice, ss_ice_ice = 2.2 × 10 = 2.2 × 1033_J/kg×k_J/kg×k

latent heat of fusion, L

latent heat of fusion, L_f_f = 333 × 10 = 333 × 1033_J/kg_J/kg

latent heat of vaporisation, L

latent heat of vaporisation, L_V_V = 2260 × 10 = 2260 × 1033_{J /kg}_{J /kg}

Heat given = Heat taken Heat given = Heat taken

By steam of 100ºC to water of 100

By steam of 100ºC to water of 100 ºC + By water from100ºC to 40ºC = By water from 0ºC to 40ºC + By iceºC + By water from100ºC to 40ºC = By water from 0ºC to 40ºC + By ice from –15ºC to ice of 0ºC + By ice of 0ºC to water to 0ºC + By water of 0ºC to 40ºC

from –15ºC to ice of 0ºC + By ice of 0ºC to water to 0ºC + By water of 0ºC to 40ºC mL mL_v_v+ mS+ mS_w_w (1100 – 40) = m (1100 – 40) = m₁₁ × S × S_w_w (40 – 0) + m (40 – 0) + m₂₂ss_ice_ice(15) + m(15) + m₂₂ × L × L_f_f + m + m₂₂SS_w_w (40–0) (40–0) m (2260 × 10 m (2260 × 1033_{+ 4200 × 60) = 0.4 ×4200 × 40 +0.1 ×2.2 × 10}_{+ 4200 × 60) = 0.4 ×4200 × 40 +0.1 ×2.2 × 10}33_{×15 + 0.1 × 333 × 10}_{×15 + 0.1 × 333 × 10}33_{+ 0.1 × 4200 × 40}_{+ 0.1 × 4200 × 40} m (2512 × 10 m (2512 × 1033_{) = 67200 + 3300 + 33300 + 16800}_{) = 67200 + 3300 + 33300 + 16800} m = m = kgkg 12560 12560 603 603 251200 251200 120600 120600

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_{= 48 g}_{= 48 g} 67. 67. 100 100 u u _v_v As for

As for questionquestion u

u + + v v = = 11000 0 ccmm ...((ii))

after displacing lens by 40 cm u and v will be after displacing lens by 40 cm u and v will be u +

u + 40, 40, v - v - 4040 for I

for Istst _condition_condition

(i) (i) u u 1 1 v v 1 1 )) u u (( 1 1 v v 1 1 f f 1 1

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uv uv 100 100 uv uv v v u u f f 1 1

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(ii) For second condition (ii) For second condition

)) 40 40 u u (( 1 1 40 40 v v 1 1 f f 1 1

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= = )) 40 40 u u )( )( 40 40 v v (( 40 40 v v 40 40 u u

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= = )) 40 40 u u )( )( 40 40 v v (( v v u u

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From (i) and (ii) From (i) and (ii)

)) 40 40 u u )( )( 40 40 v v (( 100 100 uv uv 100 100

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v v– – u u = = 4400 ...((iiii)) v + u = 100 v + u = 100

from equation (i) and (ii) from equation (i) and (ii) 2v = 140 2v = 140 v =70, u = 30 v =70, u = 30 uv uv v v u u f f 1 1

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= = 2100 2100 100 100

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f = 21 cmf = 21 cm

(9)

68.

68. _{we know that}_{we know that}

A

A₁₁VV₁₁ = A = A₂₂VV₂₂ = Q (volume flowing per second) = Q (volume flowing per second) 25 25

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× 0.6 = 1 × 0.6 = 122

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_{× v}_{× v} 2 2 v v₂₂ = = 115 5 mm//ss ...((ii)) Q = Q = AA₁₁vv₁₁==

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(5 ×10(5 ×10 – – 22₎₎22_{× 0.6 = 4.71g}_{× 0.6 = 4.71g}

force, F = rate of change of momentum force, F = rate of change of momentum F = mv

F = mv₂₂ – – mvmv₁₁ = 4.71 ×

= 4.71 × 15 – 4.71 × 0.15 – 4.71 × 0.6 6 (m = 4.7(m = 4.71, mass fl1, mass flow peow per unit r unit time)time) F = 67.9 N

F = 67.9 N

CHEMISTRY

71.

71. _{Consider the volume of the solution = x cm}_{Consider the volume of the solution = x cm}33

Then the mass of the solution

Then the mass of the solution will be = 1.13xwill be = 1.13x (mass = density × volume)

(mass = density × volume) The solution contains 18% of

The solution contains 18% of NaCl by weightNaCl by weight

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100 100 18 18 × 1.13x = 36 × 1.13x = 36 x = x = 13 13 .. 1 1 18 18 3600 3600

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= 177 cm = 177 cm33 72. 72. CCH H 33– – CCH H 2 2 – – CCOOOOHH + + CCH H 22 – – OOHH CH CH33 Propanoic Propanoic acid acid Ethanol Ethanol H H++ C CH H 33– C– CH H 22– – CCOOOOCCHH22CCHH33+ + HH22OO Ethyl propanoate Ethyl propanoate 73.

73. _{Consider that the salt contains x molecules of water .}_{Consider that the salt contains x molecules of water .}

Molecular weight of anhydrous salt = 160 g Molecular weight of anhydrous salt = 160 g

so molecular weight of hydrated salt will be = 160 + so molecular weight of hydrated salt will be = 160 + 18x g18x g Then, no. of m

Then, no. of m oles of water present in 10x gm of oles of water present in 10x gm of hydrated salt =hydrated salt =

x x 18 18 160 160 10 10

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× x× x

and weight of water present in 10 gm of hydrated salt = and weight of water present in 10 gm of hydrated salt =

x x 18 18 160 160 x x 10 10

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× 18× 18 Hydrated salt

Hydrated salt

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Anhydrous salt + Water Anhydrous salt + Water 1 100gg 66..44gg 33..66gg x x 18 18 160 160 x x 180 180

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= 3.6 = 3.6 180x = 576 + 64.8 x 180x = 576 + 64.8 x x = x = 55 74. 74. _Cr_Cr 2 2 – – 2 2 7 7 O

O + 6Fe+ 6Fe2+2+ ₊_{+ 14H}_14H++

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_6Fe_6Fe3+3+_{+ 2Cr}_{+ 2Cr}3+3+_{+ 7H}_{+ 7H} 2 2OO

Change in oxidation number of Cr is

Change in oxidation number of Cr is = 6 – 3 = 3= 6 – 3 = 3 Change in oxidation number of

Change in oxidation number of Fe is = 3 – 2 = 1Fe is = 3 – 2 = 1

75.

75. _CaCO_CaCO

3

3+ H+ H22O + COO + CO22

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Ca(HCOCa(HCO33))22

Ca(HCO