KVPY SA Stream Solution 2012

(1)

ANSWER KEY

HINTS & SOLUTIONS (YEAR-2012)

Q Q uueess .. 11 22 33 44 55 66 77 88 99 1100 11 11 1122 11 33 1144 1155 11 66 1177 1188 11 99 2200 A A nn ss .. CC BB BB AA DD BB BB BB BB BB AA CC DD AA CC BB AA CC DD BB Q Q uueess .. 2211 22 22 2233 2244 2 525 2266 2277 22 88 2299 3300 33 11 3322 33 33 3344 3355 33 66 3377 3388 33 99 4400 A A nn ss .. DD AA AA DD AA CC BB BB CC AA CC AA CC DD CC DD BB BB BB CC Q Q uueess .. 4411 44 22 4433 4444 4 545 4466 4477 44 88 4499 5500 55 11 5522 55 33 5544 5555 55 66 5577 5588 55 99 6600 A A nn ss .. BB BB CC BB BB AA DD BB AA AA CC CC AA BB DD BB DD CC CC AA Q Q uueess .. 6611 66 22 6633 6644 6 565 6666 6677 66 88 6699 7700 77 11 7722 77 33 7744 7755 77 66 7777 7788 77 99 8800 A A nn ss .. DD BB CC BB CC AA DD BB BB DD BB BB BB BB ** AA DD CC CC BB

(2)

PART-A (1 Mark)

MATHEMATICS

1. 1. _{f(x) = ax}_{f(x) = ax}22_{+ bx + c}_{+ bx + c} 10 = 4a+2b+c ... (1) 10 = 4a+2b+c ... (1) –2 –2 = 4a = 4a – 2b – 2b + c ..+ c .. ... ... ... (2)(2) 12 = 4b 12 = 4b



b = 3b = 3 2. 2. )) 75 75 .. 0 0 (( 1 1 )) 75 75 (. (. )) 1 1 (( )) 75 75 (. (. 33 33 33





= = 25 25 .. 1 1 = 4 = 4 square

square root = 2root = 2

3. 3.

10, 10

10, 10 +d , 10+2d , d+d , 10+2d , d



I, d I, d



I I 10+2d < 10+10+d 10+2d < 10+10+d d < 10 d < 10



d = 1, 2, 3, ... 9 d = 1, 2, 3, ... 9 9 triangles are possible 9 triangles are possible

4. 4. _{a = 3k}_{a = 3k} b = k b = k c = 5k – 4k = k c = 5k – 4k = k d = 6k – 5k = k d = 6k – 5k = k d d 3 3 c c 2 2 b b a a



= = kk 22kk 33kk k k 3 3



= = 22 1 1 5. 5. ]] )) 1 1 n n 2 2 (( ... ... 5 5 3 3 1 1 [[ ]] n n ... ... 2 2 1 1 [[ 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2





1 122_{+ 2}_{+ 2}22_{+ 3}_{+ 3}22_{+ ... + (2n)}_{+ ... + (2n)}22₌₌ 6 6 )) 1 1 n n 4 4 )( )( 1 1 n n 2 2 (( n n 2 2



[1 [122_{+ 3}_{+ 3}22_{... + (2n – 1)}_{... + (2n – 1)}22_{+ 2}_{+ 2}22 _[1_[122_{+ 2}_{+ 2}nn_{...+ n}_{...+ n}22_]] = = 6 6 )) 1 1 n n 4 4 )( )( 1 1 n n 2 2 (( n n 2 2



S + 4 S + 4 6 6 )) 1 1 n n 2 2 )( )( 1 1 n n (( n n

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= = 6 6 )) 1 1 n n 4 4 )( )( 1 1 n n 2 2 (( n n 2 2



S + S + 22nn((22nn



11)()(44nn



11)) – – 44nn((nn



11)()(22nn



11))

(3)

= 2n = 2n

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22nn₆₆



11

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_

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[[44nn



11



22nn



22]] = = 6 6 )) 1 1 n n 2 2 )( )( 1 1 n n 2 2 (( n n 2 2





Ratio = Ratio = 6 6 )) 1 1 n n 2 2 )( )( 1 1 n n (( n n 4 4



× × )) 1 1 n n 2 2 )( )( 1 1 n n 2 2 (( n n 2 2 6 6

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= = 22nn 11 2 2 n n 2 2

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1 1 n n 2 2 2 2 n n 2 2

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> > 10 10 00 10 10 11 200n + 200 > 202n – 101 200n + 200 > 202n – 101 2n < 301 2n < 301 n < n < 2 2 30 30 11

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maximum va maximum value = 15lue = 15 00

6. 6.

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_AD_{AD B =}_{B = 18}_{18 0}₀

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_–_–

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B B 2 2 A A

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BFC = 180BFC = 180

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– –

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B B 2 2 C C 18 18 00

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– – 2 2 A A – B + 180 – B + 180

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– – 2 2 C C – B = 180 – B = 180



18 18 00

 

2 2 C C A A

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+ + 2B2B

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3636 00



= A + C + 4B = A + C + 4B 36 36 00

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= A + B + C = A + B + C + 3B+ 3B

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B = 60B = 60

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IIFD =FD =

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IIBD BD == 2 2 B B = 30 = 30



7. 7.



+ + 2 2



+ + 2 2



= 1 = 1

 

+ + ₂₂

_

= 1 = 1



= = 1 1 2 2 1 1



= = 22



11

(4)

8. 8.



R R = = 2 2 1 1





= =_2R_2R No Noww



+ R = r + R = r 3R = r 3R = r R = r/3. R = r/3. 9. 9. _tan_tan



₌₌ AF AF AB AB ,,



x = 90x = 90



tan tan



= = 1 1 2 2 sin sin

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= = AF AF AX AX 5 5 2 2 = = AAX X ;; XXF F == 5 5 1 1 AB AB FF of of Are Are aa AXF AXF of of Are Are aa



= = )) AF AF AB AB (( 2 2 1 1 )) XF XF (( )) AX AX (( 2 2 1 1



= = 1 1 2 2 2 2 1 1 5 5 1 1 5 5 2 2 2 2 1 1



= = 5 5 1 1 10. 10.



_{RQP = 176º}_{RQP = 176º}



SPQ = 2ºSPQ = 2º



SQP = 89º (SP = PQ)SQP = 89º (SP = PQ)

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SQR = 176 – 89 = 87ºSQR = 176 – 89 = 87º 1 11.1.

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= 90 + 15× = 90 + 15× o o 2 2 1 1

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= = 2 2 19 19 55oo

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= 360 – = 360 – 2 2 19 1955 = = 2 2 19 1955 72 72 00

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= = 2 2 52 5255 Difference = Difference = 2 2 19 19 55 52 52 55

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= = 2 2 33 33 00 = 165º = 165º

(5)

12.

12. _{Let A take x}_{Let A take x}

B take

B take y hay has together is hours s together is hours ==





_



_x_x11



_y_y11



_



in of work in of work Let time be t Let time be t tt

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y y 1 1 x x 1 1 + 8 + 8 == x x 8 8 tt

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y y tt = = x x 8 8 tt

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11_x_x



_y_y11

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_



= = tt

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_y_y44..55 x x tt = = 44_y_y..55 y y x x = = tt 8 8 ;; _y_yxx = = 5 5 .. 4 4 tt tt 8 8 = = 5 5 .. 4 4 tt

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t t22_{= 36}_{= 36} t = 6 hours. t = 6 hours. 13.

13. _{Let weight of bucket be}_{Let weight of bucket be}



_{and weight of water is}_{and weight of water is}





2a + b 2a + b = = 20 20 ... (1)... (1) and 3a + 2b = 33 ... (2) and 3a + 2b = 33 ... (2)



= 7 = 7



= 6 = 6 total weight = total weight =



+ +



= 1 = 133 14. 14. _{mn = 144}_{mn = 144}

(m, n) = total 15 positive ordered pairs and ne

(m, n) = total 15 positive ordered pairs and ne gativgative ordered pairs are posse ordered pairs are poss ibleible

15. 15. 00 55, , 1100, , 1155, , ... . 4400 1 1 22, , 77, , ... . 3366 3 3 33, , 88, , ... . 3388 4 4 44, , 99, , ... . 3399

PHYSICS

16.

16. _{Momentum conservation}_{Momentum conservation}

mv = (m + m)v mv = (m + m)v v’ = v’ = 2 2 v v K.E. = K.E. = 2 2 1 1 × 2 × 2mm 2 2 2 2 v v

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= mv = mv22_/4._/4.

(6)

17.

17. _{A ba}_{A ball falls v}_{ll falls vertically dow}_{ertically downward and bounces off a}_{nward and bounces off a horizontal floor}_{horizontal floor. The}_{. The spe}_{spe ed of}_{ed of the ball}_{the ball jus}_{jus t before Down}_{t before Down}

:: mmgg––FF––mmaa₁₁ a a₁₁ = g – = g – m m F F U Upp:: mmgg++FF==mmaa₂₂ a a₂₂ = g + = g + m m F F a a₂₂ > a > a₁₁ . . 19. 19.



= 30º + 90º = 30º + 90º = 120º . = 120º . 22. 22. _{i = r = 0}_{i = r = 0} So, So,



= = 00 No dispersion. No dispersion. 23. 23.

(7)

24. 24. _IIn_n _P_P₌₌ R R 3 3 v v22 IInn QQ == R R v v 3 3 22 IInn R R == R R v v22 IInn S S == R R v v 4 4 22 S Soo,, P P < < R R < < Q Q < < SS 25.

25. _{In length 2}_{In length 2}



_{R change}_{R change}

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₂₂



_R_R



_T_T

1 1

 

TT In In dd



d d



TT 26. 26. f f _ss = mg . = mg . 27. 27. _6a_6a22_{= 4}_{= 4}



_r_r22 r r a a = = 6 6 4 4



B B_S_S = = 3 3 4 4



r r 33



_g_g B B_C_C = a = a33



_g_g C C S S B B B B = = ₃₃ 3 3 a a r r 3 3 4 4



= =



_

4 4 6 6 4 4 6 6 .. 3 3 4 4 = = 11 2 2 3 3 2 2





B B_S_S > > BB_C_C . . 28. 28. 238238_U_U 92 92



2142148484 PoPo + 6a + ne + 6a + ne U U 238 238 92 92



8484214214PoPo + 6 + 622HeHe 4 4_{+ ne}_{+ ne} S Soo,, n n = = 44

(8)

30.

30. _Du_{Du e to attract}_{e to attraction in both c}_{ion in both case be}_{ase be nd in the same}_{nd in the same direction.}_direction.

CHEMISTRY

31. 31. 22CaCa OO 22CaCaOO )) excess excess (( 22











 mole mole 2 2 1 1 40 40 20 20



molemole 2 2 1 1 mole mole 2 2 1 1

of CaO will b

of CaO will be formed i.ee formed i.e .,., 5656 2828gg 2 2 1 1





32.

32. _{It is an}_{It is an example of bromoform reaction (similar to Iodofo}_{example of bromoform reaction (similar to Iodoform reaction).}_{rm reaction).}







NaNaOHOH – – _CHBr_CHBr 3 3 + + 33. 33. 19 19 40 40_K_K++ e e – –_{s = 19 – 1 = 18}_{s = 19 – 1 = 18} N = 40 – 19 = 21 N = 40 – 19 = 21

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electroelectrons + Neutrons = 18 ns + Neutrons = 18 + 21 = 39+ 21 = 39

34.

34. NaNa₂₂O is most O is most basic Oxidbasic Oxide as e as it will foit will form NaOH on dissolving in water which is srm NaOH on dissolving in water which is s trong base.trong base.

35. 35. _{Molarity =}_{Molarity =} solution solution of of lit lit solute solute of of moles moles or or 3 3 .. 1 1 35 35 .. 0 0 = 0.269 M = 0.27 M = 0.269 M = 0.27 M 36.

36. _Den_{Den sity}_{sity ((}



_{), temperature (T) and pressu}_{), temperature (T) and pressu re (p) are intens}_{re (p) are intens ive va}_{ive variables because they donot depend upon}_{riables because they donot depend upon}

mass. mass.

37.

37. _{Potash alum is K}_{Potash alum is K}

2

2SOSO44.A.All22 (SO (SO44))33.24H.24H22OO



Empirical formula is K A Empirical formula is K A l(SOl(SO₄₄))₂₂.12H.12H₂₂OO

38. 38.

Sol. Sol.

(9)

39.

39. _{It is}_{It is example of e}_{example of e lectrophi}_{lectrophilic addi}_{lic addition reaction following the markownikov rule.}_{tion reaction following the markownikov rule.}

40.

40. _{At room te}_{At room tempe}_mperature_{rature , it is a simple acid-base}_{, it is a simple acid-base reaction r}_{reaction res}_{es ultin}_{ultin g in the formation of s}_{g in the formation of s alt.}_alt.







CHCH33 – –CHCH22 – –NHNH22 _NH_NH 3 3 + +_CH_CH 2 2CHCH33.CH.CH33COOCOO – – 41. 41. _{If Q}_{If Q} C

C < < KKCC then then reaction will move in forward direction.reaction will move in forward direction.

42.

42. _Ace_{Ace tyl salicylic acid is}_{tyl salicylic acid is common}_{common ly known}_{ly known as as}_{as as pirin.}_pirin.

43.

43. _{There is}_{There is not any acidic p}_{not any acidic proton in diethyl ether, hence it does}_{roton in diethyl ether, hence it does not exhibit strong hydrog}_{not exhibit strong hydrogen}_{en bonding.}_bonding. Or

Or

For H–b

For H–bonding in molecule highly electroonding in molecule highly electronegative elemenegative eleme nt & H nt & H should be directly cshould be directly connected. onnected. In (CIn (C₂₂HH₅₅))₂₂O,O, H is conne

H is conne cted to carbocted to carbon.n.

44.

44. _{Both A}_{Both A and B differs in}_{and B differs in position of do}_{position of double bond, hence}_{uble bond, hence they are po}_{they are positional isomers}_{sitional isomers ..}

45.

45. _(C)_(C)

PART-II

T

Two Mark

wo Mark Questions

Questions

MATHEMATICS

61. 61. c c 2 2 b b b b 2 2 a a





c c – – 2 2 b b c c – – 2 2 b b = = 2 2 2 2 2 2 c c – – b b 2 2 bc bc – – 2 2 b b ac ac 2 2 – – ab ab 2 2



= = _





c c b b 2 2 )) ac ac – – b b (( 2 2 bc bc – – ab ab 2 2 2 2 2 2



bb22₌_{= a}_ac_{c ,,} _n_nu_um_mb_be_er_{r a}_arre_{e a}_a,_{, a}_arr,_{, a}_ar_r22 (A (A)) ₂₂ ₂₂ 2 2 2 2 c c b b 2 2 b b a a 2 2



= = ₂₂ 2 2 c c ac ac 2 2 ac ac a a 2 2



= = )) c c a a 2 2 (( c c )) c c a a 2 2 (( a a



= = c c a a = = ₂₂ r r 1 1 may or may n

may or may n ot be inot be in tegetege r.r...

(B (B)) ₂₂ ₂₂ 2 2 2 2 c c 2 2 b b b b 2 2 a a



= = ₂₂ ₂₂ ₂₂ ₄₄ 2 2 2 2 2 2 r r a a 2 2 r r a a r r a a 2 2 a a



= = )) 1 1 r r 2 2 (( r r a a )) 1 1 r r 2 2 (( a a 2 2 2 2 2 2 2 2 2 2



= = ₂₂ r r 1 1 (C (C )) c c b b a a c c b b a a22 22 22









= = ₂₂ 4 4 2 2 2 2 2 2 2 2 ar ar ar ar a a r r a a r r a a a a

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



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= a = a

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2 2 4 4 2 2 r r r r 1 1 r r r r 1 1 (D (D )) b b c c a a c c b b a a22 22 22

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= = )) 1 1 r r r r (( a a )) r r r r 1 1 (( a a 2 2 4 4 2 2 2 2







= a(r

(10)

62. 62. _{z =}_{z =} xy xy 3 3 , x + 2y + 4 , x + 2y + 4



_xy_xy33 = = 99 x + 2y + x + 2y + 33((44_xy_xy



33xyxy)) = = 00 xy(x + 2y – 9) = – 12 xy(x + 2y – 9) = – 12 C – I C – I _{xy = 1, x + 2y = – 3}_{xy = 1, x + 2y = – 3} x + x + x x 2 2 + 3 + 3 = = 00 ((––1, 1, ––1, 1, 3)3) x x22_{+ 3x}_{+ 3x + 2 =}_{+ 2 = 0}_{0 –}_–1,_{1, –}_–2₂

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3 3 ,, 2 2 1 1 ,, 2 2 – – C – II C – II _{xy = –1, x + 2y = 21}_{xy = –1, x + 2y = 21}



_{x =}_{x =} 2 2 44 44 99 21 21



x + x + x x )) 1 1 (– (– 2 2 = = 33,, xx22_{– 3x – 2}_{– 3x – 2 = 0}_{= 0} x = x = 2 2 )) 2 2 (( 4 4 9 9 3 3





= = 2 2 17 17 3 3



C – III C – III _{xy = 2, x + 2y = – 6 + 9 = 3}_{xy = 2, x + 2y = – 6 + 9 = 3} x + 2. x + 2. x x 2 2 = 3 = 3



xx22_{– 3x + 4 = 0}_{– 3x + 4 = 0} D = 9 – 4.1.4 < 0 D = 9 – 4.1.4 < 0 C – IV C – IV _{xy = – 2, x + 2y – 9 = 6}_{xy = – 2, x + 2y – 9 = 6} x x + + 22y y = = 1155,, xx22_{– 15x – 4 = 0}_{– 15x – 4 = 0} x + 2 x + 2

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 

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x x 2 2 = = 1155,, 115522_{+ 4.1.4}_{+ 4.1.4} C –V C –V _x_xy_{y =}_{= 3}_3,_{, x}_{x +}_{+ 2}_2y_{y –}_{– 9}_{9 =}_{= –}_{– 4}_4,, _x_{x +}_{+ 2}_2y_{y =}_{= 5}₅ x + 2. x + 2. x x 3 3 = 5 = 5



xx22_{– 5x + 6 = 0}_{– 5x + 6 = 0}



x = 2, 3x = 2, 3 x x = = 22,, y y == 2 2 3 3 ,, z z = = 11 ((22,, 2 2 3 3 , 1) , 1) x x = = 33,, y y = = 11,, z = z = 11 ((33, , 11, , 11)) C – VI C – VI _{xy = –3, x + 2y – 9 = 4}_{xy = –3, x + 2y – 9 = 4} x + 2y = 13 x + 2y = 13 x + x + x x )) 3 3 (– (– 2 2 = 1 = 1 33 x x22_–_{– 1}₁₃_3x_{x –}_{– 6}_{6 =}_{= 0}₀ _D_{D =}_{= 1}₁₃₃22_{+ 4}_{+ 4}



_{6 = 193}_{6 = 193}

(11)

C – VII C – VII _{xy = 4}_{xy = 4} x + 2y – 9 = – 3 x + 2y – 9 = – 3 x + 2y = 6 x + 2y = 6 x + 2. x + 2. x x 4 4 = 6 = 6



xx22_{– 6x + 8 = 0}_{– 6x + 8 = 0} x = 2,4 x = 2,4 x = 2, y = 4, z = x = 2, y = 4, z = 4 4 3 3 (2, 4, (2, 4, 4 4 3 3 )) x = 4, y = 1, z = x = 4, y = 1, z = 4 4 3 3 (4, 1, (4, 1, 4 4 3 3 ))



six solution (C)six solution (C)

63. 63. AB AB = AC= AC



AB AB C ~C ~



BDBD CC BD BD AB AB = = DC DC BC BC = = BC BC AC AC cosA = cosA = x x .. x x 2 2 2 2 – – x x x x22



22 22 = = 2 2 .. 2 2 .. 2 2 2 2 x x 2 2 2 2 2 2 2 2 2 2

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

2 2 2 2 x x 4 4 x x 2 2



= = 4 4 4 4 x x 8 8 2 2



8x 8x22_{– 16 = 8x}_{– 16 = 8x}22_–_– 4 4 x x44 x x44_{= 64}_{= 64} x = x = ₂₂ ₂₂ s s == 2 2 2 2 2 2 2 2 2 2 2 2



= = ₂₂ ₂₂ + + 11 area of

area of



AB AB C =C = ((22 22



11)()(22 22



11)()(11))((11)) =

(12)

64.

64. _{Let distance b/w Pune and Mumbai be}_{Let distance b/w Pune and Mumbai be} _{speed of 1}_{speed of 1}stst_{train =}_{train =}

4 4   2 2ndnd_{train =}_{train =} 2 211 3 3   = = 7 7 2 2 distance covered by 1

distance covered by 1stst_{train in 2 hours =}_{train in 2 hours =}

4 4  



2 = 2 = 2 2   at 9:30

at 9:30 relatirelative distance to be covered =ve distance to be covered = 2 2

 

Let they meet at time t Let they meet at time t

2 2   = =

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_

tt 4 4   + + 7 7 2 2 tt 2 2   = = _tt

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 

28 28 8 8 7 7



t =t = 15 15 14 14 hours or 56 hours or 56



An An s. s. 9: 39: 3 0 0 + 56 + 56 min min = 10 = 10 : 2: 266

65.

65. _Applyin_{Applyin g py}_{g pythogorus}_{thogorus theore}_{theore m.}_m.

Path = Path = ₂₂22

_

₂₂22 ₊₊ 22 22 1 1 3 3



= = ₈_{8 +} + ₁₀₁₀

PHYSICS

66.

66. _{In parallel combination}_{In parallel combination}

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= =

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n n R R R R E E 0 0 P P₁₁ = =

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22..RR_n_n

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= = _n_n R R .. n n R R R R E E 2 2 0 0

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In serie

In serie s combinatios combinationn

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= = _R_R EE_nR_nR

0 0

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(13)

P P₂₂ = = ..nRnR nR nR R R E E 22 0 0

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B Buutt PP₁₁ = P = P₂₂ n n R R .. n n R R R R E E 2 2 0 0

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= = RR nRnR ..nRnR E E 22 0 0

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R R nR nR n n 0 0

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= =

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nRnR R R n n 0 0 R R₀₀ + nR = nR + nR = nR₀₀ + R + R R R₀₀(1 – n) = R(1 – n)(1 – n) = R(1 – n) R R R R₀₀ = 1 = 1 67. 67. R = R = g g 2 2 sin sin u u22

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= = 10 10 ºº 15 15 00 sin sin 30 3022 = 90 × = 90 × 2 2 1 1 = 45 m = 45 m COM of firecracker at 45 m from

COM of firecracker at 45 m from projectioprojection pointn point

x x_Cm_Cm = = 2 2 1 1 2 2 2 2 1 1 1 1 m m m m r r m m r r m m

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45 45 == m m r r 2 2 m m )) 27 27 (( 2 2 m m 2 2

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90 = 27 + r 90 = 27 + r ₂₂ r r ₂₂ = 90 – 27 = 64 m = 90 – 27 = 64 m If r If r ₂₂ = –27 m then = –27 m then 90 = –27 + r 90 = –27 + r ₂₂ r r ₂₂ = 117 m. = 117 m. 68. 68. _U_{Us i}_{s in}_ng_g _w_w..ff..tt K K_f_f – K – K_ii = W = W_g_g + + WW_f_f 2 2 1 1 mu mu22_–_– 2 2 1 1 mu mu22_{= mgh + W}_{= mgh + W} f f W W_f_f = – mgh = – mgh

Which is equal to energy loss in process. Which is equal to energy loss in process.

(14)

69.

69. _{Magnetic field due to wire is in}_{Magnetic field due to wire is in wards when loop mov}_{wards when loop moves}_{es toward}_{towards E curre}_{s E curre nt is clockwise.}_{nt is clockwise.}

70.

70. _{Heat gi}_{Heat given by wa}_{ven by water = 100}_{ter = 100 × 1 × 800}_{× 1 × 800 = 8000}_{= 8000 cal}_cal

Heat taken by ice = 8000 cal = m × 80 Heat taken by ice = 8000 cal = m × 80

m = 100 gm m = 100 gm

So amount of ice which does not melt = 150 – 100 = 50 gm. So amount of ice which does not melt = 150 – 100 = 50 gm.

CHEMISTRY

71.

71. _{Metal + H}_{Metal + H}

2

2SOSO44

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Metal sulphateMetal sulphate

No. of Eq. of metal = No. of eq. of me

No. of Eq. of metal = No. of eq. of me tal sulphatetal sulphate

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2 2 96 96 E E 8 8 .. 6 6 E E 2 2 ,, E = 20E = 20 Ans.Ans. 72. 72. V V 2 2 2 2 2 2 .. 0 0 V V 1 1 .. 0 0 V V ]] x x [[  _f_f

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== V V 2 2 V V 5 5 .. 0 0 = 0.25 M = 0.25 M Ans. Ans. 73. 73. 74. 74. 2 2 1 1 m m

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22_{= h}_{= h}

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_–_{– W}_W

High is the thershold freque

High is the thershold freque ncy of metal greater will be the work funncy of metal greater will be the work fun ction.ction. So, So, M M₁₁

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RRb b ;; MM₂₂

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K K ;; MM₃₃

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NNa a ;; MM₄₄

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LiLi 75. 75. HH 22 ₂₂ )) excess excess (( mole mole 2 2 44 Br Br Mn Mn KBr KBr KMnO KMnO

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No. of eq. of KMnO

No. of eq. of KMnO₄₄ = No. of eq. of Br = No. of eq. of Br ₂₂ 2 × 5 = n 2 × 5 = n_Br Br 22× 2× 2 n n_Br Br 22 = 5 mole = 5 mole