nseasr 2011 astronomy solution

8/3/2019 NSEASr 2011 Astronomy Solution

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(2) Astronomy. 1) (a) Draw a f igure indicat ing ihe relat ive posit ions ycursel f to ver ify the answer.2 ) ( b) We obt ain x~ + X 2 "" ASin (UI l + ~ ) .Now, we require XI + - Xl + X l " " 0, so thai x]

= - (XI + X 2) = - ASil1(a;J + ~ ) =AS in ( 1 W ' + ~ + 1 l ' ) . Comparing this wh l1 [hegiven expression for Xl,we get the result,

3) (d) Express : tan2 f} interms of sec ' e and obtain a quadrat ic equat ion insec O . Thisgives two values of8 sat is fying the condi tion given inoption (d) .

4) (d):5) (b} We have tan a ""~ and Ianp = ~ _NoW that t he ang lep+qCf)sO p-qcos()

between p and (~q) is ( ISO - 8). Add t he e xp re ss io ns t o gel th e r esul t6) ( d) Use the exp re ss ions f or d iff erent quanti ti es t o get t he re su lt. The physi ca l

quant ity i scalled skin depth - dis tance to which elect romagnet ic wave penet ratesi nt o a b od y.

7) (b) Red g ian t sta rs h ave low s u r fa ce t empe ra ture s . and high luminosity.8) (b}9) (b) Express tan 6Sc = tan (45(1+ 20Q) an d tan 2SQ"'"ran (4S I - 20). Use formula

fer tan (A. + 8). Also note that angles 20 an d 70 a s well a s55" an d 350 form pa i rsnf'complementary angles.

10) (c) Note t ha t u ni ts Of poa rc hen ry /r n wher eas t hose o f t:!I are farad/m. Then wecan wri te . J = H = ~ ~ J ( ~ } " ' L ) . Note , n . , unit of each of thequantities in bracket 1S 0 ]1 1 ' )1 and hence th e answer.

U) (c) T!1cspect ral type depends on the surface tempera ture of th e star. The SUIlhastempera ture of6000 K, which is typical for G class .12) ( a) Refer to the fi gu re a nd wr it e t he in iti al pot en ti al ene rgy of th e system asVI " " ' ~ [ ~ + q,q~ + ! b ! l . L ] . Simi l a rly the express ion for the f i na l po t ent i al4ns, 30 40 50ene rgy of t il e sys tem isU] =~[qlq2 + ql(h + q ' ~ h ] . Gel the dif ference4l '1 o 30 40 10between the two potential energies and compare it wuh t ile given expression 10get the answer.

13)(c) The maximum number of ecl ipses (both solar and J unar) tha t can occur in ayear i s 7. However , the combination o r solar an d lunar eclipses can be ei ther 4and 3 or5 and 2 respect ively.

l4)(c) We get An B=(6,12,IS ..... J9G}, The last term, that is, n th term in th is se tcan be written as 6 + (n -I) "" 3 96, giving" "" 6 6. This gives I1(A r-; D) =66.Aga in use n(,A V 8) = n(A) + n(O) - n(A rl 8) toget the resul t

IS)(a}16) (b) Let th e speeds of points P and Q be ! ! l _ and ~~spectively, where y= I s~n.e. , . '

and x = j cess. Obviously, 2= -/cosO~(n eg ative sign in dica tin g th e decreasedt dtofy with increase of time I) and!!:!. = -isin (j~. This gives dxfdl = tan f). Witl'1dl dt dy/dl2= ,,fj a nd -; an 3 0" = - k . w e get th e answer.


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23) (3) Refer to the figure.

oTriangles BDO and CAD can be easily proved 10b e simi t ar, so that AC : DC =SD : D B. Therefore, ---.!:._ '= _ _ : : : : : ; - , . = 3 . H en ce tile required fr ac ti on o f volumea- . 10

24)(b)25))26) (d) Wecan wri te

( X - 3 ) ' ( 5 ) ' ( 5 )( , . 'H ( 5 ) '" 1x+2 = ,1-x+2 = 1- .r+2 = 1- x+2 X(l~ 5 ]'". x+2

Taking the l im it e s r --+ c o, w e gel the result27) (a) The l at it ud e o f A m r it sa r is greeter than 23.50 N Qr th a nd hence the Sur. never

comes at the zenuh.28) (a) Expand (a - bi and similarly (b - ell, (c - d)'l and (d - a)z . Add ai l these and

note that the sum ofthese squares i s: :! O.Simpli fy the inequal ity togel the resul t.29) (a) Distance between the foci = 2 ae = = 2, s o thal "" "" I . Also the sum of the focal

d" r P" 2 "" iii Tbis ei se 1 Ii 1rsrances 0 , 15 a gtvmg c = ,+. IS gives e= -;- = J2 + I = -.30)(c) As per l ntemar iona l Ast ronomical Un ion 's conven tions , cons tel la tion

Ophiuchus occupies portion between Sccrpius and Sagit tar) us3J) (- I: ) Let a and a: 2 be the roots of the quadratic equation. Using the expressions

f or t he sum and the p roduct o f l ] 1e roots of the quadrat ic equat ion, we get 0: = I.This gives the f n a l r e su I t.

32) (b) Consider the do-t prQdUCI o r the two vectors { o + 2 5 ) and ( S a ' - 4b ) which iszero . Simplify this to get cos ()""0.5 a nd h en ce l it e r es ul t.

33))

34) (c) At Mercury's inferior conjunction if the earth is at Mercury's nodal plane,i ransi t occurs. The eer th comes at Mercury's n od al p la ne in the months Of Maya n d N o ve m be r .

35) (c) Pisces wil l be at 9 0 1 : 1 west of Gcrnini.

36)(d) We Imv


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4S)(d) Write III -6+.J3j"'6+2 @.=2.+~+2 ~, so that ,J; = J i+ II .'1 4 2 2 '1 2 '2 '12 '12S . I r: (7 /5 . . .imilerly, -vn = 'V 2 ' - V " 2 ' Thi s g ives th e r equi re d r esul t.

46) (0 :]) A t the hole , t he p re ssu re out si de t he tube ispgh Whi te I rcm inside the tubepressure is p g (/ ' - h' ) which is Jess than thai outside and hence mercury wiJ l notCOmecut at all .

47)(d) Wehavea != P i= I.Also a + p " "- I and p = I_ We can also write a 2 + a=p,l +P=_I. Take th e pr oduc tAB and l is e t he se r esul ts .

48) (d) Draw the figure yourself and check that moment of inertia of side FlS is~ ( ~ . A ts c the moment of inertia or each of thei ( fJ~)~ddin ga ll t he se One gets t he answe r.49) (c) Note that the fbcal length of the combination of lenses l$given by the relation

. . ! _ = _ !_ + _ I _ . Also use lens makers formula f er f oc al length o r each of the lensesF I, I,and get the answer.

50) (e) Use superposition theorem. Wilen de source of 6 V is: shorted, Ll1Ccurrentthrough 10ohm resistance is ~A frcmB to A. Whereas when 10 'V d e source is7sho rted, t he cur ren t happens t o be l_ A from A to B. Adding the two currents35with appropriate sense, we getthe result.

51)(n} Write (I +:r! r = I + 3x1{1 + X 1 ) + x~= 8xJ CDSJ f). Simpli fyi ng t his we gelx l i : 1+ 3(21."'( :osO) = Sees' B. Using the expression for CQS (38), we g - e t thex xresult.

52) (b) Lei ll1eupthrust acting onthe metal sphere at.OC be F f ' J = V~ p.,,,g and thai a t" F V p g (I + Y 1 \8 ) .50Cbc FYJ=V~ap ...~~g. Then. , ;: = ~p::~ = {i+r:l'io)SIIlr.e 11f1


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60) (b) The given expression " " ( 1 + cot o ? - s e c ~ (B + % ) . Also l i l L l - { B + % ) = -cotB.With til is 011 s im p li fic at io n w e g et the result

61) (el Since the charge IS s ta tionary, i ts speed is zero and he-nceno force is exertedby th e m ag ne ti c fi el d.

62) (3 ) The cent re s and r ad ii o f the two c irc les re spec ti vel y ar e (0 ,0 ). ( 3,4 ) and 2 ,7 .Ti le d is tance between t he cent re s is 5 which i s a lso t he d if fe rence between theradii . This means that the two circles touch each other internally and hence onlyo ne c om m on t an ge nt .

63) (e )Note t hat f orce const an t is inve rs el y p roporti onal t o l engt h, so that the fo rceconstants of tbe two pieces are If. =~ k and k = ~k The effective force constantin t his cas e is k' = k. +k~ = 25k This give the per iodic r ime T 'w;; ing the usual6relation, r: = 2tr . J f . .

6 ") (a ) T he C arle s ian c oo rd in at es c or re sp o nd in g t o p o la r c oo rd in at es (r~ B) o f' a p oi nta re g iv en by x = r co s e, y ~ r sin 8 s c t h at } ). --;r~ +J. From th e g i ve n e q ua ti o nwe can getr=x- 8 = . . . I = (x-3)~ ; : : : : > Xl +; = ;;r .2 - 1 6;; r. +6 4 w h ic h yields anexpression y = ~ 1 6( .v - 4 ) r ep re se nt in g a p ar ab ol a

65){b) Ler the linear mass density be A . = k(/-x). 51) t ha t th e mass element dm "". . dx-- k(l- x) dx. Th e X coo rdina te o f cen tre of m ass can be detcrm lned by u si ng th eu su al fo rm ul a. T he Y c oo rd in ate i s o b vi ou sl y z er o.

66) ( c) Re fe r t o the figure.

Th e shaded area = ! r u b - . .! . ( a ,- , y , = n a b ( I _ ! . )2 2 2 1rtim e is c o nsta nt a nd is e qu al t o H a D .T

Note t hat th e a rea swep t per unit

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67) (d) U se an expression for the area of B triangle in term s of a determinan t.Therefore,

Are.of~ABC~ ~ I :~ : : : I : ~ I ~: : : : : 1 : ~ ( O + b J : : : 1 : 0c a+b J c a+b+c I l c 1 1

68)(b) Let "(X) = 100a be th e p op ul ati on o f ti le v il lag e. L et A b e th e se t o f p erso nskn ow i n g Hind i and B be th e set o f p eo ple kno wing Eng lish . Th en , th e num ber o fp eo pl e k no w in g H in di o r E ng li sh Or both = I I (A u B ). T hu s, th e n um be r o fp eo ptew ho can understand o nly M ara th i is n{X) - 1 1 { A vB ), w hich is given to be 5 2(1 .Therefore; IOOa-n(Au8) = 5 2 ( 1 =;) I'I(AuB) ;;;;:;4S::::}1 ( A ) + n C B ) -n(Ar'lB)= 48a 0> 350+ 23a - "(A n B) = 483 . ,-, n {A rv B): lOa= 1 5 00 ( gi v en ). Thisgives (1= I S O , so th at p op ufa no n o f th e v illag e is ( X) = I D D a = 15000.

69}(c) We can wri te S -", ;~ + !+ _ !. +2 .. _+ "' !" . .. ..+_!_2 2 4 20 10 850[ I I 1 1 1 J [ 1 1 1 1 I ]'" 3 '2+"6+]2+20+ 2550 =J 'j';2+W+ 3x4 + 4)(5+ 50x51. . " I su ( I I ) .,(1 1)which can be further written as 3~,.{r+I)-3f: ;-;-:;:1 =1 .T-51 . This

can then be s impl i fled to getthe answer70) (< :) We have J . .. _ _!_=13xI4_..!...",182-1=~=~=~ (given)12! 14l 14! 14! 14! ISxl4! IS! IS!

a nd t he re fo re t he r es ul t:1. 1) (3) Fo r a tan ge nt to be parallel 10 th e X ax is, th e slop e of th e curve at it p oin t m us t

b e z er o, D if fe re nti ate t he e qu at io n a nd p ut 2.... F a ct o ri ze t h e e xp r es si o n [Q ge tdxo nl y o ne r ea l v al ue o f: . = -I a nd h en ce th e r esult.

n ){a ) Lei 5h-ol = a so th at Y-h~!_ . A dd in g l 11 e ex pr es sio ns B J1 dusing th e da ta w eoge t a + . . ! . . = 2 :: : : ; - (0 -I)~0:: : : : ; .a = I W ith th is w e get 5 J.:r~~=I~x-4 = 0 an dahence the result.

7 3) ( 1 1) G iv en thatf(x);;:: sin xcos.r = !sin 2x = - j'(.x) = cos2x. Note that Ax) is2increasing when j'(x) > 0 ~ cos2x > 0 = - x e ( D . ~ )r x e ( - - , l' [ ) Hence thec or re ct o pti on is ( d) .

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74)) We 03" w,itd" ~ 16807 c (il, n a b O o ' = 48020 .. (ii) an d n(n-I) ,; b'" =254g80 co.(iii) . Equation ( i) . .. equati on (i il g ives . !_"" 16807 =0.35. (iv)na 48020S' '1 I . ( O C ) C) 2b 48020 0875 ()um ar y. equation II -e equation III gives e n --l)a = 54880 : = _ vEquation (iv) -+ equation (v) gives . : : : . . : . . . ! . = 0.4= 1 = 5. Usi ng eqfi ), we get h = 72"and eq(iv) gives a =4.

7S}(a) With usual notations, the excess pressures for the two bubbles arep" -= ~af ]d PI != ~ respecti ve ly , obv ious ly g iv ing p~ > P8- Now, the excesss, Repressure across the interrace is P .< I - pjj = . i ! : _ where R i s the radius of curvatureRof the interrace. Substituting the values we get the answer,

76) (d)Use the express ion for the t O U i . 1 energy -- _! _ kc/ to get the value of k: and use it2to find the periodic lime.

- 'J co"(Jx)77)(a) Let 1- ,'i"'(J~)+oo"(3x)1i< H (il.Also

= 'J ,;n' (Jx) Ii< ("),oo"(3x)+Oio'(3x) ..... " Addi.,g equations (i) lind (ii), we get

" IT : 1r2 I = J{I}:u- = - = gl v en integr al [ "" -,6 127S}(b) Refer to the truth tables of NOT, OR and AND geres and verify that the

outputs come out to be I and 0 respect ively.

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79) (d } From Bohr t heory o f hydrogen atom we have In'r = "..!!_, 50 that linear2"momentum of the elect ron is mv = . . ! ! ! : _ . Use this in the express ion for de Brogl ie

2m "

wavelength it = ~to get the answer.mv80)(d) Given integral can be rewritten as

I~1 . I 5x' +5(1Dg5X5"'~ = 1 . 5 5x +(log5X5'~x = . l . 55;(5' d)1i

nseasr 2011 astronomy solution

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