HINTS & SOLUTIONS NATIONAL STANDARD EXAMINATION IN JUNIOR SCIENCE (NSEJS)

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1 NATIONAL STANDARD EXAMINATION IN JUNIOR SCIENCE (NSEJS) DATE : (a) 008, 09,.., 9997 a 008 d 0 a n 9997 a n a + (n )d (n ) (n ) 0 89 n n 90. HINTS & SOLUTIONS CODE : JS-5. (c) (c) 4. (c) Parallelogram This number should be divided by. So that result is perfect square. 5. (b) ( + y + z) 5 y( + y + z) 5 z(z + + y) 4 () () () Add (), () & () + y + z + (y + yz + z) 79 ( + y + z) 79 + y + z 7 (4) From (), (), () and (4) 5 z 9 y + y + z Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) PCCP Head Office: Plot No. A-5 [A], IPIA, Near Resonance CG Tower Contact : , SOLUTION_NSEJS_9..7 _PAGE- Toll Free : CIN: U800RJ007PLC0409

2 6. (b) p + q + r P + q + r 0 pqr 0 ( p) ( q) ( r) ( r) ( q p + pq) q p + pq r + rq + rp rpq p q r + pq + rq + rp rpq (p + q + r) + pq + rq + rp rpq (p + q + r) p + q + r + (pq + rq + rp) () (pq + rq + rp) 6 (pq + rq + rp) pq + rq + rp () Put value of () in () (c) 8. (b) 9. (b) (5) () y Now and y have same numerator but y denominator is less compare to. So y >. s 54 km/hr. t 0 sec t m 8 Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE- Toll Free : CIN: U800RJ007PTC0409

3 0. (d) a + b + c + d 4? ( a) ( b) ( c) ( b) ( c) ( d) ( c) ( d) ( a) ( a)( b)( d) d a b c ( a) ( b) ( c) ( d) ( a) ( b) ( c) ( d) ( a) ( b) ( c) ( d) ( a) ( b) ( c) ( d) 4 (a b c d) ( a)( b)( c)( d) 4 4 ( a)( b)( c)( d) 0 ( a)( b)( c)( d) 0.. (c) (49) 7(50 ) k ( ) Remainder 7. 50k 7 5. (Bonus) abc 4R s R abc (c) CI f CM() f f 40 f 00 f 00 0 f (c) + + ( ) ( ) ( ) ( ) ( ) is factor of 4 p + q () 4 p() + q 0 6 4p + q 0 4p q 6 () ( ) is factor of 4 p + q () 4 p + q 0 p q () Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) Toll Free : CIN: U800RJ007PTC0409 SOL NSEJS_9..7 _PAGE-

4 Solving () & () p 5 q 4 5. (c) 50, 50, 505,.., 599 a 50 d a n a + (n ) d (n ) 98 (n ) 49 n n S ( ) S 7, (a) A F E B D C Area of AB CF AC BE Area of AB CF AC BE Area of 68.4 BC AD 68.4 BC AD (d) a + b a + b 066 a + b (a + b) ab (a + b) 066 ab () 9 ab ab ab (b) D A B C By Ptolemy Theorem AD BC + AB DC AC BD AC AC 48,60 AC AC 0. Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE-4 Toll Free : CIN: U800RJ007PTC0409

5 9. (c) 40. (a) 5 Aug 07 Tuesday 5 Aug 0 Tuesday So after 6 years Independence day will again come on Tuesday. ( b) (m + ) (a c) (m ) (m + ) b(m + ) (m ) a c(m ) (m + ) b (m + ) (m ) a + c(m ) (m + ) + ( bm b am + a) + c(m ) for equal and opposite root coefficient of should be zero. bm b am + a 0 a b m(a + b) m a b a b. 4. (c) Let the percentage abundance of isotope and percentage abundance of isotope So Average atomic mass is 80 u X % X % 79 5 X is 8 5 X is 00 Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE-5 Toll Free : CIN: U800RJ007PTC0409

6 4. (c) Duralumin is an alloy of aluminium containing copper, manganese and magnesium. It is used for making the parts of air crafts as it is light in weight. 44. (b) Millimoles of HCl 0 0 Millimoles of NaOH So, concentration of OH [OH ] 0 4 poh 4 log 7 ph + poh 4 ph 4- poh The ph range of 8- is of weak base & it gives pale blue colour. 45. (b) Moles of HCl mole 6.5 Molarity of HCl solution is M Na + H O NaOH + H 46 g Na metal gives 80g NaOH 0.46 g is gives 0 moles of NaOH M HCl V HCl M NaOH V NaOH V HCl ml V HCl 0 L 0 ml 46. (b) Assume Caustic soda (NaOH) is a monoacidic base Calcium hydroide Ca(OH) is a diacidic base Hydrated alumina Al(OH) is a triacidic base For Neutralization with one equivalent of phosphoric acid (tribasic acid) each time (Moles base (Valnecy factor) base Equivalent of acid) The ratio of moles of bases required will be NaOH : Ca(OH) : Al(OH) : 0.5 : (d) In case (i) CO Acidic oide, MgO basic oide, N O neutral oide H O Generally it is neutral but sometimes it shows amphoteric behaviour So case (i) is correct In case (ii) SO acidic oide, NO neutral oide, CO neutral oide, Al O amphoteric oide So case (ii) is wrong In case (iii) P O 5 acidic oide, ZnO Amphoteric oide, NO neutral oide, Al O Amphoteric oide So case (iii) is wrong In case (iv) SO Acidic oide, CaO basic oide, N O Neutral oide, PbO Amphoteric oide So case (iv) is correct So correct cases are (i) & (iv) 48. (a) Weight of magnesium 4g (Given) Number of atom in magnesium 4 4 NA Weight of sulphur 4g (Given) Number of atom in sulphur 4 NA Ratio of atom in sulphur to magnesium 4N A 4 4N A 4 Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE-6 Toll Free : CIN: U800RJ007PTC0409

7 49. (c) Biology specimens are preserved in formaline solution. Formaline is (7 40%) aq. Solution of Formaldehyde or Methanal (HCHO) 50. (d) Tooth decay starts when ph of mouth is lower than 5.5. Tooth enamel is made up of calcium phosphate which does not dissolve in water, but get corroded when ph in mouth tooth is below (c) Case II Zinc is less reactive than aluminium so it will not displace aluminium. (I) Zn + CuSO 4 ZnSO 4 + Cu (II) Zn + Al (SO 4 ) No reaction (III) Zn + AgNO Zn(NO ) + Ag (IV) Zn + PbNO Zn(NO ) + Pb As per question reaction (III) will not occur, but as zinc is more reactive than silver so zinc can displace silver. 5. (c) Last discovered element in halogens is astatine Z 85 (it is a 6 th period element) The difference between 6 th & 7 th period element is of. So net halogen element will have atomic number (Z) 7 5. (c) As per Gay lussac s law : At particular temperature & pressure both SO & O occupy same volume & having same number of molecules. Suppose both contain same no. of moles '' then the ratio of their masses will be SO : O X 64 g g : So the mass of SO in flask will be twice that of oygen. 54. (b) During meteorite shower temperature of water body increases as a result ph decreases H O H + + OH K w [H + ] [OH ] as the temp increases, dissociation of water also increases. The value of K w increases & ph decreases. 55. (a) If Z 0 Electronic configuration will be, 8 Outermost shell of the element is completely filled so its valency is zero. 56. (c) O Ketone is R C R [R alkyl group] O (C H 6 O) CH C CH propan one Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE-7 Toll Free : CIN: U800RJ007PTC0409

8 57. (c) (C) Cu (s) + AgNO (aq) Cu (NO ) (aq) + Ag (s) Colourless Bluish green Final observation will be (i) Solution turns blue (ii) Silver deposite on the copper 58. (d) As this is open vessel so pressure and Volume is constant. according to ideal gas equation PV nrt n Temperature (Kelvin) n T n T assume n mole, n moles remain in vessel then n 5 mole T 7 C K ( as 5 moles of air epelled out) 59. (b) 60. (a) 00 5 T T K NaHCO Na CO + CO + H O white solid Residue When residual white powder Na CO dissolved in water it will give alkaline solution CO + H O HCO + OH When we add this solution in Alum Sol n white gelatinous ppt of Al(OH) is obtained. Number of moles of cane sugar.7 4 Number of carbon atoms present in mole cane sugar is N A Total number of carbon atoms consumed through sugar in the tea is.7 4 N A (d) t 0. sec. For block (a) displacement is same i.e 4 unit, so acceleration is zero For block (b) displacement is 6 unit same so acceleration is zero. 6. (b) I and III A q B q o t O to A velocity is constant i.e. V, also from A to B velocity is constant but < so velocity is less at AB v t 6. (c) From the definition of power of Accommodation. Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE-8 Toll Free : CIN: U800RJ007PTC0409

9 64. (d) Constant downward force of gravity only. 65. (b) i i i 5º 5º 5º v v sin i sin r A is maimum and velocity is minimum. 66. (c) a 0 <a t, b 0 <b t, density will decrease because its volume will increase. 67. (d) By flemings left hand rule. particle will turn towards left and electron will turn towards right. 68. (d) It is evaporation of water from blanket by the heat of the bo. 69. (c) Time is 50 sec. and speed increases from 0 to 88 km/hr. v u acceleration is a t m/sec v u + as s s s m 70. (a) KQ Q Electric Potential energy R As R decreases so electric potential energy increases. 7. (d) According to newton's III law of motion for every action there is equal and opposite reaction 7. (d) According to equation of continuity av constant So, a v a v Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE-9 Toll Free : CIN: U800RJ007PTC0409

10 7. (c) f 80 cm 50 cm /////////////////////////////// Since the mirror is inside the water liquid image will be formed at focus i.e. 50 cm above mirror. 74. (b) Resultant amplitude is given by A a a a,a 80 cos A a a (a) a acos (c) a 4a 4a a N K 0 V 0 V 0 I A 40 Current flowing from N to K I/ Current flowing from N to K is / A 76. (b) On the chair there will be a downward force of gravity and an upward force eerted by the flow. 77. (a) By lenz law 78. (b) Let the volume of bulb of hydrometer is V and area of cross section of rod is A For water (V + 0A) d w g mg () For liquid (V + 0A).4 g mg () For liquid (V + 0A) dg mg () From equation () and (V + 0 A) V.4 0A 0.4 V V 50A Equation in () and () (V + 0A) (V + 0A) d 50A + 0A (50A + 0A)d Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE-0 Toll Free : CIN: U800RJ007PTC0409

11 d 79. (b) g/cm 60 6 M O C 0 0 E 70 q 50 q B In COB º So 50º 80. (b) If R & R are connected in series then S R + R RR If R & R are connected in parallel then P R R S R R P RR R R R R R R A S R R RR R P RR RR If n is minimum then R R M D R RR R R RR RR R R S P So, 4 n min Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE- Toll Free : CIN: U800RJ007PTC0409

12 Corporate Office : CG Tower, A-46 & 5, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) SOL NSEJS_9..7 _PAGE- Toll Free : CIN: U800RJ007PTC0409

INDIAN ASSOCIATION OF PHYSICS TEACHERS

INDIAN ASSOCIATION OF PHYSICS TEACHERS NATIONAL STANDARD EXAMINATION IN JUNIOR SCIENCE (NSEJS) 07-08 (Total Marks : 40) Total time : 0 minutes 9--07 [CODE JS53] / 9 ONLY ONE OUT OF FOUR OPTIONS IS CORRECT.