NTSE TEST FULL TEST-2. Test Date :

Transcription

1 NTSE TEST FULL TEST- Test Date : Corporate Office : Paruslok, Boring Road Crossing, Patna-0 Kankarbagh Office : A-0, st Floor, Patrakar Nagar, Patna-0 Bazar Samiti Office : Rainbow Tower, Sai Complex, Rampur Rd., Bazar Samiti Patna-06 Call : /4/6/7

2 NTSE FULL TEST- _0--07 PAGE NO. MENTAL ABILITY. (A). (C). (D) 4. (B) 5. (B) 6. (C) 7. (A) 8. (C) 9. (C) 0. (A). (C). (A). (A) 4. (C) 5. (D) 6. (B) 7. (B) 8. (D) 9. (C) 0. (D). (B). (C). (B) 4. (B) 5. (C) 6. (B) 7. (A) 8. (A) 9. (C) 0. (A). (B). (D). (C) 4. (B) 5. (D) 6. (B) 7. (B) 8. (D) 9. (A) 40. (A) 4. (D) 4. (C) 4. (D) 44. (D) 45. (C) 46. (D) 47. (B) 48. (A) 49. (B) 50. (D) ENGLISH 5. (A) 5. (B) 5. (B) 54. (B) 55. (B) 56. (B) 57. (B) 58. (B) 59. (D) 60. (C) 6. (A) 6. (B) 6. (B) 64. (B) 65. (B) 66. (A) 67. (D) 68. (D) 69. (D) 70. (A) 7. (C) 7. (A) 7. (C) 74. (A) 75. (D) 76. (B) 77. (D) 78. (B) 79. (D) 80. (C) 8. (A) 8. (C) 8. (B) 84. (A) 85. (A) 86. (D) 87. (C) 88. (D) 89. (C) 90. (B) 9. (B) 9. (D) 9. (C) 94. (B) 95. (A) 96. (D) 97. (D) 98. (C) 99. (B) 00. (A) 0. (C) v e PHYSICS where r is position of body from the surface. r v r 7R R 8R v r R R v v e on v v e 0. (D) Total distance = =50 m Relative velocity = 0 ( 0) = 50 m/s Total dis tan ce 50 Hence t 5 sec. v (C) E mv (I) rel Corporate Office : Parus Lok Complex, Plot No.-6/7, Boring Road Crossing, Patna , Ph. No. : , 0, 54007

3 NTSE FULL TEST- _0--07 PAGE NO. 04. (B) Now v = (v + ) m/s Now E' E m v ' m v (II) From (I) & (II). mv mv or v = (v + ) or v v or v m / s v m 5 u v 5u v u f 5u u 0 or v v u 0 5 5u 0 6 5u 0 u cm 05. (B) F q 4 r q q = n (.6 x 0 9 ) n = x (A) Volume is same in both cases V AL AL 5 or A L L A L L A A L R A L A. R L L A 4 A Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

4 NTSE FULL TEST- _0--07 PAGE NO. 4 R R % change % R (C) Conceptual. 08. (B) mv qvb r mv r qb 6 mv or B T. 9 qr (C) (n m)e (4 ).E E A.5A nr R (C) B g kx kx < g For liquid N mg N N >mg hence (C) is correct. (C). (C) sinic cos 48 sin 4 Mg sin T = Ma [Newton s II law for block ] T = Ma [Newton s II law for block ] By subtracting both equations T = Mg sin Mg sinθ T = Corporate Office : Parus Lok Complex, Plot No.-6/7, Boring Road Crossing, Patna , Ph. No. : , 0, 54007

5 NTSE FULL TEST- _0--07 PAGE NO. 5. (A) The time taken by the stone to reach the lake t h 500 = 0 sec (Using g 0 Now time taken by sound from lake to the man t h sec v 40 Total time = t + t = =.5 sec. h ut gt ) 4. (B) 5. (D) 6. (D) CHEMISTRY -chloro-, -dimethyl pentane contains all the four,, and 4 carbon atoms. CH CH 4 CH CH C CH CH Cl -chloro---dimethyl pentane CH 4 H C C CH CH CH IUPAC name =, -dimethyl--butene. carbon is attached to one carbon atom. carbon is attached to two carbon atoms. carbon is attached to three carbon atoms. The hydrogen attached to carbon atom are. H CH 4 H C C C C H H CH It has one carbon atom and two hydrogen atoms. 7. (A) (i) (ii) n = 4, l = 4 p orbital n = 4, l = 0 4s orbital (iii) n =, l = d orbital (iv) n =, l = p orbital According to aufbau principle, energies of above mentioned orbitals are in the order of Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

6 NTSE FULL TEST- _0--07 PAGE NO (A) Given, Atomic number of Rb, Z = 7 Thus, its electronic configuration is [Kr] 5s. Since the last electron or valence electron enter in 5s subshell. So, the quantum numbers are n = 5, l = 0, (for s orbital) m = 0 9. (D) m l to l,s / or / The balanced chemical reaction is BaCl Na PO Ba PO 6NaCl 4 4 In this reaction, moles of BaCl combines with moles of Na po 4. Hence, 0.5 mole of BaCl require mole of Na PO 4. Since available Na PO 4 (0. mole) is less than required mole (0.), it is the limiting reactant and would determine the amount of product Ba (PO 4 ). 0. (A). (B) moles of Na PO 4 gives mole Ba (PO 4 ) 0. mole of Na PO 4 would give 0. = 0. mole Ba (PO 4 ) NaHCO Na CO H O CO g mole mole mole mole mole mole mole mole 6 x 84 x.4 = 504 gm. = 67. L Na CO + HCl NaCl + H O + CO (g) mole mole mole mole Corporate Office : Parus Lok Complex, Plot No.-6/7, Boring Road Crossing, Patna , Ph. No. : , 0, 54007

7 NTSE FULL TEST- _0--07 PAGE NO. 7 Limiting Reagent HCl is the limiting Reagent. So, From mole of HCl mole of CO gas produced. (A). (C) 4. (B) 5. (C) 6. (A) 4 mole 4 mole of CO gas = 44.8 L i.e. x.4 L Highly reactive elements are obtained by electrolysis. Roasting removes volatile impurites like water vapours and converts sulphides into oxides. ZnS + O ZnO + SO Removal of impurities from ores is known as enrichment of ore or concentration of ore. HCOOH is a monobasic acid. Base Acid Acid Base CO H O HCO OH BIOLOGY 7. (C) Cytokinin delays senescense 8. (D) 9. (B) 0. (B). (B) Fungi are eukaryotic in nature.. (A) Antibiotics are used either to kill or inhibit the growth of bacteria Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

8 NTSE FULL TEST- _0--07 PAGE NO. 8. (D) 4. (A) Glucagon increases the level of blood glucose level. 5. (A) 6. (C) 7. (D) 8. (B) 9. (D) 40. (A) 4. (A) 4. (B) 4. MATHEMATICS A E 0 B C D CED AED 0, sin 0 = 44. Let P (m, 6) divides the line segment AB joining A (,5) B in the ratio k :. k : A (, 5) P (m, 6) Applying section formula, we get the co-ordinates of P : B,5 Corporate Office : Parus Lok Complex, Plot No.-6/7, Boring Road Crossing, Patna , Ph. No. : , 0, 54007

9 NTSE FULL TEST- _0--07 PAGE NO. 9 5 k k 5 k 6 5k 0,, k k (k ) (k ) But P (m, 6) = P k 6 5k 0, (k ) (k ) k 6 m (k ) and also 5k 0 6 (k ) 5k 0 6 (k ) 5k 0 (k ) 5k 0 k 5k k 0 k = k 6 Putting k in the equation m we get : (k ) k m 0 0 k m 0 Required value of m is m =, m + = 5 (x,y) A C(,) B(4, ) x 4 x 46. y y 7 Other end = (, 7) A a h a B M b C Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

10 NTSE FULL TEST- _0--07 PAGE NO. 0 Perimeter = a + b BM = b/ b a b / h h a 4 4a b Area ABC.b. 4a b b. 4a b x + y = a (I), xy = b (II) x y x y xy x y xy x y xy a ab b 48. a ab Ans. x y b D Y C b Ar xyz a.b ab Ar (rectangle ABCD) = 4ab A x a y 4a a z B Ratio of Area Given, a+=b+=c+=d+4=a+b+c+d+5 Now, a+=a+b+c+d+5 b+=a+b+c+d+5 c+=a+b+c+d+5 d+4=a+b+c+d+5 Add all the above equations, we get (a+b+c+d)+(+++4)=4a+4b+4c+4d+0 0=a+b+c+d+0 0 0=(a+b+c+d) 0=(a+b+c+d) 0 =a+b+c+d Corporate Office : Parus Lok Complex, Plot No.-6/7, Boring Road Crossing, Patna , Ph. No. : , 0, 54007

11 NTSE FULL TEST- _0--07 PAGE NO. 50. Let O be the centre of. By symmetry O is on the perpendicular bisector of AB. Draw OE AB. Then BE = AB/ = /. If r is the radius of. We see that OB = r, and OE = r. Using Pythagoras, theorem. 5. r r Simplification gives r = /8 D C O A E B 4 cm 80 cm Area 80 4 = = 4746 cm 5. E = {,,,...4, 5 } n E 5 n(s) = 50 n E 5 PE n S S = ut 50 m 9 u h m / sec km / hr Speed= u = 40 km/hr 54. cot sec sec and 90, sec sec tan sec sec sec 0 sec, cos,, 0,, 60, 0, 60, 00 n n 55. T sin cos n T T sin cos sin cos T sin cos Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

12 NTSE FULL TEST- _0--07 PAGE NO. sin cos cos sin sin cos sin cos sin cos sin cos sin cos 56. T T sin cos cos sin T sin cos Option (A) is correct sin cos A 60 0 D B C 50 h 50 m h tan 60 AB AB h tan0 AB h AB tan h= 050 Distance b/w two planes = = 00 m 57. st three digits no s divisible by and having middle no 5 is nd No 5 rd No 455 4th No th No 754 6th No Clearly there will be six no s only 58. If x < 0 and 7 x 5x 65 x 5x 66 0 log x 5x 65 0 = (x )(x + 6) = 0 x =, 6 only x= 6 acceptable b/c x < 0 (given) 59. Let the polynomial f(x) It yields a remainder upon divisor by x i.e. f()=...() Corporate Office : Parus Lok Complex, Plot No.-6/7, Boring Road Crossing, Patna , Ph. No. : , 0, 54007

13 NTSE FULL TEST- _0--07 PAGE NO. 60. and also yields a remainder upon divisor by x i.e. f()=...() Now to find the remainder when f(x) is divided by (x )(x ) we can not use remainder theorem Let us use division algorithm Dividend = divisor quotient + remainder f(x) = (x ) (x ) q(x) +ax +b Put x=, f() =a + b a +b =...() put x=, f() =a + b a +b =...(4) by solving () and (4), we get a= and b= remainder=ax+b = x+ S S (C) 6. (C) 6.(A) 64. (C) 65. (C) 66. (C) 67. (B) 68. (D) 69. (C) 70.(C) 7. (C) 7. (D) 7. (A) 74. (A) 75. (C) 76. (C) 77.(B) 78. (D) 79. (C) 80. (C) 8. (B) 8. (C) 8. (C) 84.(A) 85. (D) 86. (A) 87. (A) 88. (C) 89. (B) 90. (B) 9.(C) 9. (D) 9. (C) 94. (B) 95. (B) 96. (A) 97. (A) 98.(A) 99. (A) 00. (D) Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7