AITS Test 02 Answer Key & Solutions (Physics, Chemistry, Botany, Zoology, 2023-24) PDF

All India Test Series (2023-24) AITS TEST - 02 DURATION DURATION:: 200 90 Minutes Minutes DATE: 24/12/2023 M. MARKS: 720 ANSWER KEY...

All India Test Series (2023-24) AITS TEST - 02 DURATION DURATION:: 200 90 Minutes Minutes DATE: 24/12/2023 M. MARKS: 720 ANSWER KEY (PHYSICS) SECTION-A 1. (3) 8. (1) 15. (4) 22. (2) 29. (4) 2. (3) 9. (2) 16. (4) 23. (4) 30. (2) 3. (1) 10. (1) 17. (1) 24. (1) 31. (1) 4. (4) 11. (1) 18. (2) 25. (2) 32. (2) 5. (3) 12. (2) 19. (1) 26. (4) 33. (1) 6. (3) 13. (3) 20. (3) 27. (1) 34. (1) 7. (3) 14. (4) 21. (2) 28. (3) 35. (4) SECTION-B 36. (4) 39. (3) 42. (4) 45. (4) 48. (1) 37. (4) 40. (4) 43. (3) 46. (1) 49. (2) 38. (1) 41. (3) 44. (1) 47. (4) 50. (1) (CHEMISTRY) SECTION-A 51. (2) 58. (4) 65. (1) 72. (1) 79. (2) 52. (3) 59. (1) 66. (3) 73. (1) 80. (1) 53. (2) 60. (2) 67. (4) 74. (2) 81. (1) 54. (4) 61. (4) 68. (2) 75. (4) 82. (3) 55. (1) 62. (4) 69. (2) 76. (3) 83. (2) 56. (2) 63. (1) 70. (1) 77. (1) 84. (3) 57. (3) 64. (2) 71. (4) 78. (1) 85. (1) SECTION-B 86. (2) 89. (2) 92. (2) 95. (3) 98. (3) 87. (2) 90. (1) 93. (2) 96. (3) 99. (3) 88. (2) 91. (1) 94. (3) 97. (1) 100. (2) (BOTANY) SECTION-A 101. (2) 108. (1) 115. (4) 122. (1) 129. (2) 102. (4) 109. (3) 116. (2) 123. (2) 130. (1) 103. (4) 110. (2) 117. (1) 124. (4) 131. (4) 104. (1) 111. (2) 118. (2) 125. (1) 132. (4) 105. (3) 112. (1) 119. (4) 126. (2) 133. (1) 106. (3) 113. (3) 120. (4) 127. (3) 134. (1) 107. (4) 114. (3) 121. (4) 128. (3) 135. (3) SECTION-B 136. (1) 139. (2) 142. (1) 145. (4) 148. (3) 137. (2) 140. (3) 143. (4) 146. (2) 149. (4) 138. (3) 141. (4) 144. (4) 147. (1) 150. (1) (ZOOLOGY) SECTION-A 151. (2) 158. (4) 165. (2) 172. (1) 179. (3) 152. (1) 159. (1) 166. (3) 173. (1) 180. (2) 153. (1) 160. (4) 167. (4) 174. (1) 181. (3) 154. (3) 161. (3) 168. (3) 175. (4) 182. (1) 155. (3) 162. (2) 169. (4) 176. (3) 183. (3) 156. (2) 163. (4) 170. (3) 177. (2) 184. (3) 157. (3) 164. (3) 171. (3) 178. (4) 185. (1) SECTION-B 186. (4) 189. (3) 192. (2) 195. (1) 198. (2) 187. (2) 190. (2) 193. (3) 196. (2) 199. (1) 188. (2) 191. (2) 194. (1) 197. (2) 200. (4)  Hints and Solution (PHYSICS) SECTION-A 8. (1) 1. (3) mv 2 T= 3= x 3y + 5 r x 5 0.36 × v 2 y = − ⇒ 36 = 3 3 2.25 1 ⇒ v = 15 m/s Slope = 3 (NEW NCERT 11th Page No. 63, 64) 2. (3) 9. (2) X2 – Y2, subtraction is not possible because of Angular velocity of particle P about point E, different dimensions. 4 4 (NEW NCERT 11th Page No. 7, 8) ωE= = rEF 2r 3. (1) Here angle is taken from displacement axis. So angle from time axis is 90° – 60° = 30° 1 Now v = tan 30° = 3 Angular velocity of particle P about point C, (NEW NCERT 11th Page No. 14) 4 4 ωC= = rFC r 4. (4) ωE 4 / 2r 1 R u 2 sin 2θ × g 2 g Ratio= =. = = cot θ ωC 4/r 2 T2 g.4u 2 sin 2 θ 2 (NEW NCERT 11th Page No. 41, 42) i.e., gT2 = 2R tan θ If T is tripled, then R becomes 9 times. 10. (1) (NEW NCERT 11th Page No. 39, 40) 0.04 × 15iˆ + 0.06 × 15 ˆj + 0.1 × 15kˆ Vcm = 5. (3) 0.2 (NEW NCERT 11th Page No. 99) W= 1 2 ( 1 ) ( ) k x22 − x12 = × 5 × 103 302 − 152 × 10−4 2 = 168.75 J 11. (1) (NEW NCERT 11th Page No. 80) ( a / 2, 3a / 2) 6. (3) 3 dy 2 y = ( t − 5) ⇒ v == 3( t − 5) dt At = t 0; = v1 75 m/s and at t = 10 sec, v2 = 75 m/s So, change in kinetic energy = Work done m1 x1 + m2 x2 + m3 x3 X cm = 1 1 m1 + m2 + m3 = mv22 − mv12 = 0 2 2 a 1× 0 + 2 × a + 3 × (NEW NCERT 11th Page No. 76, 77) = 2 = 7a 1+ 2 + 3 12 7. (3) (NEW NCERT 11th Page No. 96) P = 2mE. If E are same then P ∝ m P1 m1 9 3 ⇒ = = = P2 m2 25 5 (NEW NCERT 11th Page No. 74, 75) 12. (2) 17. (1) at = 2.5 m/s2 v = 50 m/s r = 1000 m v2 Radial acceleration, ar = r 2T = µ× 2mg 50 × 50 25 T = m1 g a= r = = 2.5 m/s2 1000 10 2m1 g = µ × 2mg The acceleration of the car is 2 2 2 m1 = µ× m = 0.4m = m =a at + ar 5 5 (NEW NCERT 11th Page No. 60) = 2 × 2.5 = m/s2 2 (NEW NCERT 11th Page No. 41, 42) 18. (2) m1 = 20 kg; m2 = 40 kg; F = 300 N ; a1 = 14ms −2 13. (3) For mass m1; F1 = m1a1 = 20 × 14 = 280 N Plastic body is defined as a body which cannot For mass m2; regain its shape after deforming force is removed. 300 – F1 = 40 × a2 Elastic limit is the limit beyond which property of elasticity vanishes. 300 – 280 = 40 × a2 (NEW NCERT 11th Page No. 169) 20 1 −2 a= 2 = ms 40 2 14. (4) (NEW NCERT 11th Page No. 54, 65) 19. (1) p2 m2v 2 ( M 2 ) ( LT −1 )2 = = = M ° L°T ° ∆L F Strain= = S m 2 rg m 2 rg ( M 2 ) L1 ( L1T −2 ) L AY S1 F1 A2Y2 (NEW NCERT 11th Page No. 7) = = 1 {= & Y1 Y2 } 2 , A1 A2=  F1 F= S2 A1Y1 F2 15. (4) S1 1 ∴ = S2 1 (NEW NCERT 11th Page No. 170) 20. (3) V2 7x + 6 y +1 =0 For negotiating curve without friction tan θ = rg 7x + 6 y =−1 =V rg tan θ 7x 6 y + =1 = 20 m/s −1 −1 72 km/h x y Intercept form + 1 = (NEW NCERT 11th Page No. 63) −1 / 7 −1 / 6 16. (4) 21. (2)  Work done on pushing the block is. = 4iˆ + 5 ˆj A  W = mgh (as no work done against friction) B = 2iˆ + 2 ˆj + 4kˆ 15 × 9.8 × 5   ⇒ W = 735 J A + B = 6iˆ + 7 ˆj + 4kˆ   (NEW NCERT 11th Page No. 70) A+ B 6iˆ + 7 ˆj + 4kˆ   = A+ B 36 + 49 + 16 6iˆ + 7 ˆj + 4kˆ = 101 (NEW NCERT 11th Page No. 33) [ 3] 22. (2) 26. (4) If the body strikes the sand floor with a velocity v, mv 2 1 Maximum tension = then Mgh = M v 2 r 2 Let F be the resisting force acting on the body. 8 × 242 ⇒ = 64 N Then, the resultant force = F – Mg 72 Using work-energy theorem (NEW NCERT (Page No. 63) 1 ( F − Mg ) x = M v2 27. (1) 2 ( F − Mg ) x =M gh The position vector of COM of the three particles is given by h    ⇒ x=  m r + m2 r2 + m3r3 F rCOM = 1 1 −1 m1 + m2 + m3 Mg (NEW NCERT 11th Page No. 78) Substituting the values, we get 23. (4)  (2)(iˆ + 5 ˆj + kˆ) + (3)(iˆ + 2 ˆj + kˆ) + (4)(2iˆ + ˆj − 2kˆ) rCOM = 1 1 2+3+ 4 m v2 = µ mg. x + kx 2 2 2  13iˆ + 20 ˆj − 3kˆ rCOM = 1 1 9 = × 2 × v 2 ( 0.5 )( 2 )(10 )( 8 ) + ( 4.5 ) 64 2 2 (NEW NCERT 11th Page No. 96) v 2 = 80 + 4.5 × 32 = 224 28. (3) v  15 m/s Given, (NEW NCERT 11th Page No. 73) I = 30 kgm2 Kr = 1500 J 24. (1) The rotational kinetic energy is given by Relation between potential energy and force is  1 2 dU ˆ dU ˆ dU ˆ K=r Iω F= − i− j− k 2 dx dy dz 1 Given U (x, y) = sin (x – y), 1500= × 30 × ω2  2 ⇒ F= − cos ( x − y ) iˆ + cos ( x − y ) ˆj ω2 =100   −π  ˆ  −π  ˆ ⇒ F π − cos  =  i + cos  j ω =10 rad/sec (O , )  2   2  2 2πf =10 rad/sec  ⇒ F π 0iˆ + 0 ˆj = 10 5 (O , ) f = Hz = Hz 2 2π π  0N ⇒ F = (NEW NCERT 11th Page No. 114, 115) (NEW NCERT 11th Page No. 77) 29. (4)      25. (2) Formula used: L = r × P = m(r × v ) Since ac = p3rt 2  Given: r = (iˆ + kˆ), v = 3(iˆ − ˆj + kˆ), m = 2 kg 2 v Substituting the values, we get = p3rt 2 r = 2[(iˆ + kˆ)] × 3[(iˆ − ˆj + kˆ)] v 2 = p 3 r 2t 2  =L 2(3iˆ − 3kˆ) 3/2 v= p rt L = 6 kg m2/s mdv (NEW NCERT 11th Page No. 106) F = = mp3/2 r dt Power = F × v = ( mp3/2r ) × ( p3/2rt ) = mp3r 2t (NEW NCERT 11th Page No. 83) [ 4] 30. (2) 1 2 1 KH = =mvx mu 2 cos 2 θ 1 2 2 Ki = mv 2 2 K = K × cos 2 45° = 1 2 Kf = m(v + 3) 2 2 (NEW NCERT 11th Page No. 72, 74) K f = 3K i 35. (4) 1 1 2 m(v + 3) 2 = mv 1 2 2 2 Potential energy U = kx i.e. U ∝ x 2 2 On solving we get, This is a equation of parabola, so graph between U 3( 3 + 1) v= and x is a parabola. 2 (NEW NCERT 11th Page No. 80) (NEW NCERT 11th Page No. 75, 75) SECTION-B 31. (1) 36. (4) MR 2 MR 2 IResultant = + ⇒ h1 = e 2 h 2 4 3MR 2 h1 (0.4) 2 ×= ⇒ = 2 0.32 m = (NEW NCERT 11th Page No. 84) 4 (NEW NCERT 11th Page No. 115, 116) 37. (4) 32. (2) In translation, v1 = 2 gh Mgl 20 × 10 × 20 2 gh ∆l = = = v2 < v1 2 AY 2 × 0.4 × 2 × 1011 In Rolling,  K2  4 × 104 1 + 2  = × 10−11  R  2× 4× 2 (NEW NCERT 11th Page No. 114) = 0.25 × 10–7 = 25 × 10–9 m 38. (1) (NEW NCERT 11th Page No. 170, 171)      (A) 4 | B |2 + | B |2 +2 × 2 | B | × | B | cos θ =| B | 33. (1)    5 | B |2 +4 | B |2 cos θ =| B | Let I be the moment of inertia of the body. Then 1 2 1 2 5 + 4cos θ =1 total KE= mv + I ω 2 2 4 cosθ = – 4 1 2 1 v2  v cosθ = – 1 or= KE mv + I 2= , ω  θ = 180° 2 2 R  R     according to energy conservation loss in KE = gain (B) | F1 + F2 |=| F1 | + | F2 | = 10 + 16 = 26 in PE.     = (C) A.B | A || B | cos θ 1 I  2  7v 2    or m + v mgh = mg =   A.B   2  R2   10 g  cos θ =       | A || B |  2 solving this, we get I = mR 2 (iˆ + ˆj ).(5kˆ) 5 cos θ = = 0 2 ×5 i.e., the body is a solid sphere (NEW NCERT 11th Page No. 114) θ = 90°   (D) A + B =iˆ + 2 ˆj + 4kˆ 34. (1)   | A+ B| = 21 1 Initial kinetic energy is K = mu 2 (NEW NCERT 11th Page No. 34, 72, 33) 2 The velocity at highest point=vx u cos θ Kinetic energy of a particle at highest point [ 5] 39. (3) 42. (4) Option A is incorrect. If the velocity is zero at an instant, then acceleration may not necessarily be zero at that instant. For e.g. when an object is thrown vertically upward, at the highest point the velocity is zero but acceleration on it due to gravity is non-zero. Option B is incorrect because if the velocity and   acceleration are in opposite direction then the  r  F  (2iˆ  3 ˆj )  ( kˆ)  3iˆ  2 ˆj object will slow down. =  (3iˆ  2 ˆj ) Option C is incorrect because if the position and (NEW NCERT 11th Page No. 106) velocity have opposite sign the particle will move towards origin. 43. (3) (NEW NCERT 11th Page No. 14, 15) ∆V P  ∆V −0.04  = −  =  40. (4) V B V 100  P = 200 atm = 200 × 1.01 × 105 Pa or N/m2 100 So B = 200 × 1.01 × 105 × = 5.05 × 1010 N/m2 0.04 (NEW NCERT 11th Page No. 173) 44. (1) dx x =3t ⇒ u x = =3 Form diagram: F. B. D dt dy y = 4t 2 ⇒ u y = = 8t dt u y 8t tan= θ = ux 3 ∴ differentiating with respect to time we get, ∑ Fy = 0 (as body moves in horizontal direction) (sec 2 θ ) dθ 8 = dt 3 (NEW NCERT 11 Page No. 65) th  64  d θ 8 ( or 1 + tan 2 θ ) dθ 8 = dt 3 ; or 1 + t 2   9  dt 3 = 41. (3) 4R 8 CM of uniform semicircular disc of radius R = dθ 3 dθ 3π or = ; at t = 3 s is dt 1 + 64 t 2 dt from the center 9 2R CM of uniform semicircular ring of radius R = dθ 8 π = rad/s dt 195 from the center (NEW NCERT 11th Page No. 35) 3R CM of solid hemisphere of radius R = from the 8 45. (4) center F ∆L R = Y× CM of hemisphere shell of radius R = from the 4A 3L 2 4 AY center =F ∆L 3L (NEW NCERT 11th Page No. 97) 4 AY Comparing with F= K ∆L , we get K = 3L (NEW NCERT 11 Page No. 170) th [ 6] 46. (1) 49. (2) Case (i) when acceleration in lift is upward. N= 1 W + Fs N =1 mg + ma Case (ii) when acceleration in lift is downward. N 2 =W − ma =mg − ma N1 3 ∴ = N2 1 mg + ma 3 (τ0 ) net = 0 = mg − ma 1 a  a ∴ N  − x  + Fa = mg mg + ma = 3mg − 3ma 2  2 4ma = 2mg a  mga mga or N  − x + = ∴ a= g 2  4 2 2 a  mga (NEW NCERT 11th Page No. 65) N  − x = 2  4 But N = mg 47. (4) a mga 1 1 1 k1k2 12k 2 12k ∴ mg − mgx = = + = or k = = 2 4 k k1 k2 k1 + k2 7k 7 a a 4a − 2a a or x = − = = (NEW NCERT 11th Page No. 59) 2 4 8 4 (NEW NCERT 11th Page No. 110) 48. (1) 50. (1) Force F = – kx = –20x At x = 15 cm =.15 m F = – 20 ×.15 = – 3 −3 Acceleration a = = −30 m/s2.1 (NEW NCERT 11th Page No. 54) Thickness = ( R2 − R1 ) ± (∆R1 + ∆R2 ) ⇒ (3.79 − 2.23) ± (0.02 + 0.01) ⇒ 1.56 ± 0.03 (NEW NCERT 11th Page No. 55) (CHEMISTRY) SECTION-A 52. (3) 51. (2) 40.6 = x (40) + (1 – x)41 Total rupees = 10 × 6.022 × 1023 = 6.022 × 1024 40.6 = 40x + 41– 41x if we spend Rs. 1,00,000 in 1 sec, x = 0.4 then Rs 6.022 × 1024 will be spent in (6.022 × 1 – x = 0.6 1024/1,00,000) = 6.022 × 1019 sec. Abundance of Z40 in nature is 40% and Abundance 6.022 × 1019 sec = 6.022 × 1019/(60×60×24×365) = of Z41 in nature is 60% 1.909 × 1012 years (NEW NCERT 11th Part-I Page. No. 17) (NEW NCERT 11th Part-I Page. No. 8) [ 7] 53. (2) 61. (4) O Δp= m × Δv 0.014 Δp = 9.1×10−28 × 4.0 ×105 × 100 this compound cannot show tautomerism. Δp 509.6 ×10−28 = (NEW NCERT 11th Part-II Page. No. 270) h 6.626 ×10−27 Hence, = Δx = Δp × 4π 509.6 ×10−28 × 4 × 3.14 54. (4) Refer theory. Δx = 0.010 cm (NEW NCERT 11th Part-I Page. No. 23) (NEW NCERT 11th Part-I Page. No. 50) 55. (1) 62. (4) Let mass of oxide = 100gm Alkyl halide cannot exhibit functional isomerism as Mass of oxygen = 26gm the functional group remain same i.e. halogens. Mass of metal = 74gm (NEW NCERT 11th Part-II Page. No. 270) Moles of oxygen = 26/16 = 1.6 Moles of metal = 74/23 = 3.2 63. (1) Empirical formula will be M2O. Z12 Z22 E n = –13.6 = –13.6 (NEW NCERT 11th Part-I Page. No. 19) n12 n 22 1 9 n 22 n2 56. (2) = ⇒ =9⇒ =3 CH2OH OCH3 n12 n 22 n12 n1 and (NEW NCERT 11th Part-I Page. No. 46) These are functional isomers. 64. (2) (NEW NCERT 11th Part-II Page. No. 270) Given: n1 = 1 and n 2 = 6. We know, Total number of spectral line 57. (3) ( n 2 - n1 )( n 2 - n1 +1) ROH + CH 3 MgBr → CH 4 + Mg ( OR )( Br ) = 2 22400 mL of CH 4 produced at STP So, total number of spectral lines = 1 mol of ROH = (= 6 − 1)( 6 − 1 + 1) 15 2.24 2 2.24 mL of CH 4 produced at STP  mo = l 22400 (NEW NCERT 11th Part-I Page. No. 44) 5.6 × 10 –3 Now molecular mass of ROH = 65. (1) 10−4 C4H10 → n-butane and isobutane are two chain ∴ mol mass of ROH = 56 isomers. (NEW NCERT 11th Part-I Page. No. 21) C5H12 → n-pentane, isopentane and neopentane are three chain isomers. 58. (4) (NEW NCERT 11th Part-II Page. No. 270) All the three structures are isomers of benzene. (NEW NCERT 11th Part-II Page. No. 270) 66. (3) 59. (1) hc hc hc = + 0.5 λ λ1 λ 2 Conc.(in ppm) = ×106 = 833.33ppm 600 1 1 1 (NEW NCERT 11th Part-I Page. No. 21) = + λ λ1 λ 2 60. (2) 1 1 1 = + CaCO3 → CaO + CO2 300 500 λ 2 100 gm of CaCO3 produces 22.4L at S.T.P. 10 gm of CaCO3 produces 2.24L CO2. λ 2 = 750 nm (NEW NCERT 11th Part-I Page. No. 21) (NEW NCERT 11th Part-I Page. No. 41) 67. (4) 73. (1) As we know, n₁ = 1 2K2CrO4 + 2HCl → 1K2Cr2O7 + 2KCl + 1H2O Number of lines = (n2 – n1 + 1) (n2 – n1)/2 =6 x + y –z = 2 n = 4. (NEW NCERT 11th Part-II Page. No. 246) Three lines are observed in the ultraviolet region of the spectrum. These corresponds to the transitions 4 74. (2) →1,3→1,2→1 So, number of lines in U.V. (Lyman +1 −2 +2 0 0 +4 Cu 2 S + 2 Fe O → 2 Cu + 2 Fe + S O 2 series) spectrum = 3. (NEW NCERT 11th Part-I Page. No. 44) FeO reduced to Fe and in Cu2S, Cu+1 reduced to Cu and S-2 oxidised to SO2. 68. (2) (NEW NCERT 11th Part-II Page. No. 237) According to Hund's rule electrons are distributed among the orbitals of subshell in such a way so as to 75. (4) give maximum number of unpaired electron with For an adiabatic process, work done is given by- parallel spin. i.e. in a subshell pairing of electron W = nCV dT will not start until and unless all the orbitals of that As the gas is monoatomic, subshell will get one electron each with same spin. 3 So, only Mn can have one unpaired electron due to CV = R 2 odd number of electrons. 3 (NEW NCERT 11th Part-I Page. No. 62) ∴ W =nRdT 2 69. (2) 3 100 = × 0.20 × 2 × ( T − 530 ) The process is an isothermal expansion. 2 Hence, q = − W 100 ⇒ T − 530 = ΔE = 0 0.6 V2 ⇒T= 696.7K W = −2.303nRTlog V1 (NEW NCERT 11th Part-II Page. No. 145)  500  76. (3) = –2.303 × 0.05 × 8.314 × 320 × log    100  2KMnO 4 + 5C2 O 2- + 4 + 16H → 2Mn 2+ + 2K + + 10CO 2 + 8H 2 O = −214.4 J KMnO 4 + 5Fe 2+ + 8H + → K + + 5Fe3+ + Mn 2+ + 4H 2 O q= +214.4, W = −214.4 (expansion work) (NEW NCERT 11th Part-I Page. No. 142) From the above equations, it is observed that 3 moles of KMnO4 will be needed to react completely 70. (1) with five moles of ferrous oxalate in acidic solution. Value of m = –1 represents one orbital. No. of equivalents of KMnO4 = No. of equivalents Therefore, maximum number of electrons will be of FeC2O4 two. Equivalents of KMnO4 = n x no. of moles = 5 × x (NEW NCERT 11th Part-I Page. No. 62) Equivalents of FeC2O4 = 3 × 1 = 3 5x = 3 71. (4) 3 x= +2 +2 +1 +4 5 2 Hg Cl2 + Sn Cl2 → Hg 2 Cl2 + Sn Cl4 (NEW NCERT 11th Part-II Page. No. 237) Hg undergoes reduction while Sn undergoes oxidation. 77. (1) (NEW NCERT 11th Part-II Page. No. 237) Refer Theory. (NEW NCERT 11th Part-I Page. No. 146) 72. (1) +7 +2 10I – + 16 H + + 2 Mn O 4– → 5I 2 + 2 Mn+ 8H 2 O (NEW NCERT 11th Part-II Page. No. 237) 78. (1) SECTION-B O O 86. (2) 5+ Cr For metal oxide M2Ox, O O Let weight of oxide = 100gm O weight of metal = 60gm KO OK OK weight of oxygen = 40gm Gram Eq. of oxygen = Gram Eq. of metal (NEW NCERT 11th Part-II Page. No. 239) 40 60 MW = ⟹ x= 16 / 2 MW / x 12 79. (2) For metal chloride MClx, The priority of –CN group is more than –CHO MW + x(35.5) = 95 group. On putting the value of x in the above equation, we (NEW NCERT 11th Part-II Page. No. 267) get MW = 24 (NEW NCERT 11th Part-I Page. No. 16) 80. (1) Internal energy is an extensive property and also a 87. (2) state function. PbO + 2HCl → PbCl2 + H₂O (NEW NCERT 11th Part-I Page. No. 144) Atomic mass of PbO = 207 + 16 = 223 Atomic mass of HCl = 36.5 81. (1) From the above chemical equation, we see that Propane-1,2,3-tricarboxylic acid is the correct HCl is the limiting reagent. IUPAC name of the given compound. 1 mole of PbO gives 1 mole of PbCl2 (NEW NCERT 11th Part-II Page. No. 267) 223 gm of PbO forms 1 mole of PbCl2. 3 gm of PbO forms 3/223 = 0.013 mole of PbCl2 82. (3) (NEW NCERT 11th Part-I Page. No. 21) 1,6-Dimethyl cyclohexene is the correct IUPAC 88. (2) name of compound given in option 3. For ABC5, (NEW NCERT 11th Part-II Page. No. 265) Overall charge = 1 + 4 + 5(–1) = 0 (NEW NCERT 11th Part-I Page. No. 17) 83. (2) In (2), 3 moles of gas becomes 1 mole of liquid. If 89. (2) greater number of molecules transforms into a small Meso-3,4-dibromohexane will have one number of molecules then entropy decreases. Also configuration as R and other S. in (2) gas molecules changes to liquid, hence (NEW NCERT 11th Part-II Page. No. 271) entropy decreases. Hence (2) has the most negative change in entropy 90. (1) (NEW NCERT 11th Part-I Page. No. 158) Z -1 Velocity of electron, v n = 2.18×106 ms (Z = 1 n 84. (3) for H ) Sulphonic group is more prior as compared to other The distance travelled by the electron in third group so 6-Methyl-4-oxohept-6-ene-1-sulphonic Bohr's orbit in 4 × 10–7s acid is the correct IUPAC name of given compound. 2.18 ×106 × 4.0 ×10−7 (NEW NCERT 11 Part-II Page. th = No. 267) = m 0.29 m 3 The circumference of third orbit 85. (1) NH2  n2  = 2πr = 2π ×  0.529×   z  OH  9 = 2π  0.529×  = 29.89×10-10 m 6-Aminocyclohex-2-ene-1-ol.  1 (NEW NCERT 11th Part-II Page. No. 265) 0.29 = Number of revolutions = 9.7 ×107 29.89 × 10−10 (NEW NCERT 11th Part-I Page. No. 46) 91. (1) 96. (3) In fully eclipsed conformation, the steric strain is In cyclic process, ∆U=0 maximum. Hence this conformation is most ∆U = Q + W unstable. This is because of the repulsion between Q = –W methyl-methyl group which are very close together. Q = Area under the circle = πr2 = π102 (NEW NCERT 11th Part-II Page. No. 271) (NEW NCERT 11th Part-I Page. No. 141) 92. (2) 97. (1) n2 Alcohol oxidized with potassium dichromate to Radius of nth Bohr’s orbit, rn = 0.529 form acetic acid which is indicated by change in z (NEW NCERT 11th Part-I Page. No. 46) colour. (NEW NCERT 11th Part-II Page. No. 247) 93. (2) Entropy of the system increases with increase in 98. (3) temperature. This is because when the temperature Ethyl-2-oxocyclohexanecarboxylate is the correct increases, the randomness of the molecule increases. IUPAC name of the given compound. (NEW NCERT 11th Part-I Page. No. 158) (NEW NCERT 11th Part-II Page. No. 267) 94. (3) 99. (3) Given that, v1 = 3v 2 pVγ = constant...(i) 1 hc  RT  γ m ( 3v 2 ) = - φ 2 For case -1,   V = constant 2 λ1  V  1 hc ⇒ TVγ −1 = constant m ( v2 ) = – φ 2 For case -2, 1/3 (given) or TV –1/3 2 λ2 But T ∝ (V) = constant hc 9hc On equating, ⇒ -φ = - 9φ 1 λ1 λ2 γ −1 = 3 9hc hc 4 ⇒ 8φ = − or γ= = 1.33 λ2 λ1 3 1  1240 1240  Cp ⇒ φ= 9× –  ⇒ =1.33 8 500 250  Cv ⇒φ =2.17eV (NEW NCERT 11th Part-I Page. No. 145) (NEW NCERT 11th Part-I Page. No. 41) 100. (2) 95. (3) 5-Bromo-2-chloromethyl-1-fluoro-1-iodopent-1- +2 −1 +6 +3 ene is the correct IUPAC name of the given FeS2 → SO 24− + Fe compound. ( n-factor = [1× ( 3 − 2 ) + 2 6 − ( −1)  = 15 (NEW NCERT 11th Part-II Page. No. 267) M So, eq. wt. of FeS2 = 15 (NEW NCERT 11th Part-I Page. No. 23) (BOTANY) SECTION-A the same time hundreds of characters can be 101. (2) considered. Cytotaxonomy that is based on Numerical taxonomy which is now easily carried out cytological information like chromosome number, using computers is based on all observable structure, behaviour and chemotaxonomy that uses characteristics. Number and codes are assigned to all the chemical constituents of the plant to resolve the characters and the data are then processed. In this confusions, are also used by taxonomists these days. way each character is given equal importance and at (NEW NCERT 11th Page No. 24) 102. (4) 106. (3) The golgi apparatus principally performs the Pteridophytes are used for medicinal purposes and function of packaging materials, to be delivered as soil-binders. Marchantia is a bryophyte. either to the intra-cellular targets or secreted outside (NEW NCERT 11th Page No. 28, 30) the cell. Materials to be packaged in the form of vesicles from the ER fuse with the cis face of the 107. (4) golgi apparatus and move towards the maturing The major pigments in the class of algae that store face. A number of proteins synthesised by their food in the form of floridean starch i.e., red ribosomes on the endoplasmic reticulum are algae are chlorophyll a, d and phycoerythrin. modified in the cisternae of the golgi apparatus (NEW NCERT 11th Page No. 27) before they are released from its trans face. Golgi apparatus is the important site of formation of 108. (1) glycoproteins and glycolipids. A flower is asymmetric (irregular) if it cannot be In animal cells, lipid-like steroidal hormones are divided into two similar halves by any vertical plane synthesised in Smooth Endoplasmic Reticulum passing through the centre, as in canna. (SER). (NEW NCERT 11th Page No. 62) (NEW NCERT 11th Page No. 95, 96) 109. (3) 103. (4) In parietal placentation, the ovules develop on the The gametes of brown algae (phaeophyceae) are inner wall of the ovary or on peripheral part. Ovary pyriform (pear-shaped) and bear two laterally is one-chambered but it becomes two chambered attached flagella. due to the formation of the false septum, e.g., (NEW NCERT 11th Page No. 27) mustard and Argemone. (NEW NCERT 11th Page No. 65) 104. (1) The red thalli of most of the red algae are 110. (2) multicellular. Some of them have complex body China rose features: Alternate phyllotaxy, superior organisation. The food is stored as floridean starch ovary, twisted aestivation, monoadelphous stamens, which is very similar to amylopectin and glycogen axile placentation. in structure. Majority of the red algae are marine (NEW NCERT 11th Page No. 61, 64) with greater concentrations found in the warmer areas. They occur in both well-lighted regions close 111. (2) to the surface of water and also at great depths in Tap roots of carrot, turnip and adventitious roots of oceans where relatively little light penetrates. sweet potato, get swollen and store food. (NEW NCERT 11th Page No. 27) Axillary buds of stems may also get modified into woody, straight and pointed thorns. Thorns are 105. (3) found in many plants such as Citrus, Bougainvillea. Bryophytes include the various mosses and Underground stems of potato, ginger, turmeric, liverworts that are found commonly growing in Zaminkand, Colocasia are modified to store food in moist shaded areas in the hills. Bryophytes are also them. They also act as organs of perennation to tide called amphibians of the plant kingdom because over conditions unfavourable for growth. these plants can live in soil but are dependent on The stems of maize and sugarcane have supporting water for sexual reproduction. They usually occur in roots coming out of the lower nodes of the stem damp, humid and shaded localities. They play an called stilt roots. important role in plant succession on bare rocks/soil. (OLD NCERT 11th Page No. 67, 68) The plant body of bryophytes is more differentiated than that of algae. It is thallus-like and prostrate or 112. (1) erect, and attached to the substratum by unicellular Depending on whether the apex gets developed into or multicellular rhizoids. They lack true roots, stem a flower or continues to grow, two major types of or leaves. They may possess root-like, leaf-like or inflorescences are defined – racemose and cymose. stem-like structures. In racemose type of inflorescences the main axis (NEW NCERT 11th Page No. 29) continues to grow, the flowers are borne laterally in an acropetal succession. (NEW NCERT 11th Page No. 61) 113. (3) 120. (4) The outside of the epidermis is often covered with a waxy thick layer called the cuticle which prevents the loss of water. Cuticle is absent in roots. Stomata are structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stoma is composed of two beans shaped cells known as guard cells which enclose stomatal pore. The guard cells possess (NEW NCERT 11th Page No. 62) chloroplasts and regulate the opening and closing of stomata. Sometimes, a few epidermal cells, in the 114. (3) vicinity of the guard cells become specialised in In mango, the pericarp is well differentiated into an their shape and size and are known as subsidiary outer thin epicarp, a middle fleshy edible mesocarp cells. and an inner stony hard endocarp. (NEW NCERT 11th Page No. 72) (NEW NCERT 11th Page No. 66) 121. (4) 115. (4) In monocot stem, the phloem parenchyma is absent, In an isobilateral leaf, the stomata are present on and water-containing cavities are present within the both the surfaces of the epidermis; and the vascular bundles. mesophyll is not differentiated into palisade and (NEW NCERT 11th Page No. 76) spongy parenchyma. (NEW NCERT 11th Page No. 77) 122. (1) The internal tissue organisation in dicot root are as follows: 116. (2) The outermost layer is epiblema. Many of the cells In a dicot stem, the outer hypodermis, consists of a of epiblema protrude in the form of unicellular root few layers of collenchymatous cells just below the hairs. The cortex consists of several layers of thin- epidermis, which provide mechanical strength to the walled parenchyma cells with intercellular spaces. young stem. The innermost layer of the cortex is called (NEW NCERT 11th Page No. 75) endodermis. (NEW NCERT 11th Page No. 74) 117. (1) In grasses, certain adaxial epidermal cells along the 123. (2) veins modify themselves into large, empty, The floral diagram and floral formula in the figure colourless cells. These are called bulliform cells. below represent the mustard plant (Family: (NEW NCERT 11th Page No. 77) Brassicaceae/Cruciferae). 118. (2) In dicot stem, a large number of vascular bundles are arranged in a ring ; the ‘ring’ arrangement of vascular bundles is a characteristic of dicot stem. Each vascular bundle is conjoint, open, and with endarch protoxylem. (NEW NCERT 11th Page No. 76) (NEW NCERT 11th Page No. 67) 119. (4) The trichomes in the shoot system are usually 124. (4) multicellular. They may be branched or unbranched Formation of synaptonemal complex takes place and soft or stiff. They may even be secretory. The during zygotene. trichomes help in preventing water loss due to The beginning of diplotene is recognised by the transpiration. dissolution of the synaptonemal complex and the tendency of the recombined homologous (NEW NCERT 11th Page No. 72) chromosomes of the bivalents to separate from each other except at the sites of crossovers. (NEW NCERT 11th Page No. 126) 125. (1) variability in the population of organisms from one Ribosomes are the granular structures first observed generation to the next. under the electron microscope as dense particles by (NEW NCERT 11th Page No. 128) George Palade (1953). They are composed of ribonucleic acid (RNA) and proteins and are not 132. (4) surrounded by any membrane. The eukaryotic As we go higher from species to kingdom, the ribosomes are 80S while the prokaryotic ribosomes number of common characteristics goes on are 70S. Each ribosome has two subunits, larger and decreasing. smaller subunits. The two subunits of 80S ribosomes (NEW NCERT 11th Page No. 08) are 60S and 40S while that of 70S ribosomes are 50S and 30S. 133. (1) (NEW NCERT 11th Page No. 98) The vertical section of a dorsiventral leaf through the lamina shows three main parts, namely, 126. (2) epidermis, mesophyll and vascular system. The Chrysophytes, Dinoflagellates and Euglenoids are epidermis which covers both the upper surface photosynthetic protists. Slime moulds are (adaxial epidermis) and lower surface (abaxial saprophytic. Protozoans are heterotrophic. epidermis) of the leaf has a conspicuous cuticle. The (NEW NCERT 11th Page No. 14, 15) abaxial epidermis generally bears more stomata than the adaxial epidermis. The latter may even lack 127. (3) stomata. The tissue between the upper and the lower All tissues on the innerside of the endodermis such epidermis is called the mesophyll. Mesophyll, which as pericycle, vascular bundles and pith constitute the possesses chloroplasts and carry out photosynthesis, stele. is made up of parenchyma. It has two types of cells (NEW NCERT 11th Page No. 74) – the palisade parenchyma and the spongy parenchyma. 128. (3) (NEW NCERT 11th Page No. 76) Agaricus (mushroom) is a member of basidiomycetes. (NEW NCERT 11th Page No. 18) 134. (1) If gynoecium is situated in the centre and other parts 129. (2) of the flower are located on the rim of the thalamus The key features of metaphase are: almost at the same level, it is called perigynous. The Spindle fibres attach to kinetochores of ovary here is said to be half inferior, e.g., plum, rose, chromosomes. peach. When a flower can be divided into two Chromosomes are moved to spindle equator and get similar halves only in one particular vertical plane, aligned along metaphase plate through spindle it is zygomorphic, e.g., pea, gulmohur, bean, Cassia. fibres to both poles. The term syncarpous is a condition when carpels are (NEW NCERT 11th Page No. 123) fused, as in mustard and tomato. When stamens are attached to the petals, they are epipetalous as in 130. (1) brinjal, or epiphyllous when attached to the perianth Lion (Panthera leo), leopard (P. pardus) and tiger as in the flowers of lily. (P. tigris) with several common features, are all (NEW NCERT 11th Page No. 64, 65) species of the genus Panthera. Solanum tuberosum is the scientific name for potato. 135. (3) (NEW NCERT 11th Page No. 06) Schwann (1839), a British Zoologist, studied different types of animal cells and reported that cells 131. (4) had a thin outer layer which is today known as the Meiosis is the mechanism by which conservation of ‘plasma membrane’. He also concluded, based on specific chromosome number of each species is his studies on plant tissues, that the presence of cell achieved across generations in sexually reproducing wall is a unique character of the plant cells. organisms, even though the process, per se, (NEW NCERT 11th Page No. 88) paradoxically, results in reduction of chromosome number by half. It also increases the genetic  SECTION-B 142. (1) 136. (1) Unlike bryophytes and pteridophytes, in In pteridophytes, the female gametophytes are gymnosperms the male and the female retained on the parent sporophytes for variable gametophytes do not have an independent free- periods. The development of the zygotes into young living existence. embryos take place within the female gametophytes. The predominant stage of the life cycle of a moss is This event is a precursor to the seed habit considered the gametophyte which consists of two stages. The an important step in evolution. first stage is the protonema stage, which develops (NEW NCERT 11th Page No. 32) directly from a spore. It is a creeping, green, branched and frequently filamentous stage. The 137. (2) second stage is the leafy stage, which develops from Roots in some genera have fungal association in the the secondary protonema as a lateral bud. form of mycorrhiza (Pinus), while in some others (NEW NCERT 11th Page No. 30, 33) (Cycas) small specialised roots called coralloid roots are associated with N2- fixing cyanobacteria. The 143. (4) stems are unbranched (Cycas) or branched (Pinus, Next to endodermis in root lies a few layers of thick- Cedrus). walled parenchymatous cells referred to as (NEW NCERT 11th Page No. 32) pericycle. Initiation of lateral roots and vascular cambium during the secondary growth takes place in 138. (3) these cells. The pith is small or inconspicuous. The Sphagnum is a moss. parenchymatous cells which lie between the xylem The pteridophytes are classified into four classes: and the phloem are called conjuctive tissue. Psilopsida (Psilotum); Lycopsida (Selaginella, (NEW NCERT 11th Page No. 74) Lycopodium), Sphenopsida (Equisetum) and Pteropsida (Dryopteris, Pteris, Adiantum). 144. (4) (NEW NCERT 11th Page No. 30, 32) The innermost layer of the cortex is called endodermis. It comprises a single layer of barrel 139. (2) shaped cells without any intercellular spaces. The The outer covering of endosperm separates the tangential as well as radial walls of the endodermal embryo by a proteinous layer called aleurone layer. cells have a deposition of water-impermeable, waxy (NEW NCERT 11th Page No. 67) material suberin in the form of casparian strips. Next to endodermis lies a few layers of thick-walled 140. (3) parenchymatous cells referred to as pericycle. (Type of aestivation) (Example) (NEW NCERT 11th Page No. 74) Sepals or petals in a whorl just Calotropis touch one another at the margin, 145. (4) without overlapping. The sub-metacentric chromosome has centromere slightly away from the middle of the chromosome The margins of sepals or petals Cassia resulting into one shorter arm and one longer arm. overlap one another but not in any In case of acrocentric chromosome the centromere particular direction. is situated close to its end forming one extremely Five petals, the largest (standard) Bean short and one very long arm, whereas the telocentric overlaps the two lateral petals chromosome has a terminal centromere. (wings) which in turn overlap the (NEW NCERT 11th Page No. 101) two smallest anterior petals (keel). (NEW NCERT 11th Page No. 64) 146. (2) A virus is a nucleoprotein and the genetic material is 141. (4) infectious. When a flower can be divided into two equal radial (NEW NCERT 11th Page No. 20) halves in any radial plane passing through the centre, it is said to be actinomorphic, e.g., mustard, Datura, chilli. Maize is a monocot plant. (NEW NCERT 11th Page No. 62, 66) 147. (1) 149. (4) Some cells in the adult animals do not appear to The M Phase represents the phase when the actual exhibit division (e.g., heart cells) and many other cell division or mitosis occurs and the interphase cells divide only occasionally, as needed to replace represents the phase between two successive M cells that have been lost because of injury or cell phases. It is significant to note that in the 24 hour death. These cells that do not divide further exit G1 average duration of cell cycle of a human cell, cell phase to enter an inactive stage called quiescent division proper lasts for only about an hour. The stage (G0) of the cell cycle. Cells in this stage remain interphase lasts more than 95% of the duration of metabolically active but no longer proliferate unless cell cycle. called on to do so depending on the requirement of (NEW NCERT 11th Page No. 121) the organism. (NEW NCERT 11th Page No. 122) 150. (1) Fusion between one large, nonmotile (static) female 148. (3) gamete and a smaller, motile male gamete is termed Universal rules of nomenclature are as follows: oogamous, e.g., Volvox, Fucus. 1. Biological names are generally in Latin and (NEW NCERT 11th Page No. 24) written in italics. They are Latinised or derived from Latin irrespective of their origin. 2. The first word in a biological name represents the genus while the second component denotes the specific epithet. 3. Both the words in a biological name, when handwritten, are separately underlined, or printed in italics to indicate their Latin origin. 4. The first word denoting the genus starts with a capital letter while the specific epithet starts with a small letter. (NEW NCERT 11th Page No. 04) (ZOOLOGY) SECTION-A 154. (3) 151. (2) Myelin sheath is responsible for faster conduction of Mammals, many terrestrial amphibians and marine impulses through the nerve fibres. The impulses in fishes mainly excrete urea and are called ureotelic the myelinated neurons jump from node to node in a animals. transmission process called saltatory conduction. (NEW NCERT, Class11th, Page No.-205) Thus, the myelination of axons increases the speed of action potential. 152. (1) (NEW NCERT, Class11th, Page No.-232) Increase in Ca++ level leads to the binding of calcium with a subunit of troponin on actin filaments and 155. (3) thereby remove the masking of active sites for List-I List-II myosin. (NEW NCERT, Class11th, Page No.-222) The process of secreting Ammonotelism ammonia 153. (1) Marine fishes Ureotelic Blood is the medium of transport for O2 and CO2. About 97 per cent of O2 is transported by RBCs in Fresh water fishes Ammonotelic the blood. The remaining 3 per cent of O2 is carried in a dissolved state through the plasma. Nearly 20- Land snails Uricotelic 25 per cent of CO2 is transported by RBCs whereas [New NCERT, Class11th, Page No.-205] 70 per cent of it is carried as bicarbonate. About 7 per cent of CO2 is carried in a dissolved state through plasma. (NEW NCERT, Class11th, Page No.-189) 156. (2) 161. (3) Each half of pectoral girdle consists of a clavicle and Muscle contains a red coloured oxygen storing a scapula. Scapula is a large triangular flat bone pigment called myoglobin. Myoglobin content is situated in the dorsal part of the thorax between the high in some of the muscles which gives a reddish second and the seventh ribs. The dorsal, flat, appearance. Such muscles are called the Red fibres. triangular body of scapula has a slightly elevated On the other hand, some of the muscles possess very ridge called the spine which projects as a flat, less quantity of myoglobin and therefore, appear expanded process called the acromion. The clavicle pale or whitish. These are the White fibres. Number articulates with this. Below the acromion is a of mitochondria are also few in them, but the amount depression called the glenoid cavity which of sarcoplasmic reticulum is high. They depend on anaerobic process for energy. articulates with the head of the humerus to form the (NEW NCERT, Class11th, Page No.-223) shoulder joint. (NEW NCERT, Class11th, Page No.-226) 162. (2) During inspiration diaphragm and external 157. (3) intercostal muscles contract which results increase The first heart sound (lub) is associated with the in the thoracic volume and causes a similar increase closure of the tricuspid and bicuspid valves whereas in pulmonary volume. An increase in pulmonary the second heart sound (dub) is associated with the volume decreases the intra-pulmonary pressure to closure of the semilunar valves. less than the atmospheric pressure which forces the (NEW NCERT, Class11th, Page No.-200) air from outside to move into the lungs, i.e., inspiration. 158. (4) (NEW NCERT, Class11th, Page No.-185, 186) When a neuron is not conducting any impulse, i.e., resting, the axonal membrane is comparatively more 163. (4) permeable to potassium ions (K+) and nearly When a stimulus is applied at a site on the polarised impermeable to sodium ions (Na+). membrane, the membrane at that site becomes freely (NEW NCERT, Class11th, Page No.-232) permeable to Na+. This leads to a rapid influx of Na+ followed by the reversal of the polarity at that site, 159. (1) i.e., the outer surface of the membrane becomes Cardiac muscle tissue is a contractile tissue present negatively charged and the inner side becomes positively charged. The polarity of the membrane at only in the heart. Cell junctions fuse the plasma that particular site is thus reversed and hence membranes of cardiac muscle cells and make them depolarised. The electrical potential difference stick together. Communication junctions across the resting plasma membrane is called the (intercalated discs) in cardiac muscle at some fusion resting potential. points allow the cells to contract as a unit, i.e., when (NEW NCERT, Class11th, Page No.-232-233) one cell receives a signal to contract, its neighbours are also stimulated to contract. 164. (3) (Old NCERT, Class11th, Page No.-105) Protonephridia or flame cells are the excretory structures in Platyhelminthes (Flatworms, e.g., 160. (4) Planaria), rotifers, some annelids and the The efferent arteriole emerging from the glomerulus cephalochordate – Amphioxus. Nephridia help to forms a fine capillary network around the renal remove nitrogenous wastes and maintain a fluid and tubule called the peritubular capillaries. A minute ionic balance. vessel of this network runs parallel to the Henle’s (NEW NCERT, Class11th, Page No.-206) loop forming a ‘U’ shaped vasa recta. (NEW NCERT, Class11th, Page No.-208) 165. (2) Asbestosis, silicosis and Emphysema are example of respiratory disorders. (NEW NCERT, Class11th, Page No.-191) 166. (3) 171. (3) During shortening of the muscle, i.e., contraction, Loose connective tissue has cells and fibres loosely the ‘I’ bands get reduced, whereas the ‘A’ bands arranged in a semi-fluid ground substance, for retain the length. The Z-lines come together, and so example, areolar tissue present beneath the skin. Often it serves as a support framework for the sarcomere shortens as well. However, the A epithelium. It contains fibroblasts (cells that produce band does not change length during muscle and secrete fibres), macrophages and mast cells. In contraction, since its length is the same as the length the dense regular connective tissues, the collagen of the thick filaments. fibres are present in rows between many parallel (NEW NCERT, Class11th, Page No.-222) bundles of fibres. (Old NCERT, Class11th, Page No.-103) 167. (4) List-I List-II 172. (1) First vertebra is the atlas and it articulates with the I Depolarization A opening and then occipital condyles. The vertebral column is phase in the closing of Na+ differentiated into cervical (7), thoracic (12), lumbar generation of an channels (5), sacral (1-fused) and coccygeal (1-fused) regions action potential starting from the skull. II Repolarization phase B opening of (NEW NCERT, Class11th, Page No.-225) in the generation of potassium gates an action potential and rushing of 173. (1) potassium Another mass of nodal tissue is seen in the lower left corner of the right atrium close to the atrio- III Absolute refracting C The sodium ventricular septum called the atrio-ventricular node phase channel remained (AVN). Right Atrioventricular valve is tricuspid open valve whereas left atrioventricular valve is mitral or IV Resting phase D All voltage gated bicuspid valve. sodium and (NEW NCERT, Class11th, Page No.-198, 199) potassium channels are closed 174. (1) (NEW NCERT, Class11 , Page No.-232, 233) th Blood group AB is universal acceptor. Blood group A contains antigen A and anti-B antibodies. (NEW NCERT, Class11th, Page No.-195) 168. (3) Bones have a hard and non-pliable ground substance 175. (4) rich in calcium salts and collagen fibres which give There are eight cranial bones i.e., parietal (2), bone its strength. Cells of cartilage are enclosed in temporal (2), frontal (1), occipital (1), ethmoid (1) small cavities within the matrix secreted by them. and sphenoid (1). Zygomatic and lacrymal bones are (Old NCERT, Class11th, Page No.-104) the facial bones whereas Ischium is a bone of pelvic girdle. 169. (4) (NEW NCERT, Class11th, Page No.-224) Conditional reabsorption of Na+ and water takes 176. (3) place in the distal convoluted tubule. Vasa recta are a series of straight capillaries in the (NEW NCERT, Class11th, Page No.-209) medulla. They lie parallel to the loop of Henle. The Henle’s loop and vasa recta play a significant role in 170. (3) producing concentrated urine. The opening between the right atrium and the right (NEW NCERT, Class11th, Page No.-210-211) ventricle is guarded by a valve formed of three muscular flaps or cusps, the tricuspid valve, whereas 177. (2) a bicuspid or mitral valve guards the opening The dorsal portion of the midbrain consists mainly between the left atrium and the left ventricle. of four round swellings (lobes) called corpora (NEW NCERT, Class11th, Page No.-198) quadrigemina.