# AIPMT Mock Test Hints & Solutions - Sri Chaitanya

Sri Chaitanya The Final Word in JEE/MedicalFoundation

HINTS & SOLUTION AIPMT MOCK TEST SECTION - I (PHYSICS) 1.

Ans. (3) The velocity-time graph for the given situation can be drawn as below. Magnitudes of slope of OA = f f and slope of BC  2 Y -1

v(ms )

O

A

B

C t

t1 D

t (s)

t2

X

E

f v  f t1  t 2 2 t 2  2t1

2.

1 2 In graph area of OAD ; s  f t1 2 From area of rectangle ABED ; s 2  (f t1 )t 1f (2t1 )2 Distance travelled in time t2 ; s3  22 or s  (f t1 )t  f t12  15s  s1  s 2  s3  15s 1 2  or (f t1 )  12s  s  f t1  2 12s (f t1 )t  t 1 or s or t1  and we get (f t1 )t1 6 2 2 1  t 1 2 s f   ft . 2  6 72 Ans. (1)  v1  5iˆ

3. 4.

5.

 v 2  5ˆj      v  v 2  v1  5ˆj  5iˆ or |  v | 5 2  1 | v | 5 2  ms 1  a  2 t 10 5 For direction, tan     1 5 1 ms 2 towards northAverage acceleration is 2 west. Ans. (1) P  mv  3.513  5.00  17.6kg  ms 1 . Ans. (3) This is the question based on impulse-momentum theorem. | F  t || change in momentum|  F  0.1 | Pf  Pi | As the ball will stop after catching; Pi  mvi  0.15  20  3  Pf  0  F  0.1  3  F  30N .

Ans. (4) This is the equilibrium of coplanar force. Hence, Fx  0  F N  Fy  0 f  mg  c  0

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6.

7.

Mocktest/AIMPT

   N  f  0  Since,  f  0   N  0 . Ans. (4) Approximate mass = 60 kg Approximate velocity = 10 ms–1 1 Approximate KE   60  100  3000J 2 KE range = 2000 to 5000 J. Ans. (3) m1u1  m 2u 2  (m1  m 2 )v

11. Ans. (3)      rF

 ˆ   (iˆ  ˆj)  (  F k) ˆ  (ˆj  k)] ˆ  F[( ˆi  k)

v  2 / 3m / s 1 2

8.

1 1  2 2 Energy loss  (0.5)  (2)  (1.5)     0.67 J  3 2 2 Ans. (4) According to the work-energy theorem, W  K 2

9.

1  v0  1 2 Case I  F  3  m    mv 0 2  2 2 where, F is the resistive force and V0 is the initial speed. Case II Let, the further distance travelled by the bullet before coming to rest is s. 1 2   F(3  s)  K f  K i   mv 0 2 1 2 1 2   mv 0 (s)   mv 0 8 2 1 (3  s)  1 or or s  1cm . 4 Ans. (4) ma 2 Moment of inertia of square plate about xy is 6 Moment of inertia about zz’ can be computed using parallel axes theorem

 F[ˆj  ˆi] 12. Ans. (3)

 2h  g h  g 1    R

d  g d  g 1   ...(ii)  R As per statement of the problem, d  2h   i.e., g 1    g 1   R R .  2h  d 13. Ans. (1) Maximum velocity v  A, (where A is the amplitude and  is the angular frequency of oscillation).  4.4  (7  10 3 )  2 / T and

or

T

7  10 3 2  22  4.4 7

= 0.01s

14. Ans. (2) Time period of the spring  m T  2    k Here k, be the force constant of spring. For the first spring  m t1  2    k1  For the second spring

...(i)

 m t 2  2   ...(ii)  k2  The effictive force constant in the series combination is kk k 1 2 k1  k 2 Time period of combination

2

Izz '

...(i)

 a   I xy  m   2  ma 2 ma 2 2ma 2    6 2 3

10. Ans. (3) According to the principle of conservation of angular momentum, in the absence of external torque, the total angular momentum of the system is constant.

 m(k1  k 2 )  T  2    k1k 2 

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4 2 m(k1  k 2 ) k1k 2 From Eqs. (i) and (ii), we obtain  m m t12  t 22  42    k k  

T2 

1

Mocktest/AIMPT

in an adiabatic process may be non-zero. 19. Ans. (1) For an adiabatic process, dQ  0 So, dU  W  nC vdT  146  103 J nfR   7  146  103 2 [f  degree of freedom]

...(iii)

2

or

1 1 t12  t 22  4 2 m     k1 k 2 

or

t12  t 22 

4 2 m(k1  k 2 k1k 2

103  f  8.3  7  146  103 2 or f  5.02  5 So, it is a diatomic gas. 20. Ans. (1) 

 t12  t 22  T 2 . 15. Ans. (2)   6 As given y  10 sin  100t  20x   4 Comparing it with y  a sin(t  kx  ) we obtain, 100rads 1 , k  20m 1

v

...(i)

1Vg 2Vg  kvT2

...(ii)

 q Linear charge density     r

16. Ans. (2) Equation of a wave y  a sin(t  k x) ...(i) Let equations of another wave may be, y  a sin(t  k x) ...(ii) y   a sin(t  k x) ...(iii) If Eq. (i) propagates with Eq. (ii), then we get y  2a cos k x sin t ...(iv) If Eq. (i) propagates with Eq. (iii), then we get y  2a sin k x cos t ...(v) After putting x = 0 in Eqs. (iv) and (v) respectively, we get y  2a sin t and y  0 Hence, (iii) is a equation of unknown wave. 17. Ans. (2) For the Carnot engine used as a refrigerator T  W  Q2  1  1  T2 

 or

Vg(1  2 ) k

21. Ans. (3)

 100   5ms 1 . k 20

It is given that,  

vT 

K.dq E   dEsin ( ˆj)   2 sin ( ˆj) r K qr E  2  d  sin ( ˆj) r r Kq  E  2  sin  d ( ˆj) r  0 q  2 2 (ˆj) 2 0r 22. Ans. (3) From FBD of sphere, using Lami’s theorem

1 10

  1

T2 T1

T2 9  T1 10 W2  90 J

So, (as W  10 J) . 18. Ans. (2) Internal energy does not change in isothermal process. S can be zero for an adiabatic process. Work done

F  tan  ...(i) mg When suspended in liquid, as  remains same, F'   tan    ...(ii) mg 1    d

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Using Eqs. (i) and (ii) F F F'  where, F'  K mg   mg  1    d F F   mg   mg K 1    d 1 1 K  2  0.8   or 1 1 d  1.6  23. Ans. (3) r1 Q r4r 2dr 4  Qr12 0 2  R E4r1  E 0 40 R 4 24. Ans. (4) W  QdV  Q(Vq  Vp )

Mocktest/AIMPT

1   2 2

  1   2   1   2      2 as (1   2 ) is negligible   2 p  1  2 26. Ans. (2) From balanced Wheatstone bridge concept, 55 20   R  220  R 80 27. Ans. (4) 28. Ans. (2) Redrawing the circuit it is a balanced Wheatstone’s bridge and no current flows in the middle resistor 

I

25. Ans. (4) Let R0 be the initial resistance of both conductors  At temperature  their rsistances will be, R 1  R 0 (1  1) and R 2  R 0 (1   2) For series combination, R s  R 1  R 2 R s0 (1   s)  R 0 (1  1)  R 0 (1   2) where R s0  R 0  R 0  2R 0  2R 0 (1   s)  2R 0  R 0( 1   2 )   2 s  1 or 2 For parallel combination, RR R  1 2 R1  R 2

R p0 (1   p) 

R 0 (1  1)R 0 (1   2) R 0 (1  1)  R 0 (1   2)

R 0R 0 R0 where, R p0  R  R  2 0 0 R0 R 2 (1  1   2  1 2 2 )  (1   p)  0 2 R 0 (2  1   2) as 1 and  2 are small quantities  1  2 is negligible 1   2 p  or 2  (1   2 ) 

20

5

10 + –

 100  (1.6  10 19 )  ( 4  10)  100  1.6  1019  14  2.24  10 16 J.

10

5V

In series R '  10   20   30  In series R ''  5   10   15  15  30  10 In parallel R  15  30 V 5  I   0.5 A . R 10 29. Ans. (1) The magnetic field in between the two wires because of each wire will be in opposite direction i  0i Bin between  0 ˆj  (  ˆj) 2x 2 (2d  x)

 0i  1 1  ˆ  ( j)  2  x 2d  x  at x = d, Bin between = 0 for x < d, Bin between = (ˆj) for x > d, Bin between = ( ˆj) Towards x, net magnetic field will add up and direction will be ( ˆj) . Towards x’, net magnetic field will add up and direction will be (ˆj) . 30. Ans. (2) The magnetic field inductions at a point P, at a distance d from O in a direction perpendicular to the plane of the wires due to currents through AOB and COD are perpendicular to each other, is 

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Hence,

Mocktest/AIMPT

B  B12  B22 1

   2I  2   2I  2  2   0 1    0 2     4 d   4 d     0 (I12  I22 ) 2d

31. Ans. (2) 32. Ans. (2) So, to enter in the region x > b r  (b  a) mv  (b  a) or Bq q(b  a)B v or . m 33. Ans. (2) 34. Ans. (2) The current at any instant is given by  I  I0 (1  e  Rt / L ) L,R I0   I0 (1  e Rt / L ) 2 1  (1  e  Rt / L ) or 2 1 2V e  Rt / L  or 2 Rt  ln 2 or L L 300  103  t  ln 2   0.693 R 2  150  0.693  103  0.10395s  0.1s 35. Ans. (2) The emf induced between the ends of the conductor 1 e  BL2 2 1   0.2  10 4  5  (1)2 2  0.5  10 4 V  5  10 5 V  50V

36. Ans. (2) Due to the flow of current,inthe same direction in two adjacent sides, an attractive magnetic force will be produced due to which the speing gets compressed. 37. Ans. (2) At t  0 , inductor behave like an infinite resistance V t  0,i  So, at R2 and at t   , inductor behaves like a conducting wire V V(R1  R 2 ) i  R eq R 1R 2 38. Ans. (4) The given circuit is under resonance as X L  X C How, power dissipated in the circuit is V2 P  242 W R 39. Ans. (3) In vacuum,  0  1 In medium,   4  4  Refractive index,     1  2 0     2 c c and wave velocity, v    2 Hence, it is clear that wavelength and velocity become half but frequency remains unchanged when the wave passes through any medium. 40. Ans. (1) At minimum deviation (   m ) A 60º r1  r2    30º (For both colours) 2 2 41. Ans. (2) Since object and image move in opposite directions, the positioning should be as shown in the figure. Object lies between focus and centre of curvature f  x  2f

Wavelength,

'

Image Object

right left

42. Ans. (3) 31  4 2

3 3 1   590 4 4 1770   442.5nm 4

2 

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Mocktest/AIMPT

condition is d sin   n ...(i) Given, d  slit  width  2  2 sin   n or 2 sin   n The maximum value of sin  is 1, hence, n  2 1  2 Thus, Eq. (i) must be satisfied by 5 integer values ie, – 2,–1,0,1,2. Hence, the maximum number of possible interference maxima is 5. 44. Ans. (2) Expression is given by 2d cosi  n B . 45. Ans. (3) Emission spectrum would rises when electron makes a jump from higher energy level to lower energy level. Frequency of emitted photon is proportional to change in energy of two energy levels, ie,  1 1   RcZ2  2  2   n1 n 2 

Chloroform & acetone form a non-ideal solution, in which A......B type interaction are more than A......A & B.......B type interaction due to H-bonding. Hence, the solution shows, negative deviation from Raoult’s Law i.e., Vmix   ve; H mix   ve  Total volume of solution = less than (30 + 50ml) or

< 80 ml 54. Ans. (2) 

55. Ans. (2) Zince blende (ZnS) has fcc structure and is an ionic crystal haing 4 : 4 co-ordination number. 56. Ans. (2) Schottky defect is caused if someof the lattice points are unoccupied. The points which are unoccupied are called vacancies or holes. The number of missing positive and negative ions is the same keeping the crystal neutral. Cations and anions are of similar size. 57. Ans. (2) d1 1 r1  ;  d 2 16 r2

CHEMISTRY 46. Ans. (4) As we know that four sodium atom are present in sodium ferrocyanide Na4[Fe(CN)6] 47. Ans. (4) KMnO 4

FeSO4

M1V1 M 2 V2 n  ; M1V1  1 M 2 V2 n1 n2 n2 

n  m  RT RT  M p    V V 

2 1 1  10    0.2 10 10 5 1

For (4) M1V1  0.02  10  5 48. Ans. (3) 49. Ans. (4)

d2 4  16  d1 1

58. Ans. (3) K.E. 

3RT 3RT 2E , Vrms   U rms  2 M M

59. Ans. (3) 7 N14  1 H1  8 O15  , Because X is the  - rays it has neither mass nor charge. 60. Ans. (2) Deflection in  - rays is large. 61. Ans. (4) According to Le-chatelier’s principle. 62. Ans. (4) AlH3 63. Ans. (3)

H 2O(l )  H 2O ( g ) :  H U ng RT

U  H ngRT : 41000  1  8.3  373

1q 1 q 7 7 1  m   2  m   2  9.6  10  4.8  10 C kg a p

50. Ans. (1) Number of radial nodes=(n-l-1) for 3s; n=3,l=0 (number of radial node=2) for 2p;n=2,l=1 (number of radial node=0) 51. Ans. (2) As stability is directly related to lattice energy and lattice energy depends on charge and size of ions. So, the order is:

=37.904 64. Ans. (1) In reaction (a) Na (s) reacts with H2O(l) to give 1

NaOH(l) and H 2 (g) . Since, the entropy of a gas is 2 higher than that of liquid and the entropy of liquid is higher than that of solid, therefore, the change in entropy of this reaction increases. 65. Ans. (3) For a given reaction

Fe2 O3  Cr2 O3  Al 2 O3  MgO.

2NO(g)  O 2(g)  2NO2(g)

Rate of reaction = k [NO]2[O2] Rate of reaction directly proportional to concentration of the compound or inversely proportional to the volume of the vessel.

52. Ans. (1) Valency of Na2S2O3 is supposed to be x, then 2 + 2x + (-6) = 0 , 2x - 4 = 0 , x = 2. 53. Ans. (2) –  –

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Mocktest/AIMPT

n v

If volume of vessel is reduced by 1/3rd of its initial value, then concentration of compound is increased by 3 times and consequently the rate of reaction by 27 times. and consequently the rate of reaction by 27 times. 66. Ans. (1) Cr2 O72  6e  Cr 3

For reduction of 1 mole of Cr2 O72  , 6 Faradays of electricity are required.  For reduction of 2.5 moles of Cr2 O 72  , 6  2.5  15 Faradays of electricity are required. 67. Ans. (4) For the cell reaction, Fe acts as cathode and Sn as anode. Hence, Eocell  E ocathode  Eoanode  0.44  (0.14)  0.30V

The negative EMF suggests that the reaction goes spontaneously in reverse direction. 68. Ans. (2) Anodic reation : H2 (P1 )  2H  Cathodic reaction : 2H   H 2 (P2 ) Ecathode



RT P2 RT P1 ln  ln 2F P1 2F P2

HI  I2  ICl  HIO4 1

0

1

7

78. Ans. (1) KO2 is used in oxygen cylinder because it absorbs CO2 and increases O2 content. Super oxides reacts with water to give H2O2 & O2 79. Ans. (1) 1

80.

82. 83.

Einf  Eanode  Ecathode P RT (H  ) 2 RT ln  ln 2 2F P1 2F (H  ) 2

Mg 2 C3  4H 2 O  CH3 C  CH  2Mg(OH) 2

77. Ans. (4)

81.

P RT RT [H  ]2  ln 2 2 ; Eanode   ln 2F [H ] 2F P1



73. Ans. (1) The triple point of any substance is that temperature and pressure at which the material can exist in all three phases (Solid, liquid and gas) in equlibrium specifically the triple point of water is 273. 16K at 611.2 Pa. 74. Ans. (1) 75. Ans. (3) 76. Ans. (2) Propyne can be prepared by the hydrolysis of magnesium carbide.

84. 85. 86.

Ionic radii  Atomic No. Ionic radius decreases from left to right in a period. Ans. (3) Lanthanide contraction takes place. Ans. (3) Cyanide process Ans. (2) Ans. (1) Follow IUPAC rule. Ans. (3) Ans. (2) Follow IUPAC Rule Ans. (2) (1) Molecule is planar..

69. Ans. (3) V O

2 5  2SO 2SO 2  O 2  3 (Catalyst)

70. Ans. (3) Gold sol is a lyophobic sol. Gold particles have very less affinity towards dispersion medium, hence its sol can be easily coagulated. 71. Ans. (2) X2 - Cl2 , Y2 - F2 , Z2 - Br2 72. Ans. (4)

87. 88. 89. 90.

(2) 6  electrons are present. Ans. (1) Ans. (4) Ans. (4) Ans. (2) Nucleophiles that are relatively weak bases such as CN- , RNH2 and X- give conjugate addition where as strong bases such as R-Li , R-MgX give direct addition.

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Sri Chaitanya The Final Word in JEE/Medical/Foundation

ANSWER KEY AIPMT MOCK TEST 1. 6. 11. 16. 21. 26. 31. 36. 41.

(3) (4) (3) (2) (3) (2) (2) (2) (2)

2. 7. 12. 17. 22. 27. 32. 37. 42.

(1) (3) (3) (2) (3) (4) (2) (2) (3)

46. 51. 56. 61. 66. 71. 76. 81. 86.

(4) (2) (2) (4) (1) (2) (2) (3) (2)

47. 52. 57. 62. 67. 72. 77. 82. 87.

(4) (1) (2) (4) (4) (4) (4) (2) (1)

91. 96. 101. 106. 111. 116. 121. 126. 131. 136. 141. 146. 151. 156. 161. 166. 171. 176.

(3) (3) (4) (2) (4) (1) (4) (2) (2) (2) (2) (2) (2) (4) (4) (3) (2) (2)

92. 97. 102. 107. 112. 117. 122. 127. 132. 137. 142. 147. 152. 157. 162. 167. 172. 177.

(1) (2) (1) (1) (3) (1) (2) (4) (2) (4) (4) (2) (4) (3) (4) (1) (1) (2)

Section- I (Physics) 3. (1) 8. (4) 13. (1) 18. (2) 23. (3) 28. (2) 33. (2) 38. (4) 43. (2) Section- II (Chemistry) 48. (3) 53. (2) 58. (3) 63. (3) 68. (2) 73. (1) 78. (1) 83. (1) 88. (4) Section-III (Biology) 93. (1) 98. (2) 103. (2) 108. (2) 113. (1) 118. (4) 123. (3) 128. (3) 133. (4) 138. (1) 143. (1) 148. (3) 153. (1) 158. (2) 163. (1) 168. (4) 173. (1) 178. (3)

4. 9. 14. 19. 24. 29. 34. 39. 44.

(3) (4) (2) (1) (4) (1) (2) (3) (2)

5. 10. 15. 20. 25. 30. 35. 40. 45.

(4) (3) (2) (1) (4) (2) (2) (1) (3)

49. 54. 59. 64. 69. 74. 79. 84. 89.

(4) (2) (3) (1) (3) (1) (1) (3) (4)

50. 55. 60. 65. 70. 75. 80. 85. 90.

(1) (2) (2) (3) (3) (3) (3) (2) (2)

94. 99. 104. 109. 114. 119. 124. 129. 134. 139. 144. 149. 154. 159. 164. 169. 174. 179.

(1) (2) (2) (4) (3) (1) (2) (4) (1) (3) (2) (2) (1) (4) (2) (4) (4) (2)

95. 100. 105. 110. 115. 120. 125. 130. 135. 140. 145. 150. 155. 160. 165. 170. 175. 180.

(3) (4) (3) (4) (2) (4) (2) (3) (4) (4) (4) (3) (4) (4) (3) (4) (1) (4)