dcx0p3on5z8dw.cloudfront.net · (1) 23/03/2020 code-b. regd. office: aakash tower, 8,pusa road, new...

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23/03/2020 CODE-B

Regd. Office: Aakash Tower, 8,Pusa Road, New Delhi-110005, Ph.011-47623456

MM : 720 TEST SERIES for NEET-2020 Time : 3.00 Hrs.

Test - 1

Answer Key

1. (3) 2. (3) 3. (2) 4. (1) 5. (3) 6. (4) 7. (4) 8. (2) 9. (2) 10. (2) 11. (2) 12. (3) 13. (1) 14. (4) 15. (2) 16. (1) 17. (4) 18. (1) 19. (3) 20. (3) 21. (1) 22. (3) 23. (4) 24. (3) 25. (3) 26. (4) 27. (4) 28. (3) 29. (4) 30. (3) 31. (3) 32. (2) 33. (4) 34. (4) 35. (1) 36. (3)

37. (1) 38. (2) 39. (3) 40. (1) 41. (2) 42. (3) 43. (4) 44. (1) 45. (1) 46. (1) 47. (1) 48. (4) 49. (3) 50. (4) 51. (3) 52. (3) 53. (3) 54. (3) 55. (4) 56. (3) 57. (2) 58. (4) 59. (2) 60. (3) 61. (4) 62. (1) 63. (2) 64. (3) 65. (3) 66. (4) 67. (2) 68. (4) 69. (4) 70. (4) 71. (1) 72. (3)

73. (3) 74. (2) 75. (1) 76. (2) 77. (4) 78. (1) 79. (4) 80. (2) 81. (4) 82. (3) 83. (2) 84. (4) 85. (4) 86. (2) 87. (3) 88. (1) 89. (2) 90. (2) 91. (4) 92. (1) 93. (1) 94. (3) 95. (2) 96. (2) 97. (3) 98. (1) 99. (3) 100. (2) 101. (2) 102. (3) 103. (3) 104. (1) 105. (1) 106. (4) 107. (3) 108. (2)

109. (3) 110. (4) 111. (3) 112. (3) 113. (3) 114. (3) 115. (4) 116. (3) 117. (1) 118. (4) 119. (1) 120. (4) 121. (3) 122. (4) 123. (3) 124. (1) 125. (2) 126. (4) 127. (1) 128. (3) 129. (1) 130. (1) 131. (1) 132. (4) 133. (1) 134. (3) 135. (4) 136. (4) 137. (2) 138. (3) 139. (4) 140. (4) 141. (2) 142. (3) 143. (1) 144. (3)

145. (4) 146 (2) 147. (1) 148. (3) 149. (2) 150. (3) 151. (2) 152. (3) 153. (2) 154. (1) 155. (3) 156. (3) 157. (2) 158. (1) 159. (3) 160. (1) 161. (3) 162. (2) 163. (2) 164. (3) 165. (4) 166. (1) 167. (2) 168. (3) 169. (4) 170. (1) 171. (2) 172. (3) 173. (4) 174. (4) 175. (1) 176. (3) 177. (2) 178. (4) 179. (2) 180. (3)

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23/03/2020 CODE-B

Regd. Office : Aakash Tower, 8,Pusa Road, New Delhi-110005, Ph.011-47623456

MM : 720 TEST SERIES for NEET-2020 Time : 3:00 Hrs.

Test - 1

Hints and Solutions

PHYSICS

1. Answer (3) For acceleration (a1) slope of v versus t graph is

a positive constant. For acceleration (–a1) slope of v versus t graph is

a negative constant. 2. Answer (3) 605 cos121xv t= π π 605 sin5yv t= − π π

( 1s) 605 cos121xv t = = π π

605 ( 1) 605= π − = − π

( 1s) 605 sin5yv t = = − π π

605 0 0= − π× =

2 2speed 605 m/sx yv v∴ = + = π

3. Answer (2) Zeroes to the left of last non-zero digit is

insignificant. 4. Answer (1) Maximum percentage errors are to be added for

multiplication and division.

1/3 2a bXc

=

100XX∆

⇒ ×

1 2100 100 1003

a b ca b c∆ ∆ ∆

= × + × + ×

1 0.3 2 1 0.9 3%3

= × + × + =

1. mÙkj (3) Roj.k (a1) ds fy, v rFkk t ds e/; vkjs[k dh <ky

/kukRed fu;rkad gS Roj.k (–a1) ds fy, v rFkk t ds e/; <ky _.kkRed

fu;rkad gS 2. mÙkj (3) 605 cos121xv t= π π 605 sin5yv t= − π π

( 1s) 605 cos121xv t = = π π

605 ( 1) 605= π − = − π ( 1s) 605 sin5yv t = = − π π

605 0 0= − π× =

2 2 605 m/s∴ = + = πx yv vpky

3. mÙkj (2) vfUre v'kwU; vad ds ck;ha rjQ ds 'kwU; vlkFkZd gSA 4. mÙkj (1) xq.kk o Òkx ds fy, vf/kdre çfr'kr =qfV dks

tksM+k tkrk gS

1/3 2a bXc

=

100XX∆

⇒ ×

1 2100 100 1003

a b ca b c∆ ∆ ∆

= × + × + ×

1 0.3 2 1 0.9 3%3

= × + × + =

Test Series for NEET-2020 Test-1 (Code-B)_(Hints & Solutions)

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5. Answer (3)

StressStrain

Y =

[Stress] [ ][ ][Strain] [ ]

FYA

= =

2[mass acceleration]

[(length) ]×

=

1

1 32

[ ] [ ][ ] [ ][ ]

M VTY MV TVT

−− −= =

6. Answer (4)

For shortest path, drift x = 0 sinr brv v⇒ = θ

10 1sin14.14 2

⇒ θ = = =r

br

vv

45⇒ θ = °

Therefore, angle w.r.t. east = ( 90 )θ + °

= 45° + 90° = 135° Time taken to cross the river

= 1 0.1h110 22

t = =×

7. Answer (4)

110 2 10 m/s2xu = × =

110 2 10 m/s2yu = × =

10 3 30 mx∴ = × =

110 3 10 92

y = × − × ×

= 30 – 45 = – 15 m

ˆ ˆr x i y j∴ = +

ˆ ˆ(30 15 ) m= −i j

8. Answer (2) x = 2t y = 3t – 6t2

5. mÙkj (3)

=Y çfrcyfoÑfr

[ ][ ][ ]

= =FYA

çfrcyfoÑfr

2[ ]

[( ) ]×

=nzO;eku Roj.k

yEckbZ

1

1 32

[ ] [ ][ ] [ ][ ]

M VTY MV TVT

−− −= =

6. mÙkj (4)

U;wure iFk ds fy, foLFkkiu x = 0 sinr brv v⇒ = θ

10 1sin14.14 2

⇒ θ = = =r

br

vv

45⇒ θ = °

blfy, iwoZ ds lkis{k dks.k = ( 90 )θ + ° = 45° + 90° = 135° unh ikj djus esa fy;k x;k le;

= 1 0.1h110 22

t = =×

7. mÙkj (4)

110 2 10 m/s2xu = × =

110 2 10 m/s2yu = × =

10 3 30 mx∴ = × =

110 3 10 92

y = × − × ×

= 30 – 45 = – 15 m

ˆ ˆr x i y j∴ = +

ˆ ˆ(30 15 ) m= −i j

8. mÙkj (2) x = 2t y = 3t – 6t2

Test-1 (Code-B)_(Hints & Solutions) Test Series for NEET-2020

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2xt =

2

3 62 4x xy∴ = × − ×

23 3 3 (1 )2 2 2

x x x x = − = −

9. Answer (2)

ˆ ˆ( ) sin cosP t B t i B t j= ω + ω

ˆ ˆcos sindP F B t i B t jdt

= = ω ω − ω ω

2 2cos sin sin cosP F B t t B t t∴ ⋅ = ω ω ω − ω ω ω

= 0

[ ]F P⇒ ⊥

10. Answer (2)

2 24

rv var r

= =

2 2 2 tv u a s= +

2 24

2 2 tv v a

r−

=× π

23

4 tv a

r⇒ =

π

2 2r ta a a∴ = +

2 22 24 34

v vr r

= +

π

2

2916

16vr

= +π

22256 9

4v

r= π +

π

11. Answer (2) From figure

tan xy

θ =

2 1 1sec d dx vdt y dt yθ

θ = × = ×

21 40 cos80

ddtθ

⇒ = × × θ

1 3 3 rad/s2 4 8

= × =

2xt =

2

3 62 4x xy∴ = × − ×

23 3 3 (1 )2 2 2

x x x x = − = −

9. mÙkj (2)

ˆ ˆ( ) sin cosP t B t i B t j= ω + ω

ˆ ˆcos sindP F B t i B t jdt

= = ω ω − ω ω

2 2cos sin sin cosP F B t t B t t∴ ⋅ = ω ω ω − ω ω ω

= 0

[ ]F P⇒ ⊥

10. mÙkj (2)

2 24

rv var r

= =

2 2 2 tv u a s= +

2 24

2 2 tv v a

r−

=× π

23

4 tv a

r⇒ =

π

2 2r ta a a∴ = +

2 22 24 34

v vr r

= +

π

2

2916

16vr

= +π

2

2256 94v

r= π +

π

11. mÙkj (2) fp= ls

tan xy

θ =

2 1 1sec d dx vdt y dt yθ

θ = × = ×

21 40 cos80

ddtθ

⇒ = × × θ

1 3 3 rad/s2 4 8

= × =


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12. Answer (3)

In UCM, speed, ,

v is constant.

13. Answer (1) Let at point P speed = u

18 3 cos60 cos30∴ ° = °u

18 3 1 18 m/s23

2

u = × =

12 182 sin30 2 1.8 s10

uTg

× ×°= = =

14. Answer (4)

OA AB BD+ +

0= + + + = +

OA AB BO OD OD

Therefore

1| | | | m2

+ + = =

OA AB BD OD

15. Answer (2) 16. Answer (1) Mean absolute error

=

| 50.1 50.0 | | 50.1 50.2 || 50.1 50.12 | | 50.1 50.08 | 0.06

4

− + − +− + −

=

17. Answer (4) 18. Answer (1) Example of dimensionless quantities having units

are solid angle and plane angle. 19. Answer (3) [B] = [3x] = [L] [P] = [MLT–1] [A] [L1/2] = [P] [B + 3x] = [MLT–1] × [L]

⇒ [A] =2 1

3/2 11/2

[ML T ] [ML T ][L ]

−−=

3/2 1

1/2 1[ML T ] [ML T ][L]

AB

−− ∴ = =

20. Answer (3)

12. mÙkj (3)

UCM esa] pky

v fu;r gS

13. mÙkj (1)

ekuk fcUnq P ij pky = u

18 3 cos60 cos30∴ ° = °u

18 3 1 18 m/s23

2

u = × =

12 182 sin30 2 1.8 s10

uTg

× ×°= = =

14. mÙkj (4)

OA AB BD+ +

0= + + + = +

OA AB BO OD OD

blfy,

1| | | | m2

+ + = =

OA AB BD OD

15. mÙkj (2)

16. mÙkj (1)

ek/; fujis{k =qfV

=

| 50.1 50.0 | | 50.1 50.2 || 50.1 50.12 | | 50.1 50.08 | 0.06

4

− + − +− + −

=

17. mÙkj (4)

18. mÙkj (1)

ek=dks okyh foekghu jkf'k;ksa ds mnkgj.k ?ku dks.k o leryh; dks.k gS

19. mÙkj (3) [B] = [3x] = [L] [P] = [MLT–1] [A] [L1/2] = [P] [B + 3x] = [MLT–1] × [L]

⇒ [A] =2 1

3/2 11/2

[ML T ] [ML T ][L ]

−−=

3/2 1

1/2 1[ML T ] [ML T ][L]

AB

−− ∴ = =

20. mÙkj (3)


(6)

x = 3t2 – 9t v = 6t – 9 ∴ v = 0⇒ t = 1.5 s

∴ Distance covered in first 3 s is = 2 × distance travelled in first 1.5 s

= 272 13.5 m4

× =

21. Answer (1)

Average speed = Total distance travelledTotal time

12 12= 3 m/s8+

=

22. Answer (3)

From starting point O time taken to have 2 m as

net displacement will be 8 s. From point A the pit is at a distance of 4 m,

therefore in 4th forward step from point A he will fall into the pit.

∴Total time taken = 8 + 4 = 12 s 23. Answer (4)

2

2590

25 25 5182 2 250 2 250 4va

x

× × = = = =× ×

m/s2

25 20 s54

vta

= = =

24. Answer (3)

N dynem cm

P Q =

5

210 dyne dyne

cm10 cmP Q×

⇒ = × 310PQ

−⇒ =

P : Q = 10–3 : 1 25. Answer (3) VA = |Slope of graph for A|

VB = |Slope of graph for B|

tan 45 1tan 60 3

A

B

VV

°∴ = =

°

26. Answer (4)

||A B

15 15 25

a b∴ = = 3, 5a b⇒ = =

x = 3t2 – 9t v = 6t – 9 ∴ v = 0⇒ t = 1.5 s ∴ çFke 3 s esa r; dh xbZ nwjh = 2 × çFke 1.5 s esa r; dh xbZ nwjh

= 272 13.5 m4

× =

21. mÙkj (1)

vkSlr pky = r; dh xbZ dyq njw h

dqy le;

12 12= 3 m/s8+

=

22. mÙkj (3)

çkjfEHkd fcUnq O ls 2 m usV foLFkkiu ds fy, fy;k

x;k le; 8 s gksxkA fcUnq A ls xM~Mk 4 m dh nwjh ij gS blfy, fcUnq A

ls vkxs dh vksj pkSFks dne esa og xM~Ms esa fxjsxk ∴ fy;k x;k le; = 8 + 4 = 12 s 23. mÙkj (4)

2

2590

25 25 5182 2 250 2 250 4va

x

× × = = = =× ×

m/s2

25 20 s54

vta

= = =

24. mÙkj (3)

Nm

=

P Q Mkbulseh

5

210

10×

⇒ = ×P QMkbu Mkbu

les h les h310P

Q−⇒ =

P : Q = 10–3 : 1 25. mÙkj (3) VA = |A ds fy, vkjs[k dh <ky|

VB = |B ds fy, vkjs[k dh <ky |

tan 45 1tan 60 3

A

B

VV

°∴ = =

°

26. mÙkj (4)

||A B

15 15 25

a b∴ = = 3, 5a b⇒ = =


(7)

27. Answer (4)

(2 1)x t= +

2(2 1)x t= +

4(2 1) 8 4v t t= + = +

Acceleration a of the particle is

8 Constantdvadt

= = =

28. Answer (3) Retardation = αv Since, =vdv adx ⇒ = −αvdv vdx

0

0

0⇒ = −

α∫ ∫x

v

dv dx

0⇒ − = −αv

x

0⇒ =αv

x

Also, = −αdv vdt

0 0

⇒ = − α∫ ∫v t

v

dv dtv

0

ln

⇒ = −α

v tv

0−α⇒ = tv v e

Therefore, v versus t graph will be an exponentially decreasing in nature.

29. Answer (4) [Angular momentum] = [Mass × velocity × distance]

= [M] × [LT–1] × [L] = [ML2T–1] [Planck’s constant] =[joule-second]

= [ML2T–2] [T] = [ML2T–1] 30. Answer (3) Let distance between poles = s

2 2 2v u as= +

2 2

2v ua

s−

=

Now, 2 2 225sv u a′ = + ×

27. mÙkj (4)

(2 1)x t= +

2(2 1)x t= +

4(2 1) 8 4v t t= + = +

d.k dk Roj.k a gS

8= = =dvadt

fu;r

28. mÙkj (3) enau = αv pw¡fd, =vdv adx ⇒ = −αvdv vdx

0

0

0⇒ = −

α∫ ∫x

v

dv dx

0⇒ − = −αv

x

0⇒ =αv

x

rFkk = −αdv vdt

0 0

⇒ = − α∫ ∫v t

v

dv dtv

0

ln

⇒ = −α

v tv

0−α⇒ = tv v e

blfy, v o t vkjs[k çÑfr esa pj?kkrkadh :i ls ?kVsxkA

29. mÙkj (4) [dks.kh; laosx] = [nzO;eku × osx × nwjh]

= [M] × [LT–1] × [L] = [ML2T–1] [Iykad fu;rkad] =[twy&lsd.M]

= [ML2T–2] [T] = [ML2T–1] 30. mÙkj (3) ekuk /kqzoksa ds e/; nwjh = s

2 2 2v u as= +

2 2

2v ua

s−

=

vc, 2 2 225sv u a′ = + ×


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2 2

2 25s v uu

s −

= + ×

2 2

2 2( )5

v uu −= +

2 2 25 2 2

5u v u+ −

= 2 23 2

5u vv +′⇒ =

31. Answer (3) 192 = 1.92 × 102 32. Answer (2)

By symmetry, they all meet at the centre of

regular hexagon. Relative velocity of A w.r.t. B vAB = vA – vB

⇒ vAB = v – vcos60° = 2 2v vv − =

Time taken to travel d with vAB

2

2AB

d d dt vv v= = =

33. Answer (4)

1ˆ3 m/smv i=

1 1

ˆ m/srmv v j= −

1 1

ˆ ˆ(3 ) m/s= −

rv i v j

2

ˆ6 m/smv i=

2 2 2

ˆ ˆsin30 cos30rmv v i v j= − − °

2 2 2r rm mv v v∴ = +

2 23ˆ ˆ ˆ62 2v vi j i= − − + 2

23ˆ ˆ6

2 2v i v j = − −

1 2r rv v=

21 2

3ˆ ˆ ˆ ˆ3 62 2vi v j i v j ⇒ − = − −

2 2

2 25s v uu

s −

= + ×

2 2

2 2( )5

v uu −= +

2 2 25 2 2

5u v u+ −

= 2 23 2

5u vv +′⇒ =

31. mÙkj (3)

192 = 1.92 × 102 32. mÙkj (2)

lefefr ls] ;s lHkh le"kV~Hkqt ds dsUnz ij feysaxsA B ds lkis{k A dk lkisf{kd osx vAB = vA – vB

⇒ vAB = v – vcos60° = 2 2v vv − =

vAB ds lkFk d r; djus esa fy;k x;k le;

2

2AB

d d dt vv v= = =

33. mÙkj (4)

1

ˆ3 m/smv i=

1 1

ˆ m/srmv v j= −

1 1

ˆ ˆ(3 ) m/s= −

rv i v j

2

ˆ6 m/smv i=

2 2 2

ˆ ˆsin30 cos30rmv v i v j= − − °

2 2 2r rm mv v v∴ = +

2 23ˆ ˆ ˆ62 2v vi j i= − − + 2

23ˆ ˆ6

2 2v i v j = − −

1 2r rv v=

21 2

3ˆ ˆ ˆ ˆ3 62 2vi v j i v j ⇒ − = − −


(9)

Comparing components we get,

21 2

33 6 and2 2v v v= − =

21

33 6 3 32 2v v⇒ − = − = × =

2 6v⇒ =

1

ˆ ˆ ˆ ˆ3 (3 3 3 ) m/sr iv i v j i j∴ = − = −

34. Answer (4)

2 2 2sin2 sinand

2u uR H

g gθ θ

= =

So, speed of projection u is required to compare these parameters.

35. Answer (1)

Least count = 1mm 0.01mm100

=

Zero error = 20 × 0.01 = 0.20 mm Linear scale reading = 25 × 0.01 = 0.25 mm Measured reading = 3 + 0.25 = 3.25 mm Absolute reading = 3.25 – 0.2 = 3.05 mm 36. Answer (3)

2 sin 2 60 sin45 6 2 s10

uTg

θ × × °= = =

At 3 2 st = particle is at the top of projectile with speed

160cos45 60 30 2 m/s2

u = ° = × =

∴Radius of curvature

2 2(30 2) 90 2 180 m

10uRg

= = = × =

37. Answer (1) Let uA = 12 m/s uB = 8 m/s

aA = –10 m/s2 aB = –10 m/s2

∴ uAB = 12 – 8 = 4 aAB = 0 m/s2

∴ SAB = uAB× t = 4t (linear equation)

38. Answer (2)

1

2

1( 2 1)

tt

=−

(for u = 0)

(for uniformly accelerated motion on a straight line)

2 1( 2 1) 3( 2 1)t t⇒ = − = −

?kVdksa dh rqyuk djus ij

21 2

33 62 2

= − =v

v vrFkk

21

33 6 3 32 2v v⇒ − = − = × =

2 6v⇒ =

1

ˆ ˆ ˆ ˆ3 (3 3 3 ) m/sr iv i v j i j∴ = − = −

34. mÙkj (4)

2 2 2sin2 sin

2θ θ

= =

u uR Hg g

rFkk

blfy, bu çkpy dh rqyuk ds fy, vko';d ç{ksi.k pky u gS

35. mÙkj (1)

vYirekad = 1mm 0.01mm100

=

'kwU; =qfV = 20 × 0.01 = 0.20 mm

js[kh; iSekuk iBu = 25 × 0.01 = 0.25 mm

ekfir iBu = 3 + 0.25 = 3.25 mm

fujis{k iBu = 3.25 – 0.2 = 3.05 mm

36. mÙkj (3)

2 sin 2 60 sin45 6 2 s10

uTg

θ × × °= = =

3 2 st = ij d.k fuEu pky ls ç{ksI; ds 'kh"kZ ij gS

160cos45 60 30 2 m/s2

u = ° = × =

∴oØrk f=T;k

2 2(30 2) 90 2 180 m

10uRg

= = = × =

37. mÙkj (1)

ekuk uA = 12 m/s uB = 8 m/s aA = –10 m/s2 aB = –10 m/s2 ∴ uAB = 12 – 8 = 4 aAB = 0 m/s2

∴ SAB = uAB× t = 4t (js[kh; lehdj.k)

38. mÙkj (2)

1

2

1( 2 1)

tt

=−

(u = 0 ds fy,)

(ljy js[kk ij ,dleku Rofjr xfr ds fy,)

2 1( 2 1) 3( 2 1)t t⇒ = − = −


(10)

39. Answer (3)

Velocity versus time graph is max 1v v at= =

10 ( )v b t t= − −

1 1( )b t t v at⇒ − = =

1( )bt t a b⇒ = +

1btt

a b⇒ =

+

a bt abtva b a b×

∴ = =+ +

∴ Average velocity

12

2 2( )av

t v v abtvt a b

× ×= = =

+

40. Answer (1)

ˆ ˆ ˆ ˆ ˆ ˆ(2 4 50 ) (5 8 5 )ˆ ˆ3 4

A B i j k i j ki j

− = + + − + += − −

( )A B−

lies in (III)rd quadrant

⇒ Angle with negative y-axis = 37° 41. Answer (2)

21 1 5 m2

= − × × = −y g

x = 10 × 1 = 10 m Where x is distance travelled by bullet in 1 s. Co-ordinate is (x, t) = (10, –5) 42. Answer (3) A = lb

= =x A lb

9.0 4.0 6.0 cm= × =

1 12 2

∆ ∆ ∆= +

x l bx l b

39. mÙkj (3)

osx rFkk le; vkjs[k gS max 1v v at= = 10 ( )v b t t= − − 1 1( )b t t v at⇒ − = = 1( )bt t a b⇒ = +

1btt

a b⇒ =

+

a bt abtva b a b×

∴ = =+ +

∴ vkSlr osx

12

2 2( )av

t v v abtvt a b

× ×= = =

+

40. mÙkj (1)

ˆ ˆ ˆ ˆ ˆ ˆ(2 4 50 ) (5 8 5 )ˆ ˆ3 4

A B i j k i j ki j

− = + + − + += − −

( )A B−

rhljs prqFkk±'k esa fLFkr gS

⇒ _.kkRed y-v{k ds lkFk dks.k = 37° 41. mÙkj (2)

21 1 5 m2

= − × × = −y g

x = 10 × 1 = 10 m tgk¡ x, 1 s esa xksyh }kjk r; dh xbZ nwjh gS funsZ'kkad gSa (x, t) = (10, –5) 42. mÙkj (3) A = lb

= =x A lb

9.0 4.0 6.0 cm= × =

1 12 2

∆ ∆ ∆= +

x l bx l b


(11)

1 0.3 1 0.12 9.0 2 4.0

= × + ×

1 1 760 80 240

= × = = 0.175 = 0.2

43. Answer (4) Slope of x-t graph gives velocity. Therefore, particle speeds up and slows down

twice and also comes to rest twice. 44. Answer (1)

Let velocity of swimmer w.r.t river is v at an angle

θ w.r.t. vertical as shown.

ˆ ˆ( sin ) cos∴ = + = − θ + θ

s sr rv v v u v i v j

( ) sintan60( ) cos

− θ∴ ° = =

θ

s x

s y

v u vv v

3 cos sinv u v⇒ θ = − θ

( 3 cos sin )v u⇒ θ+ θ =

2sin( 60 )( 3 cos sin )

u uv⇒ = =θ + °θ + θ

∴ v is minimum at 60 90θ + ° = °

30θ = °

min 2sin90 2u uv∴ = =

°

45. Answer (1)

21 10 2 5 4 20 m2× × = × =

∴From ground height = 80 – 20 = 60 m

1 0.3 1 0.12 9.0 2 4.0

= × + ×

1 1 760 80 240

= × = = 0.175 = 0.2

43. mÙkj (4)

x-t vkjs[k dh <ky osx nsrk gSA

blfy, d.k dh pky nks ckj c<+sxh rFkk ?kVsxh rFkk fojkekoLFkk esa Hkh nks ckj vk;sxk

44. mÙkj (1)

ekuk n'kkZ, vuqlkj Å/okZ/kj ds lkis{k dks.k θ ij

unh ds lkis{k rSjkd dk osx v gSA

ˆ ˆ( sin ) cos∴ = + = − θ + θ

s sr rv v v u v i v j

( ) sintan60( ) cos

− θ∴ ° = =

θ

s x

s y

v u vv v

3 cos sinv u v⇒ θ = − θ

( 3 cos sin )v u⇒ θ+ θ =

2sin( 60 )( 3 cos sin )

u uv⇒ = =θ + °θ + θ

∴ 60 90θ + ° = ° ij v U;wure gSA

30θ = °

min 2sin90 2u uv∴ = =

°

45. mÙkj (1)

21 10 2 5 4 20 m2× × = × =

∴/kjkry ls Å¡pkbZ = 80 – 20 = 60 m


(12)

CHEMISTRY

46. Answer (1)

24SO

9.6n 0.1 mol96− = =

Total electrons = 50 × 0.1 NA = 5 NA

47. Answer (1)

3 2MgCO MgO CO∆→ +

3 2MgCO COn n 1 mol1 1 4

= =

3MgCO

1w 84 21g4

∴ = × =

21% Purity 100 42%50

= × =

48. Answer (4)

% by mass density 10Mmolar mass

× ×=

98 1.8 10 18 M98

× ×= =

Now, 1 1 2 2M V M V=

1 118 V 0.2 500 V 5.55 ml⇒ = × ⇒ =

49. Answer (3) Number of lines of particular series = n2 – n1

50. Answer (4) 0KE h h= ν − ν

51. Answer (3) 52. Answer (3) 53. Answer (3)

2

34

CO11.2 10n 5 10 mol.

22.4

−−×

= = ×

Number of atoms

= 4 3A A3 5 10 N 1.5 10 N− −× × × = ×

54. Answer (3)

Mole of oxygen =

47.0568100 2

16

×=

Mass of carbon =

(100 47.05)68100 3

12

−×

=

∴ Empirical formula = 3 2C O

55. Answer (4)

46. mÙkj (1)

24SO

9.6n 0.196− = = ekys

dqy bysDVªkWu = 50 × 0.1 NA = 5 NA

47. mÙkj (1)

3 2MgCO MgO CO∆→ +

3 2MgCO COn n 1 mol1 1 4

= =

3MgCO

1w 84 21g4

∴ = × =

21 100 42%50

= × =çfr'kr 'kq)rk

48. mÙkj (4)

10M

× ×=nzO;eku çfr'kr ?kuRo

ekys j nOz ;eku

98 1.8 10 18 M98

× ×= =

vc, 1 1 2 2M V M V=

1 118 V 0.2 500 V 5.55 ml⇒ = × ⇒ =

49. mÙkj (3) fo'ks"k Js.kh dh js[kkvksa dh la[;k = n2 – n1

50. mÙkj (4) 0KE h h= ν − ν

51. mÙkj (3) 52. mÙkj (3) 53. mÙkj (3)

2

34

CO11.2 10n 5 10

22.4

−−×

= = × ekys

ijek.kqvksa dh la[;k

= 4 3A A3 5 10 N 1.5 10 N− −× × × = ×

54. mÙkj (3)

vkWDlhtu ds eksy =

47.0568100 2

16

×=

dkcZu dk nzO;eku =

(100 47.05)68100 3

12

−×

=

∴ ewykuqikrh lw= = 3 2C O

55. mÙkj (4)


(13)

56. Answer (3) 1 electron = em kg

∴ 1 kg = e

1 electronsm

∴ Mole of electrons = e

A e A

1m 1N m N

=

57. Answer (2) 2 2 22H (g) O (g) 2H O(l)+ →

t = 0, 50 ml 20 ml 0 Final 10 ml 0 ∴ Reduction in volume of gases = [50 + 20] – [10] = 60 ml 58. Answer (4) 59. Answer (2) meq. of NaOH = meq. of H2SO4

w 1 (1 2) 240

× = × ×

NaOHw 160g=

100wt of solution 160 320 g50

∴ = × =

60. Answer (3) Eq. wt of metal chloride = EM + ECl = 3 × 35.5 + 35.5 = 142 61. Answer (4) Number of neutrons = 56 – 26 = 30 62. Answer (1) 63. Answer (2)

2H 2 2

1 2

1 1R Zn n

ν = −

1 2n 3, n= = ∞

2 HH 2

R1R (1) 093

∴ ν = × × − =

64. Answer (3) Number of radial nodes = n – – 1 65. Answer (3)

h2mKE

λ =

66. Answer (4) 67. Answer (2)

2

nnr 0.53 ÅZ

=

56. mÙkj (3) 1 bysDVªkWu = em kg

∴ 1 kg = e

1m

bysDVªkWu

∴ bysDVªkWu ds eksy = e

A e A

1m 1N m N

=

57. mÙkj (2) 2 2 22H (g) O (g) 2H O(l)+ → t = 0, 50 ml 20 ml 0 vafre 10 ml 0 ∴ xSlksa ds vk;ru esa deh = [50 + 20] – [10] = 60 ml 58. mÙkj (4) 59. mÙkj (2) NaOH ds fefyrqY;kad = H2SO4 ds fefyrqY;kad

w 1 (1 2) 240

× = × ×

NaOHw 160g=

100160 320 g50

∴ = × =foy;u dk Hkkj

60. mÙkj (3) /kkrq DyksjkbM dk rqY;kadh Hkkj = EM + ECl = 3 × 35.5 + 35.5 = 142 61. mÙkj (4) U;wVªkWuksa dh la[;k = 56 – 26 = 30 62. mÙkj (1) 63. mÙkj (2)

2H 2 2

1 2

1 1R Zn n

ν = −

1 2n 3, n= = ∞

2 HH 2

R1R (1) 093

∴ ν = × × − =

64. mÙkj (3) f=T; uksMksa dh la[;k = n – – 1 65. mÙkj (3)

h2mKE

λ =

66. mÙkj (4) 67. mÙkj (2)

2

nnr 0.53 ÅZ

=


(14)

68. Answer (4)

2

n 2ZE 13.6 evn

= −

Where, n = 2, Z = 1 69. Answer (4) Lower the value of (n + ), lower will be the

energy of subshell. 70. Answer (4)

Orbital angular momentum = ( 1)+


2 224 g 22.4 L at STP12 g 11.2 L at STP

Mg(s) 2HCl(aq) MgCl (aq) H (g)+ → +

73. Answer (3) 74. Answer (2) 75. Answer (1) For ionic compounds, formula mass is used. 76. Answer (2)

NaOH2X 0.0351000218

= =+

77. Answer (4) Average atomic mass

= 1000 40 102 60 101.2100

× + ×=

78. Answer (1) 79. Answer (4) 80. Answer (2) Cr2+ : [Ar] 3d4 Mn2+ : [Ar] 3d5 Fe2+ : [Ar] 3d6 Co2+ : [Ar] 3d7 81. Answer (4) 82. Answer (3) 83. Answer (2) For H-atom, energy of orbitals depends only on

the value of n. 84. Answer (4) 85. Answer (4) Number of revolution per seconds

= 2

32

nv ZZ

2 r nnZ

∝ ∝

π

68. mÙkj (4)

2

n 2ZE 13.6 evn

= −

tgk¡, n = 2, Z = 1 69. mÙkj (4) (n + ) dk eku de gksus ij midks'k dh ÅtkZ de

gksxhA 70. mÙkj (4) d{kd dks.kh; laosx = ( 1)+ 71. mÙkj (1) 72. mÙkj (3) 2 2

24 g STP 22.4 L12 g STP 11.2 L

Mg(s) 2HCl(aq) MgCl (aq) H (g)+ → +ijij

73. mÙkj (3) 74. mÙkj (2) 75. mÙkj (1) vk;fud ;kSfxdksa ds fy, lw= nzO;eku dk mi;ksx

gksrk gSA 76. mÙkj (2)

NaOH2X 0.0351000218

= =+

77. mÙkj (4) vkSlr ijek.koh; nzO;eku

= 1000 40 102 60 101.2100

× + ×=

78. mÙkj (1) 79. mÙkj (4) 80. mÙkj (2) Cr2+ : [Ar] 3d4 Mn2+ : [Ar] 3d5 Fe2+ : [Ar] 3d6 Co2+ : [Ar] 3d7 81. mÙkj (4) 82. mÙkj (3) 83. mÙkj (2) H-ijek.kq ds fy, d{kdksa dh ÅtkZ dsoy n ds eku

ij fuHkZj djrh gSA 84. mÙkj (4) 85. mÙkj (4) çfr lsd.M çfrØe.k dh la[;k

= 2

32

nv ZZ

2 r nnZ

∝ ∝

π


(15)


% of Mg = wt. of Mg 100wt. of enzyme

×

Molecular wt. of enzyme = 24 100 2400 U1×

=

88. Answer (1)

wt. of glucose = 40 g

wt. of solution = 1.2 × 100

= 120 g

ppm of glucose = 6 540 10 3.3 10 ppm120

× = ×

89. Answer (2)

2 2

2 2

Molecules of O Mole of OMolecules of H Mole of H

=

= w 2 132 w 16

× =

90. Answer (2)

1 2

2 1

E 2000 2E 1000 1

λ= = =λ

86. mÙkj (2)

87. mÙkj (3)

Mg dk çfr'kr = Mg

100×dk Hkkj

,ta kbe dk Hkkj

,atkbe dk v.kqHkkj = 24 100 2400 U1×

=

88. mÙkj (1)

Xywdkst dk Hkkj = 40 g

foy;u dk Hkkj = 1.2 × 100 = 120 g

Xywdkst dk ppm = 6 540 10 3.3 10 ppm120

× = ×

89. mÙkj (2)

2 2

2 2

O OH H

=ds v.kq d s eksyds v.kq d s ekys

= w 2 132 w 16

× =

90. mÙkj (2)

1 2

2 1

E 2000 2E 1000 1

λ= = =λ

BOTANY

91. Answer (4) Protonema of mosses asexually reproduce by

fragmentation. 92. Answer (1) In true regeneration lost body part get

regenerated completely such as in Planaria. 93. Answer (1) Planaria - Regeneration 94. Answer (3) 95. Answer (2) 96. Answer (2) Species is the lowest taxonomic category in

hierarchy. 97. Answer (3) Panthera leo – Felidae family Canis familiaris – Canidae family Felidae and Canidae – Same order Carnivora

91. mÙkj (4)

ekWl dk izFkerarq fo[kaMu }kjk vySafxd :Ik ls tuu djrk gSA

92. mÙkj (1)

okLrfod iqujksn~Hkou esa u"V gq, 'kjhj ds Òkx iw.kZ:i ls iqu% mRiUu gks tkrs gSaA tSls % Iysukfj;k

93. mÙkj (1)

Iysukfj;k – iqujksn~Hkou

94. mÙkj (3)

95. mÙkj (2)

96. mÙkj (2)

tkfr inkuqØe esa U;wure ofxZdh Js.kh gksrk gSA

97. mÙkj (3)

isaFksjk fyvks – QsfyfM dqy

dsful Qsfefy,fjl – dsfuMh dqy

QsfyMh o dsfuMh – leku x.k dkfuZoksjk


(16)

98. Answer (1) Herbarium is not used for animals. 99. Answer (3) 100. Answer (2) First word in biological name is genus. 101. Answer (2) 102. Answer (3) Members of Oomycetes (Phycomycetes) have

coenocytic mycelia. 103. Answer (3) Conidia is commonly found in ascomycetes. 104. Answer (1) Ascospores are sexual spores. 105. Answer (1) Sporangiospores are endogenous and nonmotile. 106. Answer (4) Ascomycetes – Sac fungi 107. Answer (3) Diatoms and dinoflagellates are photosynthetic

protistans. 108. Answer (2) 109. Answer (3) Mycoplasma lack cell wall. 110. Answer (4) Some species of Micrococcus and Bacillus are

used in curing of tea leaves. 111. Answer (3) Heterocyst is responsible for N2 fixation, which

require anaerobic environment. PS-I remains active in heterocyst because it is not involved in O2 production.

112. Answer (3) 113. Answer (3) In slime moulds, spores have cellulosic cell wall. 114. Answer (3) Both archaebacteria and eukaryotes have

introns. 115. Answer (4) 116. Answer (3) Only bacteria perform anoxygenic

photosynthesis. 117. Answer (1) Mad cow disease is caused by prions.

98. mÙkj (1) tarqvksa ds fy, gjcsfj;e dk mi;ksx ugha gksrk gS 99. mÙkj (3) 100. mÙkj (2) tSfod uke esa izFke “kCn oa”k gksrk gS 101. mÙkj (2) 102. mÙkj (3) ÅekbflfVt (QkbdksekbflfVt) ds lnL;ksa esa

ladksf”kdh dodtky gksrk gSA 103. mÙkj (3) dksfufM;k lkekU; :Ik ls ,sLdksekbflfVt esa ik;k

tkrk gS 104. mÙkj (1) ,sLdksLiksj ySafxd chtk.kq gSa 105. mÙkj (1) LiksjSfUtvksLiksj vartkZr o vxfr”khy gksrs gSa 106. mÙkj (4) ,LdksekbflVht– dks"k dod 107. mÙkj (3) Mkb,Ve o Mk;uks¶ySftysV~l çdk'kla'ysf"kr

çksfVLVu gSaA 108. mÙkj (2) 109. mÙkj (3) ekbdksIykTek esa dksf”kdk fHkfÙk dk vHkko gksrk gS 110. mÙkj (4) ekbØksdksdl o cSflyl ds dqN tkfr dk mi;ksx

pk; dh ifÙk;ksa ds lalk/ku esa gksrk gSA 111. mÙkj (3) gsVjksflLV N2 fLFkjhdj.k ds fy, mÙkjnk;h gksrk gS

tks vok;oh; Ik;kZoj.k ds fy, vfuok;Z gksrk gSA PS-I gsVjksflLV esa lfØ; cuk jgrk gS D;ksafd ;g O2 ds mRiknu esa lfEefyr ughaa gksrk gSA

112. mÙkj (3) 113. mÙkj (3) voiad dod ds chtk.kqvksa esa lsY;wykst;qDr

dksf”kdkfHkfÙk gksrh gSA 114. mÙkj (3) vkdhZcSDVhfj;k o ;wdSfj;ksV~l nksuksa esa bUVªkWu gksrk gS 115. mÙkj (4) 116. mÙkj (3) dsoy thok.kq vukWDlhd`r izdk”kla”ys’k.k n”kkZrk gSA 117. mÙkj (1) eSM dkm jksx fizvkWu }kjk mRiUu gksrk gS


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118. Answer (4) Binary fission is the most common mode of

asexual reproduction in bacteria. 119. Answer (1) Chlorophyceae members contain pigments

chl. a + b. 120. Answer (4) 121. Answer (3) Red algae is red due to phycoerythrin. 122. Answer (4) Bryophytes are first embryophytes. 123. Answer (3) 124. Answer (1) 125. Answer (2) Marchantia – Male and female plants are

separate. 126. Answer (4) Pteridophytes are first tracheophytes.

Gymnosperms are first spermatophytes. 127. Answer (1) Selaginella, Salvinia and Azolla are

heterosporous. 128. Answer (3) Pteridophytes require water for fertilization. 129. Answer (1) 130. Answer (1) Egg apparatus is haploid and situated at

micropylar end. 131. Answer (1) Eucalyptus is tallest angiospermic plant, while

Sequoia is tallest gymnospermic plant. 132. Answer (4) Cycas has coralloid roots. 133. Answer (1) 134. Answer (3) 135. Answer (4) Euglenoids, cyanobacteria and members of

deuteromycetes reproduce asexually

118. mÙkj (4) f}&fo[kaMu thok.kq esa vySafxd tuu dh vfr

lkekU; fof/k gksrh gS 119. mÙkj (1) DyksjksQkblh ds lnL;ksa esa DyksjksfQy a + b o.kZd

gksrs gaAS 120. mÙkj (4) 121. mÙkj (3) yky “kSoky QkbdksbfjFkzhu ds dkj.k yky gksrk gSA 122. mÙkj (4) czk;ksQkbV~l çFke Òwz.kksn~fHkn ¼,fEcz;ksQkbV~l½ gS 123. mÙkj (3) 124. mÙkj (1) 125. mÙkj (2) ekdsZfUl;k – uj o eknk ikni i`Fkd gksrs gSa 126. mÙkj (4) VSfjMksQkbV~l izFke VªSfdvksQkbV~l gSaA vuko`rchth

izFke LiesZVksQkbV~l gSa 127. mÙkj (1) flySftuSyk] lkWfYofu;k o ,tksyk fo’kechtk.kqd

gksrs gSa 128. mÙkj (3) VSfjMksQkbV~l ds fu’kspu ds fy, ty dh

vko”;drk gksrh gSA 129. mÙkj (1) 130. mÙkj (1) vaM midj.k vxqf.kr gksrk gS rFkk chtkaM}kjh fljs

ij fLFkr gksrk gS 131. mÙkj (1) ;wdsfyIVl lcls yack vko`rchth ikni gS] tcfd

fldqvk lcls yack vuko`rchth ikni gS 132. mÙkj (4) lkbdl esa izokykHk ewy gksrs gSa 133. mÙkj (1) 134. mÙkj (3) 135. mÙkj (4) ;wfXyukW;M~l] lk;ukscSDVhfj;k o M~;wVsjksekbflfVt

ds lnL; vySafxd :Ik ls tuu djrs gSaA

.ZOOLOGY136. Answer (4) Ancylostoma i.e. Hookworm belongs to phylum

Aschelminthes. 137. Answer (2) Radula helps in feeding.

136. mÙkj (4) ,ulkbyksLVksek vFkkZr~ vadq'k Ñfe ,sLdsfYeUFkht

la?k ls lacaf/kr gksrk gSA 137. mÙkj (2) jsrhftàk ¼jsMqyk½ Òkstu djus esa lgk;d gksrk gSA


(18)

138. Answer (3)

Coelenterates exhibit tissue level of organisation. The tissues are arranged to perform the basic functions in the body.

139. Answer (4)

The given diagram is of Pleurobrachia and it belongs to phylum ctenophora.

140. Answer (4)

Chitinous exoskeleton is a universal character of arthropods.

141. Answer (2)

Chaetopleura belongs to phylum Mollusca.

142. Answer (3)

Chondrichthyes – Carcharodon, Pristis

Osteichthyes – Pterophyllum

Aves – Corvus

Mammalia – Pteropus

Amphibia – Ichthyophis

143. Answer (1)

Birds and Mammals are homeothermous.

144. Answer (3)

Lizards and snakes shed their scales as skin cast.

145. Answer (4)

Branchiostoma is a cephalochordate.

146. Answer (2)

The gills of Limulus possess many plate-like structures called lamellae. These lamellae lie parallel to each other resembling the pages of a book, hence the name book-gills.

147. Answer (1)

Tse-tse fly causes trypanosomiasis.

148. Answer (3)

Coelenterates – Physalia, Adamsia, Meandrina, Obelia, Aurelia.

149. Answer (2)

Ornithorhynchus (Platypus) is connecting link between reptiles and mammals.

150. Answer (3)

Dentalium – Tusk shell

Chaetopleura – Chiton

Pila – Apple snail

138. mÙkj (3) lhysUVsªV] Ård Lrj dk laxBu çnÆ'kr djrs gSaA

'kjhj esa vk/kkjÒwr dk;ks± dks djus ds fy, Ård O;ofLFkr gksrs gSaA

139. mÙkj (4) fn;k x;k fp= Iywjksczsfd;k dk gS rFkk ;g la?k

VhuksQksjk ls lacaf/kr gSA 140. mÙkj (4) dkbfVu ;qDr cká dadky vkFkzksZiksM~l dk ,d

lkoZf=d vfÒy{k.k gSA 141. mÙkj (2) dhVksIy;wjk eksyLdk la?k ls lacaf/kr gSA 142. mÙkj (3) dkafMªDFkht – dksjdsjksMksu] çhfLVl vksfLVDFkht – VsjksfQye ,oht ¼i{kh½ – dkoZl Lru/kkjh – Vsjksil mÒ;pj – bfDFk;ksfQl 143. mÙkj (1) i{kh rFkk Lru/kkjh lerkih gksrs gSaA 144. mÙkj (3) fNidyh rFkk liZ viuh 'kYd dks Roph; dsapqy ds

:i esa NksM+rs gSaA 145. mÙkj (4) cSafdvksLVksek ,d lsQsyksdkWMsZV gSA 146. mÙkj (2) fyeqyl ds Dykse esa ysfeyk ¼iVfydk½ uked dbZ

ifêdk tSlh lajpuk,a gksrh gSaA ;s ysfeyk ¼iVfydk½ ,d nwljs ds lekUrj gksrh gSa tks iqLrd ds iUuksa ds ln`'; gksrh gSa blfy, bls iqLr Dykse dgk tkrk gSA

147. mÙkj (1) lhlh eD[kh ds dkj.k fVªisukslksfe,fll gksrk gS 148. mÙkj (3) lhysUVªsV~l – Qkblsfy;k] ,Mefl;k] esUMjhuk]

vkscsfy;k] vksjsfy;k 149. mÙkj (2) ljhl`iksa rFkk Lru/kkfj;ksa ds chp vksjfuFkksfjadl

¼IySVhil½ ,d la;kstd dM+h gSA 150. mÙkj (3) MsUVsfy;e – jn dopj dhVksIy;wjk – dkbVu ikbyk – lsc ?kksa?kk


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151. Answer (2) In hemichordates excretory organ is proboscis

gland. 152. Answer (3) Reptiles are uricotelic animals. 153. Answer (2) Nereis possesses parapodia which helps in

swimming as well as in respiration. 154. Answer (1) Many living members of class Cyclostomata are

ectoparasites on some other fishes. 155. Answer (3) In Osteichthyes air bladder is present which

regulates buoyancy. 156. Answer (3) Nereis is dioecious but earthworms, many

tapeworms and leeches are monoecious. 157. Answer (2) Anterior head region of molluscs has sensory

tentacles. 158. Answer (1) Echinoderms e.g. Asterias and Antedon have

tube feet for locomotion 159. Answer (3) In Chondrichthyes gill slits are separate and

without operculum. 160. Answer (1) Myxine (Hag fish) is included in Division Agnatha. 161. Answer (3) Betta is fighting fish. 162. Answer (2) All vertebrates are chordates. In

cephalochordata, notochord extends from head to tail region.

163. Answer (2) In chordates central nervous system is dorsal

hollow and single. 164. Answer (3) A cloaca is not present in placental mammals or

in most bony fishes. 165. Answer (4) Poriferans exhibit cellular level of body

organisation. 166. Answer (1) Ctenoplana, Ophiura and Saccoglossus are

exclusively marine animals.

151. mÙkj (2) gsehdkMsZV~l esa 'kq.M xzafFk mRlthZ vax gksrh gSA 152. mÙkj (3) ljhl`i ;wfjd vEy mRlthZ çk.kh gaSA 153. mÙkj (2) usfjl esa iSjkiksfM;k gksrk gS tks rSjus ds lkFk lkFk

'olu esa Hkh lgk;rk djrk gSA 154. mÙkj (1) oxZ lkbDyksLVksesVk ds dbZ ltho lnL; dqN vU;

eNfy;ksa ds cká ijthoh gksrs gSaA 155. mÙkj (3) vksfLVDFkht esa ok;q dks"k mifLFkr gksrs gS tks

mRIykou dks fu;af=r djrk gSA 156. mÙkj (3) usfjl ,dfyaxkJ;h gksrs gSa ijUrq dsapq,] dbZ

QhrkÑfr rFkk tkasd mÒ;fyaxkJ;h gksrs gSaA 157. mÙkj (2) eksyLd ds vxz 'kh"kZ {ks= ij laosfnd Li'kZd gksrs gSaA 158. mÙkj (1) ,dkbuksMeZ mnkgj.k ,LVsfj;l rFkk ,aVhMksu esa

xeu ds fy, uky ikn gksrs gaSA 159. mÙkj (3) dkaMªhfDFkt esa Dykse fNnz i`Fkd ,oa çPNn jfgr

gksrs gSaA 160. mÙkj (1) feDlhu ¼gSx fQ'k½ fMohtu vXusFkk esa 'kkfey gksrk gSA 161. mÙkj (3) csVk] QkbfVax fQ'k gSA 162. mÙkj (2) lÒh d'ks:dh jTtqdh gksrs gSaA lsQsyksdkWMsZVk esa

i`"BjTtq flj ls iwaN rd QSyh jgrh gSA 163. mÙkj (2) jTtqdh esa dsUnzh; raf=dk ra= i`"Bh; [kks[kyk rFkk

,dy gksrk gSA 164. mÙkj (3) vijk Lru/kkjh ;k vf/kdka'k vfLFky eNfy;ksa esa

voLdj mifLFkr ugha gksrk gSA 165. mÙkj (4) iksjhQsjsu dksf'kdh; Lrj dk 'kjhj laxBu çnÆ'kr

djrs gSaA 166. mÙkj (1) VhuksIykuk ] vksQh;wjk rFkk lSdksXyksll dsoy leqnzh

çk.kh gSaA


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167. Answer (2) Body of hemichordates is divided into proboscis,

collar and trunk. 168. Answer (3) Ascaris is dioecious i.e. male and female are

separate. 169. Answer (4) Radial symmetry is present in coelenterates,

ctenophores and adult echinoderms. Molluscs are bilaterally symmetrical.

170. Answer (1) Mesoglea is present in diploblastic animals. 171. Answer (2) Notochord is a mesodermally derived rod-like

structure. 172. Answer (3) Obelia exhibits metagenesis. 173. Answer (4) 174. Answer (4) 175. Answer (1) Excretory structure Gills – Molluscs Malpighian tubules – Arthropods Flame cells – Platyhelminthes 176. Answer (3) Excretory system is absent in echinoderms. 177. Answer (2) Carcharodon belongs to class Chondrichthyes. 178. Answer (4) Torpedo is an electric ray. 179. Answer (2) Elephas is the largest terrestrial living animal. 180. Answer (3) Hemichordates have a rudimentary structure in

the collar region called stomochord, a structure similar to notochord.

167. mÙkj (2) gsehdkWMsZV~l dk 'kjhj 'kqaM] dkWyj rFkk o{k esa

foÒkftr gksrk gSA 168. mÙkj (3) ,Ldsfjl ,dfyaxkJ;h gksrk gS vFkkZr~ uj rFkk eknk

i`Fkd gksrs gSA 169. mÙkj (4) vjh; lefefr lhysUVsªsV~l] VhuksQksj~l rFkk o;Ld

,dkbuksMeZ esa mifLFkr gksrk gSA eksyLd f}ik'oZ lefefr dk gksrk gSA

170. mÙkj (1) ehtksfXy;k] f}dksjdh çkf.k;ksa esa mifLFkr gksrk gSA 171. mÙkj (2) 'kykdk :ih i`"BjTtq e/;Ropk ls mRiUu gksrh gSA 172. mÙkj (3) vkscsfy;k dk;kUrj.k çnÆ'kr djrk gSA 173. mÙkj (4) 174. mÙkj (4) 175. mÙkj (1) mRlthZ lajpuk Dykse – eksyLd eSyihxh ufydk – vkFkzksZiksM~l Tokyk dksf'kdk,a – IysVhgsfYeUFkht 176. mÙkj (3) ,dkbuksMeZ esa mRlthZ ra= vuqifLFkr gksrk gSA 177. mÙkj (2) dkjdsjksMksu oxZ dkaMªhfDFkt ls lacaf/kr gksrk gSA 178. mÙkj (4) VkWjihMks ,d fo|qr js gSA 179. mÙkj (2) ,fyQl lcls cM+k LFkyh; ltho çk.kh gSA 180. mÙkj (3) gsehdkWMsZV~l esa LVkseksdkWMZ uked dkWyj {ks= esa ,d

vYifodflr lajpuk gksrh gS] ,d lajpuk tks i`"BjTtq ds leku gksrh gSA

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