Solution BPKIHS Model Exam Set-III 2069-11-19

7/30/2019 Solution BPKIHS Model Exam Set-III 2069-11-19

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NAME Model Exam solution for BPKIHS Set-II I 2069-11-19

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Model Entrance Exam -2069

For

BPKIHS Set-III

Hints and Solutions

In front of Singhdurbar, PutalisadakKathmandu Tel: 01- 4231144

Date: - 2069-11-19


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Physics

1. d)

RP

2R

P

Displacement = (R)2 + (2R)2 = R 2 + 4

2. b) L = r p

=

i

2

2

j1

3

k1

1

= i (1 + 3) j (2 2) + k ( 6 2) = 4 i 0j 8k3. a)

u = 0 = aceeleration v = max = retardation v1 = 0

A t1 B t2 C v = u + t1 v1 = v t2v = 0 + t1 0 = v t2

t1 =v ............. (1) t2 =

v ............. (2)

(1) + (2), t1 + t2 =v +

v t = v

+ v =

t +

4. b) V = 2gh = 5Rg2gh = 5Rg

2gh = 5D2 g

h = 54 D

5. a) From work energy theorem,Work done by the various forces = change in kinetic energy

Fx m1gx 12 Kx

2 = 0

But Kx = m2g for just shifting m2.

Fx m1gx12m2gx = 0

F = m1 +m22 g = 0.4 1 +

22 10 = 8N

h = ?

u = 0

v = 5RgR


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6. d) WA B = m(VB VA) + (KB KA) 10 = 2 (VB VA) + 4 VB VA = 7 J/kg

7. d) Average speed =distance covered in one oscillation

time period

=

4r

T =

4r

2

=

2r

=

2Vmax

8. c) p = ghm = v = constant

vv =

B =p

vv

=gh

=2gh

B

9. b)10. a) [Torque] = [Work] = [ML2T2]11. a)12. c)

13. c) V0 =GM

R MV1V2 =

M1M2 =

M2M = 1: 2

14. d) W = Fc S cos90 = 0

Fc

90 S

15. c)

16. c) Angle of shear =rl angle of twist =

4 1031 30 = 0.12

17. c) P2V2 = P1V1 r2 = 2r1 v2 = 23v1 Pa 23v1 = (Pa + hwg) v1 8Pa = Pa + hwg

hwg = 7Pa hwg = 7 H wg h = 7H

18. a)

19. b) t =12t

5 =12 ( 15)t ............ (1) 10 =

12 (30 ) t .......... (2)

(1) (2), we get5

10= 15

30 20C


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From equation (1),12 ( 15) t = 5

1215) 86400 = 5 =

143200 = 23 10

6 /K

20. c) Q = U W nCPT = nCVT nRTCP = CV R CV = CP + R Cp < Cv

21. c) C.P =T2

T1 T2 =10 + 273

(42 + 273) (10 + 273) =28332 = 8.84

22. d) M (CP CV) = R M (525 315) = 8.31M = 0.0392 kg M = V 0.0392 = 22.4 10-3 = 1.75 kg/m3

23. c)24. b) Let S1 and S2 be distance of man from either side of valley, then,

2S1 = Vt1, 2S2 = Vt2 2(S1 + S2) = V(t1 + t2) S1 + S2 =

V(t1 + t2)2 d = S1 + S2 =

330(1 + 2)2 = 495m

25. a) Second harmonics is produced in string when string vibrates in 2loops

L = 2(2 ) = = L

26. d) Intensity level L2 L1 = logI2I1 (In bel)

3=logI2I1 103 = I2I1a

22

a12 =103 a1a2 = 110 10

27. b) Apparent depth =d1

+d

2 = d(11

+1

2 )

28. b) Dispersion 12

D2D1 =

2

2

D2D =

14 D

2 = D/4

29. c) Focal length in liquid 'fl' = f

110

7

3

5

13

5

fl= 4f

30. d)

31. a) B =0NI2a = 2 10

-6Tesla

32. c) When current is passed through solenoid, coil contracts due toattraction between turns.

33. a)34. b)35. c) Magnetic moment (M) = m 2l

'M' depends on length but pole strength (m) does not depend onlength.


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36. d) R =lA =l A

A A =VA2 (V = constant) R

1D4

37. c) I2 R3

38. c) E =1

2

C1C2

C1 + C2(V1 V2)2 =

1

4 C (2V)2 = CV2

39. d) mg = qE q =mgE ne =

mgV/d

n =mgdVe =

3.2 10-5 10 3 10-3100 1.6 10-19 =6

q = 6e

40. c) Penetrating power = Energy = hf eV = hf

41. b) By 1 and 2 decay there produced isotopes

42. a) R = R0A1/3 m1v1 + m2v2 = 0, V1

V2 =m2m1 (numerically)

A1A2 = 8

13

2

1

R

R

V1V2 =

m2m1 =

81

43. a)

44. b) =ICIE =

IE - IBIE 0.95 =

10- IB10 IB = 0.5mA

45. a)

46. b) u = 2e3 d = -e3 udd = 2e3 + -e3 + -e3 = 0

Quark combination of a neutron is udd.

47. b) En =-13.6

n2 eV

Energy emitted = E4 E2 =-13.6

42 --13.6

22 = 2.55eV

48. b) m =

2

2

0

1 C

V

m

=

2

2

0

1 C

C

m

=

49. d) t = 10T1/2 = 10 5 = 50 years approx.50. d) The threshold wavelength 5200 belongs to blue light. The

wavelength of light required for photoelectric emission should beless than 5200 i.e, infrared radiation can't eject photoelectronsbut ultraviolet radiation can eject photoelectron whatever be itspower.

Chemistry

51. b) rms =3RTM =

3R 140M at 140K


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'rms =3R 560

M at 560K 'rms =2 rms

52. b) E =hC = h

53. b) Px orbital has electron density along x-axis.

54. c) Na2311 Ne2310 +

e1 55. c) C2, N2 and F2 has no unpaired electron in their molecular orbital

configuration.56. d) Dipole forces exist only in polar molecule.57. b) IE1 for N > IF, for O due to half-filled nature in N.58. c) Eq. of metal = Eq. of oxygen

Thus,1.05

E 3.15 - 1.05

8 E = 4

59. d) BaCl2 + H2SO4

BaSO4 + 2HClMM taken 20 0.5 20 1 - -=10 20

MM found 0 10 10 10Milli mole of BaSO4 = 10

or, Mole of BaSO4 = 10-2

60. a) Fe2+Fe3+ + 1e 6e + Cr26+ 2Cr3+61. a) Cathode: H+ + e (1/2)H2

Anode: 2OHH2O + 1/2O2 + 2e62. d) It does not depend upon mass.63. d) S 2O SO2 2Cl SO42- 2BaCl BaSO4

One mole of S will give one mole of BaSO4. Thus, mole of BaSO4formed = mole of S = 8/32= 1/4

64. c) Addition of a nonvolatile solute always lowers the vapourpressure.

65. a) Tb= )21(10004.134

44.1352.01000

= 1Tb=0.156

66. d) log x/m = log K + 1n logP; this is Freundlich isotherm.

Thus slope =1n .

67. b) 2SO2(g) + O2(g) )(sPt 2SO3; phase for reactant for catalyst =2, thus heterogeneous.

68. c) Colloidal solution shows scattering of lightTyndall effect69. a) H=Ea.F.R. Ea.B.R.

70. d) r1 = K[A]1

, r2 = K[A]2

, r3 = K[A]3

. If [A] >1; r3> r2> r171. b)

normal mol.wt.exp.mol.wt


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= 1+ . For NH4Cl NH3 + HCl

Exp. mol . wt. =normal mol.wt.

2 (since =1)

72. b) The trend is based on back bonding concept.73. a) Isohydric solutions of two acids contains same [H+] ion in them.

74. b) The degree of hydrolysis of a salt of weak acid and weak base isindependent of concentration of salt.

75. c) The fact for a quantity referred as state function.76. d) S = Ssystem+ SSurrounding

S > 0, for spontaneous process.77. b) Same magnetic moment = same number of unpaired e-s =

)2( nn .Where, n = number of unpaired electronCo2+ = 3d7, 3 unpaired electronsCr2+ = 3d4, 4 unpaired electronsMn2+ = 3d5, 5 unpaired electronsFe2+ = 3d6, 4 unpaired electrons

78. d) Haemoglobin, cytochrome contain Fe.79. d) Dipole-dipole attraction give rise to more tendency for

aggregation.80. a) Due to ammonia solvated electrons.81. b) FeO + SiO2FeSiO3.82. a) FeCl2 and CuCl2 are green.

83. a) Cisplatin.84. d) Silver salts on heating gives Ag.85. d) Starch gives blue color with iodine.86. c) SN1 mechanism give rise to 50% inversion as it involves front side

as well as backside attack. This leads to racemic products.87. c) isoamylene is isopentene.88. d) SbCl5 will pull Cl- to form SbCl6- and carbocation.

89. d)

90. c) In 1,1 dichloro 1 pentene (Cl2C = CHCH2CH2CH3), one of thedoubly bonded carbon atom(C1)has a similar atoms, hence,geometrical isomerism is not possible.

91. d) C2H5NH2 2HNO C2H5OH 5PCl C2H5Cl 3NH C2H5NH2 3NH ( C2H5)2NH NH3(C2H5)3N

92. c) In Beilstein test halogens burn with green edge flame on Cu wire.93. b) LiAlH4 is weak reducing agent. It does not affect the double bond.94. c) It is non-terminal alkyne.95. d) It is Libermann nitroso reaction.

96. c)

+Anomers C6H5 CH CH3

2CH CH CH2 = CH C CH CH2 = CH C(Cl) Chloroprene

Cu2Cl2NH4Cl

Cu2Cl2


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97. a) (CH3)4N+ I-P

NaOH

, (CH3)4N+OH- + NaI(CH3)3N + CH3OH

98. d) The enzyme catalyzed conversion of starch into sugar.

99. a) Amoxicillin and ampicillin are modified antibiotics of penicillin.100. b) Osazone formation involves oxidation of two carbons.

Botany101. b) Interferons are antibody like proteinous compound produced by

certain cells of vertebrates in response to specific viral infections.102. b) The Klenfelters syndrome is trisomic condition (2n+1) of sex

chromosome in which the person with turner syndrome has 47chromosomes in 44+XXY condition.

103. b) The interchange of parts of the chromatids of a pair ofchromosome (homologous chromosomes) is called crossingover.Gene mutation involves change, loss, gain, addition or exchangeof gene between non-homologous chromosomes.

104. b) Betel (Piper) has climbing adventitious roots produced from nodesof stems.

105. b) The parenchyma is simple permanent tissues having polygonalstructure with up to 14 angles.

106. a) As compare to each other, the daughter cells produced just aftermitosis have samenumber of chromosomes and equal amount of

DNA.107. d) Glyoxisome are common character plants cells.108. b) The karyon is nucleus and karyology is the study of nucleus.

Karyogamy is fusion of compatible nuclei.109. b) Oospore is diploid spore produced through gametic copulation. It

requires motile gametes which are absent in Spirogyra.110. c) Ginkgobiloba, a gymnosperm is considered as living fossil, because

its several species has already being fossilized and very fewspecies are still found in living forms.

111. d) Majority of plant takes nitrates (NO3) form soil; limited plant cantake nitrogen in the form of nitrites (HNO2) or amino acids.112. a) Thick cuticle, reduced, palisade tissues, well developed roots,

tissues with well-developedmechanical tissues are commonadaptive character of xerophytes.

113. c) 2-3 spirally coiled male gametes are found in Dryopteris. The malegamete of Cycas has 4-5 bands of spiral coiling while Selaginellahas 2-3 spiral bandswith two flagella.

114. d) The outer seed coat or testa is produced from outer integument ofovule while fruit wall or pericarp is produced from vary wall orcarpel wall.


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115. c) Siphonogamy is formation of pollen tube during pollination. Inabsence of canal cells, pollen tube is needed for conduction ofmale gametes.

116. d) The botanical name of gram isCicerarientinum.Lathyrusodoratus,Cajanuscajan, Pisumsativa are commonly called sweet pea, chicken

pea and garden pea respectively.117. d) Shorea(sal) is typical example of wood yielding plant.118. b) The venter canal cell of archegoniumlies just above egg cell and

below neck canal cells.119. a) The cricket ball is produced from phellem of Quercus. The cricket

bat is produced from wood of Salix.120. a) Inleptosporangiatesporogenesis, single mother takes parts in the

formation of spores. It is advance character and found inDryopteris.

121. b)

The epistasis is suppressing of expression of one geneby othergene. During this process, the suppressor gene is called epistaticgene and the suppressed gene is hypostatic gene.

122. b) The sunflower belong to dicot (family Malvaceae) while Cyanodon,maize and Triticum belong to monocots (family Poaceae orGraminae).

123. c) The thallus of Volvox is colonial and show group movement calledcoenobium.

124. a) Several auxins like Indole Acetic Acid (IAA), Indole Butyric acid(IBA) and Naphthalene Acetic Acid (NAA) are also considered as

rooting hormones which can initiate adventitious roots.125. b) Potassium ion theory is related with opening and closing of

stomata. The influx of K+ ion inside guard cells causesendosmosis and stomata open while during out flux of K+ ionfrom guard cells, stomata close.

Zoology126. a) Species inhabiting different geographical areas are allopatric.

Related species which are reproductively isolated but more

physiologically similar are called sibling species.127. b) E.gingivalis lives on the surface of the teeth and gums and alsoin the gingival pockets near the basic of teeth.E.coli non-pathogenic in humans.

128. d) Exo-erythrocytic schizogony (2nd liver schizogony) occur in livercells only.

129. b) Ora serrata is the margin of the sensory retina130. d) Modulator = Hypothalamus

Effector = Islets of langerhans producing insulin, glucagon &somatostatin

131. c) Filariaworm is a viviparous nematode, larvae called microfilaria.


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132. b) Septal nephridia with complex nephrostoma, poured thenitrogenous waste into the intestine. Septal nephridia isenteronephric.

133. c) Johnston's organ lies in the second segment of antennae. In malemosquito, it helps to locate females by flight tone.

134. a) Cardiac muscles have long cylindrical, branched and uninucleatecells divided at places by intercalated disc.135. b) Nissl bodies consist of granular str. In the cyton that stains with

basic dyes and contains ribonucleo protein (RNA)136. d) Second cervicle vertebra is the axis, it has odontoid process.137. d) A sperm enters into the ovum at some point in animal

hemisphere.138. c) Endostyle of protochordate, eg:- cephalochordata is an organ

associated to feeding. It, in vertebrates, is modified into thyroid

gland.139. c) Archaeopteryx is a fossil bird which lived during upper Jurassicperiod of mesozoic era. This primitive bird is considered to be aconnecting link between reptile and birds, as it possesses bothreptilian as well as an avian characters.

140. a) Neanderthal man protruding jaw, slightly prognathus face.Cro-Magnon man Face perfectly orthognathus

141. b) Secretin stimulates pancreas possibly liver to secrete fluid baseof pancreatic juice.

142. a) Tachycardia is the condition where heart rate exceeds 90 per

minute for an average adult.143. a) Echinoderms are triploblastic, radially symmetrical often

pentamerous, larva with bilaterally symmetrical, sessile orcreeping at sea bottom.

144. a) Foramen of Monro connects lateral ventricle with dioceol (IIIventricle). Foramen of luschka present on the lateral wall ofmetacoel.

145. b) The first stage larva grow for a weak when it moults (sheds incuticle ) and becomes the second stage juvenile. The later is

infective and also known as the infective juvenile.146. b) Clitoris : the female counter part of the penis which containserectile tissue but is unconnected with urethra. Like the penis itbecomes erect under conditions of sexual stimulation, to which itis very sensitive.

147. b) Syndactyly fused digits eg:- marsupial, primatesCurved ribs = easy to climb

148. b) Diapedesis:- the movement of WBCs through narrow capillarieslike that of an Amoeba is called Diapedesis.

149. a) Stinging cells or cnidoblast cells found only in epidermis havingnematocysts.


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150. d) Cutaneous glands (mucous and poison) and chromatophores(colouration) is found in stratum spongiosum of Dermis.

English + Health + G.K

151. a 152. c 153. a 154. d 155. c 156. c 157. a 158. d

159. c 160. c 161. d 162. a 163. d 164. b 165. a 166. a

167. a 168. d 169. d 170. c 171. a 172. d 173. b 174.a

175. b 176. b 177. c 178. d 179. c 180. b 181. b 182. b

183. d 184. b 185. b 186. c. 187. b 188. a 189. d 190. c

191. b 192. d 193. c 194. b 195. d 196. a

197. b 198. c 199. a 200. a

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Solution BPKIHS Model Exam Set-III 2069-11-19