2) Kinetic Theory - Solutions

1. At Law pressure and high temperature gases behave asan ideal gas

2.

3.

4. Gas A is ideal because pV is constant for gas A so gas Aobeys Boyle’s laws for all values of pressure.

5. When air is pumped more molecules are pumped inBoyle’s law is stated for situation where number ofmolecules remain constant.

6. The minimum possible temperature on the basis ofCharle’s law is -273.150C.

7. AsIf temperature of the gas is increased 4 times thenis doubled.

8. Due to the presence of friction between the road andtyres the tyres get heated as a result of whichtemperature of air inside the tyre increase and hencepressure in tyre also increases.

9. Keeping p constant we have

10. If a gas is heated then the root mean square velocity ofits molecules is increased.

The temperature of the gas increase.11. Yes, the molar specific heat of a solid is a constant

quantity as its value is 3 cal/mol-K.12. Three because bee is free to move along x direction or y

direction or z direction.13. Degree of freedom f = 3 N - K

14. As we know mean free path

Given, and

15. The average discuss travelled by a molecule betweentwo successive collisions is known as mean free path ofthe molecule.

16. The mean free path is inversely proportional to thenumber density of the gas.

17. Argon is a monatomic gas, so

18. For 1 mole with f degrees of freedom.

Internal energy,

For n moles,

19. Brownian motion and diffusion of gases provideexperimental evidence in support of random motion ofgas molecules.

20.

21.

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23.

24. The average kinetic energies are equal because they areat the same temperature.

25. (a) (iii) table(b) Volume of the room, V = (3 × 4× 6)m3 = 72m3

Mass of air in the room = volume × density=

(72 m3) (1.3 kg/m3)=

93.6 kgObviously, this cannot be mass of a (i) a pin, (ii) apencil or a truck. It is, therefore, comparable to themass of the table.

26.Gas Molar Specific Heat, CV (= atomic

mass × cv)Helium 4.00 × 0.748 = 2.992 cal/mol KNeon 20.18 × 0.147 = 2.966 cal/mol KArgon 39.94 × 0.0760 = 3.035 cal/mol KKrypton 83.80 × 0.0358 = 3.000 cal/mol KXenon 131.3 × 0.226 = 29.673 cal/mol KWe find that CV is nearly constant and is approximatelyequal to 3 cal/mol K. It should be so on the basis ofkinetic theory of gases according to which in case of amonoatomic gas, CV = (3/2)R or CV = (3/2) × 2 cal/mol K= 3 cal/mol K.

27. All the gases listed in the table are diatomic and for suchgases CV = 5 cal/mol K. A chlorine gas moleculepossesses vibrational mode of motion and that accountsfor the slightly higher value of CV in its case.

28. (a) For the dotted line, PV/T = constant (R), i.e.,PV = RT and as such it represents the behavior of anideal gas.(b) as the gas equation PV = RT is true only athigh temperature (where one can neglectintermolecular attraction). Curve for T1 is closer to thedotted line than curve for T2.

(c) As PV = nRT,

In the present case, n (number of moles of oxygen in

1.00 × 10-3 kg) = 1/32. Thus,

= 0.26 J/K(d) No, as the value of (PV/T) depends upon the mass ofthe gas. Let m be the required mass of hydrogen. Thus,

For nR = 0.26 J/K,

Thus, m = n (2.02 g) = (0.031) (2.02 g)= 0.063 g =


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29. All gases have the same number of molecules(Avogadro’s law).As root mean square speed, , it isdifferent for all gases as is different for all gases.Neon being the lightest of the three, has the largestvalue for

30. Yes, The temperature of a gas can be increased bycompressing it. The work done in compressing the gas isconverted into its internal energy which results in anincrease in its temperature.

31. When the volume of a gas is increased, its pressuredecreases on account of the fact that : (i) the moleculeshave to travel a longer distance between impacts on thecontainer walls and (ii) these impacts are nowdistributed over a large area.

32. Self (Teacher)33. The temperature is on account of the translational

molecular motion. At absolute zero, this molecularmotion completely stops. Obviously, a temperature lessthan absolute zero is not possible.

34. According to the Dalton’s law of partial pressures, totalpressure,

35. Self (Teacher)36. Self (Teacher)37. Self (Teacher)38.

39.

40.

41.

42.

43. (a) It is a straight line.(b) It is a straight line.

44. A is an ideal gas because PV is constant at constanttemperature for an ideal gas.

45. (i) Boyle’s law(ii) Charle’s law(iii) Gay Lussac’s law.

46. Yes. This is because the mean K.E. per molecule i.e.,

depends only upon the temperature.

47. It will be doubled. ( if other factors are constant).

48. (a) (2+1) = 3 atmosphere.(b) Yes because the average translational K.E./

molecules depends upon the temperature

(c) Yes because the r.m.s. velocity depends not onlyupon temperature but also upon mass.

49. At very low temperature, Charle’s law breaks down.50. (a) The pressure of the gas increases. This is because the

pressure of a gas is proportional to the absolutetemperature of the gas if V is constant.(b) Applying boyle’s law find that the pressure increase .

51. It is defined as that temperature at which all moleculemotions cease.


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52. Since water has a high value of specific heat so it can beused as a coolant.

53. Molar specific heat at constant volume is for

monatomic gas and R for diatomic gas.

So diatomic gas has more specific heat than amonatomic gas .

54. Average speed is the arithmetic mean of the speeds ofthe molecules .

Average speed

rms speed is the root mean square sped and is definedas the square root of the mean of the squares ofdifferent speeds of the individual molecules.

rms speed

55. On pulling the piston out, the volume of the cylinder forthe given gas increases. The molecules of gas get morespace to move about. As result, lesser number ofmolecules will collide with the walls of cylinder persecond and hence less momentum is transferred to thewalls per second. Further, these collision take place onthe larger area of the walls. Due to both these reasons,the pressure decreases.

56. The basic properties of the molecules of an ideal gas are(i) zero size of the molecules and hence, zero volume ofthe molecules and (ii) no mutual intermolecular forcesbetween them . at low pressure, the volume of the givengas becomes large. Therefore the volume of themolecul.es becomes negligible in comparison to thevolume of the gas. At high temperature, the moleculeshave large K.E. and so the effect of the intermolecularforce on the motion of the molecules becomesnegligible. Hence, at low pressure and high temperature ,the real gases behave as ideal gases and gas equations isobeyed.

57. As explained in Problem 7, evaporations occurs onaccount of faster molecules escaping from the surface ofthe liquid . The liquid is therefore left with moleculeshaving lower speeds. The decreased in the averse speedof the molecules results in lowering the temperature.That is why cooling is caused by evaporation.

58.

59.

60. Gas C is ideal, because PV is constant for this gas. Itmeans the gas C obeys Boyle’s law at all pressures.

61.

62.

63.

.

64. (i) Yes, because the mean K.E. per molecules

depends only upon the temperature.

(ii) No, because for different gases, rms velocity dependsupon the mass of the gas molecules.(iii) No definite idea about pressure because masses ofthe gases are not given.

65. 273.15 K66. C . Therefore, the r.m.s. velocity Hence,

Increases in r.m.s. velocity = = 0.732 C = 73.2%

67. In the absence of forces of attractions amongst themolecules of a gas, the molecules will hit the walls of thecontainer harder. Therefore, the pressure exerted by thegas molecules would increases.

68. This is because molecules do not uninterrupted . Theyhave random motion. The scent vpour moleculesundergo a number of collisions and terrace a zig zagpath. That is why their effective displacement per unittime is low and spreading is at a much slower rate.


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69. On driving a scooter for a long time, the work doneagainst frictions is converted into heat. The gas in thetyre gest heated and hence the pressure of the gasincreases, because P T.

70. As, C2 , but C2 T ; therefore if V and T are

constants As M becomes 2 M, therefore, Pbecomes 2P.

71.

72. The average kinetic energy of gas molecules (i) isindependent of the nature of the gas.(ii) Depends upon the temperature of the gas as

=

(iii) Is independent of the volume of the gas.At absolute zero, rms velocity of the gas moleculesbecomes zero so mean K.E. per molecules of the gasbecomes zero at absolute zero.

73.

74. K.E. per molecules of a gas is directly proportional totemperature. Hence K.E. Will become four times, therms velocity is directly proportional to the square root oftemperature, hence, rms velocity becomes twice.Pressure is directly proportional to (rms velocity)2 .Hence, pressure becomes 4 times.

75.

76. Equal to more than one

77.

78. 4 hours79.

80. An object can be considered as a point object if (a) Itssize is negligible in comparison to the distance travelledby object (b) Its size is not negligible in comparison tothe distance travelled by object.

81. When an object is moving with uniform velocity along astraight line.

82. Either less than or equal to one.83. Instantaneous speed or instantaneous velocity at the

given instant.84. (a) Infinity

(b) Zero in fact it is not a piratical situation .85. The instant of time which is taken after the origin of

time (i.e. zero time) is called positive time and theinstant of time which is taken before the origin of time iscalled negative time.

86. Time taken for distance (s1) is given by,Time taken for distance (s2) is given by,Total distance coveredTotal time taken

87. .

88. Total time in going and coming back,

total distance travelled, S = 6 + 6 = 12 km ;

Average speed .

89. (i) A flying bird in the sky is a three dimensional motion(ii) A football kicked by a player is a two dimensionalmotion(iii) Earth revolving around the sun is a two dimensionalmotion(iv) The motion of the bob of a simple pendulum is onedimensional motion.

90. Speed is a scalar quantity. The speed of an object can bezero or positive but never negative. Velocity is a vectorquantity in fact velocity = speed + direction the velocityof an object can be positive zero and negative.


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91. No because the distance travelled by an object is theactual path traversed by the object during motion in agiven interval of time the displacement of an object in agiven interval of time is given by the vector dawn formthe initial position to final position.For illustrations : (i) when an object completes a circularmotion on a circle of radius r, then its displacement iszero but distance travelled is(ii) When an object moves on a triangle from A to B, B toC and C to A in a given time, its displacement in time t iszero but the distance travelled is = AB + BC + CA.

92. Average velocity = displacement/(total time taken) andaverage speed = total distance travelled/(total timetaken). If an object completes a circular path of radius rin time t, then its displacement is zero but distancetravelled by body is Therefore, the average velocityof body = zero but average speed of body

93. Since, the particles while moving with constant speed, are always at a constant distance d, they must be at

the two ends of the diameter of a circular path. Thediameter of this circular path = d. Each particle willreturn to its initial position after describing a circularpath =Time interval after which each particle will return to itsinitial position = distance travelled/speed

94. Displacement travelled in time

95. In 2 minutes 20 seconds, an athelete will complete,three complete rotations and one half rotation, i.e., itwill be at the opposite end of the diameter from the

starting point. So displacement .

96. Let u and be the velocities of two bodies. As perquestion, and or 2u = 10On solving we get and .

97. 20 m98. 9.8 m99. 1/4100. This graph is useful in following ways:

(a) To determine the velocity of the object at the giveninstant.(b) To determine the acceleration of the object.(c) To determine the total distance travelled by theobject in a given time.

101. We consider only the acceleration and not the rateof change of acceleration in a given motion of the objectbecause the basic laws of motion involve onlyacceleration and not the rate of change of acceleration.

102. Given, or where k is a constant ofproportionality.

speed or velocity of the particle,

……(i)

and acceleration of the particle,

…….(ii)From (i) and (ii) we note that acceleration of the particleis constant whereas the velocity of the particle dependsupon the time.

103. For no collision of two trains, the relative velocity offaster train w.r.t. slower train shouldbecome zero, while travelling a relative displacement dwith acceleration a.Here; S = d.As, So, or

.

There will be no collision if .

104. Acceleration is 9.8 ms-2 acting downwards andvelocity is zero.

105. Acceleration =

106. (a) Velocity is vertically upwards and acceleration isvertically downwards(b) Velocity is vertically downwards and acceleration isalso vertically downwards.

107. Velocity time graph is of the type fig. as given below


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108. It is the velocity which decides the direction ofmotion of a body. The acceleration simply tells the rateof change of velocity for example when a body is thrownvertically upwards its direction of velocity is upwardsthat is why the body goes upward where as itsacceleration is downwards.

109. The area under acceleration time graph for a giveninterval of time represents the change in speed of thebody during that interval of time.

110. Since a stone thrown vertically up with speed ureturns back the same point with the same speed henceboth the stones will strike the ground with the samespeed hence the ratio of the speeds of the two stoneswill be 1 : 1.

111. Given, SLAP

(i) The SI unit of SLAP is

(ii) Instantaneous SLAP is the value of SLAP at the giveninstant, which is the first derivative of acceleration atthe given instant and can be written as Instantaneous

SLAP

112. Yes it will be show if an object is moving withconstant speed in a circle in this case the object isacceleration but its speed neither decreases norincreases.

113. or Integrating it, we have

when then

114. Let u be the initial velocity of projection of body andbe the velocity of the same body while passing

downwards through point of projection. Thedisplacement of body s = 0. Using the relation

as, and ; wehave orIt means the final speed is independent of mass of thebody. Hence both the bodies will acquire the samespeed while passing through point of projection.

115. Acceleration,

Given, Differentiating it w.r.t. x, we

have

So, acceleration,

116. Acceleration, or

Integrating it we have

Now, velocity or

Integrating it , we have,

117.

118. 4.5 s119.

120.

121.122. Two vectors can be added only if they have the

same nature i.e. force can be added to force and not todisplacement or velocity.

123. The finite rotation about an axis is not a vectorbecause its addition to another finite rotation about adifferent axis does not obey commutative law ofaddition.

124. No. For example ; the electric current possessingboth the magnitude and direction is not a vectorquantity but is a scalar quantity.


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125. Yes. Magnitude is the same when q between andis 90o and is different if q is other than 90o.

126. The two vectors are parallel and acting in the samedirection (i.e. q = 0o).

127. (i) When three vectors are not lying in one plane,they can not produce zero resultant.(ii) When three vectors are lying in a plane and arerepresented in magnitude and direction by the threesides of a triangle taken in the same order, they canproduce zero resultant.

128. The sum of two vectors is maximum, when both thevectors are in the same direction and is minimum whenthey act in opposite direction.

129. Three forces, provided they can be represented bythe three sides of a triangle taken in the same order.

130. (i) Two only in a plane (ii) Any number of componentvectors in a plane.

131. Yes, when a vector is multiplied with a dimensionalscalar, the resultant vector will be have differencedimensions. For example, if acceleration accelerationvector is multiplied with mass (a dimensional scalar), theresultant vector has the dimensions of force.When a vector is multiplied with non-dimensional scalar,it will be a vector having dimensions as that of the givenvector.

132. Here, Ax = 2, Ay = 2 Let q be the angle which themake with x-axis then tan q = Ay/Ax = 2/2 = 1 or q =

45o.

133. Yes, it will be negative if the angle between the twovectors is between 90o to 270o.

134. As , therefore, AB cos q = AB sin q ortan q = 1 or q = p/4.

135. Area of parallelogram

(Given)

or

136. Force is a vector quantity. The two forces can beadded by laws of vectors ; parallelogram law of vectorsor triangle law of vectors which is possible if we knowthe angle between the two forces. As the angle betweenthe two given forces is not known, hence, they can notbe added.

137. Draw and from the arrow head of ,

draw , of the same length (i.e., QS = PQ) and

perpendicular to . Fig. 2(c).62. Now will

represent . Here,

or q1 = 45o

Now draw , where QT = QS. Now willrepresent

. Here,

or q2 = 45o

On measuring, the lengths of and

come out to be the same and angle betweenthem, (q1 + q2) = 45o + 45o = 90o.

138. (i) It is the product of a pure number and a velocityvector, hence the unit of product is same as that ofvelocity vector i.e. the product is a velocity of magnitude15 km h-1 towards west.(ii) It is the product of a scalar (time) and velocity vector.The unit of this product will be hour × km h-1 = km. Thusthe product is a displacement of magnitude 15 kmtowards west.

139. No ; The rectangular components of a vector hasvalues A cos q and A sin q. Since the values of cos q andsin q can never be greater than one, hence the value ofany rectangular components of a vector can never begreater than the given vector.


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140. The angle which the given velocity makes with thenorth direction is 180o - 30o = 150o. The componentvelocity along north = u cos 150o = 100 × ( ) = -86.6 km h-1.The angle which the given velocity makes with the eastdirection is The component velocityalong east .

141. (a) In this case two vectors must be in the samedirection i.e. is in the direction of . (b) In this case,

must be zero. (c) should be perpendicular to .

142. Let and ; ÐDOF =q.Fig. Resolving into two rectangular component, wehave ; B cos q along OF and B sin q along OE. Hereresultant vector is along OF.

R = B cos q. As per question,R = B/2 = B cos qor cos q = ½or q = 60o.Hence, angle between and = ÐCOD = 90o + 60o =150o.

143.

and .

144. We know that = AB cos q ; when vectors areorthogonal, q = 90o ; So, = AB cos 90o = 0When vectors are parallel, q = 0o ; So, = AB cos 0o =AB (Max).

145. 82o 49’146. 2o 52’147. 60o upstream148. 5 ms-1 ; 36o 52’ North of East149.

150. 3151. Self (teacher)152. 2s153.

154.

155.

156. When the particle describes a uniform circularmotion, its speed is constant but it has centripetalacceleration acting along the radius directed towardsthe centre of the circular path.When the particle is moving with a constant velocity,there is no change in velocity with time and hence itsacceleration is zero.

157. When a stone is dropped from a railway carriage, itwill fall vertically downwards with acceleration due togravity g. Therefore, with respect to earth, theacceleration of the stone will be g only.Inside the carriage the stone has two acceleration : (i)horizontal acceleration a, due to the motion of thecarriage and (ii) vertical acc. due to gravity g. Thus theacceleration of the stone with respect to carriage is

.

158. No ; The acceleration of a body remaining constant,the magnitude and direction of the velocity of the bodymay change. For example, under a constant acceleration(i.e. acceleration due to gravity g), the path of body inangular projection is a parabolic path.

159. (a) Vertical st. line path because w.r.t. person sittingin the train the ball has only one velocity acting vertically(b) Parabolic path because w.r.t. person outside thetrain the ball has the vertical as well as horizontalcomponent velocities.

160. A body falling freely will reach the ground earlierbecause its acceleration is g (i.e. acceleration due togravity) which is greater than the acceleration of otherbody = g sin q ; where q is the inclination of the planewith the horizontal.

161. No, because it is propelled by combustion of fueland does not move under the effect of gravity alone.

162. Time, ; Horizontal distance,

.

163. Yes164. No ; horizontal range is maximum when q = 45o and

maximum height attained by projectile is largest when q= 90o.

165. T1 = 2 u sin q/g and T2 = 2 u sin (90o - q) soT1T2 = sin q/cos q = tan q.


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166. At the maximum height of projectile, the verticalcomponent velocity becomes zero and only horizontalcomponent velocity of projectile is there.

167. Here, .

At the highest point,

168. (i) Zero. (ii) - m u cos q.169. p/6 radian per hour.170. 90o

171. It is the centripetal acceleration which isperpendicular to the direction of motion at every point,acting along the radius and directed towards the centreof the curved circular path.

172. To study the motion of a projectile, we assume that(i) there is no frictional resistance of air. (ii) The effectdue to rotation of earth and curvature of the earth isnegligible. (iii) The acceleration due to gravity isconstant in magnitude and direction at all point of themotion of projectile.

173. Horizontal range, ; Horizontal range, R

is maximum if u2 sin 2 q/g has maximum value. It will beso if sin 2 q has maximum value i.e. sin 2 q = 1

or 2 q = 90o or q = 45o.

174. A projectile when given angular projection has tworectangular component velocities, acting horizontallyand vertically. The horizontal component velocityremains constant throughout the projectile path butvertical component velocity of projectile decreases as itgoes up and becomes zero at the highest point. That iswhy, the projectile on its path, has minimum velocity atthe highest point and maximum velocity at theprojection point or at a point where it strikes thehorizontal ground during its flight.

175. In angular projection of a projectile, let h be heightattained by projectile in timings t1 and t2 respectively.

Then,

or

or total time of flight.

176. Given ; R = n H ; so,

or or q = tan-1 (4/n).

177. Let u be the velocity of ball at an instant, when itmakes an angle b with the horizontal. The horizontalcomponent velocity of the ball = u cos b.Initial horizontal component velocity of ball = u cos aAs, in angular projection of a projectile, the horizontalcomponent velocity remains unchanged, henceu cos b = u cos a or u = u cos a/cos b

178. At highest point, velocity of projectile = u cos q.K.E. of projectile at highest point

.

179. For hour hand of watch, time period, Th = 12 h ; Forearth, Te = 24 h.

Angular velocity,

or wh = 2we

180. 10.25 m181. 44.1 m ; 29.4 ms-1

182. 25 m

183. y = 2 x - 5 x2

184.


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185.

186.

187.

188.

189.

190.

191.


SaraNextGen.Com

192.

193.

194.

195.

196.

197.

198.

199.

200. Self (Teacher)201. Self (Teacher)202. Self (Teacher)203. Self (Teacher)


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204. 1.67205. Self (Teacher)206. Self (Teacher)207. 1.8 × 103 ms-1

208. 6.21 × 10-21 J209. Self (Teacher)210. Self (Teacher)211. According to the kinetic interpretation of

temperature, solute temperature mean kineticenergy of molecules .As heat is taken out, the temperature falls and hencevelocity decreases. At absolute zero, velocity of themolecules becomes zero, i.e., kinetic energy becomeszero. So no more decrease in K.E. is possible, sincetemperature can not fall further.

212. Helium is monatomic while O2 is diatomic . In case ofhelium, the supplied heat has to increase only thetranslational K.E. of the gas molecules.On the other hand, in the case of oxygen, the suppliedheat has to increase the translational, vibration androtational K.E. of gas molecules. Thus helium wouldundergo a greater temperature rise.

213. It states that in equilibrium, the total energy of thesystem is divided equally in all possible energy modeswith each mode i.e., degree of freedom having an

average energy equal to kT.

214. The K.E. of moving water is dissipated into internalenergy. The temperature of water thus increases.

215.

216.


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µ

21

217.

218.

219. (i) The moon has small gravitational force and hencethe escape velocity is small as the moon in the proximityof the earth as seen from the sun. the moon has thesame amount of heat per unit area as that of the earththough the rms speed of the air molecules is smallerthan escape velocity on the moon a significant numberof the molecules have speed greater than escapevelocity and the escape.Now rest of the molecules arrange the speeddistribution for the equilibrium temperature. Again asignificant number of molecules escape as their speedsexceed escape speed. Hence over a long time the moonhas lost most of its atmosphere. At 300 K,

for moon = 4.6 km/s(ii) As the molecules move higher their potential energyincreases and hence kinetic energy decreases and hencetemperature reduces.At greater height more volume is available and gasexpands and hence some cooling takes place.

220.


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s/km7.1\103.7

3001038.13m

Tk326

23b

rms -

-

´´´´

==n

esn

221.

222.

223.

224.

225. 2.25 R226. 5 × 10-8 m227. Self (Teacher)


SaraNextGen.Com

228.

229.

230.

231.


SaraNextGen.Com

232.

233.

234.

235.


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236. (i) 22.4 volume of one mole of a gas occupy at NTP.(ii) (a) Size of molecule of an ideal gas is negligible.(b) There is no force of attraction on repulsion amongstthe molecule of an ideal gas.(iii) At extremely low pressure and high temperature.(iv) In day to day life this concept implies that none of usis ideal or perfect. An ideal person is a theoreticalconcept however under some situation some rateperson can be taken as nearly perfect as such kinds ofperson are very limited i.e. one in million so there is noiteration between them.

237.

238.

239.

240.


SaraNextGen.Com

241. 242.

243.


SaraNextGen.Com

244.

245.

246. Self (Teacher)247. Self (Teacher)248.

249. Self (Teacher)250. 6251. 140/3

252.

253.

254.


SaraNextGen.Com

s/cm10659.4 4´

255.

256.

257.

258.

259.

260.

261.


SaraNextGen.Com

262.

263.

264.

265. It is known that one mole of any ideal gas at NTPoccupies a volume of 22.4 litre. As the cylinder has fixedcapacity of 44.8 litre it must contain 2 moles of helium

at NTP for helium (monatomic gas),

Heat required = number of moles × molar sp. Heat ×rise in temperature

266.

267.

268.

269. Self (Teacher)270. Self (Teacher)271. Self (Teacher)272.

273.


SaraNextGen.Com

.R23Cv =

\

J95.373J31.845R4515R232Q =´==´´=

274.

275.

276.

277.

278. 4649 J

279.

280.

281.


SaraNextGen.Com

282.

283.

284.

285.

286.

287.

288. The effective distance travelled by drunkard in 8steps = 5 - 3 = 2m. Therefore, he takes 40 steps to move10 metres. Now he will have to cover 5 metres more toreach the pit, for which he has to take only 5 forwardsteps. Therefore, he will have to take = 40 + 5 = 45 stepsto move 15 metres. Thus he will fall into the pit aftertaking 45 steps, i.e., after 45 seconds from the start.


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289.

290. The relative velocity will not alter at all since it isindependent of the position of bodies.

291. Let and w be the velocity of each train and windrespectively. As per question, or

i.e., velocity of each train is 3 times thevelocity of wind.

292. An inertial observes is one which is uncelebrated (i.e.a = 0) A non inertia observes is an accelerated observer.

293. Let r be the radius of semicircular path. Here,or

Diameter Magnitude of displacement =diameter

294. Self (teacher)295. Self (teacher)296. Self (teacher)297.

298. (a) (b)

299. south of east

300. 10 sec.301. 4 h 26.7 min. ;266.7 km

302.

303.

304.

305.

306.


SaraNextGen.Com

u)w(2)w( -u=+u

,w3=u

)2/r2(l p= ./lr p=;/l2r2 p==

p= /l2

.2/l2

lntDisplaceme

cetanDis p=

p=

zero;kmh48 1-

13cms1086.7 --´ 14cms102.2 --´

'3722;kmh5.6 01-

307.

308.

309.

310.

311.

312.

313.

314.

315.

316. Self (teacher)317. Self (teacher)318. Self (teacher)319.

320. 24 m321. 122.5 m322. 75 m from ground323.

324. (i) 8.36 s (ii) 7.33 s (iii) 7.82 s325. 100 m, 60 m326. (i) 170 m (ii) 125 m, -5 m/s2

327. 0.5 m/s


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sec65,ms4,ms3/2 22 --

1ms30.0 -

328. A = 25 km/h along OP lB = 10 km/h along OQAngle between and is q = 180O - 60O = 120O.

or b = 23.4o East of North.

329. Let A and B be the two forces.Then A = 3x ; B = 5x ; R = 28 N and q = 60o

Thus A/B = 3/5.Now

or or x = 28/7 = 4.Force are ; A = 3 × 4 = 12 N and B = 5 × 4 =

20 N

330.

331. Velocity :

Acceleration, when t = 3.0 s,

12.4 ms-1

If q is the angle which the direction of makes with thex-axis, then

.

332.

333.


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Ar

Br

q++= ABcoc2BAR 22

h/km8.21o120coc10252210225 =´´++=

oo

o4.23tan433.0

120coc1025120sin10

BcocAsinBtan ==

+´

=q+

q=b

q++= cosAB2BAR 22 \

o22 60cos)x5)(x3(2)x5()x3(28 ++=

x7x15x25x928 222 =++=\

jt4i3)k5jt2it3(dtd

dt)r(d 2 )))))rr

+=++==u

j4j40dtda

))rr=+=

u=

j12i3j34i3))))r

+=´+=u

\ =+=u 22 123ur

oo 76or76tan43

12tan =q===q

334. When person is at rest with respect to ground, therain is coming to him at an angle 60o with the vertical i.e.along OB, with velocity 20 ms-1, the relative velocity ofrain w.r.t. person is along OC as shown in Fig. 2(c).48.Here, ÐBOC = 60o.Velocity of rain w.r.t. ground,

Velocity of person w.r.t. ground, where, uP =OA = 20 ms-1 = CB.Velocity of rain w.r.t. person, .

(a) In DOCB, OB = CB/sin 60o =Velocity of rain w.r.t. ground,

(b) In DOCB, OC = CB/tan 60o =Velocity of rain w.r.t. person, urp = 11.55 km/h.

335. Here, andResultant vector,

.The unit vector along y direction

Required vector

336.(i)

(say)Here,

Unit vector,

(ii)say

Unit vector,

337. Here,H

ence,or 2 × 3 + 3 × (-m) + (-6) × 6 =0 or 6 - 3 m - 36 = 0 or m =-10.

338. We know that or

339. Here, ,

340. Let be perpendicular to and , then canbe taken as the cross product of and


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OBr =ur

OAp =ur

OCrp =ur

3/40)2/3/(20 ==\

h/km1.23h/km340r ==u

.h/km55.113/20 =

k7j5i3A)))r

+-= k3j4i2B)))r

-+=

k4ji5)k3j4i2()k7j5i3(BAR)))))))))rrr

+-=-+++-=+=

j)

=

\ k4j2i5)k4ji5(j)))))))

-+-=+--=

Ckj3i5)k2j2i3()k3ji2()BA(r)))))))))rr

=+-=--++-=+

35]1)3(5[C 2/1222 =+-+=

35kj3i5

CC

C)))r

) +-==

Dk5ji)k2j2i3()k3ji2(BAr)))))))))rr

=++-=---+-=-

27]51)1[(D 2/1222 =++-=

27k5ji

DD

D)))r

) ++-==

090cosABB.A90;k6jmi3B;k6j3i2A oo ==\=q+-=-+=rr)))r)))r

0=)k6+jmi3).(k6j3+i2())))))

)BA).(BA(|BA| 2 ))))))--=-

)cos1(2cos)1)(1(211B.A2B.BA.A|BA| 2 q-=q-+=-+=-))))))))

2/sin4)]2/sin21(1[2 22 q=q--=

)sin21=2cos( 2 qqQ

|2/sin2=BA|∴ q))

k5j4i3a))rr

+-= k3ji2b))rr

-+-=251546)3(5)1(4)2(3)k3ji2).(k5j4i3(b.a -=---=-+--=-+-+-=

))))))rr

312543kji

)k3ji2()k5j4i3(ba--

-=-+-´+-=´

)))))))))rr

k5ji7)83(k)910(j)512(i))))))

--=-++-+-=

Cr

Ar

Br

Cr

Ar

Br

}11)1(2{k}2211{j)}1(121{i211112kji

BAC ´--´+´-´+--´=-

=´=\)))

)))rrr

k3j3i3)3(k)3(j)3(i))))))

--=-+-+=

3327)3()3(3|C| 222 ==-+-+=\r

)kji(3

133

k3j3i3|C|

CC,isCalongvectorunit

))))))

rr

)r--=

--==\

341. Let and be the velocities of the two particlesof a pair and q be the angle between them, Fig., where

Here,The relative velocity of particle 2 w.r.t. particle 1, i.e.,

will be represented by diagonal ofparallelogram OBDC. Where ÐBOC = (180o - q). Themagnitude of the relative velocity of particle 2 w.r.t.particle 1 is given by

.The value of the relative velocity of a particle w.r.t.other particle will change with change in the value ofthe angle q. Therefore, the value of relative velocity ofparticle 2 w.r.t. particle 1 obtained above is themagnitude of relative velocity of pair of particles havingtheir velocities in the direction q and q + dq i.e. over isrange dq. As the velocities of the particles are randomlydistributed in all directors, therefore, the value of q canvery between 0 and 2 p. The magnitude of relativevelocity averaged over all the pairs is given by.

I

t means is greater than u.

342. When a body slides down along the inclined plane,the component of weight along the plane (= mg sin q)produces acceleration a in the body.

Given by,

Taking the downward motion of the body along theplane form A to B, we have,u = 0, a = g sin q, t = t, S = l, u = ?

As so

…(i)

or or

…(ii)Asu2 = u2 + 2 aS u2 = 0 + 2 (g sin q) ×l

…(iii)From (i), l µ t2, so (a) If t’ = t/2, then

or

It means in half is time of descent, the body covers onlyone-fourth of the total distance

(b) If l’ = l/2, then or

It means the body covers the half of the total distance intime 0.71 times the time of descent of body.

343. Taking motion along x-direction, we have ux = 5.0m/s ; ax = 3.0 m/s2, x = 84 m, t = ? ; ux = ?

As, or 168 =

10 t + 3 t2 or 3 t2 + 10 t - 168 = 0On solving, t = 6 sux = ux + ax t = 5.0 + 3 × 6 = 23 m/sTaking motion along y-direction, we have uy = 0, ay = 2.0m/s2, y = ?, t = 6 s ; uy = ?

As, uy =

uy + au t = 0 + 2 × 6 = 12

ms-1


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1ur

2ur

u=u=u |||| 21rr

)OB(),OA( 21 =u=urr

21ur

OD

21ur

quu-u+u=q-uu-+u+u-=u cos2)180cos(||||2|||| 22o22

22

2121

rrrr

2/sin22/sin22)cos1(2 2 qu=qú=q-u=

pppp

úûù

êëé q-

pu

=qq

pu

=qq

up

=qup

=u òòò2

0

2

0

2

0

2

02121 2/1

2/cos22

d2

sin22

d2

sin221

d21

u=u

=pu

=+--pu

=+p-pu

= 273.1)7/22(

44]1)1([

2]0coscos[

2

21ur

q=q

= singmsinmg

a

2at21

utS += 2t)sing(21

0l q+=

l/ghl2

singl2

t =q

=gh2

lt =

\

gh2singl2 =q=u

41

t2/t

l't

t'l 22

=÷øö

çèæ

=÷øö

çèæ

= 4l

'l =

2

t't

l2/l

÷øö

çèæ

=÷øö

çèæ

t71.02t

't ==

22xx t3

21

t0.584ta21

tux ´´+=\+=

,m366221

ta21

y 22y =´´==

1222y

2x ms9.251223 -=+=u+u=u

344. No, two vector of different magnitudes cannot havezero resultant. In case of three vectors, zero resultant ispossible. This occurs when the three vectors lying in aplane are represented by the three sides of a triangletaken in the same order.

345. No. The rectangular component of a vector can’t begreater than the vector itself because the rectangularcomponents of a vector are Ax = A cos q and Ay = Asin q. As sin q and cos q both are £ 1, therefore Ax and Ay

cannot be greater than A.346. Yes, null vector has a definite physical significance.347. (a) If the starting point of fly which is none corner of

room is taken as origin of coordinates, then thecoordinates of diametrically opposite corner of room are(3, 4, 5). So displacement is

and

(b) When the fly walks, then shortest distance travelledis

348. In D OPS, by vector addition,…(i)

Similarly, in D OQS, we have…(ii)

Adding (i) and (ii), we haveor

or or

orSo S is the mid point of PQ.

349. (a) Refer to Fig. , using triangle law of vectors, wehave,

or

(b)

350.

351. See Fig. ; clearly,

2 T sin q = mg or

When the rope is straight, q = 0o ;

Then, .

352. As ; so is perpendicular to .As ; so is perpendicular toAs is perpendicular to both and , so isparallel to .

353. oror ...(i)

To satisfy (i), the three possibilities can be there(i) or(ii)(iii) and are parallel to each other i.e.

, where n is a non zero real number. or

Thus if need not be equal to . Thegiven statement is true if is a zero vector or isequal to .

354. Self (teacher)355. Self (teacher)356. Self (teacher)357. Self (teacher)358. 30N, 40 N359. 49.97 kmh-1, 2 hour, 66o 25’ West of North360. (a) 53o 8’ up stream (b) 8 km h-1 (c) 6 minutes361. ms-2 South West

362. 503 N ; 783 N


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Ar

k5j4i3r)))r

++=

m750543r 222 »=++=

m4.9413543 22 =+=++=

PSOPOS +=

QSOQOS +=

QSPSOQOPOS2 +++= QSPSBAC2 +++=rrr

QSPS)BAC2( +=--rrr

QSPS0 +=

)BAC2(rrr

+=\ QSPS -=

CDBC2AB)CDBC()BCAB(BDAC ++=+++=+

ABBC2AB -+=

)ABCD( -=Q

BC2BDAC =+

CDAB)CDBC()BCAB(BDAC -=+-+=-

AB2)AB(ABBDAC =--=-

AFAEADACAB ++++

AF)DEAD(AD)DCAD(AB ++++++=

AD3)AFDC()DEAB(AD3 =++++= AO6)AO2(3 =´=

]AFDCand,DEAB[ -=-=Q

q=

sin2mg

T

¥== o0sin2mg

T

0B.A =rr

Ar

Br

0C.A =rr

Ar

Cr

CBrr

´ Br

Cr

CBrr

Ár

BCBArrrr

´=´ 0BCBA =´-´rrrr

0B)CA( =´-rrr

0CA =-rr

CArr

=0Brr

=CArr

- Br

BnCArrr

=-

BnCArrr

+=

C,BCBArrrrr

´=´ Ar

Br

Ar

BnCrr

+

22

363. (a) 5m (b) m

364. N, 45o South of West

365.

366. 1/2367. units of area

368.

369. Zero370. 2 A sin q/2371. Self (teacher)372. Let u be the horizontal velocity with which the body

is thrown horizontally. Let ux and uy be horizontal andvertical velocity with which the body strikes the groundafter 2 seconds. Let b be the angle which the directionof resultant velocity of the ball while striking the groundmake with the horizontal direction. Given, b = 45o.

Then, ( ux = u)

or or or

uy = u …(i)Taking motion of the body along vertical downwarddirection i.e. along Y axis, which is taken positivedirection, from point of projection to a point striking theground, we havey0 = 0, uy = 0, ay = 9.8 m/s2, t = 2 s, uy = ?, y = ?As, uy = uy + ay t\b uy = 0 + 9.8 × 2 = 19.6 ms-1

From (i), u = uy = 19.6 ms-1

Also,

373. Here, q = 60o, u = 392 ms-1.Total time of flight,

Horizontal

range,

Maximum height,

374. Let u be the velocity of projection from O at anangle 45o, which just clears the top A of wall of height h.Fig.Given, horizontal range = 15 + 5 = 20 m

So

or or …(i)

Let (x, y) be the coordinates of A, thenx = 15 m and y = h.Using the equation of trajectory, we have


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25

220

k4ji6)))

++-

37

úû

ùêë

é-- )kji(

31 )))

uy

xytan

u=

u

u=b Q

uyo45tan

u=

uy1

u=

2tya21tyu0yy ++=

.m6.19228.921200y =´´+´+=

s23.698.9

)2/3(39228.9

o60sin3922gsinu2T =

´´=

´´=

q=

8.9

o120sin2)392(8.9

o602sin2)392(2sing

2uR =´

=q=

m9.135788.9

)2/3(2)392(=

´=

m58808.92

)o60(sin2)392(g2

2sin2uh =´

=q

20g

2sin2u=

q

20g

o452sin2u=

´ 20g

2u=

q-q= 2cos2u

2xg21tanxy m75.3

2120

21521o45tan15h =

´´-=

375. Let the monkey stationed at A, be fired with a gunfrom O with a velocity u at an angle q with thehorizontal direction OX. Draw AC, perpendicular to OX.Let the bullet cross the vertical line AC at B after time tand coordinates of B be (x, y) w.r.t. origin O. Fig.

\b …(i)

(where OC = x)In D OAC, AC = OC tan q = x tan q …(ii)Clearly CB = y = the vertical distance travelled bythe bullet in time tTaking motion of the bullet from O to B along Y axis wehavey0 = 0, y = y, uy = u sin q, ay = - g, t = t

As,

\f0

…(iii)

\f0 AB = AC - BC = x tan q - y

[f

rom (i)]

It means the bullet will pass through the point B on

vertical line AC at a vertical distance below point

A.The distance through which the monkey falls vertically in

time .

It means the bullet and monkey will pass through thepoint B simultaneously. Therefore, the bullet will hit themonkey.

376. In Fig. Height of tower OB = 156.8 m ; u = 39.2m/s ; q = 30o Component of velocity along OX = u cos q =39.2 cos 30o =33.947 ms-1 Component of velocity alongOY = u sin q = 39.2 sin 30o = 19.6 ms-1Let t be the total time of flight (i.e. time in going from Oto D). Consider the vertical downward direction OB asthe positive direction of y axis. Taking motion of aprojectile from O to D along Y axis, we havey0 = 0, y = 156.8 m, uy = - u sin 30o = - 19.6 m/s, ay = 9.8m/s2, t = t

As,

or 156.8 = -

19.6 t + 4.9 t2

or 4.9 t2 - 19.6 t - 156.8 = 0 or t2 - 4 t - 32 =0 or t2 - 8 t + 4 t - 32 = 0or t(t - 8) + 4 (t - 8) = 0 or (t + 4) (t- 8) = 0 or t = - 4 or 8.As t = - 4 s is not possible, therefore t = 8 s.Distance from the foot of tower where it strikes theground is, BD = u cos 30o × t = 33.947 × 8 = 271.57 m.

377. Here, T = 100/7 s ; r = 12 cm

(a) Angular speed,

Linear speed, u = r w = 12 × 0.44 = 5.3 cm s-1

(b) The given motion is a uniform circular motion. Herethe direction of velocity is along the tangent to the circleat every point. The acceleration in this motion iscentripetal acceleration. It is always directed towardsthe centre of the circle. Since this direction changescontinuously at different points on circular motion,therefore, the acceleration here is not a constant vector.The magnitude of acceleration is ac = w2 r = (0.44)2 × 12 =2.3 cm s-2.


SaraNextGen.Com

q=

q=

cosux

cosuOCt

tya21tyu0yy ++=

2gt21tsinu2t)g(

21tsinu0y -q=-+q+=

÷÷ø

öççè

æ-

q´q-q=÷

øö

çèæ -q-q= 22 gt

21

cosuxsinutanxgt

21tsnutanx

2gt212gt

21tanxtanx =+q-q=

2gt21

AB2gt21t ==

2tya21tyu0yy ++=

2t8.921t)6.19(08.156 ´´+-+=

s/rad44.0)7/100()7/22(2

T2

=´

=p

=w

378. Here ; u = 20 m/s ; q = 45o, d = 25 m.Horizontal

range ; Time of

flight,

Since the goal man is already 25 m away in the directionof the ball, so to catch the ball, he is to cover a distance= 40.82 - 25 = 15.82 m, in time 2.886 s. Therefore, thevelocity of goal man to catch the ball is

379. Yes, In long jump, it matters how high one jumps. Itis explained below.For initial velocity u and angle of projection q, themaximum height,

or

and, horizontal range, 2 sin q cos

q =4 h cot qThus the span of jump depends upon (i) height hattained (ii) angle of projection, q.

380. Horizontal range

or sin 150o.

2 q = 30o or 150o or q = 15o

or75o

381. If a projectile is projected with velocity u, making anangle q with the horizontal direction, then

Horizontal range, sin 2 q

and Max. height,

Case (i) If q = a, let R = R1 and H = H1 then

…(i)

and …(ii)

Case (ii) If q = (90o - a), let R = R2 and H = H2, then

…(iii)

…(iv)From (i) and (iii) : R1 = R2

From (ii) and (iv) :

382. When a stone is going around a circular path, theinstantaneous velocity of stone is acting tangent to thecircle. When the string breaks, the centripetal forcestops to act. Due to inertia, the stone continue to movealong the tangent to circular path. That is why, the stonefillies off tangentially to the circular path.

383. Self (teacher)384. Self (teacher)385. Self (teacher)386. Self (teacher)387. Self (teacher)388. 2.0 s389. 75.57 m/s, 23o33’ with horizontal390. 5o 43’391. 4 u2/5 g392.

393. 8 p rad. s-1 ; 1257.1 cm s-1

394. 439 km h-2

395. 0.86 ms-2

396. 30o


SaraNextGen.Com

m82.40o452sin8.9

2202sing

2uR =´=q=

s886.2o45sin8.9

220gsinu2T =´=

q=

1ms483.5886.282.15 -==u

g2

2sin2uh q=

q= 2sin

h2g

2u

q=q= 2sin

h22sing

2uR

102sin230

g2sin2u45 q

=q

==

o30sin21

303010452sin ==

´´

=q

R

2uR =

g2

2sin2uH q=

g2sin2u

1R a=

a= 2sing2

2u1H

a=a-=a-

= 2sing

2u)2o180sin(g

2ug

)o90(2sin2u2R

a=a-= 2cosg2

2u)o90(2sing2

2u2H

g2

2u)2cos2(sing2

2u2H1H =q+q=+

3/2

397. 10 ms-1, 53o 8’398. 60o

399.

400. Self (Teacher)401. Self (Teacher)402.

403.

404. 4.862 cal/mol K405. Self (Teacher)406. Self (Teacher)407. Self (Teacher)408. Self (Teacher)409. Self (Teacher)410. Self (teacher)411. Self (teacher)412.

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428. Angular displacement is a vector quantity providedangle q is small because the commutative law of vectoraddition for large angle is not valid, where as for smallangles, the law is valid i.e. q1 + q2 ¹ q2 +q1, but Dq1 + Dq2

= Dq2 + Dq1. The same is explained below.Consider a book with front page F lying in the plane ofpaper. It can be rotated about two mutuallyperpendicular axes 1 and 2 as shown in Fig. 2(d).25(a).(i) Let the book be rotated trough an angle q1(= 90o) inthe clockwise direction about axis 1. It takes the positionas shown in Fig. 2(d).25(b). On further rotating the bookthrough an angle q2(= 90o) in anticlockwise directionabout axis 2, the book occupies the position as shown inFig. 2(d).25(c). (ii) From the initial position of book,[Fig2(d).25(a)], if the book is rotated through q1 (= 90o)in the anticlockwise direction first about the axis 2, itoccupies the position as shown in Fig. 2(d).25(d). Onfurther rotating the book through q2(= 90o) in clockwisedirection about axis 1, the book occupies the positionshown in Fig. 2(d).25(e).From above we note that the final positions of the bookshown in Fig. 2(d).25(c) and Fig. 2(d).25(e) are not thesame. Hence q1 + q2 ¹ q2 + q1.This shows that q1 and q2 are not vectors as they do notobey the commutative law of vector addition. If thebook is rotated through a smaller angle Dq1 and Dq2 (say2o or 3o), the final positional of the book in the two casediscussed above would almost be the same. As Dq ® 0,the final position of the book will becomeIndistinguishable, henceDq1 + Dq2 = Dq2 + Dq1.


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2) Kinetic Theory - Solutions