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MHT-CETTRIUMPHPHYSICSHINTS TO MULTIPLE CHOICE QUESTIONS,EVALUATION TESTS&MHT-CET 2019 (6th May, Afternoon) PAPER

CONTENTSr. No.TextbookChapterNo.Chapter NamePage No.Std. XI11Measurements122Scalars and Vectors1234Force2645Friction in Solids and Liquids5158Refraction of Light6869Ray Optics84712Magnetic Effect of Electric Current99813Magnetism113Std. XII91Circular Motion121102Gravitation142113Rotational ace Tension232157Wave Motion244168Stationary Waves260179Kinetic Theory of Gases and Radiation2801810Wave Theory of Light3091911Interference and Diffraction3182012Electrostatics3372113Current Electricity3662214Magnetic Effect of Electric Current3852315Magnetism4042416Electromagnetic Induction4102517Electrons and Photons4342618Atoms, Molecules and Nuclei4452719Semiconductors4702820Communication SystemsthMHT-CET 2019 (6 May, Afternoon) Paper481486

TextbookChapter No.01MeasurementsHints10.Classical Thinkingo13.Temperature is a fundamental quantity.26. 1 dyne 10 5 N, 1 cm2 10 4 m2103 dyne/cm2 103 10 5/10 4 N/m2 102 N/m2ORUsing quick conversion for pressure,1 dyne/cm2 0.1 N/m2103 dyne/cm2 103 0.1 102 N/m2 57. 57 57 rad 60 180 60 b Radius of earth 6.4 106 mDistance of the moon from the earth,6.4 106 60 180bs 3.86 108 m s 57 11. 12.Torque [M1L2T 2],Angular momentum [M1L2T 1]So mass and length have the same dimensions.13.According to Poiseuille’s formula,πPr 4 8l (dV / dt) [ ] 15.[Dipole moment] [M0L1T1A1][Electric flux] [M1L3T 3A 1][Electric field] [M1L1T 3A 1]Critical Thinking2.3.Physical quantity (M) Numerical value (n) Unit (u)If physical quantity remains constant thenn 1/u n1u1 n2u2.Because in S.I. system, there are sevenfundamental quantities.mass pressure m (F / A) F V Adensity(m / V) F (A s) F s workA6. m mv kg sec 7.Curie disintegration/second8. 21.-59.Y 16.17.Bxt is unitless.Unit of B is m 1s 1.F Ldyne 10 N. 0.1 N/m2cm 2 10-4 m 2A DLDistance of sun from earth, s 1.5 1011 mAngular diameter of sun,1920 1920 rad 1920 3600 180 60 60 Diameter of sun, D s 1920 1.5 1011 3600 180D 1.4 109 m d 100 %Percentage error d 0.01 100 % 1.03 0.97%1.Parallactic angle, 57 23.[M1L-1T-2 ][L4 ] [M1L 1T 1]131[L ][L / T ]1 2Li energy stored in an inductor2 [M1L2T 2]The dimension of a quantity is independent ofchanges in its magnitude.1 c velocity of light 0 0 M1 L1T 2 mg 1 1 1 [L1T 1] r L M T L1 1

MHT-CET Triumph Physics (Hints)24.From F at bt21 2 M1LT Fa [M1L1T 3] T1 t M1L1T 2 F [M1L1T 4]b 2 2 Tt 25. 26. 27. 28.2 M1L1/ 2 T 2 a 1 1 4 [L 1/2 T2]b M L T [M1L1T 2] [L2]a [L1T 1]b [M1L 3]c [L2a] [LbT b] [McL 3c] [Mc L2a b 3c T b]Comparing powers of M, L and T,c 1, 2a b 3c 1, b 2b 22a 2 3(1) 12a 2a 14π 2 a xLet T2 y zG M4 2 being pure number is dimensionless.[M 0 L1T 0 ]x[M0L0T2] [M-1L3T-2 ] y [M1L0 T 0 ]z [M0L0T2] [Lx] [M 1L3T 2] y [M1] zComparing powers of M, L and Ty z 0,x 3y 0 and 2y 2y 1Substituting value of y,z 1, x 34π 2 a 3Thus, T2 GMT PaDbSc[M0L0T1] [M1L 1T 2]a [M1L 3T0]b [M1L0T 2]cComparing powers of M, L, Ta b c 0, a 3b 0 and 2a 2c 131and c 1.Solving, a - , b 22In the given wave equation x denotesæ xödisplacement. Thus ççç has dimensions of T.è vøHence from the principle of homogenity k hasdimensions of T.30.F a x M1L1T 2 Fa [M1L1/2T 2] L1/ 2 xbt2 F M1L1T 2 F [M1L1T 4]b 2 t T 2 29. a t2bxa [t2] [T2]T2P bxP T T2b Px M1L 1T 2 L1 2 T 4 M1 M1 a2 [T ] 4 [M1T 2]b T 31. By principle of dimensional homogeneity a V 2 [P] [a] [P] [V2] [M1L 1T 2] [L6] [M1L5T 2]Dimensions of b are same as that of V,[b] [L3] M1L5T 2 a M1L2T 2 3 b L 32.Let G cxgypzSubstituting dimensions,[M 1L3T 2] [M0L1T 1]x [M0L1T 2]y [M1L 1T 2]zComparing powers of M, L, T 1 z,x y z 3 and x 2y 2z 2Solving, x 0, y 233.Acceleration due to gravity g g [L1T 2]a 1, b 21st systemL1 1 cm 10 5 km1T1 1 s min60abst22nd systemL2 1 kmT2 1 min1 10 5 km 1 /60 min n L 1 T 1 980 L2 T 2 1 km 1 min 980 10 5 3600 35.28 km min 2 2

Chapter 01: Measurements39.The number of significant figures in all of thegiven number is 4.41.A vernier calliper has a least count 0.01 cm.Hence measurement is accurate only uptothree significant figures.42. In multiplication or division, final resultshould retain the same number of significantfigures as there are in the original number withthe least significant figures.Area of rectangle 6 12 72 m243.am 20.17 21.23 20.79 22.07 21.785am 21.21 a1 21.21 20.17 1.04 a2 21.21 21.23 0.02 a3 0.42 a4 0.86 a5 0.57 am 45.Δa1 Δa 2 Δa 3 Δa 4 Δa 5549.Maximum possible error in measurement of T L L 2 %2T T L (0.1 2 3) % 6.1%50.T 2π l / g T2 4π2l/g g % error in l 51. d 0.00547.54.Avogadro number (N) represents the numberof atoms in 1 gram mole of an element. i.e., ithas the dimensions of mole 1.55.As the graph is a straight line , P Q, orP constant.P Constant Q i.e.,QCompetitive ThinkingP 3. 4% 2 2% 8%48. m 2 v Percentage error in K.E %v m (0.75 2 1.85)% 4.45%I 2 Rt4.2 H I R t % 100 2 % Error,Rt H I 2 2 1 1 6%[Energy] [M1L2T 2] [M1L1T 1] [L1] [T 1] [P1A1/2T 1] FF 2Also maximum error in pressure (P) F l P 100 2 100 100 Fl P max V l b h 100 100%Vbh l 0.02 0.01 0.02 100% 13.12 7.18 4.16 0.77%53. 4 r 100 0.1% and V r 3r3 V%Percentage error in volume V3 r 0.3% r 0.1 100 0.2%and error in T 2 100 % error in g % error in l % error in T 0.1 0.2 0.3 %H 100 % 25% 0.020 46.1mm0.1 100 0.1% 100 100100cm52.1.04 0.02 0.42 0.86 0.57 0.585 100 %Percentage error d 4 2 lT2 The van der Waals equation for ‘n’ moles ofthe gas is, n 2a P [V nb] nRTV2 VolumePressurecorrectioncorrectionF V 2 FlV Fl 4PV 2 A 2 2a 2 nn2nnThus, S.I.units of a is N m4/mol2.3

MHT-CET Triumph Physics (Hints)4. 12.From Van der Waal’s equation,nb has dimensions of volume.Vb nThus, S.I. units of b is m3/mol.[x] [bt2]metrexmunit of b 2 22thr(hour)13.Energy force distance, so if both are increasedby 4 times then energy will increase by 16 times.14.1 dyne 10 5 N and 1 cm 10 2 m 1 dyne/cm 10 3 N/m108 dyne/cm 105 N/mRC is the time constant of RC circuit and L is the time constant of LR circuit. Hence, R L both RC and have the dimensions of time R Alternate method:coulombRC ohm farad ohm voltcoulomb coulombvolt voltampereampere second [T]L henry ohm secondNow, Rohmohm second [T]Lhave the dimensions of time.Both RC andR 15.16.[ 0 L] [C] X Gm1m 2r2Fr 2 G m1m 2F M1L1T 2 L2 [M 1L3T 2][G] 2 M 22.C [K] 26.F vF kv[W] M1L2 T 2 1 2 [M T ]22[x ] L k M1L1T 2 F 1 1 [M1L0T 1]v LT 28.R PV M1L 1T 2 L3 1 2 2 1 [M L T θ ] T 29.F xd [x] F d 1/ 21/2 M1L1T 2 M1L 3T0 M 3/ 2 L 1/ 2 T 2 30.F XLinear densityLinear density is mass per unit length 31. 32.11 0 0 2C 0 0[C] [M0L1T–1] 1 0 –2 2 C2 [M L T ] 4W 0 LV C V Q Currentttt20.1 2Kx225. M1 [M1L1T 2] 1 [X] L 2 0 –2 [X] [M L T ]The van der Waals equation for ‘n’ moles ofthe gas is, n 2a P 2 [V nb] nRTV VolumePressurecorrectioncorrectionF V 2 FlV Fl 4PV 2 A 2 2a 2 nn2nn4 Fl a 2 [M1L5T 2mol 2] n 0 [ 0] 33.q1 q 24 Fr 2A2 T2 [M 1 L 3 T4 A2](M1L1T 2 ) L2Electric Field [M1L1T 2 ]Force [A1T1 ]Charge[E] [M1L1T 3A 1]

Chapter 01: Measurements34. 35.36.Capacitance (C) Charge(Q)Voltage(V)But, Voltage (V) Work (W)Charge(Q)C 39. QQ2 WWQ40. Q 2 C [M 1L 2T2Q2]2 2 ML T Magnetic flux BA,where, B magnetic flield, A area 41.42. 0 1 0 1 M L T A 37.Energy density is given by U 1 B22 0Also,Energy density B2 [ML 1T 2]2 0[M1L2 T 2 ][M1L2 T 1 ]2EJ 2 [M0L0T0]M 5G 2[M1 ]5 [M 1L3T 2 ]2The dimensions of angle are [M0L0T0].45.[G] [M 1 L3 T 2][c] [M0 L1 T 1][h] [M1 L2 T 1]Now, let the relation between given quantitiesand length be,L Gx cy hz[L1] [M 1 L3 T 2]x [M0 L1 T 1]y [M1 L2 T 1]zWe get, x z 0i.e., z x (i)3x y 2z 1.(ii) 2x y z 0.(iii)y 3x.[from (i) and (iii)]number of turns currentmetre.( [Current] Ampere [A])c [T]v [L1 T 1 ]a [L1T 2]t[T1 ]b v(t c) [L1T 1] T1 [L1]Y Now,[Area] [A] [L2]Magnetic intensity H nI 2 A L L [LA] m2 1kg 2 33smsDimension [M1 L0T 3]44. BA Area magnetic intensity B H Wm2F A cos(Bx) C cos(Dt)But, 2 x 2 t F A cos C cos T on comparing we get,2 metre 1B 2 second 1and, D Tmetresecond 1 D i.e. velocity 1metresecond B B,H Units of solar constant :43.where, H magnetic intensity M 0 L1T 1 M1L1T 3 A 1 kg[ 0E2] [ 0] [E]2 [M 1L 3T4A2] [M1L1T 3A 1]2 [M1L1T 2A0]OR1ε0 E2 u2where u is energy density and has dimensions[M1L1T 2]A[H] L Drift velocity vd Electricfield E M 1L0T2A1 Permeability Mobility Energy [ML2 T 2 ] Volume[L3 ] X[M 1L 2 T 4 A 2 ] [M 3L 2T8A4]3Z2[M1L0 T 2 A 1 ]25

MHT-CET Triumph Physics (Hints)Substituting the value in equation (ii),3x 3x 2z 1 i.e., z 12Substituting this value we get,x 48. 1 3and y 22Ghc3/2T krxρySzTime (T) [L0M0T1]Radius (r) [L1M0T0]Density ( ) [L 3 M1T0]Surface tension (S) [L0M1T 2][T1] k[L]x[ML 3]y[MT 2]z T1 kLx 3y My z T 2z1 2z 1 z ;21y z 0 y z ;23x 3y 0 x 3y ;2L 46. r 3SLet the physical quantity formed of thedimensions of length be given as,T kr3/2 1/2 S 1/2 k 47.49.50. e2 .(i)[L] [c] [G] 4 0 Now,Dimensions of velocity of light [c]x [LT 1]xDimensions of universal gravitational constant[G]y [L3T 2M 1]ye2 1 L 2 G c 0 6VIU[V] [M1L2T 3A 1] using V qRA[ ] [M1L3T 3A 2] using l1[ ] [M 1L 3T3A2] using [R] [M1L2T 3A 2] using R PVNTS.I. unit: J K 1 [M1 L2 T 2 K 1]Boltzmann constant (kB) 12F dv A dx Ns [M1 L 1 T 1]m2Water equivalent is the mass of water that willabsorb or lose same quantity of heat as that ofthe substance for the same change intemperature.S.I. unit : kg [M1 L0 T0]Coefficient of thermal conductivity (K)Q At x S.I. unit : J/ m s K [M1 L1 T 3 K 1]S.I. unit :z dimensionless,k Z[M1L2 T 2 K 1 K1 ] [α] [M1L1T 2][L1 ] And P 1 1 2 [M L T ] [β] 1 1 2 P [M L T ] [β] [M0L2T0]y e2 3 2 zDimensions of [ML T ] 40 Substituting these in equation (i)[L] [LT 1]x [M 1L3T 2]y [ML3T 2]z Lx 3y 3z M y z T x 2y 2zSolving for x, y, zx 3y 3z 1 y z 0x 2y 2z 0Solving the above equation,11x 2, y , z 22 Zshould bek Coefficient of viscosity ( ) zxIn the given equation,53.30 VSD 29 MSD291 VSD MSD30L.C. 1 MSD – 1 VSD1 29 0.5 1 MSD 30 30 1 minute

Chapter 01: Measurements54.20 VSD 16 MSD1 VSD 0.8 MSDLeast count MSD – VSD 1 mm – 0.8 mm 0.2 mmLeast count of vernier 1 M.S.D. – 1 V.S.D. 29 0.5 0.5 30 0.5 30Reading of vernier M.S. reading V.S. reading L.C.0.5 58.5 9 30 58.65 Main scale0.8 mm0055. 56.57.58.59.1mm10For a given vernier callipers,1 MSD 5.15 5.10 0.05 cm2.45 0.049 cm1 VSD 50L.C 1 MSD 1VSD 0.001 cmThus, the reading 5.10 (0.001 24) 5.124 cm diameter of cylinder 5.124 cmAs per the question, the measured value is3.50 cm. Hence the least count must be0.01 cm 0.1 mmFor vernier scale, where the 10 divisions invernier scale matches with 9 division in mainscale and main scale has 10 divisions in 1 cm1 MSD 1 mm and 9 MSD 10 VSD,Least count 1 MSD – 1 VSD 0.1 mmHence, correct option is (B).One main scale division, 1 M.S.D. x cmOne vernier scale division,(n 1) x1 V.S.D. nLeast count 1 M.S.D. – 1 V.S.D.nx nx x x cm.nn1mm100 0.01 mmDiameter Main scale reading (Divisions oncircular scale least count)1 0 52 0.52 mm100 Diameter 0.052 cm.61.A 4πr2 Fractional error A 100 2 0.3% 0.6%A4 3πr3Δr 100% error in volume 3 r0.1 100 3 5.3 63.Volume of sphere (V) 64.R 65.Given that: P V R V I IRVI 3 3 6%a 3b2cd a 3 100 a 3 1% 3% b error contributed by b 2 100 b 2 2% 4% c error contributed by c 100 3%c d error contributed by d 100 4% d Percentage error in P is given as,Dp 100 (error contributed by a) (errorpcontributed by b) (error contributed by c) (error contributed by d) 3% 4% 3% 4% 14%error contributed by aLeast count of screw gauge 30 VSD 29 MSD29MSD1 VSD 30 A2 r Ar 7

MHT-CET Triumph Physics (Hints)66. 67.a 2 b2cPercentage error is given by, x 2 a 2 b c xabc (2 2) (2 3) 4 4 6 4 14 x% 14%xCorrectedGiven: x Least count 71.PitchNo.of div.in circular scale69. 70.0.5 0.05 mm50Actual measurement0.5 0.05 mm 2 0.5 mm 25 50 1 mm 0.25 mm 0.05 mm 1.20 mmZero error 5 Main Scale Reading (MSR) 0.5 mmCircular Scale Division (CSD) 25thNumber of divisions on circular scale 50Pitch of screw 0.5 mm0.5 0.01 mmLC of screw gauge 50zero error 5 LC –0.05 mmzero correction 0.05 mmObserved reading 0.5 mm (25 0.01) mm 0.75 mmCorrected reading 0.75 mm 0.05 mm 0.80 mmLeast count of screw gauge 0.001 cm 0.01 mmMain scale reading 5 mm,Zero error – 0.004 cm – 0.04 mmZero correction 0.04 mmObservedMainscale (Division least count)readingreadingObserved reading 5 (25 0.01) 5.25 mm8Observed zeroreadingreadingcorrectionCorrected reading 5.25 0.04 5.29 mm 0.529 cm0.5Least count 0.01 mm50Diameter of ball D 2.5 mm (20) (0.01)D 2.7 mmMM V 4 D 3 3 2 M D 3 D max M0.550 0.01 mmActual reading 0.01 35 3 3.35 mmTaking error into consideration 3.35 0.03 3.38 mm. 68. 0.01 2% 3 100% 2.7 max 3.1% 72.Pressure (P) Force (F)Area (A)F. ( Area length2)L2Percentage error in pressure is given by, P F L 100 100 2 100PFL 4% 2 (3%) 4% 6% 10%massm 3Density ( ) Volume l (for cube V l3) 73.74.Percentage relative error in density will be, m l 100 100 3 100l m 1.5 (3 1) 1.5 3 4.5%Least count of both instrument0.5mm 5 10 3mm d l 1004MLgY ld 2 l d Y 2 ld Y maxError due to l measurement 0.5 /100mm 2%0.25mm ll

Chapter 01: MeasurementsError due to d measurement,0.52 d0.5 / 100100 2 2%0.5mmd0.2575. 76. 77.We have;lT 2 gSquaring2 l T2 4 g 78. lg 4 2TFractional error in g is g l T 2gTl 2 g L T 2 gL T L T 2 g g T LTime for 20 oscillations 40 s40Time for 1 oscillation 20T 2s4 2 L 4(3.14) 2 0.98 9.68 m/s2g T2 2 2 LgLT2% Accuracy in determination of g, L T g 100 2 100 100 LTg L t 100 2 100 Lt0.11 100 2 100 2090100 200 0.5 2.22200 90 2.72 3%100 100 125 145 T 0.8 % 0.7 % 1.5 %T79.g % error in g 4 2lT2 g 100g l T 100 2 100 T l 0.1 0.1 g 9.68 2 2 98 0.1 g 9.68 0.1 98 Given : T 2 D 1.25 10–2 m; h 1.45 10–2 mThe maximum permissible error in D D 0.01 10–2 mThe maximum permissible error in h h 0.01 10–2 mg is given as a constant and is errorless.rhgdhg 103 N/m 103 N/mT 24 T d h % errorTdh T d h 100 100 100Tdh 0.01 10 2 0.01 10 2 100 21.45 10 2 1.25 1081. g 4 2.EI 0.1 0.1 100 2 100 1.406%64 16 EII 0.1 0.1 100 2 100 1.406%64 16 EIII 0.1 0.1 100 2 100 2.72%20 9 M1L2 ML2 [M1L2T 2A 2]1 1 2Q2 A T These are the dimensions of unit Henry.82. x2 batFrom principle of homogeneity, ‘b’ will havethe dimensions of x2[b] [L2] .(i)Also,[P] [M1L2T 3][t] [T1][L2 ][b][a] 1 2 3[P][t] [M L T ][T1 ]Given: P 9

MHT-CET Triumph Physics (Hints) [a] [M 1T2][L2 ][b] [M1L2T–2][a] [M 1T 2 ] Torsional constant K [K] [ ][K] [M1L2T 2]From (iii) and (iv),[b] [K][a]85.[T] [M1T 2][ ] [M1L 1T 1][ ] [M1L 3]From the options given, T M1T 2 [L1T 1] [v] M1L 1T 1 v 86.Planck constant (h) is related to angularnh Lmomentum (L) as,2 [h] [L] [mvr]Moment of inertia I mr2h mvr v I mr 2rBut 2 f h [2 f ] [f ] I .(iii) .(iv)mvt83.F ma m Ft [m] [F1 V 1 T1] v 84. .(ii)Ftv FEE [Surface tension] 2 L L (VT) 2 [E1V–2T–2]T Evaluation Test1.A given dimensional formula may representtwo or more physical quantities. But given aphysical quantity, it has unique dimensions 6.2.[A] [M1L1T 2][mass][m] [length] [M1L 1T0][A] [M1L1T 2 ][B] [m] [M1L 1T 0 ] [M0L2T 2]This is a dimensional formula for latent heat.3.An instrument is said to have a high degree ofprecision if measured value remainsunchanged over number of readings repeated.Here readings are constant upto threesignificantfigures.Henceaveragemeasurement is precise. But, as zero error isnot considered readings are inaccurate.5.Capacitance, C [C] 10qqq2 V work / charge W[C1 ]2 [M 1L 2T2C2][M1L2 T 2 ]7.Two full turns of circular scale covers distanceof 1 mm. Hence one full turn will coverdistance of 0.5 mm.0.5L.C. of given instrument 50Diameter Zero error MSR CSR LC0.5 0.02 4 37 50 4.39 mmIn the expression, U A xx2 BB must have the dimensions of x2 i.e., [L2]Dimensions of A 8.1 2 22Ux 2 M L T L L1/ 2x [M1L7/2T 2]AB [M1L7/2T 2] [L2] [M1L11/2T 2]The number 37800 has three significant digitsbecause the terminal zeros in a numberwithout a decimal point are not significant. Allzeros occuring between two non-zero digitsare significant.

Chapter 01: Measurements9. A B [D] [E] 10.11. CD E C C C [A] [B] D E D E [A] [B] [M0LT 1] C 0 1 D [A] [M LT ] 00 C M LT [D] [E] 1 0 1 [T] LT M LT A2BC1/3 D3Taking logarithm of both sides,1log X 2 log A log B log C 3log D3Partially differentiating, X A B 1 C D 2 3XAB3 CD APercentage error in A 2 2 1% 2%A B 3%Percentage error in B B1 C 14 4% %Percentage error in C 3 C 33 DPercentage error in D 3 3 5% 15%DThe minimum percentage error is contributedby C. Hence the correct choice is (C).Given X E [M1L2T 2], G [M 1L3T 2], I [M1L1T 1] GI 2 M [M 1L3T 2 ][M1L1T 1 ]2 [M1 ] 2 [M1L2T 2 ]2 E [M0L1T0]This is the dimension of wavelength.12.Mere dimensional correctness of an equationdoes not ensure its physical correctness. Adimensionally correct equation may or maynot be physically correct but a dimensionallyincorrect equation is definitely incorrect.13.Percentage error r 3 100r0.4 3 100 19.35%6.2Nearest answer is option (C).14.Here [N] [M0L 2T 1][n1] [n2] [M0L 3T0]and [z1] [z2] [M0L1T0][N]Hence [D] [z1][n1 ][M 0 L 2T 1 ] [M0L1T0][M 0 L 3T 0 ] [M0L2T 1] 15. As zero of circular scale is above the referenceline of graduation, zero correction is positiveand zero error is negativeZero error 4 10 3 cm16.Relative velocity is defined as the time rate ofchange of relative position of one object withrespect to another. It is not the ratio of similarquantities.17.M.S.D. 3.48 cm, V.S.D. 6L.C. 0.01 cmobserved internal diameter of calorimeterD0 M.S.D. (V.S.D. L.C.) 3.48 (6 0.01) 3.48 0.06D0 3.54 cmzero error 0.03 cmSince, zero error is negative, it is added intoobserved reading.Corrected internal diameter,D D0 zero errorD 3.54 0.03 3.57 cm 18. M.S.D. 6.4 cm,

MHT-CET Triumph Physics (Hints) 4. From Van der Waal’s equation, nb has dimensions of volume. b = V n 12 2 Thus, S.I. units of b is m 3/mol. 12. [x] = [bt2] unit of b = 2 x t = 2 metre (hour) = 2 m hr 13. Energy = force distance, so if both are increased by 4 times then energy will increase by 16 times. 14. 1 dyne = 10 5 N and 1 cm = 10 2 m