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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
The diameter of a ball was measured five times with the aid of a micrometer whose absolute error `(Deltad_("inst"))= pm 0.01 mm`. The results of measuring the diameter of the ball are `d_(1)=5.27 mm, d_(2)=5.30 mm, d_(3)=5.28 mm, d_(4)=5.32 mm` and `d_(5)=5.28 mm`. Find (a) mean value of ball diameter (b) mean absolute error (c) result of measurement (d) relative error (e) persentage error. |
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Answer» (a) The mean value of the ball diameter, i.e., `d_(m)=(5.27+5.30+5.28+5.32+5.28)/(5)=5.29 mm` (b) The absolute error in the measurements are : `Deltad_(1)=|d_(m)-d_(1)|=0.02 mm` `Deltad_(2)=|d_(m)-d_(2)|=0.01 mm` `Deltad_(3)=|d_(m)-d_(3)|=0.01 mm` `Deltad_(4)=|d_(m)-d_(4)|=0.03 mm` `Deltad_(5)=|d_(m)-d_(5)|=0.01 mm` Mean absolute error, `Deltad_("mean")=(0.02+0.01+0.01+0.03+0.01)/(5)` `cong 0.02 mm`. (c) Since the mean absolute error `(Deltad_("mean"))` is greater than the instrumental error `(Delta d_("inst"))`, the result of measurement is `d=d_(m) pm Deltad_("mean")=(5.29 pm 0.02) mm` (As a rule, we take either `Deltad_("mean")` or `Deltad_("inst")`, depending upon which of these errors is greater) (d) Relative error, `(Deltad_("mean"))/d_(m)=0.02/5.29=0.004` (e) Percentage error, `(Deltad_("mean"))/d_(m)xx100=0.004xx100=0.4 %` |
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| 52. |
Four physical quantities are listed in column I. Their values are listed in column II in random order. Match the column correctly. |
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Answer» 1. At room temperature thermal energy of a gas molecule is, `E=3/2 Kt=3/2xx1.38xx10^(-23)xx300=6.21xx10^(-21) J` or `E=(6.21xx10^(21)//1.6xx10^(-19))=0.039 eV` So, thermal energy of air molecule is of the order of 0.02 eV, i.e., `[a] rarr [e]` 2. Energy of the order of MeV is associates with nuclear phenomena such as fission, fusion, radioactivity or binding energy. So binding energy per nucleon is of the order of MeV, i.e., `[b] rarr [h]` 3. As X-rays are produced by kV potential difference across the X-ray tube, so energy of X-rays is of the order of keV, i.e., `[c] rarr [g]` 4. As atoms get ionised by few eV so the energy of light emitted by atoms is of the order of few eV, i.e., `[d] rarr [f]` |
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| 53. |
There is a concept that if unit of a physical quantity is large then the multiplier of the unit becomes less while expressing any fixed magnitude. If in a new system of unit, unit of time is 0.5 second and the unit of length is 4 metre. Answer the following questions. Unit of speed becomes (compare to its initial value) :A. one eighthB. eight timesC. halfD. double |
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Answer» Correct Answer - b `nv =` constant Unit of speed is `m s^(-1)` Hence, `(4xxm)/(s//2)=8 ms^(-1)` |
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| 54. |
In British engineering system the unit of mass is slug while the unit of force is force is pound. How many kg are in a slug? |
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Answer» We know that, 1 slug is that mass of a body which moves with an acceleration of `1 ft//s^(2)` under the action of a force of one pound, i.e., 1 slug = 1b force`//(ft//s^(2))` but ib force `=[g] xx 1b` (mass) or 1b force `=[32.2 ft//s^(2)] xx 1b` (mass) So, 1 slug `=[32.3 ft//s^(2)] 1b` (mass) `//[ft//s^(2)]` or 1 slug `=32.2 1b` (mass) `=32.2 xx 0.4536 kg` 1 slug `=14.59 kg` |
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| 55. |
When a body is weighed successively in the two pans of a physical balance with unequal arms, the apparent masses are found to be `M_(1)` and `M_(2)`. Show that the length of the arms are in the ratio `sqrt(M_(1)): sqrt(M_(2))`. |
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Answer» Let `l_(1)` and `l_(2)` be the lengths of left and right arms of the balance. Then, `l_(1)M=l_(2)M_(1) rArr l_(1)/l_(2)=M_(1)/M` and `l_(1)M_(2)=l_(2)M rArr l_(1)/l_(2)=M/M_(2)` or `l_(1)/l_(2)l_(1)/l_(2)=M_(1)/MM/M_(2)=M_(1)/M_(2)` `:. l_(1)/l_(2)=sqrt(M_(1)/M_(2))` |
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| 56. |
In an experiment to determine the specific heat of a metal,a `0.20 kg` block of the metal at `150 .^(@) C` is dropped in a copper calorimeter (of water equivalent `0.025 kg`) containing `150 cm^3` of water at `27 .^(@) C`. The final temperature is `40.^(@) C`. The specific heat of the metal is. |
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Answer» Let specific heat of brass `=C_(1)` Mass of brass `m_(1)=0.2 kg` Fall in temp. of brass `Delta theta_(1)=150-40` `=110^(@)C=110 K` Mass of water `m_(2)=0.150 kg` Rise in temp. of water `Delta theta_(2)=40-27` `=13^(@)C=13 K` Water equivalent of calorimeter `w=0.025 kg` According to the principle of caloriemetry, Heat lost = Heat gained `m_(1)C_(1)Delta theta_(1)=m_(2)C_(2)Delta theta_(2)+wC_(2)Delta theta_(2)` `C_(1)=((m_(2)+w)C_(2)Delta theta_(2))/(m_(1)Delta theta_(1))` ...(i) `=((0.150+0.025)xx4.2xx10^(3)xx13)/(0.2xx110)` `=(0.175xx4.2xx10^(3)xx13)/(0.2xx110)` `=0.434xx10^(3)J kg^(-1) K^(-1)` From eqn. (i), the maximum permissible error in sp. heat is given by `(deltaC_(1))/C_(1)=(deltam_(2))/((m_(2)+omega))+(delta(Deltatheta_(2)))/(Delta theta_(2))+(Delta m_(1))/m_(1)+(delta(Delta theta_(1)))/(Deltatheta_(1))` `=0.001/0.175+0.2/13+0.001/0.200+0.2/110` `=0.0057+0.0153+0.005+0.0018` `=0.0278` `:. DeltaC_(1)=0.0278xx0.434xx10^(3)J kg^(-1) K^(-1)` `=0.012xx10^(3) J kg^(-1) K^(-1)` thus, specific heat of brass is `C_(1)=(0.43 pm 0.01)xx10^(3) J kg^(-1) K^(-1)` |
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| 57. |
The dimensions of gravitational constant G are :A. `[MLT^(-2)]`B. `[ML^(3)T^(-2)]`C. `[M^(-1)L^(3)T^(-2)]`D. `[M^(-1)LT^(-2)]` |
| Answer» Correct Answer - c | |
| 58. |
E, m, L, G denote energy mass, angular momentum & gravitation constant respectively. The dimensions of `(EL^(2))/(m^(5)G^(2))` will be that of :A. lengthB. massC. timeD. angle |
| Answer» Correct Answer - d | |
| 59. |
The velocity v of a particle at time t is given by `v=at+b/(t+c)`, where a, b and c are constants. The dimensions of a, b, c are respectively :-A. `[LT^(-2)[, [L]` and `[T]`B. `[L^(2)], [T]` and `[LT^(2)]`C. `[LT^(2)], [LT]` and `[L]`D. `[L], [Lt]` and `[T^(2)]` |
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Answer» Correct Answer - a Dimension of at = Dimension of v Dimension of c = Dimension of t Dimension of `b/(t+c)=` Dimension of v |
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| 60. |
The speed `(v)` of ripples on the surface of waterdepends on surface tension `(sigma)`, density `(rho)` and wavelength `(lambda)`. The square of speed `(v)` is proportional toA. `sigma/(rho lambda)`B. `rho/(sigma lambda)`C. `lambda/(sigma rho)`D. `rho lambda sigma` |
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Answer» Correct Answer - a `v prop sigma^(x)rho^(y)lambda^(z)` `v=ksigma^(x)rho^(y)lambda^(z)` `[LT^(-1)]=[MT^(-2)]^(x)[ML^(-3)]^(y)[L]^(z)` `x+y=0" "," "-2x=-1" "," "-3y+z=1` `x=1/2" "y=-1/2" "z=-1/2` Hence, `v=(sigma^(1//2))/(rho^(1//2)lambda^(1//2))` `v^(2) equiv (sigma/(rho lambda))` |
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| 61. |
The velocity of the waves on the surface of water is proptional to `lambda^(a) rho^(beta) g^(gamma)` where `lambda` = waver length, `rho` = density and `g` =m acceleration due to gravity. Which of the following relation is correct?A. `alpha = beta = gamma`B. `beta = gamma ne alpha`C. `gamma = alpha ne beta`D. `alpha ne beta ne gamma` |
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Answer» Correct Answer - c `v prop lambda^(alpha) rho^(beta) g^(gamma)` `[LT^(-1)]=[L]^(alpha)[ML^(-3)]^(beta)[LT^(-2)]^(gamma)` `[LT^(-1)]=[L^(alpha-3beta+gamma)M^(beta)T^(-2gamma)]` Comparing, `beta=0, gamma=1/2, alpha-3 beta+gamma=1` `alpha-0+1/2=1` `alpha=1/2` `:. gamma = alpha ne beta` |
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| 62. |
The velocity of water wave `v` may depend on their wavelength `lambda`, the density of water `rho` and the acceleration due to gravity `g`. The method of dimensions gives the relation between these quantities asA. `V^(2) prop g^(-1) gamma^(-1)`B. `V^(2) prop g lambda`C. `V^(2) prop g lambda rho`D. `V^(2) prop g^(-1) lambda^(-3)` |
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Answer» Correct Answer - b See question 38 above. `v prop sqrt(g lambda)rArr v^(2) prop g lambda` |
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| 63. |
In the formula `X=3YZ^(2)`, X and Z have dimensions of capacitance and magnetic induction respectively. What are the dimensions of Y in MKSQ system? |
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Answer» As `r=CV, C=q/V=q^(2)/W [as V=W/q]` So, `[X] rarr [C]=[Q^(2)/(ML^(2)T^(-2))]` `=[M^(-1)L^(-2)T^(2)Q^(2)]` and `F= Bil sin theta," "[B]=[F/(il)]` So, `[Z] rarr [B]=[(MLT^(-2))/(QT^(-1)L)]=[MT^(-1)Q^(-1)]` But as it is given `X= 3YZ^(2), " "i.e., Y=X//(3Z^(2))` So, `[Y]=([M^(-1)L^(-2)T^(2)Q^(2)])/([MT^(-1)Q^(-1)]^(2)) rarr[M^(-3)L^(-2)T^(4)Q^(4)]` |
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| 64. |
Given `F = (a//t) + bt^2` where F denotes force and t time. The diamensions of a and b are respectively:A. `[MLT^(-1)]` and `[MLT^(-4)]`B. `[LT^(-1)]` and `[T^(-2)]`C. `[T]` and `[T^(-2)]`D. `[LT^(-2)]` and `[T^(-2)]` |
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Answer» Correct Answer - a `[a/t]=[F]` `[b/t^(2)]=[F]` |
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| 65. |
Column-1 gives three physical quantities. Select the appropriate units for the choices given in column-II. Some of the physical quantities may have more than one choice. `{:("Column-I",,"Column-II"),("Capacitance",,"ohm-second"),("Inductance",,"Coulomb"^(2)//"Joule"),("Magnetic induction",,"coulomb"//"volt"),(,,"newton"//"ampere metre"),(,,"volt-second"//"ampere"):}` |
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Answer» (I) `q=CV i.e., [C]=[q//V]` so `[C]=[M^(-1)L^(-2)T^(4)A^(2)]` `U=1/2Li^(2), i.e., [L]=[U//i^(2)]` so `[L]=[ML^(2)T^(-2)A^(-2)]` `F=Bil sin theta, i.e., [B]=[F//il]` so `[B]=[MT^(-2)A^(-1)]` (II) Now the dimensions from given units are `ohmxx sec equiv [R][T] rarr [ML^(2)T^(-3)A^(-2)][T]` `=[ML^(2)T^(-2)A^(-2)]` `"coul"^(2)-"joule"^(-1) equiv [q^(2)/W] rarr ([A^(2)T^(2)])/([ML^(2)T^(-2)])` `=[M^(-1)L^(-2)T^(4)A^(2)]` `"coul (volt)"^(-1) equiv[q^(2)/W] rarr ([A^(2)T^(2)])/([ML^(2)T^(-2)])=[M^(-1)L^(-1)T^(4)A^(2)]` `"newton (amp-m)"^(-1) equiv [F/(il)] rarr ([MLT^(-2)])/([AL])=[MT^(-2)A^(-1)]` `"volt sec (amp)"^(-1) equiv [(WxxT)/(qxxA)] rarr [(ML^(2)T^(-2)T)/(ATA)]` `=[ML^(2)T^(-2)A^(-2)]` Comparing dimensions of II with I, we find that : Capacitance has unit `"coulomb"^(2)"-joule"^(-1)`, and coulomb `("volt")^(-1)`, inductance has units ohm-sec and volt-sec `("ampere")^(-1)`, and magnetic induction has unit newton `("ampere m")^(-1)`. |
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| 66. |
In an experiment on determining density of a rectangular block, the dimensions of the block are mesured with vernier callipers `(V.C = 0.01 cm)` and its mass is measured with a beam balance `(L.C. = 0.1 g)`. The measured values are : Mass`(m) = 39.3` g Length`(l) = 5.12` cm Breadth`(b) = 2.56` cm Thickness`(t)=0.37` cm Calculate density of the block with permissible limits of error. |
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Answer» Density `(rho)= ("Mass")/("Volume")=m/(lxxbxxt)` `=39.3/(5.12xx2.56xx0.37)g cm^(-3)` `=8.1037 g cm^(-3)` `=8.1 g cm^(-3)` (rounding off the result) The maximum relative error in the density is given by `(Delta rho)/rho=(Delta m)/m+(Delta l)/(l)+(Delta b)/b+(Delta t)/t` `=0.1/39.3+0.01/5.12+0.01/2.56+0.01/0.37` `=0.0024+0.0019+0.0039+0.027` `=0.0358` `:. " "Delta rho =0.0358 rho=0.0358xx8.1037 cong 0.3 g cm^(-3)` Thus, the density of the solid is `(8.1 pm 0.3)g cm^(-3)` Note : The main contribution to the relative error comes from the measurement of thickness (t). Hence, precision in the measurement of density can be increased by measuring thickness with screw gauge (L.C. = 0.001 cm) instead of vernier callipers (V.C. = 0.01 cm). experimentalists keep such facts in mind while carrying out different measurements in a given experiment. |
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| 67. |
Match the units/dimensions in Column -I with the physical quantities /expressions in column -II. |
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Answer» Correct Answer - `a rarr p, q, r; b rarr s; c rarr p, q, r; d rarr s` `{:(Q=MxxcxxDeltaT,c rarr" Specific heat"),(Q=MxxL,L rarr" Latent heat"):}` Mean square velocity `=[LT^(-1)]^(2)=[L^(2)T^(-2)]` |
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| 68. |
Indicate which pair of physical quantities gives below has not the same units and dimensions?A. Momentum and impulseB. Torque and angular momentumC. Acceleration and gravitational field strengthD. Pressure and modulus of elasticity |
| Answer» Correct Answer - b | |
| 69. |
If mass, length and acceleration is taken as base quantities in a system then dimension of length in dimensional formula of energy is : |
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Answer» Correct Answer - 1 `E=[ML^(2)T^(-2)]=[MLLT^(-2)]=[MLA]` where `A=[LT^(-2)]=` Acceleration `:.` Dimensions of `[L]` is 1 in dimensional formula for energy. |
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| 70. |
If L, C and R represent the physical quantities indutance, capacitance and resistance respectively, the conbinations which have the dimensions of frequency are :A. `(1//RC)`B. `(R//L)`C. `(1//sqrt(LC))`D. `(C//R)` |
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Answer» Correct Answer - a, b, c Time constant `= RC` Time constant `= L//R` `:.` Frequency `=I/(RC)`, Frequency `=R/L` Frequency `=1/(2pi) sqrt(1/(LC))` |
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| 71. |
Which physical quantities have same dimensions ?A. Force and powerB. Torque and energyC. Torque and powerD. Force and Torque |
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Answer» Correct Answer - b Torque and energy both have same dimension `[ML^(2)T^(-2)]`. |
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| 72. |
Which one of the following pair of quantities has same dimension ?A. Force and work doneB. Momentum and impulseC. Pressure and forceD. Surface tension and stress |
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Answer» Correct Answer - b According to impulse momentum theorem Impulse = Change in linear momentum Hence, both the quentities will have same dimension. |
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| 73. |
Check the correctness of the relation `c = (1)/(sqrt(mu_0 in_0))` where the symbols have their usual meaning. |
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Answer» Correct Answer - Correct both dimensionally and numerically Principle of homogeneity (Dimensions of each term on both side of an equation must be same). |
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| 74. |
The dimensions of self-inductance are :A. `[MLT^(-2)A^(-2)]`B. `[ML^(2)T^(-1)A^(-2)]`C. `[ML^(2)T^(-2) A^(-2)]`D. `[ML^(2)T^(-2)A^(2)]` |
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Answer» Correct Answer - c Energy stored in a coil `U=(1//2)Li^(2)` |
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| 75. |
The SI unit of inductance, the henry can be written as :A. weber/ampereB. volt -second/ampereC. joul/`("ampere")^(2)`D. ohm-second |
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Answer» Correct Answer - a, b, c, d Inductance, `L=-e/(((di)/(dt)))` unit `=("volt-second")/("ampere")` or joule`//("ampere")^(2)` `phi=Li` Hence, unit of L weber/ampere. |
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| 76. |
In terms of potential difference V, electric current I, permitivity `epsi_(0)`, permeability `mu_(0)` and speed of light c, the dimensionally correct equations (s) is (are) :A. `mu_(0)I_(2)=epsiV^(2)`B. `mu_(0)I=mu_(0)V`C. `I=epsi_(0)V`D. `mu_(0)cI=epsi_(0)V` |
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Answer» Correct Answer - a, c `[V]=[M^(1)L^(2)T^(-3)A^(-1)]` `[I]=[A]` `[c]=[L^(1)T^(-1)]` `[epsi_(0)]=[M^(-1)L^(-3)T^(4)A^(2)]` `[mu_(0)]=[M^(-1)L^(1)T^(-2)A^(-2)]` Option (a) : `[mu_(0)I^(2)]=[M^(1)L^(1)T^(-2)]` `[E_(0)V^(2)]=[M^(1)L^(1)T^(-1)]` Option (b) : `[epsi_(0)I]=[M^(1)L^(-3)T^(4)A^(3)]` `[mu_(0)V]=[M^(2)L^(3)T^(-5)A^(-3)]` Option (c) : `[I]=[A]` `[epsi_(0)cV]=[A]` Option (d) : `[mu_(0)cI]=[M^(1)L^(2)T^(-3)A^(-1)]` `[epsi_(0)V]=[L^(-1)TA]` Options (a) and (c) correct. |
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| 77. |
Let `[epsi_(0)]` denote the dimensional formula of the permittivity of the vacuum and `[mu_(0)]` that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current :A. `[epsi_(0)]=[M^(-1)L^(-3)T^(2)I]`B. `[epsi_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`C. `[mu_(0)]=[MLT^(-2)I^(-2)]`D. `[mu_(0)]=[ML^(2)T^(-1)I]` |
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Answer» Correct Answer - b, c Since, `F=1/(4pi epsi_(0))(q_(1)q_(2))/r^(2)` Hence, `epsi_(0)=((q_(1)q_(2))/(4pi Fr^(2)))` and `q=I xxt` Hence, `epsi_(0) rarr [M^(-1)L^(-3)T^(4)I^(2)]` `B=mu_(0)/(2pi) I/r` `mu_(0)=((Bxx2pixxr)/I)` Hence, `mu_(0) rarr [ML^(2)T^(-1)]` |
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