Questions
The quantities x = \(\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}},(y)=\frac{E}{B}\) and z = \(\frac{1}{CR}\) are defined where C-capacitance, R-Resistance, /-length, E-Electric field, B-magnetic field and ε0 , μ0 , -free space permittivity and permeability respectively. Then :
(a) x, y and z have the same dimension.
(b) Only x and z have the same dimension.
(c) Only x and y have the same dimension.
( d) Only y and z have the same dimension.
Answer/Explanation
Ans
(a) x, y and z have the same dimension.
We know that Speed of light, c = \(\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}=x\) Also, c =\(\frac{E}{B}=y\)
Time constant, τ = Rc = t
∴ z = \(\frac{l}{Rc}=\frac{l}{t}\)= Speed
Question
Dimensional formula for thermal conductivity is (here K denotes the temperature :
(a) MLT-2K (b) ML -2 K-2
(c) MLT-3K (d) ML -3K-1
Answer/Explanation
Ans
(d) ML -3K-1
From formula, \(\frac{dQ}{dt}=kA\frac{dT}{dx}\)
⇒ k = \(\frac{\frac{dQ}{dt}}{\tfrac{(A)dT}{dx}}\), [k] = \(\frac{ML^2T^-3}{L^2KL^-1}\)=[MLT-3K-1
Question
A quantity x is given by (IF2/WL4) in terms of moment of inertia l, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:
(a) planck’s constant (b) force constant
(c) energy density (d) coefficient of viscosity
Answer/Explanation
Ans
(c) energy density
Dimension of Force F = M1L1T 2
Dimension of velocity V= L1T1
Dimension of work= M1L2T2
Dimension of length = L
Moment of inertia = ML2
∴ x = \(\frac{IFv^2}{WL^4}=\frac{(M^1L^2)(M^1L^1T^{-2})(L^1T^{-2})^2}{(M^1L^2)(L^4)}\)
= \(\frac{(M^1L^{-2}T^{-2})}{L^3}=M^1L^{-1}T^{-2}\) =
Question
Amount of solar energy received on the earth’s surface per unit area per unit time is defined a solar constant. Dimension of solar constant is :
(a) ML2T -2 (b) ML0T-3
(c) M2L0T-1 (d) MLT-2
Answer/Explanation
Ans
(b) ML0T-3
Solar constant = \(\frac{Energy}{Time Area}\)
Dimension of Time = T
Dimension of Area = L2
Question
If speed V, area A and force F are chosen as fundamental units, then the dimension of Young’s modulus will be :
(a) FA2V-1 (b) FA2V-3 (c) FA2V-2 (d) FA-1Vo
Answer/Explanation
Ans
(d) FA-1Vo
Young’s modulus, Y = \(\frac{stress}{strain}\)
⇒ Y
Question
If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is :
(a) [P2AT-2] (b) [PA-1 T-2] (c) [PA1/2T -1] (d) [P1/2AT-1]
Answer/Explanation
Ans
(c) [PA1/2T -1]
Energy, E α AaTbPc
or, E = kAaTbPc… (i)
where k is a dimensionless constant and a, b and c are the exponents.
Dimension of momentum, P = M1L1T-1
Dimension of area, A = L2
Dimension of time, T = T1
Putting these value in equation (i), we get
M1L2T-2 = McL2a+cTb-c
by comparison
c=l
2a+c=2
b-c=-2
c=1,a=1/2,b=-1 :. E=A1/2T-1P
Question
Which of the following combinations has the dimension of electrical resistance ( ∈0 is the permittivity of vacuum
and μ0 is the permeability of vacuum)?
(a) \(\sqrt{\frac{\mu _{o}}{\varepsilon _{o}}}\)
(b) \(\frac{\mu _{o}}{\varepsilon _{o}}\)
(c) \(\sqrt{\frac{\varepsilon _{o}}{\mu _{o}}}\)
(d) \(\frac{\varepsilon _{o}}{\mu _{o}}\)
Answer/Explanation
Ans
(a) \(\sqrt{\frac{\mu _{o}}{\varepsilon _{o}}}\)
\(\sqrt{\frac{\mu _{o}}{\varepsilon _{o}}}=\sqrt{\frac{\mu _{o}^2}{\varepsilon _{o}\mu _{o}}}=\mu _{o}c\)
μoc → MLT-2A-2 × LT-1
ML2T-3A-2
Question
In the formula X = 5YZ2, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units ?
(a) [M-3 L-2 8A4] (b) [M-1 L-2 4A2] (c) [M-2 Lo -4A-2] (d) [M-2 L-2 6A3]
Answer/Explanation
Ans
(a) [M-3 L-2 8A4]X = 5YZ2
⇒ Y α \(\frac{X}{Z^2}\) …..(i)
X = Capacitance = \(\frac{Q}{V}=\frac{Q^2}{W}=\frac{[A^2T^2]}{[ML^2T^{-2}]}\)
X = [M-1L-2T4A2]
Z = B = \(\frac{F}{IL}\) ∴ [F = ILB]
Z = [MT-2A-1]
Y = \(\frac{[M^{-1}L^{-2}T^4A^2]}{[MT^{-2}A^{-1}]^2}\)
Y = [M-3L-2T8A4
Question
In SI units, the dimensions of \(\sqrt{\frac{\epsilon _{o}}{\mu _{o}}}\) is :
(a) A-1TML3 (b) AT2M-1L-1
(c) AT-3ML3\2 (d) A2T3 M-1L-2
Answer/Explanation
Ans
(d) A2T3 M-1L-2
\([\sqrt{\frac{\varepsilon _{o}}{\mu _{o}}}]=\sqrt{\frac{\varepsilon _{o}^2}{\mu _{o}\varepsilon _{o}}}=[\frac{\varepsilon _{o}}{\sqrt{\mu _{o}\varepsilon _{o}}}]=\varepsilon _{o}E[LT^{-1}]\chi [\varepsilon _{o}]\)\([\therefore \frac{1}{\sqrt{\mu _{o}\varepsilon _{o}}}=C]\)
\(\therefore F=\frac{q^2}{4\pi \varepsilon _{o}r^2}\)
⇒[εo] = \(\frac{[AT]^2}{[MLT^{-2}]\chi [L^2]}=[{A^2M^{-1}L^{-3}T^4}]\)
Question
Let /, r, c and v represent inductance, resistance, capacitance and voltage, respectively. The dimension of \(\frac{l}{rcv}\) in SI units will be:
(a) [LA-2] (b) [A-1] (c) [LTA] (d) [LT2]
Answer/Explanation
Ans
(b) [A-1]
As we know,
\([\frac{l}{r}]=[T] and [cv]=[AT]\)
Question
The force of interaction between two atoms is given by F = αβexp\((-\frac{x^2}{\alpha kT})\); where x is the distance, k is the Boltzmann constant and T is temperature and α and β are two constants. The dimensions of β is:
(a) M0L2T4 (b) M2LT-4
(c) ML-2 (d) M2L2 -2
Answer/Explanation
Ans
(b) M2LT-4
Force of interaction between two atoms, F = αβe \((-\frac{x^2}{\alpha kT})\) Since exponential terms are dimensionless
∴ \([\frac{x^2}{\alpha kT}]=M^oL^oT^o\)
⇒ \(\frac{L^2}{[\alpha ]ML^2T^{-2}}=M^oL^oT^o\) ⇒[α] = M-1T2
Inductance \(\frac{L}{R}=t\) and R =\(\frac{v}{l}\) ohm-sec, volt-second (ampere)-1
Magnetic Induction F = lIB⇒B=\(\) newton (ampere-meter)
Question
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young’s modulus will be:
(a) V-2A2 F-2 (b) V-2A2F
(c) V-4A-2F (d) V-4A2F
Answer/Explanation
Ans
(d) V-4A2F
Let [Y] = [V]a [F]b [A]c
[ML-1T2] = [LT-1]a [MLT-2]b [LT-2]c
[ML-1Y-2] = [MbLa+b+c T-a-2b-2c]
Comparing power both side of similar terms we get,
b = 1, a+ b + c = – l, -a -2b -2c = – 2
solving above equations we get:
a=-4, b= 1, c=2
so [Y] = [V-4FA2] = [V-4A2
Question
A quantity f is given by f =\(\sqrt{\frac{hc^5}{G}}\) where c is speed of light, G 1miversal gravitational constant and h is the Planck’s constant. Dimension of/is that of:
(a) area (b) energy
(c) momentum (d) volume
Answer/Explanation
Ans
(b) energy
Dimension of [h] = [ML2T-1]
[C] = [LT-1]
[G] = [M-1L3T-2]
Hence dimension of
\([\sqrt{\frac{hC^5}{G}}]=\frac{[ML^2T^{-1}].[L^5T^{-5}]}{[M^{-1}L^3T^{-2}]}\)
= [ML2T-2
Question
Expression for time in terms of G (universal gravitational constant), h (Planck’s constant) and c (speed oflight) is roportional to:
(a)\(\sqrt{\frac{hc^2}{G}}\)
(b)\(\sqrt{\frac{c^3}{Gh}}\)
(c)\(\sqrt{\frac{Gh}{c^5}}\)
(d)\(\sqrt{\frac{Gh}{c^3}}\)
Answer/Explanation
Ans
(c)\(\sqrt{\frac{Gh}{c^5}}\)
Let t α Gx hy Cz
Dimensions of G = [M-1L3Y-2],
h = [ML2T-1] and C = [LT-1]
[Tl= [M-1L3T-2]×[ML2T-1]y[LT-1]z
[M0L0T1] = [M-x+y L3x+2y+z T-2x-y-z]
By comparing the powers of M, L, T both the sides
-x+y=O x=y
3x + 2y + z = 0 ⇒ Sx + z = 0 ….. (i)
-2x-y-z= 1 ⇒ 3x+z=-1 ….. (ii)
Solving eqns. (i) and (ii),
x = y = \(-x+y=O x=y
3x + 2y + z = 0 ⇒ Sx + z = 0 ….. (i)
-2x-y-z= 1 ⇒ 3x+z=-l ….. (ii)
Solving eqns. (i) and (ii),
x = y = \(\frac{1}{2}\), z = – \(\frac{5}{2}\)
Question
The dimensions of stopping potential V0 in photoelectric effect in units of Planck’s constant ‘h ‘, speed of light ‘c’ and Gravitational constant ‘G’ and ampere A is:
(a) h1/3G2/3c1/3 A -1 (b) h2/3c5/3G1/3A -1
(c) h-2/3 e-1/3 G4/3 A-1 (d) h2G3/2c1/3 A-1
Answer/Explanation
Ans
(None)
Stopping potential (V0) α hxLy Gz Cr ]
Here,h = Planck’s constant = ML2T-1I= current= [A]
G = Gravitational constant= [M-1L3T-2] and c = speed of light = [LT-1]
V0 = potential= [ML2T-3A-1]:. [ML2T3A-1]=[ML2T-1]x [A]y[M-1L3T2]z[LT-1]rMx-z; L2x+3z+r; T-x-2z-r; Ay
Comparing dimension of M, L, T, A, we get
y=-1,x=0,z=-1,r=5
:. Vo α h0 I-1G-1C
Question
The dimensions of \(\frac{B^2}{2\mu _{o}}\), where B is magnetic field and μo is the magnetic permeability of vacuum, is:
(a) MLT-2 (b) ML2 – 1 (c) ML2 – 2 (d)ML-1T-2
Answer/Explanation
Ans
(d)ML-1T-2
The quantity \(\frac{B^2}{2\mu _{o}}\) is the energy density of magnetic field.
⇒\([\frac{B^2}{2\mu _{o}}]=\frac{Energy}{Volumn}=\frac{Force\chi displacement}{(displacement)^3}\)
Question
The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants G, h and c. Which of the following correctly gives the Planck length?
(a) G2hc (b) \((\frac{Gh}{c^3})^{1/2}\) (c) \(\frac{1}{G^2h^2c}\) (d) Gh2c3
Answer/Explanation
Ans
(b) \((\frac{Gh}{c^3})^1/2\)
Plank length is a unit of length, lp = 1.616229 x 1 o-35 m
lp
Question
A physical quantity P is described by the relation p = a1/2 b2 c3 d -4 If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1 %, 3% and 5%, then the relative error in P will be :
(a) 8% (b) 12% (c) 32% (d) 25%
Answer/Explanation
Ans
(a) 8%
Let mass, related as M α TxCyhz
M1 L0T0 = (T1)x (L1T-1)y (M1L2T-1)z
M1L0T0 = Mz Ly+ 2z + Tx-y-z
z=l
y+2z=0
y=-2
x-y-z=0
x+2-1=0
x=-1
M = [T-1 C-2 h1
Question
In the following ‘I’ refers to current and other symbols have their usual meaning, Choose the option that corresponds to the dimensions of electrical conductivity :
(a) M-1L-3T3I (b) M-1L-3T3I2
(c) M-1L3T3I (d) ML-3T 3I2
Answer/Explanation
Ans
(b) M-1L-3T3I2
We know that resistivity r = \(\frac{RA}{1}\)
Conductivity = \(\frac{1}{resistivity}=\frac{1}{RA}\)
= \(\frac{1I}{VA}(QV=RI)\)
=\(\frac{[L][I]}{\frac{[ML^2T^{-2}]}{[I][T]}}\chi [L^2]\) \(\because v=\frac{w}{q}=\frac{w}{it}\)
= [M-1 L– 3T3][I2]= [M-1L-3 T3I2
Question
If electronic charge e, electron mass m, speed of light in vacuum c and Planck’s constant hare taken as fundamental quantities, the permeability of vacuum μ0 can be expressed in units of:
(a) \((\frac{h}{me^2})\)
(b)\((\frac{hc}{me^2})\)
(c)\((\frac{h}{ce^2})\)
(d)\((\frac{mc^2}{he^2})\)
Answer/Explanation
Ans
(c)\((\frac{h}{ce^2})\)
Let μ0 related with e, m, c and h as follows.
μo = keamb cchd
[MLT-2A-2] = [AT]a [M]b [LT-1] c[ML2T -1]
= [Mb+d Lc+2d Ta-c-d Aa]
On comparing both sides we get
a= – 2 … (i)
b + d = I … (ii)
c + 2d = l … (iii)
a – c – d = -2 … (iv)
By equation (i), (ii), (iii) & (iv) we get,
a = – 2, b = 0, c = – 1, d = 1
Question
If the capacitance of a nanocapacitor is measured in terms of a unit ‘u’ made by combining the electric charge ‘e’, Bohr radius ‘a0‘, Planck’s constant ‘h’ and speed of light ‘c’ then:
(a) \(u=\frac{e^2h}{a_{o}}\)
(b)\(u=\frac{hc}{e^2a_{o}}\)
(c)\(u=\frac{e^2c}{ha_{o}}\)
(d)\(u=\frac{e^2a_{o}}{hc}\)
Answer/Explanation
Ans
(d)\(u=\frac{e^2a_{o}}{hc}\)
Let unit ‘u’ related with e, a0, h and c as follows.
[u] = [e]a [a0] [h]c [C]d
Using dimensional method,
[M-1L-2T+4A+2] = [A1T1]a[L]b[ML2T-1]c[LT-1]d
[M-1L-2T+4A+2] = [Mc Lb+2c+d Ta-c-d Aa]
a = 2, b = 1, c = – 1, d = – 1
Question
From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is:
(a)\(\frac{ch}{2\pi \varepsilon _{o}^2}\)
(b)\(\frac{e^2}{2\pi \varepsilon _{o}Gm_{e}^2}\)
(c)\(\frac{\mu _{o}\varepsilon _{o}G}{c^2he^2}\)
(d)\(\frac{2\pi \sqrt{\mu _{o}\varepsilon _{o}}h}{ce^2G}\)
Answer/Explanation
Ans
(b)\(\frac{e^2}{2\pi \varepsilon _{o}Gm_{e}^2}\)
The dimensional formulae of
e = [ MOLOT1A1]
Eo = [ M-1L3T4A2]
G=[ M-1L3T-2] and me=[ M1LOTO]
Now, \(\frac{e^2}{2\pi \varepsilon _{o}Gm_{e}^2}\)
\(=\frac{[M^0L^0T^1A^1]^2}{2\pi [M^{-1}L^{-1}T^4A^2][M^{-1}L^3T^{-2}][M^1L^0T^0]^2}\)
\(=\frac{[T^2A^2]}{2\pi [M^{-1-1+2}L^{-3+3}T^{4-2}A^2]}\)
\(=\frac{[T^2A^2]}{2\pi [M^0L^0T^2A^2]}=\frac{1}{2\pi }\)
Question
In terms of resistance R and time T, the dimensions of ratio \(\frac{\mu }{\varepsilon }\) of the permeabilityμ and permittivity E is:
(a) [RT-2] (b) [R2T-1] (c) [R2] (d) [R2T2]
Answer/Explanation
Ans
(c) [R2]
Dimensions of μ= [MLT-2A-2]
Dimensions of E = [M-1L-3T4A-2]
Dimensions of R = [ML2T-3A-2]
\(\therefore \frac{Dimensions of \mu }{Dimensions of \epsilon }=\frac{[MLT^{-2}A^{-2}]}{[M^{-1}L^{-3}T^4A^2]}\)
= [M2L4T-6A-4] = [R2
Question
Let [ ∈0 ] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A= electric current, then:
(a)∈0 = [M-1 L-3 T2 A]
(b) ∈0 = [M-1 L-3 T4 A2]
(c) ∈0 = [M1 L2 T1 A2]
(d) ∈0 = [M1 L2 T1 A]
Answer/Explanation
Ans
(b) ∈0 = [M-1 L-3 T4 A2]
As we know, \(F = \frac{1q_1q_2}{4\pi\varepsilon _{0} R^2}\)
⇒ \(\varepsilon _{0} = \frac{1q_1q_2}{4\pi FR^2}\)
Question
The dimensions of angular momentum, latent heat and capacitance are, respectively.
(a) ML2T1A2, L2T-2, M-1L-2T2
(b) ML2T-2, L2T2, M-1L-2T4A2
(c) ML2T-1, L2T-2, ML2TA2
(d) ML2T-1, L2T-2, M-1L-2T4A2
Answer/Explanation
Ans
(d) ML2T-1, L2T-2, M-1L-2T4A2
Angular momentum= m x v x r = ML2 T-1
Latent heat \(L = \frac{Q}{m}= \frac{ML^2}T^{-2}{M}=L^2T^{-2}\)
Question
Which of the following set have different dimensions?
(a) Pressure, Young’s modulus, Stress
(b) EMF, Potential difference, Electric potential
(c) Heat, Work done, Energy
(d) Dipole moment, Electric flux, Electric field
Answer/Explanation
Ans
d) Dipole moment, Electric flux, Electric field
Electric flux ΦE = E .S
∴ Dimensionally ΦE
Question
Pressure depends on distance as, \(P = \frac{\alpha }{\beta }exp(-\frac{\alpha z}{k\Theta })\), where α, β are constants, z is distance, k is Boltzman’s constant and Θ is temperature. The dimension of β are
(a) M0L0T0
(b) M-1L-1T-1
(c) M°L2T°
(d) M-1L1T2
Answer/Explanation
Ans
(c) M°L2T°
The power of an exponent is a number or constant.
Therefore, dimensionally \(\frac{\alpha z}{k\theta }=[M^oL^oT^o]\)
\(\therefore \alpha =\frac{k\theta }{z}\)
\(\therefore \alpha =\frac{[ML^2T^{-2}\theta ^{-1}][\theta ]}{[L]}=[MLT^{-2}]\)
As \(k=\frac{joule}{kelvin}=[ML^2T^{-2}\theta ^{-1}]\) and \(z=[L]\)
And, dimensionally \(p=\frac{\alpha }{\beta }\Rightarrow \beta = \frac{\alpha }{P}\)
\(\therefore [\beta ]=\frac{MLT^{-2}}{ML^{-1}T^{-2}}=[M^oL^2T^o]\)
Question
The dimension of \((\frac{1}{2})\varepsilon_{o}E^2 \) (εo : permittivity of free
space, E electric field)
(a) MLT-1 (b) ML2T-2
(c) ML-1T-2 (d) ML2T-1
Answer/Explanation
Ans
(c) ML-1T-2
Here \((\frac{1}{2})\varepsilon _{o}E^2\) represents, energy density i.e., energy per unit volume.
Question
To find the distance d over which a signal can be seen clearly in foggy conditions, a railways-engineer uses dimensions and assumes that the distance depends on the mass density p of the fog, intensity (power/area) S of the light from the signal and its frequency f The engineer finds that d is proportional to s1/n. The value of n is
Answer/Explanation
Ans
(3)
Let d ∝ px Sy fz = Kp x Sy fz where K is a dimensionless constant.
M0L1T° = MxL-3xMyT-yT-z
Equating dimensions both sides
MoL1T° = Mx+y L-3xT-y-z
:. x + y = 0, – 3x = 1
:. x=-\(\frac{1}{3}\) and y= \(\frac{1}{3}\)
Question
The dimension of electrical conductivity is ……………. .
Answer/Explanation
Ans
[M-1L-3T3 A2]
Conductivity, σ = \(\frac{J}{E}=\frac{i/A}{F/q}=\frac{i^2t}{FA}\)
= [M-1L-3T3 A2
Question
The equation of state for real gas is given by \((p+\frac{a}{V^2})(V-b)=RT\). The dimenAsions of the constant α is ……………….. .
Answer/Explanation
Ans
[M-1L-3T3A2]
Here, \([\frac{\alpha }{V^2}]=[P]\Rightarrow [\alpha ]=[PV^2]=\frac{MLT^{-2}}{L^2}L^6=ML^5T^{-2}\)
Question
In the formula X = 3 YZ2, X and Z have dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are ____ , ____ .
Answer/Explanation
Ans
[M-3L-2T4Q4]
[X] = [C] = [M-1 L-2 T2 Q2]
[Z] = [B] = [MT-1 Q-1] \((\because Y=\frac{X}{3Z^2}given)\)
Question
Planck’s constant has dimension ____ .
Answer/Explanation
Ans
[ML2T-1]
Using E = hv ⇒ Planck’s constant,
MCQs with One or More Than One Correct
Question
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity X as follows:
[position] = [Xα]; [speed] = [Xβ]; [acceleration] =[Xp]; [linear momentum] = [Xq]; [force] =[Xp]. Then
(a) a+p=2β (b) p+q-r=β
(c) p-q+r = α (d) p+q+r=β
Answer/Explanation
Ans
(a) a+p=2β
(b) p+q-r=β
Given position, L = [Xα]
Speed, LT-1 = [Xβ]
Acceleration, LT-2 = [XP]
Linear momentum, MLT-1 = [Xq]
Force, MLT-2 = [Xr]
\(\frac{Position}{Speed}= time\), T = \(-\frac{[\chi ^{\beta }]}{\chi ^{\beta }}=\chi ^{\alpha -\beta }\)
Acceleration = \(\frac{Speed}{Time}=\frac{\chi ^{\beta }}{\chi ^{\alpha -\beta }}= X^p\)
xα-β+p =Xβ
:. α+p= 2β
Hence option (a) is correct.
Force = linear momentum/time
\([X^r]=\frac{[X^q]}{X^{\alpha -\beta }}\)
⇒ r = q + β – α ⇒ r = q + β – (2β – p)
⇒ r = q – β + p ⇒ p + q – r = β
Question
Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimensions of L, which of the following statement(s) is/are correct?
(a) The dimension of force is L-2
(b) The dimension of linear momentum is L-1
(c) The dimension of energy is L-2
(d) The dimension of power is L-5
Answer/Explanation
Ans
(a) The dimension of force is L-2
(b) The dimension of linear momentum is L-1
(c) The dimension of energy is L-2
According to question, dimensions of angular momentum
[mvr] = M°L0T° and of mass [m] = M°L0T0
ML2T-1 = MLOT° ⇒ T = L2
Momentum \(p=mv=\frac{mvr}{r}=\frac{M^oL^oT^o}{L}=L^{-1}\)
Energy \(E=\frac{1}{2}mv^2=\frac{1(mv)^2}{2m}=L^{-2}\)
Power \(P=\frac{E}{t}=\frac{L^{-2}}{T}=\frac{L^{-2}}{L^2}=L^{-4}\)
Question
A length-scale (l) depends on the permittivity(ε) of a dielectric material. Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge ( q) carried by each of the particles. Which of the following expression(s) for I is(are) dimensionally correct?
(a) \(l=\sqrt{(\frac{nq^2}{\varepsilon n^{2/3}k_BT})}\)
(b) \(l=\sqrt{(\frac{\varepsilon k_BT}{nq^2})}\)
(c) \(l=\sqrt{(\frac{q^2}{\varepsilon n^{2/3}k_BT})}\)
(d) \(l=\sqrt{(\frac{q^2}{\varepsilon n^{1/3}k_BT})}\)
Answer/Explanation
Ans
(b) \(l=\sqrt{(\frac{\varepsilon k_BT}{nq^2})}\)
(d) \(l=\sqrt{(\frac{q^2}{\varepsilon n^{1/3}k_BT})}\)
Dimensions of [n] = [L-3], [q] = [AT]
\([\frac{q^2}{\varepsilon }]=[Fr^2]=[U\chi r]=[ML^3T^{-2}]\)
[kBT] = [U] = [ML2T-2]
Dimension of l = [L]
(a) R.H.S. = \(\sqrt{(\frac{q^2}{\varepsilon })\chi \frac{1}{(k_BT)}\chi n}\)
\(=\sqrt{[U\chi r]\chi \frac{1}{[U]}\chi n}=\sqrt{[n]\chi [r]}=\sqrt{[L^{-3}][L]}=[L^{-1}]\)
(b) R.H.S. = \(\sqrt{\frac{\varepsilon (k_BT)}{nq^2}}=\sqrt{\frac{(k_BT)}{n(q^2/\varepsilon )}}=\sqrt{\frac{[U]}{[n][U\chi r]}}\)
\(=\sqrt{\frac{1}{[n][r]}}=\sqrt{\frac{1}{[L^{-3}][L]}}=[L]\)
(c) R.H.S. = \(\sqrt{(\frac{q^2}{\varepsilon })\frac{1}{(k_BT)}\chi \frac{1}{n^{2/3}}}\)
\(=\sqrt{[U\chi r]\chi \frac{1}{[U]}\frac{1}{[L^{-2}]}}=[L^{3/2}]\)
(d) R.H.S = \(\sqrt{(\frac{q^2}{\varepsilon })\chi\frac{1}{k_BT}\chi \frac{1}{n^{1/3}} }=\sqrt{[U\chi r]\chi \frac{1}{[U]}\chi \frac{1}{n^{1/3}}}=\sqrt{[r]\chi \frac{1}{[n^{1/3}]}}\)
Question
In terms of potential difference V, electric current I, permittivity ε0, permeability μ0 and speed of light c, the dimensionally correct equation(s) is(are)
(a) μ0I2 = ε0V2
(b) μoI=μoV
(c) I=ε0cV
(d) μ0cI = ε0V
Answer/Explanation
Ans
(a) μ0I2 = ε0V2
(c) I=ε0cV
Using, \(C=\frac{1}{\sqrt{\mu _o\varepsilon _o}}\) and \(R=\sqrt{\frac{\mu_o}{\varepsilon _{o}}}\)
(a) \(\mu _oI^2=\varepsilon _oV^2\Rightarrow \frac{\mu _o}{\varepsilon _o}=\frac{V^2}{I^2}=R^2=\sqrt{(\frac{\mu _o}{\varepsilon _o})^2}\)
(b) \(\varepsilon _oI=\mu _oV\Rightarrow \frac{\mu _o}{\varepsilon _o}=\frac{I}{V}=\frac{1}{R}(\because v =RI))
Dimensionally incorrect.
(c) I=εoCV
\(\therefore \frac{1}{\varepsilon _oC}=\frac{V}{I}=R\therefore \frac{1}{\varepsilon _o\frac{1}{\sqrt{\mu _o\varepsilon _o}}}=R\)
Dimensionally correct.
(d) μoCI=εoV
Question
Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are)
(a)\(M\alpha \sqrt{c}\) (b) \(M\alpha \sqrt{G}\)
(c) \(L\alpha \sqrt{h}\) (d) \(L\alpha \sqrt{G}\)
Answer/Explanation
Ans
(a)\(M\alpha \sqrt{c}\)
(b) \(M\alpha \sqrt{G}\)
(c) \(L\alpha \sqrt{h}\)
As Planck’s constant h, speed of light c and gravitational constant G are used as basic units for length L and Mass M so, L α hx cy G2………..(i)
and Mα hP cq Gr ……… (ii)
Dimensions of [h] = [M L2 T-1], [c] = [L T-1]
[G] = [M-1 L3 T-2]
Using principle of homogeneity of dimensions
For eqn. (i)
[MoLTo]=[Mx L2x T-x][Ly T-y][M-z L3z T-2z]
MoLTo= M(x-z)L(2x + y + 3z) T(-x-y- 2z)
On comparing powers from both sides, we get
x-z= 0, 2.x+ y + 3z= 1, -x-y-2z = 0
On solving these eqns., we get
\(x=\frac{1}{2},y=-\frac{3}{2},z=\frac{1}{2}\)
\(\therefore L=K\sqrt{\frac{hG}{c^3}}; K\) is a constant
In the same way solving eqn. (ii) we get,
Question
The pairs of physical quantities that have the same dimensions is (are):
(a) Reynolds number and coefficient of friction
(b) Curie and frequency of a light wave
(c) Latent heat and gravitational potential
(d) Planck’s constant and torque
Answer/Explanation
Ans
(a) Reynolds number and coefficient of friction
(b) Curie and frequency of a light wave
(c) Latent heat and gravitational potential
Reynold’s number
= Coefficient of friction= [M°L0To] i.e., dimensionless.
Curie is the unit of radioactivity (number of atoms decaying per second) and frequency also has the unit per second i.e., [M°L0T-1].
Latent heat \(= \frac{Q}{m}\) and Gravitation potential = \(\frac{W}{m}\)
Dimensionally same [MoL2T-2
Question
The dimensions of the quantities in one (or more) of the following pairs are the same. Identify the pair (s)
(a) Torque and Work
(b) Angular momentum and Work
(c) Energy and Young’s modulus
(d) Light year and Wavelength
Answer/Explanation
Ans
(a) Torque and Work
(d) Light year and Wavelength
Torque, τ = F × r × sin θ
and work done, W = F × d × cos θ
Dimensionally, θ = W = [ML2T -2]
Match the Following (Based viewed on Desktop/Laptop and ipad)
Question
Match List I with List II and select the correct answer using the codes given below the lists:
List I List II
P. Boltzmann constant 1. [ML2T-1]
Q. Coefficient of viscosity 2. [ML -1T-1]
R. Planck constant 3. [ML-3K-1]
S. Thermal conductivity 4. [ML2T-2K-1]
Codes:
P Q R S
(a) 3 1 2 4
(b) 3 2 1 4
(c) 4 2 1 3
(d) 4 1 2 3
Answer/Explanation
Ans
(c) 4 2 1 3
Boltzmannn constant \([K_B]=\frac{u}{\theta } (\because u=\frac{1}{2}K_BT))
\(\therefore [K_B]=ML^2T^{-2}K^{-1}\)
Coefficient of viscosity \([\eta ]=\frac{F}{6\pi rv}=\frac{MLT^{-2}}{L\chi T^{-1}}=ML^{-1}T^{-1})
Planck constant, \(h=\frac{E}{v}=\frac{ML^2T^{-2}}{T^{-1}}=ML^{-2}T^{-1}\)
Thermal conductivity \(K_{conductivity}=\frac{Hl}{tA\Delta T}\)
Question
Match the physical quantities given in column I with dimensions expressed in terms of mass (M), length (L), time (T), and charge (Q) given in column II and write the correct answer against the matched quantity in a tabular form in your answer book.
Column I Column II
Angular momentum ML2T-2
Latent heat ML2Q-2
Torque ML2T-1
Capacitance ML3T-1Q-2
Inductance M-1L-2T2Q2
Resistivity L2T-2
Answer/Explanation
Ans
Angular Momentum= mvr = [ML2T-1]
Latent heat [L] = Q/m [L] = [L2T-2]
Torque,;= F × r [τ] = [ML2T-2]
Capacitance \(C=\frac{1}{2}=\frac{q^2}{u};[C]=[M^{-1}L^{-2}T^2Q^2]\)
Inductance \(L=\frac{2u}{i^2};[L]=[ML^2Q^{-2}]\)
Passage
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [ E] and [ B] stand for dimensions of electric and magnetic fields respectively, while [ε0] and [μ0] stand for d1mens10ns of the permittivity and permeability of free space respectively. [L] and [1] are dimensions of length and time respectively. All the quantities are given in SI units.
Question
The relation between [ E] and [ B] is
(a) [E] = [B][L][T] (b) [E] = [B] [L]-1 [T]
(c) [E] = [B] [L] [T]-1 (d) [E]=[B] [L]-1 [T]-1
Answer/Explanation
Ans
(c) [E] = [B] [L] [T]-1
Using, \(C = \frac{E}{B}\) where C = speed of light
∴ E = CB = LT-1
Question
The relation between [ ε0] and [μ0] is
(a) [μo] = [ε0] [L]2 [T]-2 (b) [μ0] = [εo] [L]-2 [T]2
(c) [μo] = [εo]-1 [L]2 [T]-2 (d) [μo] = [εo]-1 [L]-2 [T]2
Answer/Explanation
Ans
(d) [μo] = [εo]-1 [L]-2 [T]2
We know that
\(C=\frac{1}{\sqrt{\mu _o\varepsilon _o}}\therefore C^2=\frac{1}{\mu _o\varepsilon _o}\)
Subjective Problems
Question
Write the dimensions of the following in terms of mass, time, length and charge
(i) magnetic flux
(ii) rigidity modulus
Answer/Explanation
Ans
(i) Magnetic Flux \(\phi =BA=\frac{F}{qv}A=[M^1L^2T^{-1}Q^{-1}]\)
Question
A gas bubble, from an explosion under water, oscillates with a period T proportional to padbEc. Where ‘P’ is the static pressure, ‘d’ is the density of water and ‘E’ is the total energy of the explosion. Find the values of a, band c.
Answer/Explanation
Ans
As per question, T α PadbEc or [T] = [P]a[d]b[E]c
or [M°LoT1] = [ML-1T-2]a [ML-3]b [ML2T-2]c
or MoLoT1 = Ma + b + c L-a -3b + 2c T 2a – 2c
a+b+c=0 ……(i)
-a-3b +2c=0 ……(ii)
-2a-2c= 1 …….(iii)
Solving, eqns. (i), (ii) and (iii) we get