IIT-JEE Advanced Physics:Dimensions of Physical Quantities: Important Questions

IIT JEE advanced PhysicsIIT JEE Advanced Chemistry IIT JEE Advanced Maths

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^-1V^o

Answer/Explanation

Ans
(d) FA^-1V^o
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) [PA^1/2T ^-1] (d) [P^1/2AT^-1]

Answer/Explanation

Ans
(c) [PA^1/2T ^-1]
Energy, E α A^aT^bP^c
or, E = kA^aT^bP^c… (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 = M^cL^2a+cT^b-c
by comparison
c=l
2a+c=2
b-c=-2
c=1,a=1/2,b=-1 :. E=A^1/2T^-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) \(\sqrt{\frac{\mu _{o}}{\varepsilon _{o}}}\)
\(\sqrt{\frac{\mu _{o}}{\varepsilon _{o}}}=\sqrt{\frac{\mu _{o}^2}{\varepsilon _{o}\mu _{o}}}=\mu _{o}c\)
μ_oc → 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 L^o ^-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^-3ML^3\2 (d) A²T³ M^-1L^-2

Answer/Explanation

Ans
(d) A²T³ M^-1L^-2\([\sqrt{\frac{\varepsilon _{o}}{\mu _{o}}}]=\sqrt{\frac{\varepsilon _{o}^2}{\mu _{o}\varepsilon _{o}}}=[\frac{\varepsilon _{o}}{\sqrt{\mu _{o}\varepsilon _{o}}}]=\varepsilon _{o}E[LT^{-1}]\chi [\varepsilon _{o}]\)\([\therefore \frac{1}{\sqrt{\mu _{o}\varepsilon _{o}}}=C]\)
\(\therefore F=\frac{q^2}{4\pi \varepsilon _{o}r^2}\)
⇒[ε_o] = \(\frac{[AT]^2}{[MLT^{-2}]\chi [L^2]}=[{A^2M^{-1}L^{-3}T^4}]\)

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^-4Force 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] = [M^bL^a+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 G^x h^y C^z
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 L^3x+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) h^1/3G^2/3c^1/3 A^-1 (b) h^2/3c^5/3G^1/3A ^-1
(c) h^-2/3 e^-1/3 G^4/3 A^-1 (d) h²G^3/2c^1/3 A^-1

Answer/Explanation

Ans
(None)
Stopping potential (V₀) α h^xL^y G^z C^r ]
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]^rM^x-z; L^2x+3z+r; T^-x-2z-r; A^y
Comparing dimension of M, L, T, A, we get
y=-1,x=0,z=-1,r=5
:. V_o α h⁰ I^-1G^-1C

Question

The dimensions of \(\frac{B^2}{2\mu _{o}}\), where B is magnetic field and μ_o is the magnetic permeability of vacuum, is:
(a) MLT^-2 (b) ML^{2 – 1} (c) ML^{2 – 2} (d)ML^-1T^-2

Answer/Explanation

Ans
(d)ML^-1T^-2The 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, l_p = 1.616229 x 1 o^-35 m
l_p

Question

A physical quantity P is described by the relation p = a^1/2 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 T^xC^yh^z
M¹ L⁰T⁰ = (T¹)^x (L¹T^-1)^y (M¹L²T^-1)^z
M¹L⁰T⁰ = M^z L^{y+ 2z} + T^x-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^-3T³I²We 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^{– 3}T³][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.
μ_o = ke^am^b c^ch^d
[MLT^-2A^-2] = [AT]^a [M]^b [LT^-1] ^c[ML²T ^-1]
= [M^b+d L^c+2d T^a-c-d A^a]
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] = [M^c L^b+2c+d T^a-c-d A^a]
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 = [ M^OL^OT¹A¹]
E_o = [ M^-1L³T⁴A²]
G=[ M^-1L³T^-2] and m_e=[ M¹L^OT^O]
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^-1L^-3 T² A]
(b) ∈₀= [M^-1L^-3T⁴ A²]
(c) ∈₀= [M¹ L² T¹ A²]
(d) ∈₀ = [M¹ L² T¹ A]

Answer/Explanation

Ans
(b) ∈₀= [M^-1L^-3T⁴ 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 Φ_E = E .S
∴ Dimensionally Φ_E

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 \((\frac{1}{2})\varepsilon_{o}E^2 \) (ε_o : 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^-2Here \((\frac{1}{2})\varepsilon _{o}E^2\) 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 s^1/n. The value of n is

Answer/Explanation

Ans
(3)
Let d ∝ p^x S^y f^z = Kp ^x S^y f^z where K is a dimensionless constant.
M⁰L¹T° = M^xL^-3xM^yT^-yT^-z
Equating dimensions both sides
M^oL¹T° = M^x+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] = [X^α]; [speed] = [X^β]; [acceleration] =[X^p]; [linear momentum] = [X^q]; [force] =[X^p]. 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 = [X^α]
Speed, LT^-1 = [X^β]
Acceleration, LT^-2 = [X^P]
Linear momentum, MLT^-1 = [X^q]
Force, MLT^-2 = [X^r]
\(\frac{Position}{Speed}= time\), T = \(-\frac{[\chi ^{\beta }]}{\chi ^{\beta }}=\chi ^{\alpha -\beta }\)
Acceleration = \(\frac{Speed}{Time}=\frac{\chi ^{\beta }}{\chi ^{\alpha -\beta }}= X^p\)
x^α-β+p =X^β
:. 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^-2According to question, dimensions of angular momentum
[mvr] = M°L⁰T° and of mass [m] = M°L⁰T⁰
ML²T^-1 = ML^OT° ⇒ 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 (k_B), 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]
\([\frac{q^2}{\varepsilon }]=[Fr^2]=[U\chi r]=[ML^3T^{-2}]\)
[k_BT] = [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) μ_oI=μ_oV
(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=ε_oCV
\(\therefore \frac{1}{\varepsilon _oC}=\frac{V}{I}=R\therefore \frac{1}{\varepsilon _o\frac{1}{\sqrt{\mu _o\varepsilon _o}}}=R\)
Dimensionally correct.
(d) μ_oCI=ε_oV

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 h^x c^y G²………..(i)
and Mα h^P c^q G^r ……… (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)
[M^oLT^o]=[M^x L^2x T^-x][L^y T^-y][M^-z L^3z T^-2z]
M^oLT^o= 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°L⁰T^o] 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 [M^oL²T^-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) [μ_o] = [ε₀] [L]² [T]^-2 (b) [μ₀] = [ε_o] [L]^-2 [T]²
(c) [μ_o] = [ε_o]^-1 [L]² [T]^-2 (d) [μ_o] = [ε_o]^-1 [L]^-2 [T]²

Answer/Explanation

Ans
(d) [μ_o] = [ε_o]^-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 p^ad^bE^c. 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 P^ad^bE^c or [T] = [P]^a[d]^b[E]^c
or [M°L^oT¹] = [ML^-1T^-2]^a [ML^-3]^b [ML²T^-2]^c
or M^oL^oT¹ = M^{a + 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

Questions

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Question

Resources

Members

Company