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Units and Dimensions Previous Year Questions With Solutions

1.0Introduction

Units and dimensions are key concepts in physics and engineering, used to measure and express physical quantities. A unit is a standard of measurement (e.g., meter for length, kilogram for mass), while dimensions describe the physical nature of a quantity, expressed in terms of basic quantities like length, mass, and time. The International System of Units (SI) includes seven base units, and derived units are combinations of these. Dimensions ensure equations are consistent and help with unit conversions and the formulation of physical laws. Understanding units and dimensions is essential for accurate and universal communication in science and technology.

JEE Main Previous Year Solved Questions on Units and Dimensions
JEE Adv Previous Year Solved Questions on Units and Dimensions

2.0Key Concepts to Remember

Dimensional Equation:The equation formed by equating a physical quantity to its dimensional formula is known as a dimensional equation.

$M a ss \to [M^{1} L^{0} T^{0}]$

$L e n g t h \to [M^{0} L^{1} T^{0}]$

$T im e \to [M^{0} L^{0} T^{1}]$

$T e m p er a t u re \to [M^{0} L^{0} T^{0} K^{1}]$

$C u rre n t \to [M^{0} L^{0} T^{0} A^{1}]$

$Lu min o u s I n t e n s i t y \to [M^{0} L^{0} T^{0} c d^{1}]$

$A m o u n t o f S u b s t an ce \to [M^{0} L^{0} T^{0} m o l^{1}]$

3.0Rule of Dimensions

Only the same physical quantities can be added or subtracted.

A + B = C – D

Dimensionless Quantities

Dimensionless Quantities are:

• Ratio of physical quantities with same dimensions.

• All mathematical constants.

• All standard mathematical functions and their inputs (exponential, logarithmic, trigonometric & inverse trigonometric).

4.0Important Formulas

Conversion Between System of Units	$n_{2} = n_{1} (\frac{M _{1}}{M _{2}})^{a} (\frac{L _{1}}{L _{2}})^{b} (\frac{T _{1}}{T _{2}})^{c}$
Errors in Measurement	Error = True value – Measured value
Absolute Error (a)	$Δ a = a_{T} - a$
Mean Absolute Erroram $Δ a = a_{T} - a$	$(Δ a)_{m} = \frac{1}{n} \sum_{i = 0}^{n} ∣ a_{i} ∣$
Relative or Fractional Error	$R e l a t i v e E rror = \frac{M e an A b so l u t e E rror}{T r u e Va l u e}$
Percentage Error	$P erce n t a g e E rror = \frac{M e an A b so l u t e E rror}{T r u e Va l u e} \times 100$
Addition or Subtraction of Quantities:	$Δ X = Δ A + Δ B$
Multiplication or Division of Quantities:	$\frac{Δ X}{X} = \pm (\frac{Δ A}{A} + \frac{Δ B}{B})$
Maximum Fractional Error	$1. If X = A^{n} then \frac{Δ X}{X} = n (\frac{Δ A}{A}) 2. If X = A^{p} B^{q} C^{r} then \frac{Δ X}{X} = [p (\frac{Δ A}{A}) + q (\frac{Δ B}{B}) + r (\frac{Δ C}{C})] 3. If X = \frac{A ^{p} B ^{q}}{C ^{r}} then \frac{Δ X}{X} = [p (\frac{Δ A}{A}) + q (\frac{Δ B}{B}) + r (\frac{Δ C}{C})]$
Vernier Callipers	$L . C = M - V = (\frac{b - a}{b}) M$
Measurement with Vernier Calliper	$R e a d in g = MSR + (V S co in c i d e d d i v i s i o n w i t h MS \times L C)$
Positive Zero Error Correction	READING = (Main Scale Reading) + (Vernier Scale Reading) – ZE Positive zero error = VS Coincided division with MS × LC
Negative Zero Error Correction	Negative zero error = –(Total VSD – VS Coincided division with MS) × LC
Least Count of Screw Gauge	$L e a s t co u n t o f scre w g a ug e = \frac{P i t c h ( P )}{T o t a l n o . o f d i v i s i o n s o n t h e c i rc u l a r sc a l e ( N )}$

5.0Past Year Questions with Solutions on Units And Dimension:JEE (Mains)

Q-1.The dimension of , where B is magnetic field and µ₀ is the magnetic permeability of vacuum, is:

$(1) [M L^{- 1} T^{- 2}] (2) [M L^{2} T^{- 1}] (3) [M L T^{- 2}] (4) [M L^{2} T^{- 2}]$

Solution: Ans(1)

Magnetic energy stored per unit volume is

$\frac{B ^{2}}{2 μ _{0}} Dimensions is [M L^{- 1} T^{- 2}]$

Q-2.The dimension 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 :

$[(1)] h^{2} G^{3/2} c^{1/3} A^{- 1} [(2)] h^{- 2/3} c^{- 1/3} G^{4/3} A^{- 1} [(3)] h^{1/3} G^{2/3} c^{1/3} A^{- 1} [(4)] h^{2/3} c^{5/3} G^{1/3} A^{- 1}$

Solution: Ans(Bonus)

$V_{0} = h^{x} c^{y} G^{z} A^{w}$

$\frac{M L ^{2} T ^{- 2}}{A T} = (M L^{2} T^{- 1})^{x} (L T^{- 1})^{y} (M^{- 1} L^{3} T^{- 2})^{z} A^{w}$

w=-1

x-z=1

2x+y+3x=2

-x-y-2z=-3

$2 x = 0 \Rightarrow x = 0$

z=-1,y=5

$V_{0} = h^{0} c^{5} G^{- 1} A^{- 1}$

Q-3.A quantity f is given by $f = \frac{h c ^{5}}{G}$ where c is the speed of light, G universal gravitational constant and h is Planck's constant. Dimension of f is that of :

(1) Momentum (2) Area (3) Energy (4) Volume

Solution:Ans(3)

$[h] = M^{1} L^{2} T^{- 1}, [c] = L^{1} T^{- 1}, [G] = M^{- 1} L^{3} T^{- 2}$

$[f] = \frac{M ^{1} L ^{2} T ^{- 1} \cdot L ^{5} T ^{- 5}}{M ^{- 1} L ^{3} T ^{- 2}} = M^{1} L^{2} T^{- 2}$

Q-4.If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be:

$(1) F A^{- 1} V^{0} (2) F A^{2} V^{- 1} (3) F A^{2} V^{- 3} (4) F A^{2} V^{- 2}$

Solution:(1)

$Y = F^{x} A^{y} V^{z}$

$M^{1} L^{- 1} T^{- 2} = [M L T^{- 2}]^{x} [L^{2}]^{y} [L T^{- 1}]^{z}$

$= M^{x} L^{x + 2 y + z} T^{- 2 x - z}$

Comparing power of ML and T

$x = 1, y = - 1, z = 0$

$Y = F A^{- 1} V^{0}$

Q-5.If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is :

$[(1)] [P A^{- 1} T^{- 2}] [(2)] [P^{1/2} A^{2} T^{- 1}] [(3)] [P^{2} A T^{- 2}] [(4)] [P^{1/2} A T^{- 1}]$

Solution: Ans(2)

$L e t [E] = [P]^{x} [A]^{y} [T]^{z}$

$M^{1} L^{2} T^{- 2} = [M L T^{- 1}]^{x} [L^{2}]^{y} [T]^{z}$

$M^{x} L^{x + 2 y} T^{- x + z}$

x=1

x+2y=2

1+2y=2

$y = \frac{1}{2}$

-x+z=-2

-1+z=-2

z=-1

$[E] = [P A^{1/2} T^{- 1}]$

Q-6.Amount of solar energy received on the earth's surface per unit area per unit time is defined as a solar constant. Dimension of solar constant is:

$[(1)] [M L^{2} T^{- 2}] (2) [M L T^{- 2}] (3) [M^{2} L^{0} T^{- 1}] (4) [M L^{0} T^{- 3}]$

Solution:(4)

$S = \frac{P}{A} = \frac{M L ^{2} T ^{- 3}}{L ^{2}} = [M L^{0} T^{- 3}]$

Q-7. A quantity x is given by $(\frac{I F v ^{2}}{W L ^{4}})$ in terms of moment of inertia I, force F, velocity v , work W and Length L. The dimensional formula for x is same as that of:

(1) Planck's constant (2) Force constant (3)Coefficient of viscosity (4)Energy density

Solution:(4)

$x = \frac{I F v ^{2}}{W L ^{4}}$

$x = \frac{[ M L ^{2} ] [ M L T ^{- 2} ] [ L T ^{- 1} ] ^{2}}{[ M L ^{2} T ^{- 2} ] [ L ^{4} ]} = \frac{M ^{2} L ^{5} T ^{- 4}}{M L ^{6} T ^{- 2}} = M L^{- 1} T^{- 2}$

$[Energy Density] = [\frac{E}{V}] = [\frac{M L ^{2} T ^{- 2}}{L ^{3}}] = [M L^{- 1} T^{- 2}]$

Q-8.The quantities and are defined where C-capacitance, R-Resistance, l-length, E-Electric field, B-magnetic field and Î₀, m₀,-free space permittivity and permeability respectively. Then :

(1) Only x and y have the same dimension

(2) x, y and z have the same dimension

(3) Only x and z have the same dimension

(4) Only y and z have the same dimension

Solution:Ans(2)

$x = \frac{1}{μ _{0} ϵ _{0}} = speed \Rightarrow [x] = [L T^{- 1}] y = \frac{E}{B} = speed \Rightarrow [y] = [L T^{- 1}] z = \frac{l}{RC} = \frac{l}{τ} \Rightarrow [z] = [L T^{- 1}]$

(2) x, y and z have the same dimension

Q-9. Given a charge q, current I and permeability of vacuum $μ_{0}$ . Which of the following quantities has the dimension of momentum ?

$(1) q I / μ_{0} (2) q μ_{0} I (3) q^{2} μ_{0} I (4) q μ_{0} / I$

Solution:Ans(2)

Q=AT

I=A

$μ_{0} = [M L T^{- 2} A^{- 2}]$

$Q^{x} μ_{0}^{y} I^{z} = [A T]^{x} [M L T^{- 2} A^{- 2}]^{y} [A]^{z} = M L T^{- 1}$

$M L T^{- 1} = [M]^{y} [L]^{y} [T]^{x - 2 y} [A]^{- 2 y + z + x}$

now y=1

x-2y=-1

-2y+z=0

x=y=z=1

Q-10.If O and O are the permeability and permittivity of free space, respectively, then the dimension of $(\frac{1}{μ _{0} ϵ _{0}})$ is :

$(1) L / T^{2} (2) L^{2} / T^{2} (3) T^{2} / L (4) T^{2} / L^{2}$

Solution: Ans(2)

$c = \frac{1}{μ _{0} ϵ _{0}} \Rightarrow \frac{1}{μ _{0} ϵ _{0}} = c^{2} \Rightarrow [c^{2}] = [L^{2} T^{- 2}]$

Q-11.The equation for real gas is given by $(P + \frac{a}{V ^{2}}) (V - b) = RT$ , where P,V,T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of $a b^{- 2}$ is equivalent to that of :

(1) Planck's constant (2) Compressibility (3) Strain (4) Energy density

Solution: Ans(4)

$(P + \frac{a}{V ^{2}}) (V - b) = RT$

$∴ \Rightarrow [a] = [P] [V^{2}] = [M L^{- 1} T^{- 2}] [L^{6}] = [M L^{5} T^{- 2}]$

$[b] = [V] = L^{3}$

$\Rightarrow [a b^{- 2}] = [M L^{5} T^{- 2}] [L^{- 6}] = [M L^{- 1} T^{- 2}]$

Dimensions of Energy Density

Q-12.Match List–I with List–II

List–I	List–II
Coefficient of viscosity	$M L^{- 1} T^{- 1}$
Intensity of wave	$[M L^{0} T^{- 3}]$
Pressure gradient	$[M L^{- 2} T^{- 2}]$
Compressibility	$[M^{- 1} L^{1} T^{2}]$

Choose the correct answer from the options given below :

(1) (A)–(I), (B)–(IV), (C)–(III), (D)–(II)

(2) (A)–(IV), (B)–(I), (C)–(II), (D)–(III)

(3) (A)–(IV), (B)–(II), (C)–(I), (D)–(III)

(4) (A)–(II), (B)–(III), (C)–(IV), (D)–(I)

Solution:Ans(2)

A. Coefficient of Viscosity = $[M L^{- 1} T^{- 1}]$

B. Intensity $I = [M L^{0} T^{- 3}]$

C.Pressure Gradient = $[M L^{- 2} T^{- 2}]$

D=Compressibility K= $[M^{- 1} L^{1} T^{2}]$

Q-13. A physical quantity C is related to four other quantities p, q, r and s as follows $C = \frac{p ^{2}}{q r ^{3/5} s ^{1/2}}$ . The percentage errors in the measurement of p, q, r and s are 1%, 2% 3% and 2% respectively.The percentage error in the measurement of C will be _________%.

Solution: Ans(15)

$c = \frac{p ^{2}}{q ^{1} r ^{3/5} s ^{1/2}} = p^{2} q^{- 1} r^{- 3/5} s^{- 1/2}$

$(\frac{Δ C}{C})_{max} = ∣2 \frac{Δ p}{p} + \frac{Δ q}{q} + \frac{3}{5} \frac{Δ r}{r} + \frac{1}{2} \frac{Δ s}{s}$

=1+2 ✕ 2+3 ✕ 3+12 ✕ 2%=15 %

Q-14.Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a timeT = 4 seconds . If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after

Solution: Ans(8)

$T \propto m^{x} G^{y} a^{z}$

$\Rightarrow T \propto M^{x} [M^{- 1} L^{3} T^{- 2}]^{y} [L]^{z}$

$\Rightarrow T \propto M^{x - y} L^{3 y + z} T^{- 2 y}$

x-y=0x=y

$- 2 y = 1 \Rightarrow y = - \frac{1}{2}, x = - \frac{1}{2}$

$3 y + z = 0 \Rightarrow z = - 3 y = \frac{3}{2}$

$T \propto m^{- 1/2} G^{- 1/2} a^{3/2}$

$T \propto (\frac{a ^{3}}{m})^{1/2} \Rightarrow T = 4 \times (\frac{2 ^{3}}{2})^{1/2} = 8 s$

Q-15.Given below are two statements :

Statement (I) : The dimensions of Planck’s constant and angular momentum are the same.

Statement (II) : In Bohr’s model electrons revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck’s constant.

In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Both Statement I and Statement II are correct

(2) Statement I is incorrect but Statement II is correct

(3) Statement I is correct but Statement II is incorrect

(4) Both Statement I and Statement II are incorrect

Solution:Ans(3)

E=hf

$[M L^{2} T^{- 2}] = [h] [T^{- 1}]$

$[h] = [M L^{2} T^{- 1}]$

$[L] = [m v r] [M L^{2} T^{- 1}]$

$L = \frac{nh}{2 π}$

L is integral multiple of $\frac{h}{2 π}$

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Units and Dimensions Previous Year Questions With Solutions

1.0Introduction

Units and dimensions are key concepts in physics and engineering, used to measure and express physical quantities. A unit is a standard of measurement (e.g., meter for length, kilogram for mass), while dimensions describe the physical nature of a quantity, expressed in terms of basic quantities like length, mass, and time. The International System of Units (SI) includes seven base units, and derived units are combinations of these. Dimensions ensure equations are consistent and help with unit conversions and the formulation of physical laws. Understanding units and dimensions is essential for accurate and universal communication in science and technology.

JEE Main Previous Year Solved Questions on Units and Dimensions
JEE Adv Previous Year Solved Questions on Units and Dimensions

2.0Key Concepts to Remember

Dimensional Equation:The equation formed by equating a physical quantity to its dimensional formula is known as a dimensional equation.

$M a ss \to [M^{1} L^{0} T^{0}]$

$L e n g t h \to [M^{0} L^{1} T^{0}]$

$T im e \to [M^{0} L^{0} T^{1}]$

$T e m p er a t u re \to [M^{0} L^{0} T^{0} K^{1}]$

$C u rre n t \to [M^{0} L^{0} T^{0} A^{1}]$

$Lu min o u s I n t e n s i t y \to [M^{0} L^{0} T^{0} c d^{1}]$

$A m o u n t o f S u b s t an ce \to [M^{0} L^{0} T^{0} m o l^{1}]$

3.0Rule of Dimensions

Only the same physical quantities can be added or subtracted.

A + B = C – D

Dimensionless Quantities

Dimensionless Quantities are:

• Ratio of physical quantities with same dimensions.

• All mathematical constants.

• All standard mathematical functions and their inputs (exponential, logarithmic, trigonometric & inverse trigonometric).

4.0Important Formulas

Conversion Between System of Units	$n_{2} = n_{1} (\frac{M _{1}}{M _{2}})^{a} (\frac{L _{1}}{L _{2}})^{b} (\frac{T _{1}}{T _{2}})^{c}$
Errors in Measurement	Error = True value – Measured value
Absolute Error (a)	$Δ a = a_{T} - a$
Mean Absolute Erroram $Δ a = a_{T} - a$	$(Δ a)_{m} = \frac{1}{n} \sum_{i = 0}^{n} ∣ a_{i} ∣$
Relative or Fractional Error	$R e l a t i v e E rror = \frac{M e an A b so l u t e E rror}{T r u e Va l u e}$
Percentage Error	$P erce n t a g e E rror = \frac{M e an A b so l u t e E rror}{T r u e Va l u e} \times 100$
Addition or Subtraction of Quantities:	$Δ X = Δ A + Δ B$
Multiplication or Division of Quantities:	$\frac{Δ X}{X} = \pm (\frac{Δ A}{A} + \frac{Δ B}{B})$
Maximum Fractional Error	$1. If X = A^{n} then \frac{Δ X}{X} = n (\frac{Δ A}{A}) 2. If X = A^{p} B^{q} C^{r} then \frac{Δ X}{X} = [p (\frac{Δ A}{A}) + q (\frac{Δ B}{B}) + r (\frac{Δ C}{C})] 3. If X = \frac{A ^{p} B ^{q}}{C ^{r}} then \frac{Δ X}{X} = [p (\frac{Δ A}{A}) + q (\frac{Δ B}{B}) + r (\frac{Δ C}{C})]$
Vernier Callipers	$L . C = M - V = (\frac{b - a}{b}) M$
Measurement with Vernier Calliper	$R e a d in g = MSR + (V S co in c i d e d d i v i s i o n w i t h MS \times L C)$
Positive Zero Error Correction	READING = (Main Scale Reading) + (Vernier Scale Reading) – ZE Positive zero error = VS Coincided division with MS × LC
Negative Zero Error Correction	Negative zero error = –(Total VSD – VS Coincided division with MS) × LC
Least Count of Screw Gauge	$L e a s t co u n t o f scre w g a ug e = \frac{P i t c h ( P )}{T o t a l n o . o f d i v i s i o n s o n t h e c i rc u l a r sc a l e ( N )}$

5.0Past Year Questions with Solutions on Units And Dimension:JEE (Mains)

Q-1.The dimension of , where B is magnetic field and µ₀ is the magnetic permeability of vacuum, is:

$(1) [M L^{- 1} T^{- 2}] (2) [M L^{2} T^{- 1}] (3) [M L T^{- 2}] (4) [M L^{2} T^{- 2}]$

Solution: Ans(1)

Magnetic energy stored per unit volume is

$\frac{B ^{2}}{2 μ _{0}} Dimensions is [M L^{- 1} T^{- 2}]$

Q-2.The dimension 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 :

$[(1)] h^{2} G^{3/2} c^{1/3} A^{- 1} [(2)] h^{- 2/3} c^{- 1/3} G^{4/3} A^{- 1} [(3)] h^{1/3} G^{2/3} c^{1/3} A^{- 1} [(4)] h^{2/3} c^{5/3} G^{1/3} A^{- 1}$

Solution: Ans(Bonus)

$V_{0} = h^{x} c^{y} G^{z} A^{w}$

$\frac{M L ^{2} T ^{- 2}}{A T} = (M L^{2} T^{- 1})^{x} (L T^{- 1})^{y} (M^{- 1} L^{3} T^{- 2})^{z} A^{w}$

w=-1

x-z=1

2x+y+3x=2

-x-y-2z=-3

$2 x = 0 \Rightarrow x = 0$

z=-1,y=5

$V_{0} = h^{0} c^{5} G^{- 1} A^{- 1}$

Q-3.A quantity f is given by $f = \frac{h c ^{5}}{G}$ where c is the speed of light, G universal gravitational constant and h is Planck's constant. Dimension of f is that of :

(1) Momentum (2) Area (3) Energy (4) Volume

Solution:Ans(3)

$[h] = M^{1} L^{2} T^{- 1}, [c] = L^{1} T^{- 1}, [G] = M^{- 1} L^{3} T^{- 2}$

$[f] = \frac{M ^{1} L ^{2} T ^{- 1} \cdot L ^{5} T ^{- 5}}{M ^{- 1} L ^{3} T ^{- 2}} = M^{1} L^{2} T^{- 2}$

Q-4.If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be:

$(1) F A^{- 1} V^{0} (2) F A^{2} V^{- 1} (3) F A^{2} V^{- 3} (4) F A^{2} V^{- 2}$

Solution:(1)

$Y = F^{x} A^{y} V^{z}$

$M^{1} L^{- 1} T^{- 2} = [M L T^{- 2}]^{x} [L^{2}]^{y} [L T^{- 1}]^{z}$

$= M^{x} L^{x + 2 y + z} T^{- 2 x - z}$

Comparing power of ML and T

$x = 1, y = - 1, z = 0$

$Y = F A^{- 1} V^{0}$

Q-5.If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is :

$[(1)] [P A^{- 1} T^{- 2}] [(2)] [P^{1/2} A^{2} T^{- 1}] [(3)] [P^{2} A T^{- 2}] [(4)] [P^{1/2} A T^{- 1}]$

Solution: Ans(2)

$L e t [E] = [P]^{x} [A]^{y} [T]^{z}$

$M^{1} L^{2} T^{- 2} = [M L T^{- 1}]^{x} [L^{2}]^{y} [T]^{z}$

$M^{x} L^{x + 2 y} T^{- x + z}$

x=1

x+2y=2

1+2y=2

$y = \frac{1}{2}$

-x+z=-2

-1+z=-2

z=-1

$[E] = [P A^{1/2} T^{- 1}]$

Q-6.Amount of solar energy received on the earth's surface per unit area per unit time is defined as a solar constant. Dimension of solar constant is:

$[(1)] [M L^{2} T^{- 2}] (2) [M L T^{- 2}] (3) [M^{2} L^{0} T^{- 1}] (4) [M L^{0} T^{- 3}]$

Solution:(4)

$S = \frac{P}{A} = \frac{M L ^{2} T ^{- 3}}{L ^{2}} = [M L^{0} T^{- 3}]$

Q-7. A quantity x is given by $(\frac{I F v ^{2}}{W L ^{4}})$ in terms of moment of inertia I, force F, velocity v , work W and Length L. The dimensional formula for x is same as that of:

(1) Planck's constant (2) Force constant (3)Coefficient of viscosity (4)Energy density

Solution:(4)

$x = \frac{I F v ^{2}}{W L ^{4}}$

$x = \frac{[ M L ^{2} ] [ M L T ^{- 2} ] [ L T ^{- 1} ] ^{2}}{[ M L ^{2} T ^{- 2} ] [ L ^{4} ]} = \frac{M ^{2} L ^{5} T ^{- 4}}{M L ^{6} T ^{- 2}} = M L^{- 1} T^{- 2}$

$[Energy Density] = [\frac{E}{V}] = [\frac{M L ^{2} T ^{- 2}}{L ^{3}}] = [M L^{- 1} T^{- 2}]$

Q-8.The quantities and are defined where C-capacitance, R-Resistance, l-length, E-Electric field, B-magnetic field and Î₀, m₀,-free space permittivity and permeability respectively. Then :

(1) Only x and y have the same dimension

(2) x, y and z have the same dimension

(3) Only x and z have the same dimension

(4) Only y and z have the same dimension

Solution:Ans(2)

$x = \frac{1}{μ _{0} ϵ _{0}} = speed \Rightarrow [x] = [L T^{- 1}] y = \frac{E}{B} = speed \Rightarrow [y] = [L T^{- 1}] z = \frac{l}{RC} = \frac{l}{τ} \Rightarrow [z] = [L T^{- 1}]$

(2) x, y and z have the same dimension

Q-9. Given a charge q, current I and permeability of vacuum $μ_{0}$ . Which of the following quantities has the dimension of momentum ?

$(1) q I / μ_{0} (2) q μ_{0} I (3) q^{2} μ_{0} I (4) q μ_{0} / I$

Solution:Ans(2)

Q=AT

I=A

$μ_{0} = [M L T^{- 2} A^{- 2}]$

$Q^{x} μ_{0}^{y} I^{z} = [A T]^{x} [M L T^{- 2} A^{- 2}]^{y} [A]^{z} = M L T^{- 1}$

$M L T^{- 1} = [M]^{y} [L]^{y} [T]^{x - 2 y} [A]^{- 2 y + z + x}$

now y=1

x-2y=-1

-2y+z=0

x=y=z=1

Q-10.If O and O are the permeability and permittivity of free space, respectively, then the dimension of $(\frac{1}{μ _{0} ϵ _{0}})$ is :

$(1) L / T^{2} (2) L^{2} / T^{2} (3) T^{2} / L (4) T^{2} / L^{2}$

Solution: Ans(2)

$c = \frac{1}{μ _{0} ϵ _{0}} \Rightarrow \frac{1}{μ _{0} ϵ _{0}} = c^{2} \Rightarrow [c^{2}] = [L^{2} T^{- 2}]$

Q-11.The equation for real gas is given by $(P + \frac{a}{V ^{2}}) (V - b) = RT$ , where P,V,T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of $a b^{- 2}$ is equivalent to that of :

(1) Planck's constant (2) Compressibility (3) Strain (4) Energy density

Solution: Ans(4)

$(P + \frac{a}{V ^{2}}) (V - b) = RT$

$∴ \Rightarrow [a] = [P] [V^{2}] = [M L^{- 1} T^{- 2}] [L^{6}] = [M L^{5} T^{- 2}]$

$[b] = [V] = L^{3}$

$\Rightarrow [a b^{- 2}] = [M L^{5} T^{- 2}] [L^{- 6}] = [M L^{- 1} T^{- 2}]$

Dimensions of Energy Density

Q-12.Match List–I with List–II

List–I	List–II
Coefficient of viscosity	$M L^{- 1} T^{- 1}$
Intensity of wave	$[M L^{0} T^{- 3}]$
Pressure gradient	$[M L^{- 2} T^{- 2}]$
Compressibility	$[M^{- 1} L^{1} T^{2}]$

Choose the correct answer from the options given below :

(1) (A)–(I), (B)–(IV), (C)–(III), (D)–(II)

(2) (A)–(IV), (B)–(I), (C)–(II), (D)–(III)

(3) (A)–(IV), (B)–(II), (C)–(I), (D)–(III)

(4) (A)–(II), (B)–(III), (C)–(IV), (D)–(I)

Solution:Ans(2)

A. Coefficient of Viscosity = $[M L^{- 1} T^{- 1}]$

B. Intensity $I = [M L^{0} T^{- 3}]$

C.Pressure Gradient = $[M L^{- 2} T^{- 2}]$

D=Compressibility K= $[M^{- 1} L^{1} T^{2}]$

Q-13. A physical quantity C is related to four other quantities p, q, r and s as follows $C = \frac{p ^{2}}{q r ^{3/5} s ^{1/2}}$ . The percentage errors in the measurement of p, q, r and s are 1%, 2% 3% and 2% respectively.The percentage error in the measurement of C will be _________%.

Solution: Ans(15)

$c = \frac{p ^{2}}{q ^{1} r ^{3/5} s ^{1/2}} = p^{2} q^{- 1} r^{- 3/5} s^{- 1/2}$

$(\frac{Δ C}{C})_{max} = ∣2 \frac{Δ p}{p} + \frac{Δ q}{q} + \frac{3}{5} \frac{Δ r}{r} + \frac{1}{2} \frac{Δ s}{s}$

=1+2 ✕ 2+3 ✕ 3+12 ✕ 2%=15 %

Q-14.Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a timeT = 4 seconds . If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after

Solution: Ans(8)

$T \propto m^{x} G^{y} a^{z}$

$\Rightarrow T \propto M^{x} [M^{- 1} L^{3} T^{- 2}]^{y} [L]^{z}$

$\Rightarrow T \propto M^{x - y} L^{3 y + z} T^{- 2 y}$

x-y=0x=y

$- 2 y = 1 \Rightarrow y = - \frac{1}{2}, x = - \frac{1}{2}$

$3 y + z = 0 \Rightarrow z = - 3 y = \frac{3}{2}$

$T \propto m^{- 1/2} G^{- 1/2} a^{3/2}$

$T \propto (\frac{a ^{3}}{m})^{1/2} \Rightarrow T = 4 \times (\frac{2 ^{3}}{2})^{1/2} = 8 s$

Q-15.Given below are two statements :

Statement (I) : The dimensions of Planck’s constant and angular momentum are the same.

Statement (II) : In Bohr’s model electrons revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck’s constant.

In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Both Statement I and Statement II are correct

(2) Statement I is incorrect but Statement II is correct

(3) Statement I is correct but Statement II is incorrect

(4) Both Statement I and Statement II are incorrect

Solution:Ans(3)

E=hf

$[M L^{2} T^{- 2}] = [h] [T^{- 1}]$

$[h] = [M L^{2} T^{- 1}]$

$[L] = [m v r] [M L^{2} T^{- 1}]$

$L = \frac{nh}{2 π}$

L is integral multiple of $\frac{h}{2 π}$

Join ALLEN!

Units and Dimensions Previous Year Questions With Solutions

1.0Introduction

2.0Key Concepts to Remember

3.0Rule of Dimensions

4.0Important Formulas

5.0Past Year Questions with Solutions on Units And Dimension:JEE (Mains)

Table of Contents

Join ALLEN!

Units and Dimensions Previous Year Questions With Solutions

1.0Introduction

2.0Key Concepts to Remember

3.0Rule of Dimensions

4.0Important Formulas

5.0Past Year Questions with Solutions on Units And Dimension:JEE (Mains)

Table of Contents