JEE Advanced Numericals Unit & Dimension - Physics for JEE Main & Advanced

 Page 1


 
    
 
1  
 
   
UNIT & DIMENSION 
 
1. Young's modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational 
constant G, Planck's constant h and the speed of light c, as Y = c
a
 h
ß
 G
?
. Which of the following is the 
correct option ?   
 (A) a = 7, ß = –1, ? = –2        (B) a = –7, ß = –1, ? = –2    
 (C) a = 7, ß = –1, ? = 2     (D) a = –7, ß = 1, ? = –2 
2. In a particular system of units, a physical quantity can be expressed in terms of the electric charge e, 
electron mass m
e
, Planck's constant h, and Coulomb's constant
0
1
k
4
=
p?
,  where ?
0
  is the permittivity of 
vacuum. In terms of these physical constants, the dimension of the magnetic field is [B] = [e]
a
 [m
e
]
ß
 [h]
?
 
[k]
d
. The value of a + ß + ? + d is ________.    
3. A physical quantity S
?
 
is defined as 
( )
0
S E B/ = × µ
? ??
, where E
?
is electric field, B
?
is magnetic field and 
0
µ is 
the permeability of free space. The dimensions of S
?
 
are the same as the dimensions of which of the 
following quantity (ies) ?     
 (A) 
Energy
charge × current
 (B) 
Force
Length × Time
 (C) 
Energy
Volume
 (D) 
Power
Area
 
4. 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] = [ ???? a
]; [speed] = [ ???? ß
]; [acceleration] =[ ???? p
]; [linear momentum] = [ ???? q
]; 
[force] = [ ???? r
]. Then -     
 (A) a + ???? = 2 ß  (B) ???? + ???? – ???? = ß 
 (C) ???? - ???? + ???? = a  (D) ???? + ???? + ???? = ß 
5. Let us consider a system of units in which mass and angular momentum are dimensionless. If length has 
dimension of L, which of the following statement (s) is/are correct ?   
 (A) The dimension of force is L
–3
 (B) The dimension of energy is L
–2
 
 (C) The dimension of power is L
–5
 (D) The dimension of linear momentum is L
–1
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units.   
 (There are two questions based on Paragraph "X", the question given below is one of them) 
6. 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
 
JEE	Advanced	Numericals
Page 2


 
    
 
1  
 
   
UNIT & DIMENSION 
 
1. Young's modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational 
constant G, Planck's constant h and the speed of light c, as Y = c
a
 h
ß
 G
?
. Which of the following is the 
correct option ?   
 (A) a = 7, ß = –1, ? = –2        (B) a = –7, ß = –1, ? = –2    
 (C) a = 7, ß = –1, ? = 2     (D) a = –7, ß = 1, ? = –2 
2. In a particular system of units, a physical quantity can be expressed in terms of the electric charge e, 
electron mass m
e
, Planck's constant h, and Coulomb's constant
0
1
k
4
=
p?
,  where ?
0
  is the permittivity of 
vacuum. In terms of these physical constants, the dimension of the magnetic field is [B] = [e]
a
 [m
e
]
ß
 [h]
?
 
[k]
d
. The value of a + ß + ? + d is ________.    
3. A physical quantity S
?
 
is defined as 
( )
0
S E B/ = × µ
? ??
, where E
?
is electric field, B
?
is magnetic field and 
0
µ is 
the permeability of free space. The dimensions of S
?
 
are the same as the dimensions of which of the 
following quantity (ies) ?     
 (A) 
Energy
charge × current
 (B) 
Force
Length × Time
 (C) 
Energy
Volume
 (D) 
Power
Area
 
4. 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] = [ ???? a
]; [speed] = [ ???? ß
]; [acceleration] =[ ???? p
]; [linear momentum] = [ ???? q
]; 
[force] = [ ???? r
]. Then -     
 (A) a + ???? = 2 ß  (B) ???? + ???? – ???? = ß 
 (C) ???? - ???? + ???? = a  (D) ???? + ???? + ???? = ß 
5. Let us consider a system of units in which mass and angular momentum are dimensionless. If length has 
dimension of L, which of the following statement (s) is/are correct ?   
 (A) The dimension of force is L
–3
 (B) The dimension of energy is L
–2
 
 (C) The dimension of power is L
–5
 (D) The dimension of linear momentum is L
–1
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units.   
 (There are two questions based on Paragraph "X", the question given below is one of them) 
6. 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
 
JEE	Advanced	Numericals
 
  
    
 
2 
 
   
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units. 
 (There are two questions based on Paragraph "X", the question given below is one of them) 
7. The relation between [?
0
] and [µ
0
] is :-    
 (A) [µ
0
] = [?
0
][L]
2
[T]
–2
  (B) [µ
0
] = [?
0
][L]
–2
[T]
2
 
 (C) [µ
0
] = [?
0
]
–1
[L]
2
[T]
–2
  (D) [µ
0
] = [?
0
]
–1
[L]
–2
[T]
2 
8. A length-scale ( ?) depends on the permittivity (e) 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 expressions(s) for ? is(are) dimensionally correct? 
 
 (A)
2
B
nq
kT
? ?
=
? ?
e
? ?
? (B)
B
2
kT
nq
e ? ?
=
? ?
? ?
? (C)
2
2/3
B
q
n kT
? ?
=
? ?
? ?
e
? ?
? (D) 
2
1/3
B
q
n kT
? ?
=
? ?
? ?
e
? ?
? 
9. In terms of potential difference V, electric current I, permittivity e
0
, permeability µ
0
 and speed of light c, 
the dimensionally correct equation(s) is(are)    
 (A) µ
0
I
2
 = e
0
V
2
 (B) e
0
I = µ
0
V (C) I = e
0
cV (D) µ
0
cI = e
0
V  
10. 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) ? Mc (B) ? MG (C) ? L h (D) ? LG  
11. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer 
uses dimensional analysis and assumes that the distance depends on the mass density ? 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.    
Page 3


 
    
 
1  
 
   
UNIT & DIMENSION 
 
1. Young's modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational 
constant G, Planck's constant h and the speed of light c, as Y = c
a
 h
ß
 G
?
. Which of the following is the 
correct option ?   
 (A) a = 7, ß = –1, ? = –2        (B) a = –7, ß = –1, ? = –2    
 (C) a = 7, ß = –1, ? = 2     (D) a = –7, ß = 1, ? = –2 
2. In a particular system of units, a physical quantity can be expressed in terms of the electric charge e, 
electron mass m
e
, Planck's constant h, and Coulomb's constant
0
1
k
4
=
p?
,  where ?
0
  is the permittivity of 
vacuum. In terms of these physical constants, the dimension of the magnetic field is [B] = [e]
a
 [m
e
]
ß
 [h]
?
 
[k]
d
. The value of a + ß + ? + d is ________.    
3. A physical quantity S
?
 
is defined as 
( )
0
S E B/ = × µ
? ??
, where E
?
is electric field, B
?
is magnetic field and 
0
µ is 
the permeability of free space. The dimensions of S
?
 
are the same as the dimensions of which of the 
following quantity (ies) ?     
 (A) 
Energy
charge × current
 (B) 
Force
Length × Time
 (C) 
Energy
Volume
 (D) 
Power
Area
 
4. 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] = [ ???? a
]; [speed] = [ ???? ß
]; [acceleration] =[ ???? p
]; [linear momentum] = [ ???? q
]; 
[force] = [ ???? r
]. Then -     
 (A) a + ???? = 2 ß  (B) ???? + ???? – ???? = ß 
 (C) ???? - ???? + ???? = a  (D) ???? + ???? + ???? = ß 
5. Let us consider a system of units in which mass and angular momentum are dimensionless. If length has 
dimension of L, which of the following statement (s) is/are correct ?   
 (A) The dimension of force is L
–3
 (B) The dimension of energy is L
–2
 
 (C) The dimension of power is L
–5
 (D) The dimension of linear momentum is L
–1
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units.   
 (There are two questions based on Paragraph "X", the question given below is one of them) 
6. 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
 
JEE	Advanced	Numericals
 
  
    
 
2 
 
   
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units. 
 (There are two questions based on Paragraph "X", the question given below is one of them) 
7. The relation between [?
0
] and [µ
0
] is :-    
 (A) [µ
0
] = [?
0
][L]
2
[T]
–2
  (B) [µ
0
] = [?
0
][L]
–2
[T]
2
 
 (C) [µ
0
] = [?
0
]
–1
[L]
2
[T]
–2
  (D) [µ
0
] = [?
0
]
–1
[L]
–2
[T]
2 
8. A length-scale ( ?) depends on the permittivity (e) 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 expressions(s) for ? is(are) dimensionally correct? 
 
 (A)
2
B
nq
kT
? ?
=
? ?
e
? ?
? (B)
B
2
kT
nq
e ? ?
=
? ?
? ?
? (C)
2
2/3
B
q
n kT
? ?
=
? ?
? ?
e
? ?
? (D) 
2
1/3
B
q
n kT
? ?
=
? ?
? ?
e
? ?
? 
9. In terms of potential difference V, electric current I, permittivity e
0
, permeability µ
0
 and speed of light c, 
the dimensionally correct equation(s) is(are)    
 (A) µ
0
I
2
 = e
0
V
2
 (B) e
0
I = µ
0
V (C) I = e
0
cV (D) µ
0
cI = e
0
V  
10. 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) ? Mc (B) ? MG (C) ? L h (D) ? LG  
11. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer 
uses dimensional analysis and assumes that the distance depends on the mass density ? 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.    
 
    
 
 
   
3  
SOLUTIONS 
 
1. Ans. (A) 
Sol.   Y c hG
a ß ?
= 
 
1 2 1 2 1 13 2
ML T (LT ) (ML T ) (M L T )
- - - a - ß - - ?
= 
 1 = ß – ? ...(1) 
 –1 = a + 2ß + 3? ...(2) 
 –2 = – a – ß – 2? ...(3) 
 –3 = ß + ? 
   1 = ß – ? 
 –2 = 2ß           ? ß = –1, ? = –2 
 –1 = a – 2 – 6       ?  a = 7 
2. Ans. (4) 
Sol. ( )
e
B e m hk
ß
a ?d
= 
 
[ ] [ ]
[ ]
[ ] [ ]
e
B e m hk
ß ?
a d
=
 
[ ] [ ] [ ] [ ] [ ]
1 –2 –1 2 –1 3 –2 –4
M T A AT m ML T ML A T
? d
a ß
=
1 –2 –1 2r 3 – –4 –2
MT A m L T A
ß+?+d + d a ? d a d
= 
 Compare : ß + ? + d = 1; 2? + 3d = 0,  
a – ? – 4d = –2, a – 2d = – 1 
 On solving a = 3, ß = 2, ? = –3, d = 2  
 a + ß + ? + d = 4 
3. Ans. (B, D) 
Sol. 
0
1
S [E B] = ×
µ
? ??
 
 S is poynting vector denotes flow of energy per 
unit area per unit time 
 
2
watt
S
m
=
?
 
 Hence B, D are correct 
4. Ans. (A, B) 
Sol. Given,  
L = x
a
   
....(1) 
 LT
–1
 = x
ß
   ....(2) 
 LT
–2
 = x
p
   ....(3) 
 MLT
–1
 = x
q
   ....(4) 
 MLT
–2
 = x
r
   ....(5) 
 
(1)
(2)
 ?  T = x
a–ß 
  
 From (3) 
  
a
a-ß
=
p
2( )
x
x
x
 
 ?  a + p = 2ß     (A) 
 From (4)  
  M = x
q–ß
  
 From (5)  ?   x
q
 = x
r
  x
a–ß
  
 ? a + r – q = ß ....(6) 
 Replacing value 'a' in equation (6) from (A) 
  2ß – p + r – q = ß 
 ? p + q – r = ß (B) 
 Replacing value of 'ß' in equation (6) from (A)
 2a + 2r – 2q = a + p 
  a = p + 2q – 2r 
5. Ans. (A, B, D) 
Sol. Mass = M
0
L
0
T
0
 
 MVr = M
0
L
0
T
0
 
 M
0
1
1 00 0
1
L
. L M LT
T
= 
 L
2
 = T
1
   ....(1) 
 Force = M
1
L
1
T
–2
 (in SI) 
 = M
0 
L
1
 L
–4
 (In new system from equation (1)) 
 = L
–3
 
 Energy = M
1
L
2
T
–2 
 (In SI) 
 = M
0
L
2
L
–4
  (In new system from equation (1)) 
 = L
–2 
 
Power = 
Energy
Time
 
 = M
1
L
2
T
–3
 (in SI) 
 = M
0
L
2
L
–6
  (In new system from equation (1)) 
 = L
–4
 
 Linear momentum = M
1
L
1
T
–1
 (in SI) 
 = M
0
L
1
L
–2 
 (In new system from equation (1)) 
 = L
–1 
6. Ans. (C) 
Sol. We have 
E
B
C
= 
 ? [B] = 
[ ]
[ ]
[ ]
1 1
E
EL T
C
-
= 
 ? [E] = [B] [L][T
–1
]
 
Page 4


 
    
 
1  
 
   
UNIT & DIMENSION 
 
1. Young's modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational 
constant G, Planck's constant h and the speed of light c, as Y = c
a
 h
ß
 G
?
. Which of the following is the 
correct option ?   
 (A) a = 7, ß = –1, ? = –2        (B) a = –7, ß = –1, ? = –2    
 (C) a = 7, ß = –1, ? = 2     (D) a = –7, ß = 1, ? = –2 
2. In a particular system of units, a physical quantity can be expressed in terms of the electric charge e, 
electron mass m
e
, Planck's constant h, and Coulomb's constant
0
1
k
4
=
p?
,  where ?
0
  is the permittivity of 
vacuum. In terms of these physical constants, the dimension of the magnetic field is [B] = [e]
a
 [m
e
]
ß
 [h]
?
 
[k]
d
. The value of a + ß + ? + d is ________.    
3. A physical quantity S
?
 
is defined as 
( )
0
S E B/ = × µ
? ??
, where E
?
is electric field, B
?
is magnetic field and 
0
µ is 
the permeability of free space. The dimensions of S
?
 
are the same as the dimensions of which of the 
following quantity (ies) ?     
 (A) 
Energy
charge × current
 (B) 
Force
Length × Time
 (C) 
Energy
Volume
 (D) 
Power
Area
 
4. 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] = [ ???? a
]; [speed] = [ ???? ß
]; [acceleration] =[ ???? p
]; [linear momentum] = [ ???? q
]; 
[force] = [ ???? r
]. Then -     
 (A) a + ???? = 2 ß  (B) ???? + ???? – ???? = ß 
 (C) ???? - ???? + ???? = a  (D) ???? + ???? + ???? = ß 
5. Let us consider a system of units in which mass and angular momentum are dimensionless. If length has 
dimension of L, which of the following statement (s) is/are correct ?   
 (A) The dimension of force is L
–3
 (B) The dimension of energy is L
–2
 
 (C) The dimension of power is L
–5
 (D) The dimension of linear momentum is L
–1
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units.   
 (There are two questions based on Paragraph "X", the question given below is one of them) 
6. 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
 
JEE	Advanced	Numericals
 
  
    
 
2 
 
   
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units. 
 (There are two questions based on Paragraph "X", the question given below is one of them) 
7. The relation between [?
0
] and [µ
0
] is :-    
 (A) [µ
0
] = [?
0
][L]
2
[T]
–2
  (B) [µ
0
] = [?
0
][L]
–2
[T]
2
 
 (C) [µ
0
] = [?
0
]
–1
[L]
2
[T]
–2
  (D) [µ
0
] = [?
0
]
–1
[L]
–2
[T]
2 
8. A length-scale ( ?) depends on the permittivity (e) 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 expressions(s) for ? is(are) dimensionally correct? 
 
 (A)
2
B
nq
kT
? ?
=
? ?
e
? ?
? (B)
B
2
kT
nq
e ? ?
=
? ?
? ?
? (C)
2
2/3
B
q
n kT
? ?
=
? ?
? ?
e
? ?
? (D) 
2
1/3
B
q
n kT
? ?
=
? ?
? ?
e
? ?
? 
9. In terms of potential difference V, electric current I, permittivity e
0
, permeability µ
0
 and speed of light c, 
the dimensionally correct equation(s) is(are)    
 (A) µ
0
I
2
 = e
0
V
2
 (B) e
0
I = µ
0
V (C) I = e
0
cV (D) µ
0
cI = e
0
V  
10. 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) ? Mc (B) ? MG (C) ? L h (D) ? LG  
11. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer 
uses dimensional analysis and assumes that the distance depends on the mass density ? 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.    
 
    
 
 
   
3  
SOLUTIONS 
 
1. Ans. (A) 
Sol.   Y c hG
a ß ?
= 
 
1 2 1 2 1 13 2
ML T (LT ) (ML T ) (M L T )
- - - a - ß - - ?
= 
 1 = ß – ? ...(1) 
 –1 = a + 2ß + 3? ...(2) 
 –2 = – a – ß – 2? ...(3) 
 –3 = ß + ? 
   1 = ß – ? 
 –2 = 2ß           ? ß = –1, ? = –2 
 –1 = a – 2 – 6       ?  a = 7 
2. Ans. (4) 
Sol. ( )
e
B e m hk
ß
a ?d
= 
 
[ ] [ ]
[ ]
[ ] [ ]
e
B e m hk
ß ?
a d
=
 
[ ] [ ] [ ] [ ] [ ]
1 –2 –1 2 –1 3 –2 –4
M T A AT m ML T ML A T
? d
a ß
=
1 –2 –1 2r 3 – –4 –2
MT A m L T A
ß+?+d + d a ? d a d
= 
 Compare : ß + ? + d = 1; 2? + 3d = 0,  
a – ? – 4d = –2, a – 2d = – 1 
 On solving a = 3, ß = 2, ? = –3, d = 2  
 a + ß + ? + d = 4 
3. Ans. (B, D) 
Sol. 
0
1
S [E B] = ×
µ
? ??
 
 S is poynting vector denotes flow of energy per 
unit area per unit time 
 
2
watt
S
m
=
?
 
 Hence B, D are correct 
4. Ans. (A, B) 
Sol. Given,  
L = x
a
   
....(1) 
 LT
–1
 = x
ß
   ....(2) 
 LT
–2
 = x
p
   ....(3) 
 MLT
–1
 = x
q
   ....(4) 
 MLT
–2
 = x
r
   ....(5) 
 
(1)
(2)
 ?  T = x
a–ß 
  
 From (3) 
  
a
a-ß
=
p
2( )
x
x
x
 
 ?  a + p = 2ß     (A) 
 From (4)  
  M = x
q–ß
  
 From (5)  ?   x
q
 = x
r
  x
a–ß
  
 ? a + r – q = ß ....(6) 
 Replacing value 'a' in equation (6) from (A) 
  2ß – p + r – q = ß 
 ? p + q – r = ß (B) 
 Replacing value of 'ß' in equation (6) from (A)
 2a + 2r – 2q = a + p 
  a = p + 2q – 2r 
5. Ans. (A, B, D) 
Sol. Mass = M
0
L
0
T
0
 
 MVr = M
0
L
0
T
0
 
 M
0
1
1 00 0
1
L
. L M LT
T
= 
 L
2
 = T
1
   ....(1) 
 Force = M
1
L
1
T
–2
 (in SI) 
 = M
0 
L
1
 L
–4
 (In new system from equation (1)) 
 = L
–3
 
 Energy = M
1
L
2
T
–2 
 (In SI) 
 = M
0
L
2
L
–4
  (In new system from equation (1)) 
 = L
–2 
 
Power = 
Energy
Time
 
 = M
1
L
2
T
–3
 (in SI) 
 = M
0
L
2
L
–6
  (In new system from equation (1)) 
 = L
–4
 
 Linear momentum = M
1
L
1
T
–1
 (in SI) 
 = M
0
L
1
L
–2 
 (In new system from equation (1)) 
 = L
–1 
6. Ans. (C) 
Sol. We have 
E
B
C
= 
 ? [B] = 
[ ]
[ ]
[ ]
1 1
E
EL T
C
-
= 
 ? [E] = [B] [L][T
–1
]
 
 
  
    
 
  
 
4 
 
7. Ans. (D) 
Sol. We have, 
 
00
1
C =
µ?
 
 
[ ]
2
00
1
C
? ?
? =
? ?
µ?
? ?
 
 
[ ][ ]
2 2
0 0
1
LT
-
? =
µ?
 ? [µ
0
] = [?
0
]
–1
[L]
–2
[T]
2 
8. Ans. (B, D) 
Sol. We know,  
 
2
2
0
1q
F
4 r
=
pe
 
2
2
0
q
(Fr )4 ? = p
e
  
 So, dimension 
 
2
2
0
q
dim (Fr ) =
e
 = MLT
–2
× L
2
 = ML
3
T
–2
 
 Similarly; 
B
3
E K T
2
= ?  
dim (K
B
T) = dim(Energy) = ML
2
T
–2
 
(A)  
2 –3 3 –2
2 –2
B
nq L ML T 1
kT L ML T
×
= =
e
 
(B)  
( )
2 –2 3
2 –2 2
E vol ML T L
L
Fr MLT L
× ×
= =
×
 
(C)   
2 2/3 2 2 2
3 3/2
2 2
Fr (vol) MLT L L
LL
(K ) ML T
-
-
××
= = =
e
 
(D)  
2 1/3 2 2
2 2
Fr (vol) MLT L L
L
Energy ML T
-
-
×
= = 
 [? dimension n = 
3
1
dim L
vol
-
? ?
=
? ?
? ?
] 
9. Ans. (A, C)  
Sol. Using =
µ?
00
1
C &  
µ
=
?
0
0
R we can check the correctness. 
(A)  µ
0
I
2
 = ?
0
V
2
 
 
µ
= =
?
2
2 0
2
0
V
R
I
 
 ? R
2
 = R
2
 correct 
(B)  ?
0
I = µ
0
V 
 
?
=
µ
0
0
V
I
 
  =
2
1
R
R
not correct 
(C)  I = ?
0
cV 
 
?
=?=
µ?
0
0
00
I
c
V
 
 
?
= =
µ
0
0
1 1
R R
 correct 
(D)  µ
0
CI = ?
0
V 
 
µ
=
?
0
0
C V
I
 
 
µ
=
? µ?
0
0 00
1
R 
 ?
µ
= ? =
?? ?
0
0 0 0
1 R
R R incorrect 
10. Ans. (A, C, D)  
Sol. [h] = [ML
2
T
–1
] 
 [c] = [LT
–1
] 
 [G] = [M
–1
L
3
T
–2
] 
 For unit of length 
 Let L ? h
x 
c
y
 G
z
 
 
? From principle of homogeneity 
 [LHS] = [RHS] 
 ? [M
0
LT
0
] = [ML
2
T
–1
]
x
 [LT
–1
]
y
 [M
–1
L
3
T
–2
]
z
 
 M
0
 L T
0 
=  M
x–z
 L
2x + y + 3z 
 T
–x–y–2z
 
 comparing we get 
 x – z = 0 
 2x + y + 3z = 1 
 –x – y – 2z = 0 
 Solving we get,  x = 
1
2
, y = 
–3
2
, z = 
1
2
 
 L = 
3
hG
c
 
 For unit of mass 
 Let M ? h
x
 c
y
 G
z 
 
Solving in similar manner as above 
 We get x = 
1
2
, y = 
1
2
, z = 
–1
2
 
 M = 
hc
G
 
 
Page 5


 
    
 
1  
 
   
UNIT & DIMENSION 
 
1. Young's modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational 
constant G, Planck's constant h and the speed of light c, as Y = c
a
 h
ß
 G
?
. Which of the following is the 
correct option ?   
 (A) a = 7, ß = –1, ? = –2        (B) a = –7, ß = –1, ? = –2    
 (C) a = 7, ß = –1, ? = 2     (D) a = –7, ß = 1, ? = –2 
2. In a particular system of units, a physical quantity can be expressed in terms of the electric charge e, 
electron mass m
e
, Planck's constant h, and Coulomb's constant
0
1
k
4
=
p?
,  where ?
0
  is the permittivity of 
vacuum. In terms of these physical constants, the dimension of the magnetic field is [B] = [e]
a
 [m
e
]
ß
 [h]
?
 
[k]
d
. The value of a + ß + ? + d is ________.    
3. A physical quantity S
?
 
is defined as 
( )
0
S E B/ = × µ
? ??
, where E
?
is electric field, B
?
is magnetic field and 
0
µ is 
the permeability of free space. The dimensions of S
?
 
are the same as the dimensions of which of the 
following quantity (ies) ?     
 (A) 
Energy
charge × current
 (B) 
Force
Length × Time
 (C) 
Energy
Volume
 (D) 
Power
Area
 
4. 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] = [ ???? a
]; [speed] = [ ???? ß
]; [acceleration] =[ ???? p
]; [linear momentum] = [ ???? q
]; 
[force] = [ ???? r
]. Then -     
 (A) a + ???? = 2 ß  (B) ???? + ???? – ???? = ß 
 (C) ???? - ???? + ???? = a  (D) ???? + ???? + ???? = ß 
5. Let us consider a system of units in which mass and angular momentum are dimensionless. If length has 
dimension of L, which of the following statement (s) is/are correct ?   
 (A) The dimension of force is L
–3
 (B) The dimension of energy is L
–2
 
 (C) The dimension of power is L
–5
 (D) The dimension of linear momentum is L
–1
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units.   
 (There are two questions based on Paragraph "X", the question given below is one of them) 
6. 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
 
JEE	Advanced	Numericals
 
  
    
 
2 
 
   
 
PARAGRAPH "X" 
 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 [?
0
] and [µ
0
] stand for 
dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of 
length and time respectively. All the quantities are given in SI units. 
 (There are two questions based on Paragraph "X", the question given below is one of them) 
7. The relation between [?
0
] and [µ
0
] is :-    
 (A) [µ
0
] = [?
0
][L]
2
[T]
–2
  (B) [µ
0
] = [?
0
][L]
–2
[T]
2
 
 (C) [µ
0
] = [?
0
]
–1
[L]
2
[T]
–2
  (D) [µ
0
] = [?
0
]
–1
[L]
–2
[T]
2 
8. A length-scale ( ?) depends on the permittivity (e) 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 expressions(s) for ? is(are) dimensionally correct? 
 
 (A)
2
B
nq
kT
? ?
=
? ?
e
? ?
? (B)
B
2
kT
nq
e ? ?
=
? ?
? ?
? (C)
2
2/3
B
q
n kT
? ?
=
? ?
? ?
e
? ?
? (D) 
2
1/3
B
q
n kT
? ?
=
? ?
? ?
e
? ?
? 
9. In terms of potential difference V, electric current I, permittivity e
0
, permeability µ
0
 and speed of light c, 
the dimensionally correct equation(s) is(are)    
 (A) µ
0
I
2
 = e
0
V
2
 (B) e
0
I = µ
0
V (C) I = e
0
cV (D) µ
0
cI = e
0
V  
10. 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) ? Mc (B) ? MG (C) ? L h (D) ? LG  
11. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer 
uses dimensional analysis and assumes that the distance depends on the mass density ? 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.    
 
    
 
 
   
3  
SOLUTIONS 
 
1. Ans. (A) 
Sol.   Y c hG
a ß ?
= 
 
1 2 1 2 1 13 2
ML T (LT ) (ML T ) (M L T )
- - - a - ß - - ?
= 
 1 = ß – ? ...(1) 
 –1 = a + 2ß + 3? ...(2) 
 –2 = – a – ß – 2? ...(3) 
 –3 = ß + ? 
   1 = ß – ? 
 –2 = 2ß           ? ß = –1, ? = –2 
 –1 = a – 2 – 6       ?  a = 7 
2. Ans. (4) 
Sol. ( )
e
B e m hk
ß
a ?d
= 
 
[ ] [ ]
[ ]
[ ] [ ]
e
B e m hk
ß ?
a d
=
 
[ ] [ ] [ ] [ ] [ ]
1 –2 –1 2 –1 3 –2 –4
M T A AT m ML T ML A T
? d
a ß
=
1 –2 –1 2r 3 – –4 –2
MT A m L T A
ß+?+d + d a ? d a d
= 
 Compare : ß + ? + d = 1; 2? + 3d = 0,  
a – ? – 4d = –2, a – 2d = – 1 
 On solving a = 3, ß = 2, ? = –3, d = 2  
 a + ß + ? + d = 4 
3. Ans. (B, D) 
Sol. 
0
1
S [E B] = ×
µ
? ??
 
 S is poynting vector denotes flow of energy per 
unit area per unit time 
 
2
watt
S
m
=
?
 
 Hence B, D are correct 
4. Ans. (A, B) 
Sol. Given,  
L = x
a
   
....(1) 
 LT
–1
 = x
ß
   ....(2) 
 LT
–2
 = x
p
   ....(3) 
 MLT
–1
 = x
q
   ....(4) 
 MLT
–2
 = x
r
   ....(5) 
 
(1)
(2)
 ?  T = x
a–ß 
  
 From (3) 
  
a
a-ß
=
p
2( )
x
x
x
 
 ?  a + p = 2ß     (A) 
 From (4)  
  M = x
q–ß
  
 From (5)  ?   x
q
 = x
r
  x
a–ß
  
 ? a + r – q = ß ....(6) 
 Replacing value 'a' in equation (6) from (A) 
  2ß – p + r – q = ß 
 ? p + q – r = ß (B) 
 Replacing value of 'ß' in equation (6) from (A)
 2a + 2r – 2q = a + p 
  a = p + 2q – 2r 
5. Ans. (A, B, D) 
Sol. Mass = M
0
L
0
T
0
 
 MVr = M
0
L
0
T
0
 
 M
0
1
1 00 0
1
L
. L M LT
T
= 
 L
2
 = T
1
   ....(1) 
 Force = M
1
L
1
T
–2
 (in SI) 
 = M
0 
L
1
 L
–4
 (In new system from equation (1)) 
 = L
–3
 
 Energy = M
1
L
2
T
–2 
 (In SI) 
 = M
0
L
2
L
–4
  (In new system from equation (1)) 
 = L
–2 
 
Power = 
Energy
Time
 
 = M
1
L
2
T
–3
 (in SI) 
 = M
0
L
2
L
–6
  (In new system from equation (1)) 
 = L
–4
 
 Linear momentum = M
1
L
1
T
–1
 (in SI) 
 = M
0
L
1
L
–2 
 (In new system from equation (1)) 
 = L
–1 
6. Ans. (C) 
Sol. We have 
E
B
C
= 
 ? [B] = 
[ ]
[ ]
[ ]
1 1
E
EL T
C
-
= 
 ? [E] = [B] [L][T
–1
]
 
 
  
    
 
  
 
4 
 
7. Ans. (D) 
Sol. We have, 
 
00
1
C =
µ?
 
 
[ ]
2
00
1
C
? ?
? =
? ?
µ?
? ?
 
 
[ ][ ]
2 2
0 0
1
LT
-
? =
µ?
 ? [µ
0
] = [?
0
]
–1
[L]
–2
[T]
2 
8. Ans. (B, D) 
Sol. We know,  
 
2
2
0
1q
F
4 r
=
pe
 
2
2
0
q
(Fr )4 ? = p
e
  
 So, dimension 
 
2
2
0
q
dim (Fr ) =
e
 = MLT
–2
× L
2
 = ML
3
T
–2
 
 Similarly; 
B
3
E K T
2
= ?  
dim (K
B
T) = dim(Energy) = ML
2
T
–2
 
(A)  
2 –3 3 –2
2 –2
B
nq L ML T 1
kT L ML T
×
= =
e
 
(B)  
( )
2 –2 3
2 –2 2
E vol ML T L
L
Fr MLT L
× ×
= =
×
 
(C)   
2 2/3 2 2 2
3 3/2
2 2
Fr (vol) MLT L L
LL
(K ) ML T
-
-
××
= = =
e
 
(D)  
2 1/3 2 2
2 2
Fr (vol) MLT L L
L
Energy ML T
-
-
×
= = 
 [? dimension n = 
3
1
dim L
vol
-
? ?
=
? ?
? ?
] 
9. Ans. (A, C)  
Sol. Using =
µ?
00
1
C &  
µ
=
?
0
0
R we can check the correctness. 
(A)  µ
0
I
2
 = ?
0
V
2
 
 
µ
= =
?
2
2 0
2
0
V
R
I
 
 ? R
2
 = R
2
 correct 
(B)  ?
0
I = µ
0
V 
 
?
=
µ
0
0
V
I
 
  =
2
1
R
R
not correct 
(C)  I = ?
0
cV 
 
?
=?=
µ?
0
0
00
I
c
V
 
 
?
= =
µ
0
0
1 1
R R
 correct 
(D)  µ
0
CI = ?
0
V 
 
µ
=
?
0
0
C V
I
 
 
µ
=
? µ?
0
0 00
1
R 
 ?
µ
= ? =
?? ?
0
0 0 0
1 R
R R incorrect 
10. Ans. (A, C, D)  
Sol. [h] = [ML
2
T
–1
] 
 [c] = [LT
–1
] 
 [G] = [M
–1
L
3
T
–2
] 
 For unit of length 
 Let L ? h
x 
c
y
 G
z
 
 
? From principle of homogeneity 
 [LHS] = [RHS] 
 ? [M
0
LT
0
] = [ML
2
T
–1
]
x
 [LT
–1
]
y
 [M
–1
L
3
T
–2
]
z
 
 M
0
 L T
0 
=  M
x–z
 L
2x + y + 3z 
 T
–x–y–2z
 
 comparing we get 
 x – z = 0 
 2x + y + 3z = 1 
 –x – y – 2z = 0 
 Solving we get,  x = 
1
2
, y = 
–3
2
, z = 
1
2
 
 L = 
3
hG
c
 
 For unit of mass 
 Let M ? h
x
 c
y
 G
z 
 
Solving in similar manner as above 
 We get x = 
1
2
, y = 
1
2
, z = 
–1
2
 
 M = 
hc
G
 
 
 
    
 
 
   
5  
11. Ans. (3) 
Sol. L = (I)
n
(d)
y
(f)
z
 
 L = (M
1
L
0
T
–3
)
x
(M
1
L
–3
)
y
(T
–1
)
2
 
 L = M
x + y
L
–3y
T
–3×–2
  
 –3y = 1 x + y = 0 
 y = 
1
3
- 
1
x
3
- = 0 
 
1
x
3
= 
 L = (I)
1/3
(d)
–1/3
(f)
2 
 n = 3