1. If the dimensions of a physical quantity are given by Ma Lb Tc, then the physical quantity will be :
   

1. 1. Velocity if a = 1, b = 0, c = – 1 ​

   
   
   

   2. ​Acceleration if a = 1, b = 1, c = – 2

   
   
   

   3. Force if a = 0, b = – 1, c = – 2 ​

   
   
   

   4. ​Pressure if a = 1, b = – 1, c = – 2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Pressure =	[MLT-2]	= ML-1T-2
   	[L2]

   
   ⇒ a = 1, b = – 1, c = – 2.

   Correct Option: D

   Pressure =	[MLT-2]	= ML-1T-2
   	[L2]

   
   ⇒ a = 1, b = – 1, c = – 2.

---

2. Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would be​

1. 1. ML²T ^{– 2}

   
   
   

   2. ML²T ^{– 1}I ^{– 1}

   
   
   

   3. ML²T ^{– 3}I ^{– 2}

   
   
   

   4. ML²T ^{– 3}I ^{– 1}

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Dimensions of Resistance, =	[V]	=	[ML2T-3I-1]	= [ML2T-3I-2]
   	[I]		[I]

   Correct Option: C

   Dimensions of Resistance, =	[V]	=	[ML2T-3I-1]	= [ML2T-3I-2]
   	[I]		[I]

---

---

3. Which two of the following five physical parameters have the same dimensions?​​
   [2008]
   ​(A)​Energy density ​
   (B)​Refractive index ​
   (C)​Dielectric constant ​
   (D)​Young’s modulus ​
   (E)​Magnetic field
   

1. 1. (B) and (D) ​

   
   
   

   2. ​(C) and (E)

   
   
   

   3. (A) and (D)

   
   
   

   4. ​(A) and (E)

   
   
   

---

1. View Hint View Answer Discuss in Forum

   [Energy density] =	[Work done]
   	[Volume]

   
   

   =	ML2T-2	= ML-1T-2
   	L3

   
   

   [Young's Modulus] =		F	×	l
   		A		∆l

   
   

   =	MLT-2	.	L	= ML-1T-2
   	L2		L

   
   

   Correct Option: C

   [Energy density] =	[Work done]
   	[Volume]

   
   

   =	ML2T-2	= ML-1T-2
   	L3

   
   

   [Young's Modulus] =		F	×	l
   		A		∆l

   
   

   =	MLT-2	.	L	= ML-1T-2
   	L2		L

   
   

---

4. The velocity v of a particle at time t is given by v = at +	b	, where a, b and c are constant.
   	t + c

   The dimensions of a, b and c are respectively​​
   

1. 1. ​L², T and LT² ​

   
   
   

   2. LT², LT and L

   
   
   

   3. L, LT and T² ​

   
   
   

   4. ​LT^{– 2}, L and T

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Dimension of  a. t = dimension of velocity​ ​
   a .t = L T^-1 ⇒ a = LT^-2
   Dimension of c = dimension of t
   ​(two physical quantity of same dimension can only be added)
   So, dimension of c = T
   

   Dimension of	b	= Dimension of v
   	t + c

   
   

   	b	= L T-1 ⇒ b.T-1 = LT-1 ⇒ b = L
   	T + T

   
   So, answer is LT^-2, L & T
   

   Correct Option: D

   Dimension of  a. t = dimension of velocity​ ​
   a .t = L T^-1 ⇒ a = LT^-2
   Dimension of c = dimension of t
   ​(two physical quantity of same dimension can only be added)
   So, dimension of c = T
   

   Dimension of	b	= Dimension of v
   	t + c

   
   

   	b	= L T-1 ⇒ b.T-1 = LT-1 ⇒ b = L
   	T + T

   
   So, answer is LT^-2, L & T
   

---

---

5. The dimension of	1	ε0 E2 , where ε0 is permittivity of free space
   	2

   and E is electric field, is :​​​​
   

1. 1. ​ML² T ^{– 2}

   
   
   

   2. ML ^{– 1}T ^{– 2}

   
   
   

   3. ML²T ^{– 1} ​

   
   
   

   4. MLT ^{– 1}

   
   
   

---

1. View Hint View Answer Discuss in Forum

   	1	ε0 E2 represents energy density i.e., energy per unit volume.
   	2

   
   

   ⇒			1	ε0 E2		=	ML2T-2	= ML-1T-2
   			2				L3

   Correct Option: B

   	1	ε0 E2 represents energy density i.e., energy per unit volume.
   	2

   
   

   ⇒			1	ε0 E2		=	ML2T-2	= ML-1T-2
   			2				L3

---