QBManager

Units and Measurements

139490 The physical quantity having the dimensions $[M^{- 1} L^{- 3} T^{3} A^{2}]$ is

1 resistance

2 resistivity

3 electrical conductivity

4 electromotive force

Explanation:

C Resistivity $(ρ) = \frac{RA}{l}$
$R = \frac{V}{I} and V = \frac{W}{Q}$
So, $ρ = \frac{W}{QI} \cdot \frac{A}{l}$
dimension of $(ρ) = \frac{[{ML}^{2} T^{- 2}] [L^{2}]}{[AT] [A] [L]} = [{ML}^{3} A^{- 2} T^{- 3}]$
Electrical conductivity $σ = \frac{1}{ρ}$
$= [M^{- 1} L^{- 3} T^{3} A^{2}]$
Resistance $= \frac{resistivity \times length}{area}$
$= [{ML}^{3} T^{- 3} A^{- 2}] \frac{[L]}{[L^{2}]} = [{ML}^{2} T^{- 3} A^{- 2}]$
Electromotive force $=$ resistance $\times$ current
$= [{ML}^{2} T^{- 3} A^{- 2}] [A]$
$= [{ML}^{2} T^{- 3} A^{- 1}]$

J and K-CET-2012

Units and Measurements

139491 The dimension of $\frac{a}{b}$ in the equation $p = \frac{a - t^{2}}{b x}$ , where $p$ is pressure, $x$ is distance and $t$ is time is

1

[{LT}^{- 3}]

2

[{ML}^{3} T^{- 1}]

3

[M^{2} {LT}^{- 3}]

4

[{MT}^{- 2}]

Explanation:

D Dimension of pressure $[P] = \frac{[F]}{[A]}$
$= \frac{[{MLT}^{- 2}]}{[L^{2}]}$
$[P] = [M^{+ 1} L^{- 1} T^{- 2}]$
Dimension of Distance $(x) = [L]$
Dimension of Time $(t) = [T^{1}]$
$P = \frac{a - t^{2}}{bx} = \frac{a}{bx} - \frac{t^{2}}{bx}$
Dimension of $P =$ Dimension of $\frac{a}{bx}$
$[P] = [{ML}^{- 1} T^{- 2}]$
$[{ML}^{- 1} T^{- 2}] = [\frac{a}{b}] \frac{1}{[L]}$
The dimension of $[\frac{a}{b}] = [{MT}^{- 2}]$

JCECE-2011

Units and Measurements

139492 $[{ML}^{- 1} T^{- 1}]$ stand for dimension of

1 work

2 torque

3 linear momentum

4 coefficient of viscosity

Explanation:

D (i) Dimension of work
$W = f.d$
$W = [{MLT}^{- 2}] [L]$
$= [{ML}^{2} T^{- 2}]$
(ii) Dimension of torque
$T = f \times r$
$= [{MLT}^{- 2}] [L]$
$T = [{ML}^{2} T^{- 2}]$
(iii) Linear momentum
$P = m . v .$
$= [M] [{LT}^{- 1}]$
$= [MLT^{- 1}]$
(iv) Dimension of coefficient of viscosity
$F = 6 π η rv$
$η = \frac{F}{6 π r \cdot v}$
$= \frac{[{MLT}^{- 2}]}{[L] [{LT}^{- 1}]} = [{ML}^{- 1} T^{- 1}]$

UPSEE - 2013

Units and Measurements

139493 If $p$ represents radiation pressure, $c$ represents speed of light and $S$ represents radiation energy striking unit area per sec. The non-zero integers $x, y, z$ such that $p^{x} S^{y} c^{z}$ is dimensionless are

1

x = 1, y = 1, z = 1

2

x = - 1, y = 1, z = 1

3

x = 1, y = - 1, z = 1

4

x = 1, y = 1, z = - 1

Explanation:

C Given,
$P = radiation pressure$
$C = speed of light$
$S = radiation energy$
$x, y$ and $z$ are non zero intezers.
$[P^{x} S^{y} C^{z}] = [M^{0} L^{0} T^{0}]$
The dimension of $P = [{ML}^{- 1} T^{- 2}]$
The dimension of $S = [{MT}^{- 3}]$
The dimension of $C = [{LT}^{- 1}]$
Putting the dimension in equation (i)
${[{ML}^{- 1} T^{- 2}]}^{x} {[{MT}^{- 3}]}^{y} \cdot {[{LT}^{- 1}]}^{z} = [M^{0} L^{0} T^{0}]$
$[M^{x + y} \cdot L^{- x + z} \cdot T^{- 2 x - 3 y - z}] = [M^{0} L^{0} T^{0}]$
$x + y = 0$
$z - x = 0$
$- 2 x - 3 y - z = 0$
From equation (iii)
$x = z$
From equation (ii)
$∴ z = - y$
By solving, we get,
$x = 1, y = - 1, z = 1$
Hence, option (c) is correct.

AIPMT 1992

Units and Measurements

139494 The dimensional formula for permeability of free space, $μ_{0}$ is

1

[{MLT}^{- 2} A^{- 2}]

2

[M L - 1 T^{- 2} A^{2}]

3

[{ML}^{- 1} T^{- 2} A^{2}]

4

[{MLT}^{- 2} A^{- 1}]

Explanation:

A The dimensional formula for permeability of free space,
$μ_{0} = \frac{2 π \times force \times distance}{current \times current \times length}$
Dimension of force $= [{MLT}^{- 2}]$
Dimension of current $= [A]$
Dimension of length and distance $= [L]$
$∴ μ_{0} = \frac{[{MLT}^{- 2}] [L]}{[A] [A] [L]}$
$μ_{0} = [{MLT}^{- 2} A^{- 2}]$

UPCPMT-2007

Units and Measurements

139490 The physical quantity having the dimensions $[M^{- 1} L^{- 3} T^{3} A^{2}]$ is

1 resistance

2 resistivity

3 electrical conductivity

4 electromotive force

Explanation:

C Resistivity $(ρ) = \frac{RA}{l}$
$R = \frac{V}{I} and V = \frac{W}{Q}$
So, $ρ = \frac{W}{QI} \cdot \frac{A}{l}$
dimension of $(ρ) = \frac{[{ML}^{2} T^{- 2}] [L^{2}]}{[AT] [A] [L]} = [{ML}^{3} A^{- 2} T^{- 3}]$
Electrical conductivity $σ = \frac{1}{ρ}$
$= [M^{- 1} L^{- 3} T^{3} A^{2}]$
Resistance $= \frac{resistivity \times length}{area}$
$= [{ML}^{3} T^{- 3} A^{- 2}] \frac{[L]}{[L^{2}]} = [{ML}^{2} T^{- 3} A^{- 2}]$
Electromotive force $=$ resistance $\times$ current
$= [{ML}^{2} T^{- 3} A^{- 2}] [A]$
$= [{ML}^{2} T^{- 3} A^{- 1}]$

J and K-CET-2012

Units and Measurements

139491 The dimension of $\frac{a}{b}$ in the equation $p = \frac{a - t^{2}}{b x}$ , where $p$ is pressure, $x$ is distance and $t$ is time is

1

[{LT}^{- 3}]

2

[{ML}^{3} T^{- 1}]

3

[M^{2} {LT}^{- 3}]

4

[{MT}^{- 2}]

Explanation:

D Dimension of pressure $[P] = \frac{[F]}{[A]}$
$= \frac{[{MLT}^{- 2}]}{[L^{2}]}$
$[P] = [M^{+ 1} L^{- 1} T^{- 2}]$
Dimension of Distance $(x) = [L]$
Dimension of Time $(t) = [T^{1}]$
$P = \frac{a - t^{2}}{bx} = \frac{a}{bx} - \frac{t^{2}}{bx}$
Dimension of $P =$ Dimension of $\frac{a}{bx}$
$[P] = [{ML}^{- 1} T^{- 2}]$
$[{ML}^{- 1} T^{- 2}] = [\frac{a}{b}] \frac{1}{[L]}$
The dimension of $[\frac{a}{b}] = [{MT}^{- 2}]$

JCECE-2011

Units and Measurements

139492 $[{ML}^{- 1} T^{- 1}]$ stand for dimension of

1 work

2 torque

3 linear momentum

4 coefficient of viscosity

Explanation:

D (i) Dimension of work
$W = f.d$
$W = [{MLT}^{- 2}] [L]$
$= [{ML}^{2} T^{- 2}]$
(ii) Dimension of torque
$T = f \times r$
$= [{MLT}^{- 2}] [L]$
$T = [{ML}^{2} T^{- 2}]$
(iii) Linear momentum
$P = m . v .$
$= [M] [{LT}^{- 1}]$
$= [MLT^{- 1}]$
(iv) Dimension of coefficient of viscosity
$F = 6 π η rv$
$η = \frac{F}{6 π r \cdot v}$
$= \frac{[{MLT}^{- 2}]}{[L] [{LT}^{- 1}]} = [{ML}^{- 1} T^{- 1}]$

UPSEE - 2013

Units and Measurements

139493 If $p$ represents radiation pressure, $c$ represents speed of light and $S$ represents radiation energy striking unit area per sec. The non-zero integers $x, y, z$ such that $p^{x} S^{y} c^{z}$ is dimensionless are

1

x = 1, y = 1, z = 1

2

x = - 1, y = 1, z = 1

3

x = 1, y = - 1, z = 1

4

x = 1, y = 1, z = - 1

Explanation:

C Given,
$P = radiation pressure$
$C = speed of light$
$S = radiation energy$
$x, y$ and $z$ are non zero intezers.
$[P^{x} S^{y} C^{z}] = [M^{0} L^{0} T^{0}]$
The dimension of $P = [{ML}^{- 1} T^{- 2}]$
The dimension of $S = [{MT}^{- 3}]$
The dimension of $C = [{LT}^{- 1}]$
Putting the dimension in equation (i)
${[{ML}^{- 1} T^{- 2}]}^{x} {[{MT}^{- 3}]}^{y} \cdot {[{LT}^{- 1}]}^{z} = [M^{0} L^{0} T^{0}]$
$[M^{x + y} \cdot L^{- x + z} \cdot T^{- 2 x - 3 y - z}] = [M^{0} L^{0} T^{0}]$
$x + y = 0$
$z - x = 0$
$- 2 x - 3 y - z = 0$
From equation (iii)
$x = z$
From equation (ii)
$∴ z = - y$
By solving, we get,
$x = 1, y = - 1, z = 1$
Hence, option (c) is correct.

AIPMT 1992

Units and Measurements

139494 The dimensional formula for permeability of free space, $μ_{0}$ is

1

[{MLT}^{- 2} A^{- 2}]

2

[M L - 1 T^{- 2} A^{2}]

3

[{ML}^{- 1} T^{- 2} A^{2}]

4

[{MLT}^{- 2} A^{- 1}]

Explanation:

A The dimensional formula for permeability of free space,
$μ_{0} = \frac{2 π \times force \times distance}{current \times current \times length}$
Dimension of force $= [{MLT}^{- 2}]$
Dimension of current $= [A]$
Dimension of length and distance $= [L]$
$∴ μ_{0} = \frac{[{MLT}^{- 2}] [L]}{[A] [A] [L]}$
$μ_{0} = [{MLT}^{- 2} A^{- 2}]$

UPCPMT-2007

Units and Measurements

139490 The physical quantity having the dimensions $[M^{- 1} L^{- 3} T^{3} A^{2}]$ is

1 resistance

2 resistivity

3 electrical conductivity

4 electromotive force

Explanation:

C Resistivity $(ρ) = \frac{RA}{l}$
$R = \frac{V}{I} and V = \frac{W}{Q}$
So, $ρ = \frac{W}{QI} \cdot \frac{A}{l}$
dimension of $(ρ) = \frac{[{ML}^{2} T^{- 2}] [L^{2}]}{[AT] [A] [L]} = [{ML}^{3} A^{- 2} T^{- 3}]$
Electrical conductivity $σ = \frac{1}{ρ}$
$= [M^{- 1} L^{- 3} T^{3} A^{2}]$
Resistance $= \frac{resistivity \times length}{area}$
$= [{ML}^{3} T^{- 3} A^{- 2}] \frac{[L]}{[L^{2}]} = [{ML}^{2} T^{- 3} A^{- 2}]$
Electromotive force $=$ resistance $\times$ current
$= [{ML}^{2} T^{- 3} A^{- 2}] [A]$
$= [{ML}^{2} T^{- 3} A^{- 1}]$

J and K-CET-2012

Units and Measurements

139491 The dimension of $\frac{a}{b}$ in the equation $p = \frac{a - t^{2}}{b x}$ , where $p$ is pressure, $x$ is distance and $t$ is time is

1

[{LT}^{- 3}]

2

[{ML}^{3} T^{- 1}]

3

[M^{2} {LT}^{- 3}]

4

[{MT}^{- 2}]

Explanation:

D Dimension of pressure $[P] = \frac{[F]}{[A]}$
$= \frac{[{MLT}^{- 2}]}{[L^{2}]}$
$[P] = [M^{+ 1} L^{- 1} T^{- 2}]$
Dimension of Distance $(x) = [L]$
Dimension of Time $(t) = [T^{1}]$
$P = \frac{a - t^{2}}{bx} = \frac{a}{bx} - \frac{t^{2}}{bx}$
Dimension of $P =$ Dimension of $\frac{a}{bx}$
$[P] = [{ML}^{- 1} T^{- 2}]$
$[{ML}^{- 1} T^{- 2}] = [\frac{a}{b}] \frac{1}{[L]}$
The dimension of $[\frac{a}{b}] = [{MT}^{- 2}]$

JCECE-2011

Units and Measurements

139492 $[{ML}^{- 1} T^{- 1}]$ stand for dimension of

1 work

2 torque

3 linear momentum

4 coefficient of viscosity

Explanation:

D (i) Dimension of work
$W = f.d$
$W = [{MLT}^{- 2}] [L]$
$= [{ML}^{2} T^{- 2}]$
(ii) Dimension of torque
$T = f \times r$
$= [{MLT}^{- 2}] [L]$
$T = [{ML}^{2} T^{- 2}]$
(iii) Linear momentum
$P = m . v .$
$= [M] [{LT}^{- 1}]$
$= [MLT^{- 1}]$
(iv) Dimension of coefficient of viscosity
$F = 6 π η rv$
$η = \frac{F}{6 π r \cdot v}$
$= \frac{[{MLT}^{- 2}]}{[L] [{LT}^{- 1}]} = [{ML}^{- 1} T^{- 1}]$

UPSEE - 2013

Units and Measurements

139493 If $p$ represents radiation pressure, $c$ represents speed of light and $S$ represents radiation energy striking unit area per sec. The non-zero integers $x, y, z$ such that $p^{x} S^{y} c^{z}$ is dimensionless are

1

x = 1, y = 1, z = 1

2

x = - 1, y = 1, z = 1

3

x = 1, y = - 1, z = 1

4

x = 1, y = 1, z = - 1

Explanation:

C Given,
$P = radiation pressure$
$C = speed of light$
$S = radiation energy$
$x, y$ and $z$ are non zero intezers.
$[P^{x} S^{y} C^{z}] = [M^{0} L^{0} T^{0}]$
The dimension of $P = [{ML}^{- 1} T^{- 2}]$
The dimension of $S = [{MT}^{- 3}]$
The dimension of $C = [{LT}^{- 1}]$
Putting the dimension in equation (i)
${[{ML}^{- 1} T^{- 2}]}^{x} {[{MT}^{- 3}]}^{y} \cdot {[{LT}^{- 1}]}^{z} = [M^{0} L^{0} T^{0}]$
$[M^{x + y} \cdot L^{- x + z} \cdot T^{- 2 x - 3 y - z}] = [M^{0} L^{0} T^{0}]$
$x + y = 0$
$z - x = 0$
$- 2 x - 3 y - z = 0$
From equation (iii)
$x = z$
From equation (ii)
$∴ z = - y$
By solving, we get,
$x = 1, y = - 1, z = 1$
Hence, option (c) is correct.

AIPMT 1992

Units and Measurements

139494 The dimensional formula for permeability of free space, $μ_{0}$ is

1

[{MLT}^{- 2} A^{- 2}]

2

[M L - 1 T^{- 2} A^{2}]

3

[{ML}^{- 1} T^{- 2} A^{2}]

4

[{MLT}^{- 2} A^{- 1}]

Explanation:

A The dimensional formula for permeability of free space,
$μ_{0} = \frac{2 π \times force \times distance}{current \times current \times length}$
Dimension of force $= [{MLT}^{- 2}]$
Dimension of current $= [A]$
Dimension of length and distance $= [L]$
$∴ μ_{0} = \frac{[{MLT}^{- 2}] [L]}{[A] [A] [L]}$
$μ_{0} = [{MLT}^{- 2} A^{- 2}]$

UPCPMT-2007

Units and Measurements

139490 The physical quantity having the dimensions $[M^{- 1} L^{- 3} T^{3} A^{2}]$ is

1 resistance

2 resistivity

3 electrical conductivity

4 electromotive force

Explanation:

C Resistivity $(ρ) = \frac{RA}{l}$
$R = \frac{V}{I} and V = \frac{W}{Q}$
So, $ρ = \frac{W}{QI} \cdot \frac{A}{l}$
dimension of $(ρ) = \frac{[{ML}^{2} T^{- 2}] [L^{2}]}{[AT] [A] [L]} = [{ML}^{3} A^{- 2} T^{- 3}]$
Electrical conductivity $σ = \frac{1}{ρ}$
$= [M^{- 1} L^{- 3} T^{3} A^{2}]$
Resistance $= \frac{resistivity \times length}{area}$
$= [{ML}^{3} T^{- 3} A^{- 2}] \frac{[L]}{[L^{2}]} = [{ML}^{2} T^{- 3} A^{- 2}]$
Electromotive force $=$ resistance $\times$ current
$= [{ML}^{2} T^{- 3} A^{- 2}] [A]$
$= [{ML}^{2} T^{- 3} A^{- 1}]$

J and K-CET-2012

Units and Measurements

139491 The dimension of $\frac{a}{b}$ in the equation $p = \frac{a - t^{2}}{b x}$ , where $p$ is pressure, $x$ is distance and $t$ is time is

1

[{LT}^{- 3}]

2

[{ML}^{3} T^{- 1}]

3

[M^{2} {LT}^{- 3}]

4

[{MT}^{- 2}]

Explanation:

D Dimension of pressure $[P] = \frac{[F]}{[A]}$
$= \frac{[{MLT}^{- 2}]}{[L^{2}]}$
$[P] = [M^{+ 1} L^{- 1} T^{- 2}]$
Dimension of Distance $(x) = [L]$
Dimension of Time $(t) = [T^{1}]$
$P = \frac{a - t^{2}}{bx} = \frac{a}{bx} - \frac{t^{2}}{bx}$
Dimension of $P =$ Dimension of $\frac{a}{bx}$
$[P] = [{ML}^{- 1} T^{- 2}]$
$[{ML}^{- 1} T^{- 2}] = [\frac{a}{b}] \frac{1}{[L]}$
The dimension of $[\frac{a}{b}] = [{MT}^{- 2}]$

JCECE-2011

Units and Measurements

139492 $[{ML}^{- 1} T^{- 1}]$ stand for dimension of

1 work

2 torque

3 linear momentum

4 coefficient of viscosity

Explanation:

D (i) Dimension of work
$W = f.d$
$W = [{MLT}^{- 2}] [L]$
$= [{ML}^{2} T^{- 2}]$
(ii) Dimension of torque
$T = f \times r$
$= [{MLT}^{- 2}] [L]$
$T = [{ML}^{2} T^{- 2}]$
(iii) Linear momentum
$P = m . v .$
$= [M] [{LT}^{- 1}]$
$= [MLT^{- 1}]$
(iv) Dimension of coefficient of viscosity
$F = 6 π η rv$
$η = \frac{F}{6 π r \cdot v}$
$= \frac{[{MLT}^{- 2}]}{[L] [{LT}^{- 1}]} = [{ML}^{- 1} T^{- 1}]$

UPSEE - 2013

Units and Measurements

139493 If $p$ represents radiation pressure, $c$ represents speed of light and $S$ represents radiation energy striking unit area per sec. The non-zero integers $x, y, z$ such that $p^{x} S^{y} c^{z}$ is dimensionless are

1

x = 1, y = 1, z = 1

2

x = - 1, y = 1, z = 1

3

x = 1, y = - 1, z = 1

4

x = 1, y = 1, z = - 1

Explanation:

C Given,
$P = radiation pressure$
$C = speed of light$
$S = radiation energy$
$x, y$ and $z$ are non zero intezers.
$[P^{x} S^{y} C^{z}] = [M^{0} L^{0} T^{0}]$
The dimension of $P = [{ML}^{- 1} T^{- 2}]$
The dimension of $S = [{MT}^{- 3}]$
The dimension of $C = [{LT}^{- 1}]$
Putting the dimension in equation (i)
${[{ML}^{- 1} T^{- 2}]}^{x} {[{MT}^{- 3}]}^{y} \cdot {[{LT}^{- 1}]}^{z} = [M^{0} L^{0} T^{0}]$
$[M^{x + y} \cdot L^{- x + z} \cdot T^{- 2 x - 3 y - z}] = [M^{0} L^{0} T^{0}]$
$x + y = 0$
$z - x = 0$
$- 2 x - 3 y - z = 0$
From equation (iii)
$x = z$
From equation (ii)
$∴ z = - y$
By solving, we get,
$x = 1, y = - 1, z = 1$
Hence, option (c) is correct.

AIPMT 1992

Units and Measurements

139494 The dimensional formula for permeability of free space, $μ_{0}$ is

1

[{MLT}^{- 2} A^{- 2}]

2

[M L - 1 T^{- 2} A^{2}]

3

[{ML}^{- 1} T^{- 2} A^{2}]

4

[{MLT}^{- 2} A^{- 1}]

Explanation:

A The dimensional formula for permeability of free space,
$μ_{0} = \frac{2 π \times force \times distance}{current \times current \times length}$
Dimension of force $= [{MLT}^{- 2}]$
Dimension of current $= [A]$
Dimension of length and distance $= [L]$
$∴ μ_{0} = \frac{[{MLT}^{- 2}] [L]}{[A] [A] [L]}$
$μ_{0} = [{MLT}^{- 2} A^{- 2}]$

UPCPMT-2007

Units and Measurements

139490 The physical quantity having the dimensions $[M^{- 1} L^{- 3} T^{3} A^{2}]$ is

1 resistance

2 resistivity

3 electrical conductivity

4 electromotive force

Explanation:

C Resistivity $(ρ) = \frac{RA}{l}$
$R = \frac{V}{I} and V = \frac{W}{Q}$
So, $ρ = \frac{W}{QI} \cdot \frac{A}{l}$
dimension of $(ρ) = \frac{[{ML}^{2} T^{- 2}] [L^{2}]}{[AT] [A] [L]} = [{ML}^{3} A^{- 2} T^{- 3}]$
Electrical conductivity $σ = \frac{1}{ρ}$
$= [M^{- 1} L^{- 3} T^{3} A^{2}]$
Resistance $= \frac{resistivity \times length}{area}$
$= [{ML}^{3} T^{- 3} A^{- 2}] \frac{[L]}{[L^{2}]} = [{ML}^{2} T^{- 3} A^{- 2}]$
Electromotive force $=$ resistance $\times$ current
$= [{ML}^{2} T^{- 3} A^{- 2}] [A]$
$= [{ML}^{2} T^{- 3} A^{- 1}]$

J and K-CET-2012

Units and Measurements

139491 The dimension of $\frac{a}{b}$ in the equation $p = \frac{a - t^{2}}{b x}$ , where $p$ is pressure, $x$ is distance and $t$ is time is

1

[{LT}^{- 3}]

2

[{ML}^{3} T^{- 1}]

3

[M^{2} {LT}^{- 3}]

4

[{MT}^{- 2}]

Explanation:

D Dimension of pressure $[P] = \frac{[F]}{[A]}$
$= \frac{[{MLT}^{- 2}]}{[L^{2}]}$
$[P] = [M^{+ 1} L^{- 1} T^{- 2}]$
Dimension of Distance $(x) = [L]$
Dimension of Time $(t) = [T^{1}]$
$P = \frac{a - t^{2}}{bx} = \frac{a}{bx} - \frac{t^{2}}{bx}$
Dimension of $P =$ Dimension of $\frac{a}{bx}$
$[P] = [{ML}^{- 1} T^{- 2}]$
$[{ML}^{- 1} T^{- 2}] = [\frac{a}{b}] \frac{1}{[L]}$
The dimension of $[\frac{a}{b}] = [{MT}^{- 2}]$

JCECE-2011

Units and Measurements

139492 $[{ML}^{- 1} T^{- 1}]$ stand for dimension of

1 work

2 torque

3 linear momentum

4 coefficient of viscosity

Explanation:

D (i) Dimension of work
$W = f.d$
$W = [{MLT}^{- 2}] [L]$
$= [{ML}^{2} T^{- 2}]$
(ii) Dimension of torque
$T = f \times r$
$= [{MLT}^{- 2}] [L]$
$T = [{ML}^{2} T^{- 2}]$
(iii) Linear momentum
$P = m . v .$
$= [M] [{LT}^{- 1}]$
$= [MLT^{- 1}]$
(iv) Dimension of coefficient of viscosity
$F = 6 π η rv$
$η = \frac{F}{6 π r \cdot v}$
$= \frac{[{MLT}^{- 2}]}{[L] [{LT}^{- 1}]} = [{ML}^{- 1} T^{- 1}]$

UPSEE - 2013

Units and Measurements

139493 If $p$ represents radiation pressure, $c$ represents speed of light and $S$ represents radiation energy striking unit area per sec. The non-zero integers $x, y, z$ such that $p^{x} S^{y} c^{z}$ is dimensionless are

1

x = 1, y = 1, z = 1

2

x = - 1, y = 1, z = 1

3

x = 1, y = - 1, z = 1

4

x = 1, y = 1, z = - 1

Explanation:

C Given,
$P = radiation pressure$
$C = speed of light$
$S = radiation energy$
$x, y$ and $z$ are non zero intezers.
$[P^{x} S^{y} C^{z}] = [M^{0} L^{0} T^{0}]$
The dimension of $P = [{ML}^{- 1} T^{- 2}]$
The dimension of $S = [{MT}^{- 3}]$
The dimension of $C = [{LT}^{- 1}]$
Putting the dimension in equation (i)
${[{ML}^{- 1} T^{- 2}]}^{x} {[{MT}^{- 3}]}^{y} \cdot {[{LT}^{- 1}]}^{z} = [M^{0} L^{0} T^{0}]$
$[M^{x + y} \cdot L^{- x + z} \cdot T^{- 2 x - 3 y - z}] = [M^{0} L^{0} T^{0}]$
$x + y = 0$
$z - x = 0$
$- 2 x - 3 y - z = 0$
From equation (iii)
$x = z$
From equation (ii)
$∴ z = - y$
By solving, we get,
$x = 1, y = - 1, z = 1$
Hence, option (c) is correct.

AIPMT 1992

Units and Measurements

139494 The dimensional formula for permeability of free space, $μ_{0}$ is

1

[{MLT}^{- 2} A^{- 2}]

2

[M L - 1 T^{- 2} A^{2}]

3

[{ML}^{- 1} T^{- 2} A^{2}]

4

[{MLT}^{- 2} A^{- 1}]

Explanation:

A The dimensional formula for permeability of free space,
$μ_{0} = \frac{2 π \times force \times distance}{current \times current \times length}$
Dimension of force $= [{MLT}^{- 2}]$
Dimension of current $= [A]$
Dimension of length and distance $= [L]$
$∴ μ_{0} = \frac{[{MLT}^{- 2}] [L]}{[A] [A] [L]}$
$μ_{0} = [{MLT}^{- 2} A^{- 2}]$

UPCPMT-2007