CONCEPT:
EXPLANATION:
i.e. \(Density = \frac{{Mass}}{{Volume}}\)
Now,
Dimensional formula of mass = [M]
Dimensional formula of volume = [L^{3}]
\(Density = \frac{{\left[ M \right]}}{{\left[ {{L^3}} \right]}}\)
∴ Density = ML^{-3} T^{0}
∴ Dimensional formula of density is [ML^{-3} T^{0}].
Sl. No. |
Quantity |
Common Symbol |
SI Unit |
Dimension |
1 |
Velocity |
v, u |
ms-1 |
LT-1 |
2 |
Acceleration |
a |
ms-2 |
LT-2 |
3 |
Force |
F |
Newton (N) |
M L T-2 |
4 |
Momentum |
p |
Kg-ms-1 |
M L T-1 |
5 |
Gravitational constant |
G |
N-m2Kg-2 |
L3 M-1 T-2 |
6 |
Torque |
τ |
N-m |
M L2 T-2 |
7 |
Bulk Modulus |
B |
Nm2 |
M L-1 T-2 |
8 |
Energy |
E, U, K |
joule (J) |
M L2 T-2 |
9 |
Heat |
Q |
joule (J) |
M L2 T-2 |
10 |
Pressure |
P |
Nm-2 (Pa) |
M L-1 T-2 |
11 |
Electric Field |
E |
Vm-1, NC-1 |
M L I-1 T-3 |
12 |
Potential (voltage) |
V |
V, JC-1 |
M L2 I-1 T-3 |
13 |
Magnetic Field |
B |
Tesla (T), Wb m-1 |
M I-1T-2 |
14 |
Magnetic Flux |
ΦB |
Wb |
M L2 I-1 T-2 |
15 |
Resistance |
R |
Ohm (Ω) |
M L2 I-2 T-3 |
16 |
Electromotive Force |
E |
Volt (V) |
M L2 I-1 T-3 |
17 | Density | ρ, d | kg/m^{3} | ML-3 T0 |
The distance covered by a particle in time 't' is given by \(s = at + bt^2\) where 'a' and 'b' are two constants. The dimensional formula of 'b' is:
Concept:
Principle of homogeneity of dimensions:
Explanation:
Given:
s = at + bt^{2}
From the principle of dimensional homogeneity, the left-hand side of the equation dimensionally equal to the right-hand side of the equation. So, the dimension of (s), (at) and (bt^{2}) is the same.
The dimension formula of distance (s) = [L]
For the first term:
∴ [L] = [a] [T]
\(⇒ \left[ a \right] =\frac{[L]}{[T]}=[LT^{-1}]\)
For the second term:
⇒ [L] = [b] [T^{2}]
\(⇒ \left[b \right] =\frac{[L]}{[T^2]}=[LT^{-2}]\)
The dimensional formula of 'b' is [M^{0 }L^{1} T^{-2}].
CONCEPT:
Work done (W) = F.s cos θ
Where F is force, s is displacement and θ is the angle between F and s.
Dimensional formula of work (W) = [ML2T-2]
Energy (E) = Work done (W)
Dimensional formula of Energy (E) = dimensional formula of work done (W) = [ML2T-2]
EXPLANATION:
CONCEPT:
P = F/A
Acceleration (a) = (v- u)/t
Where v and u are final and initial velocities respectively that have the dimension [M0L1T-1] and t = time that has dimension [M0L0T1].
It multiplies to give a dimensional formula of [M0L1T-2].
Force (F) = ma
The mass and acceleration have the dimensions of [M1L0T0] and [M0L1T-2] respectively, it multiplies to give the dimension of [M1L1T-2].
EXPLANATION:
Pressure (P) = F/A
F = force that has the dimension of [M1L1T-2] and A = area that has the dimension of [M0L2T0].
Dimensions of P = [F]/[A] = [M1L-1T-2].
CONCEPT:
EXPLANATION:
\(\Rightarrow Speed (v) = \frac{Distance (d)}{time (t)}\)
As we know, the dimension formula of distance (d) = [L]
The dimension formula of time (t) = [t]
∴ The dimension formula of speed is
\(\Rightarrow v =\frac{[L]}{[T]}=[M^0LT^{-1}]\)
So option 2 is correct.
Important Points
Quantity |
Unit |
Dimension |
Pressure |
Pascal |
[ML-1T-2] |
Stress |
Pascal |
[ML-1T-2] |
Velocity |
m/s |
[LT-1] |
Speed |
m/s |
[LT-1] |
Force | Newton | [MLT-2] |
Impulse | Newton-second | [MLT-1] |
Work | Joule | [ML2T-2] |
Energy | Joule | [ML2T-2] |
Capacitance (C) |
Coulomb/volt or Farad |
[M-1L2T4A2] |
Resistivity or Specific resistance (ρ) |
Ohm-meter |
[ML3T-3A-2] |
Electric current (I) |
Ampere |
[A] |
Electric charge (q) |
Coulomb |
[AT] |
Inductance (H) |
henry |
[ML2T-2A-2] |
Concept:
Explanation:
From the definition of acceleration,
\(A = \frac{{dv}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}}\)
\(\smallint dv = \smallint A\;dt\)
V = A × T
\(\frac{x}{T} = A\;T\)
⇒ x = A T^{2}
To write in the form of Dimensions
⇒ x = [F^{0}AT^{2}]
Explanation:
Vacuum permittivity:
\({F_c} = \frac{1}{{4\pi {_0}}} \times \frac{{{q_1}{q_2}}}{{{r^2}}}\)
Where ϵ_{0} is the permittivity of vacuum.
Dimension will be [M^{-1}L^{-3}T^{4}I^{2}].
Vacuum permeability:
μ_{0} = 4 π × 10^{-7} N/A^{2}
Dimension will be [MLT^{-2}I^{-2}]
Dimensions for \(\dfrac{1}{μ_0 ϵ_0}\) will be
\(\frac{1}{{\left[ {{{\rm{M}}^{ - 1}}{{\rm{L}}^{ - 3}}{{\rm{T}}^4}{{\rm{I}}^2}} \right] \times \left[ {{{\rm{M}}^1}{{\rm{L}}^1}{{\rm{T}}^{ - 2}}{{\rm{I}}^{ - 2}}} \right]}} = \left[ {{{\rm{L}}^2}{{\rm{T}}^{ - 2}}} \right]\)
The correct answer is Specific resistance.
Key Points
Concept:
Surface tension:
Explanation:
we know that,
\(Surface\;tension = \frac{{Force}}{{length}}\)
Dimensional formula for surface tension is -
\(Surface\;tension = \frac{{Force}}{{length}}=\frac{[ML{T}^{-2}]}{[L]}=[M{T}^{-2}]\)
CONCEPT:
EXPLANATION:
i.e. \(Density = \frac{{Mass}}{{Volume}}\)
Now,
Dimensional formula of mass = [M]
Dimensional formula of volume = [L^{3}]
\(Density = \frac{{\left[ M \right]}}{{\left[ {{L^3}} \right]}}\)
∴ Density = ML^{-3} T^{0}
∴ Dimensional formula of density is [ML^{-3} T^{0}].
Sl. No. |
Quantity |
Common Symbol |
SI Unit |
Dimension |
1 |
Velocity |
v, u |
ms-1 |
LT-1 |
2 |
Acceleration |
a |
ms-2 |
LT-2 |
3 |
Force |
F |
Newton (N) |
M L T-2 |
4 |
Momentum |
p |
Kg-ms-1 |
M L T-1 |
5 |
Gravitational constant |
G |
N-m2Kg-2 |
L3 M-1 T-2 |
6 |
Torque |
τ |
N-m |
M L2 T-2 |
7 |
Bulk Modulus |
B |
Nm2 |
M L-1 T-2 |
8 |
Energy |
E, U, K |
joule (J) |
M L2 T-2 |
9 |
Heat |
Q |
joule (J) |
M L2 T-2 |
10 |
Pressure |
P |
Nm-2 (Pa) |
M L-1 T-2 |
11 |
Electric Field |
E |
Vm-1, NC-1 |
M L I-1 T-3 |
12 |
Potential (voltage) |
V |
V, JC-1 |
M L2 I-1 T-3 |
13 |
Magnetic Field |
B |
Tesla (T), Wb m-1 |
M I-1T-2 |
14 |
Magnetic Flux |
ΦB |
Wb |
M L2 I-1 T-2 |
15 |
Resistance |
R |
Ohm (Ω) |
M L2 I-2 T-3 |
16 |
Electromotive Force |
E |
Volt (V) |
M L2 I-1 T-3 |
17 | Density | ρ, d | kg/m^{3} | ML-3 T0 |