Question and Answer on Unit and Dimension
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- What are fundamental and derived quantities?
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Fundamental Quantities:
Fundamental quantities are the basic physical quantities that do not depend on any other quantity for their definition.
The seven SI fundamental quantities are: length, mass, time, electric current, temperature, amount of substance, and luminous intensity.
Derived Quantities:
Derived quantities are obtained by combining fundamental quantities through mathematical relations.
Examples include speed, force, energy, pressure, and density.
2. What are the fundamental units in SI system?
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| Fundamental Quantity | SI Unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric Current | ampere | A |
| Temperature | kelvin | K |
| Amount of Substance | mole | mol |
| Luminous Intensity | candela | cd |
3. See the units and dimension of some common physical quantites.
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| Physical Quantity | SI Unit | Dimension |
|---|---|---|
| Velocity | m s-1 | [L T-1] |
| Acceleration | m s-2 | [L T-2] |
| Area | m2 | [L2] |
| Volume | m3 | [L3] |
| Density | kg m-3 | [M L-3] |
| Force | newton (N) | [M L T-2] |
| Energy | joule (J) | [M L2 T-2] |
| Power | watt (W) | [M L2 T-3] |
| Pressure | pascal (Pa) | [M L-1 T-2] |
4. Using dimensional analysis, prove 1N=105dyne and 1J=107 erg
5. What are the limitations of dimensional analysis?
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What are the limitations of dimensional analysis?
- Dimensional analysis cannot determine the numerical (dimensionless) constants in an equation, such as 1/2, 2, π, etc.
- It cannot identify functions involving trigonometric, exponential, or logarithmic terms. For example, sinθ, ex, log x cannot be derived using dimensions.
- It cannot be used when the physical quantity depends on more than three variables, because we can form only a limited number of independent dimensional equations.
- Dimensional analysis cannot distinguish between quantities that share the same dimensions (e.g., force and weight, or torque and energy).
- Equations that are dimensionally correct may still be physically incorrect.
- It cannot be applied when a physical quantity is the sum or difference of two or more terms (e.g., s = ut + ½at² includes addition of different terms).
6. If the energy (E), velocity (v) and force (F) be taken as fundamental quantities, then the dimensions of mass will be:
(a) Fv-2
(b) Fv-1
(c) Ev-2
(d) Ev2
If the energy (E), velocity (v) and force (F) be taken as fundamental quantities, then the dimensions of mass will be: Solution:Show Answer
(a) Fv-2
(b) Fv-1
(c) Ev-2
(d) Ev2
We know that Energy (E) = Force (F) × distance, and distance = velocity × time.
Therefore, E = F v t ⇒ t = E / (Fv).
Also, Force F = mass × acceleration = m (v / t) ⇒ m = F t / v.
Substituting t = E / (Fv), we get:
m = (F × E / (Fv)) / v = E / v2.
Hence, the dimension of mass = E v-2.
7. Taking force F, length L and time T to be the fundamental quantities, find the dimensions of (a) mass, (b) density, (c) pressure, (d) momentum, (e) energy
Solution: Given: (a) Mass: (b) Density: (c) Pressure: (d) Momentum: (e) Energy: Final Answers:Show Answer
Force F = mass × acceleration = m × (L T-2)
⟹ m = F L-1 T2
[mass] = F L-1 T2
Density = mass / volume = (F L-1 T2) / L3
⟹ [density] = F L-4 T2
Pressure = Force / Area = F / L2
⟹ [pressure] = F L-2
Momentum = mass × velocity = (F L-1 T2) × (L T-1)
⟹ [momentum] = F T
Energy = Force × distance = F × L
⟹ [energy] = F L
(a) [M] = F L-1 T2
(b) [ρ] = F L-4 T2
(c) [P] = F L-2
(d) [p] = F T
(e) [E] = F L
Some Numericals on Unit and Dimension
1.
The velocity v of a particle at time t is given by
v = a t + b / (t + c),
where a, b and c are constants. The dimensions of a, b and c are respectively:
(a) L2, T and L T2
(b) L T2, L T and L
(c) L, L T and T2
(d) L T-2, L and T
Solution:
We know that velocity (v) has the dimension of L T-1. First term: a t → [a][t] = L T-1 Second term: b / (t + c) Also, since (t + c) is an addition, [c] must have the same dimension as t. Final Answers: Correct option: (d) L T-2, L and TShow Answer
Given: v = a t + b / (t + c)
Since both terms on the right-hand side have the dimension of velocity,
the dimension of each term must be the same as v.
⟹ [a] = L T-2
⟹ [b] / [t] = L T-1
⟹ [b] = L T0 × T = L
⟹ [b] = L
⟹ [c] = T
[a] = L T-2, [b] = L, [c] = T
2.
If x = a t + b t2, where x is the distance travelled by the body in kilometers while t is the time in seconds, then the unit of b is:
(a) km/s
(b) km·s
(c) km/s2
(d) km·s2
Solution:
Given: x = a t + b t2
The two terms on the right-hand side must have the same unit as x (i.e., km).
For the first term: a t → unit of a × s = km
For the second term: b t2 → unit of b × s2 = km Final Answer: Correct option: (c) km/s2Show Answer
Here, x → distance → unit = kilometer (km)
t → time → unit = second (s)
⟹ unit of a = km/s
⟹ unit of b = km/s2
The unit of b is km/s2.
3. A force is given by F = a t + b t2 where t is the time. The dimensions of a and b are:
(a) MLT-4 and MLT
(b) MLT-1 and MLT0
(c) MLT-3 and MLT-4
(d) MLT-3 and MLT0
Solution:
Given F = a t + b t2
For the first term: a t → [a][t] = M L T-2
For the second term: b t2 → [b][t2] = M L T-2 Final Answer: Correct option: (c)Show Answer
The dimension of Force [F] = M L T-2.
⟹ [a] = M L T-3
⟹ [b] = M L T-4
Dimensions of a and b are MLT-3 and MLT-4.
4. For one mole of gas, Van der Waals’ equation is (P + a / V2)(V – b) = nRT.
Find the dimensions of a and b.
Since terms added together must have the same dimensions, the term We know: Also, since Final answer:Show Answer
Solution
a / V2 must have the same dimensions as pressure P.[P] = M L-1 T-2
[V] = L3
So,
[a / V2] = [P]
⇒ [a] = [P] · [V]2
⇒ [a] = (M L-1 T-2) · (L3)2
⇒ [a] = M L5 T-2
V - b is a difference, b must have same dimensions as V:[b] = [V] = L3
[a] = M L5 T-2[b] = L3
