## Chapter 1 Topic B Dimension Synopsis

Dimension of physical quantity is the power or index to which fundamental units like mass, length, time, etc are raised to express that quantity.

The dimensional formula is the relationship between a physical quantity and the dimension of its fundamental units.

**Example:** Dimensional formula of force = MLT^{-2}.

Dimensional equation is the equation by which any physical quantity is expressed in terms of dimensional formula.

**Example:** Dimensional equation of force is [F]= MLT^{-2}.

1. The value of a constant in a dimensional equation can not be determined by using dimensional analysis.

**Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment**

**Example:** The time period of a simple pendulum is \(T=k \sqrt{\frac{1}{g}}\)

Where l is length, g is acceleration due to gravity and k is a constant.

We can not determine the value of the constant k by using dimensional analysis.

2. The relation containing a constant which is not dimensionless can not be established by dimensional analysis.

**Example:** We can not find the law of mm2 from dimensional gravitation \(F=G \frac{m_1 m_2}{d^2}\) analysis because G is not dimensionless.

## Chapter 1 Topic B Dimension Short And Long Answer Type Questions

**Question 1. What do you mean by the dimension of a physical quantity? What is dimensional formula?**

**Answer:**

- Dimension of a physical quantity is the power or index to which fundamental units like mass, length, time, etc are raised to express that quantity.
- Dimensional formula is the relationship between a physical quantity and the dimension of its fundamental units.

**Question 2. What is dimensional equation?**

**Answer:**

Dimensional equation is the equation by which any physical quantity is expressed in terms of dimensional formula.

**Question 3. How are the dimensional formulae of fundamental physical quantities expressed**

**Answer:**

There are seven fundamental units in SI. These are length, mass, time, temperature, electric current, luminous intensity, and quantity of matter, units of which are considered as fundamental units.

These dimensional formulas are denoted as L, M, T, Θ, I, J, and N, respectively.

**Question 4. Write down the dimensional formula and dimensional equation of velocity.**

**Answer:**

Velocity is the rate of change of displacement with time.

∴ \(\text { unit of velocity }=\frac{\text { unit of displacement }}{\text { unit of time }}\)

= \(\text { unit of length } \times(\text { unit of time })^{-1}\)

∴ dimension of velocity is 1 in length and -1 in time.

Dimensional formula of speed is LT^{-1}.

If velocity is expressed by v, then dimensional equation of velocity is [v] = LT^{-1}.

**Question 5. Write down the dimensional formula and dimensional equation of force.**

**Answer:**

Force = mass x acceleration

∴ unit of force

= unit of mass x unit of acceleration

= \(\text { unit of mass } \times \frac{\text { unit of length }}{(\text { unit of time })^2}\)

= \(\text { unit of mass } \times \text { unit of length }\) x \(\times(\text { unit of time })^{-2}\)

∴ dimension of force is 1 in mass; 1 in length and -2 in time.

Dimensional formula of force is MLT^{-2}.

If force is expressed by F, then dimensional equation of force is [F] = MLT^{-2}.

**Question 6. Write down the dimensional formula and dimensional equation of a physical quantity x whose dimension is 0 in mass, 1 in length, and -2 in time.**

**Answer: **

Since the dimension of the quantity is 0 in mass, 1 in length, and -2 in time, the dimensional formula of the quantity is M^{0}LT^{-2}.

Dimensional equation of the quantity is [x] = M^{0}LT^{-2}.

**Question 7. The dimensional equation of a physical quantity x is [x] = M ^{-2}L^{3}T^{-1}. Write down the dimension and the dimensional formula of the quantity.**

**Answer:**

Here, [x] = M^{-2}L^{3}T^{-1}.

∴ Dimension of the quantity is -2 in mass, 3 in length, and -1 in time.

Dimensional formula of the quantity is M^{-2}L^{3}T^{-1}

**Question 8. Show that plane angle is a dimensionless physical quantity.**

**Answer:**

Magnitude of an angle in radian

\((\theta)=\frac{\text { length of arc }(s)}{\text { radius of the circle }(r)}\)∴ Dimentional formula of angle

= \(\frac{\text { Dimensional formula of length of arc }}{\text { Dimensional formula of radius }}\)

= \(\frac{L}{L}=L^0\)

∴ Plane angle is a dimensionless physical quantity.

## Chapter 1 Topic B Dimension Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which of the following quantities has a unit but no dimension?

- Strain
- Atomic weight
- Angle
- None of the above

Answer: 3. Angle

Question 2. Dimensional formula of force is

- MLT
^{-3} - M
^{2}LT^{-2} - ML
^{-1}T^{-2} - MLT
^{-2}

Answer: 4. MLT^{-2}

Question 3. Solid angle has

- Both dimension and unit
- Only dimension but no unit
- Only unit but no dimension
- Neither dimension nor unit

Answer: 3. Only unit but no dimension

Question 4. Strain has

- Both dimension and unit
- Only dimension but no unit
- Only unit but no dimension
- Neither dimension nor unit

Answer: 4. Neither dimension nor unit

Question 5. Dimension of time in the dimensional formula of power is

- -1
- -2
- -3
- -4

Answer: 3. -3

Question 6. Dimension of absolute humidity is ML^{-3}. It is similar to the dimension of

- Mass
- Density
- Specific gravity
- Area

Answer: 2. Density

Question 7. Dimensional formula of a dimensionless physical quantity is

- M
^{0}L^{0}T^{0} - MLT
- M
^{2}LT^{-2} - None of these

Answer: 1. M^{0}L^{0}T^{0}

Question 8. Dimensional formula of weight is

- MLT
^{-2} - ML
^{-1}T^{-2} - M
^{-1}LT^{-2} - MLT
^{-3}

Answer: 1. MLT^{-2}

Question 9. Which two physical quantities of the following have same dimensional formula?

- Velocity, speed
- Displacement, work done
- Force, momentum
- Velocity, acceleration

Answer: 1. Velocity, speed

Question 10. Dimensionless physical quantity is

- Mass
- Specific heat
- Weight
- Atomic weight

Answer: 4. Atomic weight

Question 11. ML^{-1}T^{-2} is dimensional formula of

- Acceleration
- Density
- Force
- Pressure

Answer: 4. Pressure

Question 12. Dimensional formula of surface tension is

- MLT
^{-2} - MT
^{-2} - MT
^{-1} - LT
^{-1}

Answer: 2. MT^{-2}

Question 13. In A=B+C equation

- Dimension of A and C are same but that of B is different
- Dimension of A, B, and C are equal
- A, B, and C all have different dimension
- Dimension of a and b are same but that of c is different

Answer: 2. Dimension of A, B, and C are equal

Question 14. in (P+a/V^{2}) (V-b) = RT equation dimensional formula is

- MLT
^{-2} - ML
^{2}T^{-1} - ML
^{5}T^{-2} - ML
^{3}T^{-2}

Answer: 3. ML^{5}T^{-2}

## Chapter 1 Topic B Dimension Answer In Brief

**Question 1. Give example of two physical quantities which do not have any dimension or unit.**

**Answer:** Atomic mass and specific gravity are two physical quantities which do not have any dimension or unit.

**Question 2. Give an example of a dimensionless physical quantity which has unit.**

**Answer:** Angle is a physical quantity which is dimensionless but has unit (radian).

**Question 3. Write dimension of acceleration.**

**Answer:** Acceleration has dimension 1 in length, 0 in mass, and -2 in time.

**Question 4. Give an example of a physical quantity which have unit but no dimension.**

**Answer:** Angle is such a physical quantity which have unit (radian) but no dimension.

**Question 5. Give example of two physical quantities which have no unit and dimension.**

**Answer:** Atomic weight and specific density are two such physical quantities which have no unit and dimension.

**Question 6. Write dimensional formula of velocity.**

**Answer:** Dimensional formula of velocity is LT^{-1}.

**Question 7. Which physical quantity have dimensional formula ML ^{-1}T^{-2}?**

**Answer:** Pressure is a physical quantity which have dimensional formula ML^{-1}T^{-2}. (strain also have dimensional formula ML^{-1}T^{-2}).

**Question 8. A physical quantity have unit °F ^{-1}. Write dimensional formula of the physical quantity.**

**Answer:** Dimensional formula of the physical quantity is K^{-1}.

## Chapter 1 Topic B Dimension Fill In The Blanks

Question 1. _____________ of two sides of a correct equation or relation is always same.

Answer: Dimension

Question 2. Angle is a ____________ physical quantity.

Answer: Dimensionless

Question 3. Nuclear density is a dimensionless quantity. Its dimensional formula is __________

Answer: M^{0}L^{0}T^{0}

Question 4. All the terms on the two sides of a physical equation must have the _______ dimension.

Answer: Same

Question 5. Dimension of atomic weight is _______

Answer: M^{0}L^{0}T^{0}

Question 6. _________ is such a physical quantity whose dimension is T^{-1}

Answer: Frequency

Question 7. Pressure and _________ both have the dimensional formula.

Answer: Stress

## Chapter 1 Topic B Dimension State Whether True Or False

Question 1. Dimensional formula is the relationship between a physical quantity and the dimension of its fundamental units.

Answer: True

Question 2. Dimensional formula of kinetic energy is ML^{2}T^{-2}.

Answer: True

## Chapter 1 Topic B Dimension Numerical Examples

**Key Information:**

In any mathematical equation involving physical quantities, each term on either side of the equation must have the same dimension.

**Question 1. The two arms of the balance beam of a common balance are equal but the masses of the two scale pans are different. When a body is weighed first in the left pan and then in the right pan, 10 g and 10.2 g are obtained respectively as masses. What is the real mass of the body?**

**Answer: **

If the masses of the body are m_{1} and m_{2} in the two cases, m_{1} = 10g and m_{2} = 10g.

∴ Real mass of the body,

\(m=\frac{m_1+m_2}{2}=\frac{10 \mathrm{~g}+10.2 \mathrm{~g}}{2}=10.1 \mathrm{~g}\)**Question 2. In P = \(\frac{W}{t}\) equation p, W, and t are power, work done, and time. Find dimensional formula of P.**

**Answer:**

Here P = \(\frac{W}{t}\)

∴ Dimensional formula of P is

\([P]=\left[\frac{W}{t}\right]=\frac{M L^2 \mathrm{~T}^{-2}}{\mathrm{~T}}=\mathrm{ML}^2 \mathrm{~T}^{-3}\)**Question 3. Buoyancy of a body is equal to the weight of the displaced liquid by the body. By using this relation find dimensional formula of buoyancy.**

**Answer:**

Buoyancy = weight of the displaced liquid

= v.ρ.g[where v is the volume of the displaced liquid of density ρ and g is acceleration due to gravity]

∴ Dimensional formula of buoyancy = [vρg]=L^{3}.ML^{-3}.LT^{-2} = MLT^{-2}

**Question 4. Velocity of a particle is \(v=a t^2+\frac{b}{t+c}\) where t is time. Find dimensional formula of a, b, and c.**

**Answer:**

Here, the equation is \(v=a t^2+\frac{b}{t+c}\)

∴ Dimensional formula of the left hand side is [v] = LT^{-1}

Now [at^{2}] = [a]T^{+2} = LT^{-1}

∴ [a] = LT^{-3}

Again, \(\frac{[b]}{[t]+[c]}=\mathrm{LT}^{-1} \text { or, } \frac{[b]}{\mathrm{T}}=\mathrm{LT}^{-1}\)

∴ Dimensions of a, b, and c are LT^{-3}, L, and T.

**Question 5. Vander Waals equation is \(\left(p+\frac{a}{v^2}\right)(v-b)\) = RT. Find dimension of a and b.**

**Answer:**

\(\frac{a}{v^2}\) is added to pressure. So its dimension would be the same as that of pressure P.

∴ \({[P]=\left[\frac{a}{v^2}\right] \text { or, } \mathrm{ML}^{-1} \mathrm{~T}^{-2}=\frac{[a]}{\left(\mathrm{L}^3\right)^2}}\)

∴ \({[a]=M L^{6-1} \mathrm{~T}^{-2}=\mathrm{ML}^5 \mathrm{~T}^{-2}}\)

Again b is subtracted from v. So its dimension would be the same as that of volume v.

∴ [v] = [b]

or, [b] = L^{3}

**Question 6. In \(s=\frac{1}{2} a t^2\) equation, s, a and t are displacement, acceleration, and time. By using dimensional analysis check whether the equation is correct or not.**

**Answer:**

The equation is \(s=\frac{1}{2} a t^2\)

∴ Dimension of the left hand side is = [s] = L and dimension of the right-hand side is

= \(\left[\frac{1}{2} a t^2\right]=\mathrm{LT}^{-2} \cdot \mathrm{T}^2=\mathrm{L}\)

∴ Dimension of the L.H.S= dimension of the R.H.S.

∴ The equation is dimensionally correct.

**Question 7. In the K = It + mt ^{2} + nt^{3} equation t and K are expressed in second and m/s respectively. Find SI unit of I and n, where l, m, and n are three physical quantities.**

**Answer:**

According to the principle of dimensional homogeneity, in any mathematical expression or equation involving physical quantities, each term on either side of the equation must have the same dimension.

Here, dimension of K is = [K] = LT^{-1}

∴ Dimension of It is [It] = [K] or, [I]T=LT^{-1}

∴ [l] = LT^{-2}

∴ Unit of in SI is m · s^{-2}.

Again, dimension of nt^{3} is [nt^{3}] = [n]T^{3} = LT^{-1}

∴ \([n]=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^3}=\mathrm{LT}^{-4}\)

∴ Unit of n in Sl is m · s^{-4}.

**Question 8. Dimensional formula of three physical quantities A, B, and C are MLT ^{-2}, ML^{2}T^{-3}, and LT^{-1} respectively. Show that the equation A = \(\frac{B}{C}\) is dimensionally correct.**

**Answer:** For the given equation, dimensional formula of the left hand side is

[A] = MLT^{-2} and dimensional formula of the right-hand side is

= \(\left[\frac{B}{C}\right]=\frac{\mathrm{ML}^2 \mathrm{~T}^{-3}}{\mathrm{LT}^{-1}}=\mathrm{MLT}^{-2}\)

∴ \({[A]=\left[\frac{B}{C}\right] . \quad therefore[A]=\left[\frac{B}{C}\right]}\) equation is dimensionally correct.

**Question 9. Time period of a simple pendulum is T = \(2 \pi \sqrt{\frac{1}{g}}\), where l is effective length of the pendulum and g is acceleration due to gravity. Using dimensional analysis verify whether the equation is correct or not.**

**Answer: **

Here time period of a simple pendulum is T = \(2 \pi \sqrt{\frac{1}{g}}\)

∴ Dimension of the left hand side of the equation is [T] = T and dimension of the right hand side is

= \(2 \pi \sqrt{\frac{1}{g}}=\left[L^{\frac{1}{2}} \cdot L^{-\frac{1}{2}} T^{+\frac{2}{2}}\right] \text {, as }[g]=L T^{-2}\)

= \(L^{\frac{1}{2}}-\frac{1}{2} \cdot T^{+1}=T\)

∴ T= \(2 \pi \sqrt{\frac{1}{g}}\) equation is dimensionally correct.