Important Short Questions With Answers – Physical Quantities | Class 11 Physics

Q.1 Find the dimensional formula of Power (P).

Solution:

Power, P = $\frac{Work\text{ }done}{time}$

$\therefore$ Dimensional formula of power = $\frac{[ML{}^{2}{{T}^{-2}}]}{[T]}$ = [ML²T ^-3]

 Note: The dimensional formula of power is [ML²T ^-3] and dimensions of specific heat capacity are 0 in mass, 2 in length and –2 in time.

Q.2 Find the dimensional formula of specific heat capacity (S).

Solution:

Heat loss/gain, dQ = $mS\Delta \theta $

Or,   S = $\frac{dQ}{m\Delta \theta }$

$\therefore$ Dimensional formula of specific heat capacity = $\frac{[M{{L}^{2}}{{T}^{-2}}]}{[M][K]}$ = [M⁰L²T ^-2K^-1]

Note: The dimensional formula of specific heat capacity is [M⁰L²T ^-2K^–1] and dimensions of specific heat capacity are 0 in mass, 2 in length, –2 in time and –1 in temperature.

Q.3 Find the dimensional formula of Pressure (P).

Solution:

Pressure, P =  $\frac{Force}{area}$ ,

$\therefore$ Dimensional formula of pressure = $\frac{[ML{{T}^{-2}}]}{[{{L}^{2}}]}$ = [ML^–1T ^-2]

Q.4 Find the dimensional formula of surface tension (T).  

Solution:

Surface tension, T = $\frac{Force}{length}$

$\therefore$ Dimensional formula of surface tension = $\frac{[ML{{T}^{-2}}]}{[L]}$ = [ML⁰T ^-2]

Q.5 Find the dimensional formula of coefficient of viscosity ($\eta$).

Solution:

Viscous force, F = 6$\pi \eta$rv_t where ‘r’ is radius and ‘v_t’is terminal velocity.

Or, $\eta$ = $\frac{F}{6\pi r{{v}_{t}}}$

$\therefore$ Dimensional formula of coefficient of viscosity ($\eta$) = $\frac{[ML{{T}^{-2}}]}{[L][L{{T}^{-1}}]}$ = [ML^-1T ^-1]

Q.6 Find the dimensional formula of specific latent heat capacity (L).

Solution:

Heat loss/gain, dQ = mL       

Or, L = $\frac{dQ}{m}$

$\therefore$ Dimensional formula of specific latent heat capacity (L) = $\frac{[M{{L}^{2}}{{T}^{-2}}]}{[M]}$ = [M⁰L²T^-2]

Q.7 Find the dimensional formula of gas constant ‘R’ from PV = nRT or PV = RT.

Solution:                      

We have, PV= nRT

R = $\frac{PV}{nT}$                 

= $\frac{F}{A}$$\frac{V}{nT}$

= $\frac{[ML{{T}^{-2}}]{{[L]}^{3}}}{{{[L]}^{2}}[K]}$ [ $\because$ n is a number and has no dimension]

$\therefore$ The dimensional formula of

R = [M L² T ^-2 K^-1]

Q.8 Find the dimensions of Planck’s constant ‘h’ at from where ‘$\lambda$’ is wavelength and ‘p’ is momentum of photon.

Solution:                      

The given equation is $\lambda$ = $\frac{h}{p}$ 

Or, h = $\lambda$×p

$\therefore$ Dimensional formula of h = [L] [MLT ^-1] = [ML²T ^-1]

Q.9 The energy of a photon is given by E = hf. Find the dimension and unit of Plank’s constant h, where f is the frequency of radiation.

Solution:                      

The given relation is E = hf

Or, h = $\frac{E}{f}$ = $\frac{E}{1/T}$ $\because$ frequency, f = $\frac{1}{tim{{e}^{{}}}period}$

$\therefore$ The dimensional formula of Plank’s constant (h) = [ML²T^–2] [T] =[ML²T^–1] 

The dimensions of Plank’s constant (h) are 1 in mass, 2 in length and –1 in time.

Again, h = $\frac{E}{f}$, so the unit of Plank’s constant is Joule sec^–1 ( J/S ).

Q.10 Find the dimensional formula of Gravitational constant (G).

Solution:                      

We have the gravitational force, F = G $\frac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$

[where, m_1, m₂ are the masses of two bodies and d is the distance between them.]

Or, G = $\frac{F{{d}^{2}}}{{{m}_{1}}{{m}_{2}}}$

$\therefore$The dimensional formula of universal gravitational constant (G) = $\frac{[ML{{T}^{-2}}]{{[L]}^{2}}}{[M][M]}$

= [M^–1L³T ^-2]

Q.11 Check the correctness of v² = u² + 2as using dimensional analysis.

Solution:                      

The dimension of L.H.S., v²  = [M⁰LT^-1]²

= [M⁰L²T^-2]

The dimension of R.H.S.,

u² + 2as = [M⁰LT^-1]² + 2 [M⁰LT^-2] [L]

= [M⁰L²T^-2] + 2[M⁰L²T^-2]

= 3[M⁰L²T^-2]

= [M⁰L²T^-2] $\because$ 3 has no dimension.

Here, dimensional formula of L.H.S. = The dimensional formula of R.H.S.

$\therefore$ The given relation v² = u² + 2as is dimensionally correct.

Q.12 Check the correctness of formula T = 2$\pi$$\sqrt{\frac{l}{{g}}}$ using dimensional analysis.

Solution:                      

The dimensional formula of L.H.S., T = [T]

The dimensional formula of R.H.S., 2$\pi$$\sqrt{\frac{l}{g}}$ = $\sqrt{\frac{[L]}{[L{{T}^{-2}}]}}$ $\because$ 2$\pi$ has no dimension.

= $\sqrt{\frac{1}{[{{T}^{-2}}]}}$

= $\sqrt{[{{T}^{2}}]}$

= [T]

Here, dimensional formula of L.H.S. = Dimensional formula of R.H.S.

$\therefore$ The given relation T = 2$\pi$$\sqrt{\frac{l}{g}}$ is dimensionally correct.

Q.13 A student writes $\sqrt{\frac{\mathbf{R}}{\mathbf{2GM}}}$for escape velocity. Check the correctness of the formula by using dimensional method.

Solution:                      

The given formula is escape velocity,

V_e = $\sqrt{\frac{R}{2GM}}$

The dimensional formula of L.H.S.,

V_e = [LT^–1]

The dimensional formula of R.H.S.,

$\sqrt{\frac{R}{2GM}}$ = $\sqrt{\frac{[L]}{[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}][M]}}$ ,  $\sqrt{\frac{1}{2}}$ has no dimension.

= $\sqrt{\frac{[{{T}^{2}}]}{[{{L}^{2}}]}}$

= [L^–1T]

Here, dimensional formula of L.H.S. $\ne$ dimensional formula of R.H.S.

$\therefore$ The given formula for escape velocity, V_e= $\sqrt{\frac{R}{2GM}}$is dimensionally incorrect.

Q.14 Using dimensional analysis, check the correctness of formula t = 2$\pi$$\sqrt{\frac{{m}}{{k}}}$, where ‘t’ be the time period, ‘m’ is the mass and ‘k’ is the force per unit displacement.

Solution:                      

The dimensional formula of L.H.S., t = [T]

The dimensional formula of R.H.S., 2$\pi$$\sqrt{\frac{m}{k}}$ =  2$\pi$$\sqrt{\frac{m}{F/l}}$

= $\sqrt{\frac{[M][L]}{[ML{{T}^{-2}}]}}$ ($\because$2$\pi$ has no dimension.)

= $\sqrt{\frac{1}{[{{T}^{-2}}]}}$

= $\sqrt{[{{T}^{2}}]}$

= [T]

Here, dimensional formula of L.H.S. = Dimensional formula of R.H.S.

$\therefore$ The given relation t = 2$\pi$$\sqrt{\frac{m}{k}}$ is dimensionally correct.

Q.15 Using dimensional analysis, check the correctness of the relation h = $\frac{\mathbf{2Tcos}\theta }{r\rho g}$, where ‘h’ is height, ‘T’ is surface tension, ‘r’ is the radius and ‘${\rho }$’ is density of liquid.

Solution:                      

The dimensional formula of L.H.S., h = [L]

The dimensional formula of R.H.S.,$\frac{2Tcos\theta }{r\rho g}$ = $\frac{[M{{L}^{0}}{{T}^{-2}}]}{[L][M{{L}^{-3}}][L{{T}^{-2}}]} $here 2 and cos$\theta$ have no dimension.

= $\frac{1}{[{{L}^{-1}}]}$

= [L]

Here, dimensional formula of L.H.S. = Dimensional formula of R.H.S.

$\therefore$ The given relation, h = $\frac{2Tcos\theta }{r\rho g}$ is dimensionally correct.

Q.16 In one of the printed documents the unit of universal gravitational constant is given as Nmkg^–2. Check its correctness from dimensional analysis.

Solution:                      

We have the gravitational force, F = G $\frac{{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$, where m_1, m₂ are the masses of two bodies and ‘d’ is the distance between them.

Or, G = $\frac{F{{d}^{2}}}{{{m}_{1}}{{m}_{2}}}$

$\therefore$ The dimensional formula of universal gravitational constant (G) = $\frac{[ML{{T}^{-2}}]{{[L]}^{2}}}{[M][M]}$          

= [M^–1L³T^–2]

But, the dimensional formula of given unit Nmkg^–2

= [MLT^–2] [L] [M^–2]

= [M^–1L²T^–2]

Here, the dimensional formula of G $\ne$the dimensional formula of given unit.

So, the given unit is not the unit of G.

Q.17 A student writes an expression for the momentum (p) of a body of mass (m) with total energy (E) and considering the duration of time (t) as p =$\sqrt{\frac{{2mE}}{{t}}}$ . Check its correctness using dimensional analysis.

Solution:                      

The given relation is p = $\sqrt{\frac{2mE}{t}}$

The dimensional formula of L.H.S, p = mv = [MLT^-1]

Dimensional formula of R,H.S, $\sqrt{\frac{2mE}{t}}$= $\sqrt{\frac{[M][M{{L}^{2}}{{T}^{-2}}]}{[T]}}$       [$\because$2 has no dimension.]

= $\sqrt{\frac{[{{M}^{2}}{{L}^{2}}{{T}^{-2}}]}{[T]}}$

= $\frac{[ML{{T}^{-1}}]}{[{{T}^{1/2}}]}$

= [MLT ^{-1 -1/2}]

= [MLT ^-3/2]

Here, the dimensional formula of L.H.S $\ne$ the dimensional formula of R.H.S.

$\therefore$ The given expression p =$\sqrt{\frac{2mE}{t}}$ is dimensionally incorrect.

Q.18 Check dimensionally the correctness of the Stoke’s formula, F = 6 $\pi \eta$rv, where symbols have their usual meanings. Ans: Correct

Solution:                      

The given formula is F = 6$\pi \eta$rv

The dimensional formula of L.H.S.,

F = [MLT ^-2]

The dimensional formula of R.H.S.,

6$\pi \eta$rv =  [ML^-1T ^-1] [L] [LT ^-1]

[$\because$ 6$\pi$ has no dimension.]

= [MLT ^-2]

Here, the dimensional formula of L.H.S = the dimensional formula of R.H.S.

$\therefore$ The given formula F = 6$\pi \eta$rv is dimensionally correct.

Q.19 Convert 10 ergs into joules using dimensional analysis.

Solution:                      

Here, ‘ergs’ and ‘joules’ are the units of energy in CGS and SI system.

The dimensional formula of energy = [ML²T^-2]

Given system (CGS)	New system (SI)
n1 = 10	n2 = ?
M1 = 1g	M2 = 1kg
L1 = 1cm	L2 = 1m
T1 = 1 sec	T2 = 1 sec

We have, n₂ = n_{1 ${{\left[ \frac{{{M}_{1}}}{{{M}_{2}}} \right]}^{a}}$${{\left[ \frac{{{L}_{1}}}{{{L}_{2}}} \right]}^{b}}$${{\left[ \frac{{{T}_{1}}}{{{T}_{2}}} \right]}^{c}}$}

where,  a =1 , b = 2 & c = –2 $\because$ dimensional formula of energy, E = [ML²T^-2]

$\therefore$  n₂ = 10  _{${{\left[ \frac{1g}{1kg} \right]}^{1}}$${{\left[ \frac{1cm}{1m} \right]}^{2}}$${{\left[ \frac{1sec}{1sec} \right]}^{-2}}$}

Or, n₂ = 10 ${{\left[ \frac{1g}{1000g} \right]}^{1}}$${{\left[ \frac{1cm}{100cm} \right]}^{2}}$[1]^–2

Or,   n₂  = 10×$\frac{1}{1000}$×$\frac{1}{10000}$

Or, n₂ = 1×10^–6

Since, n₁u₁ = n₂u₂

$\therefore$ 10 ergs = 10^–6 Joules.

Q.20 The density of gold is 19.3 gcm^-3.  Express this value in SI unit using dimensional analysis.

Solution:                                   

Here, gcm^-3 is the unit of density in CGS system.

The dimensional formula of density

= [ML^–3T⁰]

Given system (CGS)	New system (SI)
n1 = 19.3	n2 = ?
M1 = 1g	M2 = 1kg
L1 = 1cm	L2 = 1m
T1 = 1 sec	T2 = 1 sec

We have, n₂ = n_{1 ${{\left[ \frac{{{M}_{1}}}{{{M}_{2}}} \right]}^{a}}$${{\left[ \frac{{{L}_{1}}}{{{L}_{2}}} \right]}^{b}}$${{\left[ \frac{{{T}_{1}}}{{{T}_{2}}} \right]}^{c}}$}

where,  a =1 , b = –3 & c = 0 $\because$ dimensional formula of density = [ML³T⁰]

$\therefore$  n₂ = 19.3  _{${{\left[ \frac{1g}{1kg} \right]}^{1}}$${{\left[ \frac{1cm}{1m} \right]}^{-3}}$${{\left[ \frac{1sec}{1sec} \right]}^{0}}$}

Or, n₂ = 19.3 ${{\left[ \frac{1g}{1000g} \right]}^{1}}$${{\left[ \frac{1cm}{100cm} \right]}^{-3}}$×1

Or, n₂  = 19.3×$\frac{1}{1000}$×1000000

Or, n₂ = 19.3×1000 = 19300

Since, n₁u₁ = n₂u₂

$\therefore$ 19.3 gcm^-3 = 19300 kgm^-3

Q.21 Is dimensionally correct relation be always a correct physical relation? Justify your answer. What about dimensionally wrong equation?

Solution:                      

No, the dimensionally correct relation may not be a correct physical relation.

For eg. v² = u² + 5as

The given relation is dimensionally correct but physically incorrect. But dimensionally wrong relation can never be a correct physical relation.

For eg. v = u + at²

The given relation v = u + at² is both dimensionally and physically incorrect.

Detailed answer of above question:

Solution:  No, the dimensionally correct relation may not be a correct physical relation.

For eg. v² = u² + 5as

The dimension of L.H.S., v²  = [M⁰LT^-1]² = [M⁰L²T^-2]

The dimension of R.H.S., u² + 2as

= [M⁰LT^-1]² + 2 [M⁰LT^-2] [L]

= [M⁰L²T^-2] + 2[M⁰L²T^-2]

= 3[M⁰L²T^-2]

= [M⁰L²T^-2]

The given relation is dimensionally correct but physically incorrect. But dimensionally wrong relation can never be a correct physical relation.

For eg. v = u + at²

The dimension of L.H.S. = [M⁰LT^-1]

The dimension of R.H.S.

= [M⁰LT^-1] + [M⁰LT^-2] [T]²

= [M⁰LT^-1] + [M⁰LT⁰]

The given relation v = u + at² is dimensionally incorrect and physically incorrect as well.

Q.22 What are the limitations of dimensional analysis?

Solution:                      

Following are the limitations of dimensional analysis:

1. It does not give information about a physical quantity whether it is vector or scalar quantity.

2. It gives no information about dimensionless constant.

3. It is not applicable to derive a formula having more than one term either in L.H.S. or in R.H.S.

4. It cannot differentiate between two different physical quantities having same dimensions.

5. It is not applicable to trigonometric function, logarithmic function, and exponential function etc.

Q.23 Taking force length and time as fundamental quantity find the dimensional formula of density and gravitational constant. 

Solution:                      

According to the question, taking force, length and time as fundamental quantity, the dimensional formula of a physical quantity can be expressed as [F^a L^b T^c] instead of [M^a L^b T^c].

Now,

(i) Density, $\rho$ =$\frac{m}{v}$ $\because$ F = ma $\therefore$ m = $\frac{F}{a}$

$\rho$ = $\frac{F}{a}$× $\frac{1}{v}$

$\therefore$ The dimensional formula of density, $\rho$ = $\frac{[F]}{[L{{T}^{-2}}]{{[L]}^{3}}}$= [FL^-4T²]

(ii) Again F = G

Or, G = $\frac{F{{d}^{2}}}{{{m}_{1}}{{m}_{2}}}$= $\frac{F{{d}^{2}}}{F/a\times F/a}$ = $\frac{F\times {{d}^{2}}\times {{a}^{2}}}{F\times F}$

The dimensional formula of Universal Gravitational Constant,

G = $\frac{{{[L]}^{2}}{{[L{{T}^{2}}]}^{2}}}{[F]}$

= [F^-1L⁴T^-4]

Q.24 In Van der Waal’s equation, (P + $\frac{{a}}{{{{V}}^{{2}}}}$) (v–b) = RT, find the dimensional formula for ‘a’ and ‘b’.

Solution:                      

The given Van der Waal’s equation is

(P + $\frac{a}{{{V}^{2}}}$) (v–b) = RT

The dimensional formula of  $\frac{a}{{{V}^{2}}}$ = The dimensional formula of  ‘P’

Or, The dimensional formula of ‘a’ = The dimensional formula of P × V²

Or, The dimensional formula of ‘a’ = The dimensional formula of $\frac{F}{A}$ × V²

$\therefore$ The dimensional formula of ‘a’ = $\frac{F}{a}$ $\frac{[ML{{T}^{-2}}]}{{{[L]}^{2}}}$× [L³]² = [ML⁵T^-2]

Similarly,

The dimensional formula of ‘b’ = The dimensional formula of ‘V’

$\therefore$ The dimensional formula of ‘b’ = [L³]

= [M⁰L³T⁰]

Q.25 The force ‘F’ is given in terms of time ‘t’ and displacement ‘x’ by equation F = AsinBx + CsinDt. What is the dimension of  $\frac{\mathbf{D}}{\mathbf{B}}$?

Solution:                      

We know angle has no dimension i.e. [M⁰L⁰T⁰].

So we can write, 

the dimensional formula of Bx = The dimensional formula of Dt

The dimensional formula of $\frac{D}{B}$ = The dimensional formula of $\frac{x}{t}$

= $\frac{[L]}{[T]}$= [LT^–1] = [M⁰LT^–1]

Alternative method:

We know angle has no dimension i.e. [M⁰L⁰T⁰].

So we can write, 

The dimensional formula of Bx = [M⁰L⁰T⁰]

Or, the dimensional formula of B = $\frac{[{{M}^{0}}{{L}^{0}}{{T}^{0}}]}{dimensional\text{ }formula\text{ }of\text{ x}}$

= $\frac{[{{M}^{0}}{{L}^{0}}{{T}^{0}}]}{[L]}$= [L^–1]

Similarly, 

The dimensional formula of Dt = [M⁰L⁰T⁰]

Or, the dimensional formula of D = $\frac{[{{M}^{0}}{{L}^{0}}{{T}^{0}}]}{dimensional\text{ }formula\text{ }of\text{ }t}$

= $\frac{[{{M}^{0}}{{L}^{0}}{{T}^{0}}]}{[T]}$= [T ^–1]

Now,

The dimensional formula of $\frac{D}{B}$ = [LT ^–1]

= [M⁰LT ^–1]

Q.26 The displacement ‘y’ is expressed as y = a +bt + ct² + dt³, where ‘t’ is time. Find the dimensions of a, b, c and d. 

Answer:                        

According to the principle of homogeneity, for any correct physical relation, the dimensions of each term on L.H.S. are equal to the dimensions of each term on R.H.S.

So, dimensional formula of ‘a’ = dimensional formula of ‘y’

$\therefore$ Dimensional formula of a = [L]

Similarly, dimensional formula of bt = dimensional formula of y

Or, dimensional formula of b = $\frac{dimensional\text{ }formula\text{ }of\text{ }y}{dimensional\text{ }formula\text{ }of\text{ }t}$

$\therefore$ The dimensional formula of b =$\frac{[L]}{[T]}$

= [LT^–1] = [M⁰LT^–1]

Also, dimensional formula of ct² = dimensional formula of y

Or, dimensional formula of c = $\frac{dimensional\text{ }formula\text{ }of\text{ }y}{dimensional\text{ }formula\text{ }of\text{ }{{t}^{2}}}$

$\therefore$ The dimensional formula of c =$\frac{[L]}{[{{T}^{2}}]}$

= [LT^–2]

= [M⁰LT^–2]

And dimensional formula of dt³ = dimensional formula of y

Or, dimensional formula of d = $\frac{dimensional\text{ }formula\text{ }of\text{ }y}{dimensional\text{ }formula\text{ }of\text{ }{{t}^{3}}}$

$\therefore$ The dimensional formula of d =$\frac{[L]}{[{{T}^{3}}]}$ = [LT^–3]

= [M⁰LT^–3]

Q.27 Differentiate between precise and accurate measurements.

Solution:                      

Precise measurement:

It refers to the closeness of measured values to each other. In other words the limit or resolution to which a physical quantity is measured by the instrument is called precision. It is associated with small random errors.

– Less the least count of an instrument, more is the preciseness in the measurement.

– A precise measurement is not necessarily an accurate measurement.

Accurate measurement:

It refers to the closeness of measured values to the true value. It is associated with small symmetric uncertainties.

– Good accuracy means the reading or mean of set of readings is very close to the true value.

– Greater the number of significant figures, greater is the accuracy.

Example: Let us consider a man is 70 kg (true value).

1. The set of measurements 90 kg, 80 kg, 60 kg and 50 kg is accurate but not precise.

2. The set of measurements 85.23kg, 85.25kg, 85.26kg is precise but not accurate.

3. The set of measurements 69.81kg, 69.94kg, 70.00kg, 75.11kg is both accurate and precise.

Q.28 The length of the rod is exactly 1cm. An observer records the readings as 1cm, 1.0cm, 1.00cm, or 1.000cm, which is the most accurate measurement?

Solution:                      

Accuracy increases with increase in number of significant figures. Therefore, 1.000 is the most accurate as it has greatest number of significant figure (four) and the least count of the instrument that measures 1.000cm is less than all others.

Q.29 The diameter of a steel rod is given as 56.47 ± 0.02 mm. What does it mean?

Solution:                      

Here, ± 0.02 mm indicates the least count of the instrument. In other words, the maximum error that can be made by this instrument is ± 0.02 mm.  The diameter of the steel rod is given as 56.47 ± 0.02 mm. This means the true diameter of the rod lies between 56.45 mm and 56.49 mm.

Q.30 What do you mean by significant figure?

Solution:                      

The meaningful digits in a number is called the significant figures. The significant figures in measurement of a physical quantity gives the extent to which measurement is more reliable.

Let us consider 2.51mm is the thickness of a glass plate measured by an instrument of least count 0.01mm, there are 3 significant figures in the above number. This measurement is reliable upto second place after decimal. Greater the number of significant figures, greater is the accuracy. Among the measurements 1cm, 1.0cm, 1.00cm and 1.000cm, the measurement 1.000cm is the most accurate as it has greatest number of significant figure (four) and the least count of the instrument that measures 1.000cm is less than all others.

Q.31 The length of rod is exactly 1 cm. An observer records the readings as 1.0 cm, 1.00 cm, and 1.000 cm, which is the most accurate measurement?

Solution:                      

Accuracy increases with increase in number of significant figures. Therefore, 1.000 is the most accurate as it has greatest number of significant figure (four) and the least count of the instrument that measures 1.000cm is less than all others.

Q.32 A student writes an expression of the force causing a body of mass (m) to move in a circular motion with a velocity (v) as F = mv². Use the dimensional method to check its correctness.

Solution:                      

The given equation is F = mv²

The dimensional formula of L.H.S.,

F = [MLT^–2]

The dimensional formula of R.H.S.,

mv² = [M] [LT^–1]² = [ML²T^–2]

Here, dimensional formula of L.H.S. $\ne$ Dimensional formula of R.H.S.

$\therefore$ The given relation, F = mv² is dimensionally incorrect.

Q.33 Name any two physical quantities which have the same dimensions. Can a quantity have unit but no dimension? Explain.

Solution:                      

This is one of the limitations of dimensional analysis that it cannot differentiate between two different physical quantities having same dimensions i.e. two different physical quantities can have same dimensions. For example; kinetic energy and torque are different physical quantities yet have same dimensions.

Kinetic Energy, K.E.= $\frac{1}{2}$mv²

$\therefore$ Dimensional formula of K.E. = [M] [LT^-1]² = [ML²T^-2]

Torque, $\tau$= Force × perpendicular distance

$\therefore$ Dimensional formula of torque = [MLT^-2] [L] = [ML²T^-2]

Yes, physical quantity can have unit but no dimension. For example, angle.

The unit of angle is radian, degree, grade but no dimension.

Any angle ($\theta$) can be expressed as $\theta$= $\frac{l}{r}$

Dimension of $\theta$ = $\frac{[L]}{[L]}$ = no dimension i.e. [M⁰L⁰T⁰].

Also Read: Physical Quantities Notes Class 11