**Class – 11 Physics, Chapter – 3 : DIMENSION ANALYSIS**

* Dimension :* Dimensions of the derived unit may be defined as the powers to which the fundamental units of the mass, length and time etc must be raised to represent that unit.

Example : speed = distance / time

= [ L ] / [ T ]

= [ M^{0}LT^{-1} ]

Here speed is said to be possessed zero dimension in mass , 1 dimension in length and (-1) dimension in time

Force = mass * acceleration

=[MLT^{-2}]

Here, force is said to possess one dimension in mass , one dimension in length and (-2) dimension in time

**Dimensional formula :** It may be defined as an expression which shows how and which of the fundamental units are required to represent the unit of a physical quantity.

[M^{0}LT^{-1}] is the dimensional formula of speed

[MLT^{-2}] is the dimensional formula of force

[ML^{2}T-^{2}] is the dimensional formula of work

**Dimensional equation :** The equation obtained by equating the physical quantity with its dimensional formula is called as dimensional equation.

Examples :

Speed =[M^{0}LT^{-1}]

Force =[MLT^{-2}]

Work =[ML^{2}T^{-2}]

**The principle of homogeneity of dimensions :**

It states that “ A given physical relation is dimensionally correct if the dimensions of the various terms on either sides of the relation are the same “ .

**Some important points for solving the problems in the board exam and competition exam :**

- The formula of a physical quantity may be different but its dimensions are the same example : area of rectangle = length * breadth =[M
^{0}L^{2}T^{0}]

Area of square =(side)^{2}=[M^{0}L^{2}T^{0}]

Area of circle = ⊓r^{2} = [ M^{0}L^{2}T^{0 }]

Area of triangle = ½ * base * height = [ M^{0}L^{2}T^{0 }]

2) Work, all types of energy, torque, couple, moment of force, all have the same

dimension i.e.[ ML^{2}T^{-2} ]

3) The physical quantity which is the ratio of two similar physical quantities ( as for

example relative density, strain, refractive index, and relative permeability etc.)

has no dimension .

4) Some important relations :

- Latent heat = heat energy /mass
- Charge = current * time
- Capacitance = charge / electric potential
- Resistance = voltage / current
- Specific heat = (heat * energy ) / ( mass * temperature )
- Electric potential = electric work /charge
- Electric intensity =force /charge
- Gas constant = ( pressure * volume ) / temperature

5) some important formula :

- F = IBL ⟹ B = F/IL
- F = GMm/r
^{2 }⟹ G = Fr^{2}/Mm - R = ρL/A ⟹ ρ = RA/L
- F = Qq/4⊓∊r
^{2 }⟹ ∊ = Qq/4⊓r^{2}F

6) Some important physical quantities having same dimension

- [ M
^{0}L^{0}T^{-1}] = frequency, angular frequency, angular velocity, velocity gradient - [ ML
^{-1}T^{-2 }] = pressure ,stress modulus of elasticity energy density - [ ML
^{2}T^{-1}] = angular momentum , planck’s constant - [ MLT
^{-2}] = thrust ,force ,weight ,energy density - [ MLT
^{-1}] = momentum , impulse - [ M
^{0}L^{0}T ] = L/R,(LC)1/2 ,and RC - [ML
^{0}T^{0}] = mass and inertia

Where L= inductance of the coil

R=resistance of the coil

C=capacitance of the coil

**solved examples **

**Question 1: what are the uses of dimensional equation ?**

**Answers :** Uses of dimensional equation are given below .

- Dimensional equation is used to check the dimensional correctness of a given physical relation .
- Dimensional equation is used to convert a physical quantity from one system of unit to another system of unit
- Dimensional equation is used to establish a relationship between different physical quantities.

**Question 2 : Give two drawbacks of dimensional analysis.**

**Answers :** The drawbacks of dimensional analysis are given as

- The method does not tell us where the equation is wrong.
- If the dimension had been the same on each side of the equation, we would know only that it might be correct , for the method does not provide a check on numerical factors .

**Question 3 : The wavelength λ associated with a moving electron depends upon its mass m, its velocity v, and planck’s constant h . prove dimensionally that λ ∝ h/mv**

**( HPSEB 2004 )**

**Solution: **

Let *λ *∝ m^{a} ————————–(1)

*λ *∝ v^{b} —————————(2)

*λ *∝ h^{c} —————————-(3)

On combining all above three equations , w have

*λ *∝ m^{a}h^{c}v^{b}

=km^{a}h^{c}v^{b} ——————————————(4)

Where k is proportionality constant

Writing down dimension on both sides , we get

[ M^{0}LT^{0 }] = k[ M ]^{a}[ LT^{-1 }]^{b}[ ML^{2}T^{-1 }]^{c}

⇒ [ M^{0}LT^{0 }] = k[ M^{a+c}L^{b+2c}T^{-b-c }]

By equating the dimensions on both sides we get

a + c = 0 ⇒ a = – c

– b – c = 0 ⇒ b = – c

And b + 2c = 1

⇒ – c + 2c = 1

⇒ c = 1

a = – 1

b = – 1

Now putting the values of a,b and c in the equation (4)

λ = km^{-1}v^{-1}h^{1}

⇒ λ= kh/mv

⇒ *λ *∝ h/mv

( proved )

**Question 4 :**** If the velocity of sound in a gas depends on its elasticity and density, derive the relation for the velocity of sound in a medium by the method of the dimension .**

**Solution :** Let v, E and p represent the velocity, elasticity and density respectively.

According to problem

v ∝ E^{a}—————————–(1)

v ∝ p^{b}——————————(2)

On combining above two equations (1) &(2) we get

v ∝ E^{a} p^{b}——————————-(3)

⇒ v = kE^{a} p^{b}

Where k is proportionality constant

By writing down dimensions on both sides, we get

[ M^{0}LT^{-1} ] = k[ ML^{-1}T^{-2}]^{a}[ML^{-3}]^{b}

[ M^{0}LT^{-1 }] = k[M^{b+a}L^{-a-3b}T^{-2a}-2a ]

By equating the dimensions on both sides , we get

a + b = 0 ⟹ a = – b

And -2a = – 1 ⟹ a = ½

Now putting a = ½ & b = – ½ in the equation ( 3 ), we get

v ∝ E^{1/2} p^{-1/2}

v ∝ √(p/E)

*The above expression is the required relation.*

**Question 5: Check the dimensional correctness of the relation τ=Iα where is torque , I is the moment of inertia and is the angular acceleration .( PSEB 1990 )**

**Solution :** we have

[ τ ] = [ ML^{2}T^{-2 }]

[ I ] = [ ML^{2} ]

[ α ] = [ T^{-2} ]

LHS = τ

= [ ML^{2}T^{-2} ]

RHS = Iα

=[ ML^{2} ][ T^{-2} ]

= [ ML^{2}T^{-2} ]

Since LHS = RHS

*Therefore , the given equation is dimensionally correct .*

**Question 6 : The value of G in cgs system is 6.67*10 ^{-8} dyn-cm^{2}-g^{-2}. Calculate its value in SI units**

**Solution :**

n_{1 }= 6.67*10^{-8} n_{2}=?

M_{1 }= 1gm M_{2 }= 1kg

L_{1 }= 1 cm L_{2 }= 1m

T_{1 }= 1 sec T_{2 }= 1sec

[ G ] = [ M^{-1}L^{3}T^{-2 }]

n_{2 }= n_{1}( M_{1 }/ M_{2})^{-1 }* ( L_{1 }/ L_{2})^{3 }* ( T_{1 }/ T_{2 })^{-2}

Now, putting the values of M_{1}, M_{2}, L_{1}, L_{2}, T_{1}, T_{2 } and n_{1}, we get

n_{2 }= 6.67 * 10^{-8 }* ( 1 gm /1 kg )^{-1 }* ( 1 cm / 1 m)^{3 }* ( 1 sec / 1 sec)^{-2}

= 6.67 *10^{-8 }* ( gm / 11000 gm)^{-1 }* ( 1 cm / 100 cm)^{3 }* ( 1 )^{-2}

= 6.67 * 10^{-8 }* ( 10^{-3})^{-1} * 10^{-6 }* 1

= 6.67 * 10^{-11}

This the value of G in SI units.

**Question 7 : check the dimensional consistency in the following relation **

**s = ut + 1 / 2at ^{2}**

**Where symbols have their usual meaning.**

**Solution : **

[ S ] = [ L ]

[ u ] = [ LT^{-1} ]

[ t ] = [ T ]

[ a ] = [ LT^{-2} ]

Putting the dimensional formulae of all the terms in the LHS & RHS respectively.

LHS = S

= [ L ]

RHS = ut + 1 / 2at2

= [ LT^{-1} ] *[ T ] + [ LT^{-2} ] * [ T^{2} ]

= [ L ] + [ L ]

= [ L ]

*The given equation is dimensionally correct.*

**Question 8: Consider the equation 1/2mv ^{2}=mgh. Here m is the mass of the body, v is the velocity, g is the acceleration due to gravity and h is the height. Check the consistency of the this equation .**

**Solution **: 1 / 2mv^{2} = mgh

Putting the dimensional formulae on LHS & LHS respectively

LHS = 1 / 2mv^{2}

= [ M ][ LT^{-1} ]^{2}

= [ ML^{2}T^{-2} ]

RHS = mgh

= [ M ][ LT^{-2} ][ L ]

= [ ML^{2}T^{-2} ]

*Hence, it is dimensionally correct.*

**Question 9 : Find the value of force of 100 dynes on a system based on meter ,kg and minute as fundamental units.**

**Solution :**

n_{1} = 100 dynes n_{2} =?

M_{1 }= 1gm M_{2} = 1kg

L_{1} = 1 cm L_{2} = 1m

T_{1} = 1 sec T_{2} = 1 minute

[ F ] = [ MLT^{-2} ]

n_{2} = n_{1}( M_{1}/M_{2 }) * ( L_{1} / L_{2} ) * ( T_{1 }/ T_{2} )^{-2}

= 100 ( 1 gm / 1 kg) * ( 1 cm / 1 m) * ( 1 sec / 1 minute)^{-2}

= 100(1 gm /1000 gm) * (1 cm / 100 cm) * (1 sec / 60 sec)^{-2}

= 100 * 10^{-3} * 10^{-2} * 60 * 60

= 3.6 new units

This is the required value of 100 dynes in the given system of units.

**Question 10 : If the escape velocity depends on acceleration due to gravity of the planet and the radius of the planet, establish dimensional relation between them.**

**Solution :**

Let v, g and r represent the escape velocity, acceleration due to gravity and the radius of the planet.

v ∝ g^{a}

v ∝ r^{b}

On combining the above two equations , we get

v ∝ g^{a}r^{b}—————————–(1)

v = kg^{a}r^{b}

Where k is proportionality constant

Putting down the dimensional formulae of each term on both sides we get

[ LT^{-1 }] = k[ LT^{-2} ]^{a}[ L ]^{b}

[ LT^{-1} ] = k[ L^{2b} + bT^{-2a} ]

By equating the dimensions on both sides we get

-2a = – 1 ⟹ a = 1/ 2

And a + b = 1

Putting the value of a , we get

⟹ 1 / 2 + b = 1

⟹ b = 1-1 / 2

b = 1 /2

Now the putting the values of a and b in equation (1), we get

v ∝ g1/2r-½

v ∝ √( g/r )

This is the required dimensional relation.

**Question 11: In the equation C = A + B, can the dimensions of C, B and A be different?**

**Answer :** According to principal of homogeneity of dimensions, the dimensions of C, B and A can not be different. It should be same.

**Question 12 : Will the dimension of physical quantity change with system of units.**

**Answers :** No, dimension of a physical quantity is independent of the system of units.

*Multiple choice questions ( MCQ ) *

*( Both for medical and engineering )*

**Question 1 : The dimensions of electric current are **

** (a) [ M ^{0}L^{0}T^{-1}Q ] (b) [ ML^{2}T^{-1}Q ]**

** (c) [ M ^{2}LT^{-1}Q ] (d) [ M^{2}L^{2}T^{-2} ]**

** [ Pb PMT 2001 ]**

** Answer :** (a) [ M** ^{0}**L

**T**

^{0}**Q ]**

^{-1} **Explanation :** Electric current = amount of charge / time

= Q / t

= [ M** ^{0}**L

**T**

^{0}**Q ]**

^{-1}**Question 2 :** The SI unit of gravitational potential is

(a) J (b) Jkg^{-1}

(c) J-kg (d) J-kg^{2}

** [AFMC- pune 2001]**

**Answers:** (b) Jkg^{-1}

**Explanation :** Gravitational potential = work /mass

SI unit of gravitational potential = SI unit of work / SI unit of mass = Jkg^{-1}

**Question 3 : Which of the following pair does not have similar dimensions **

** (a) Stress and pressure (b) Angle and strain **

** (c) Tension and surface tension (d) Planck’s constant and angular momentum**

** [AIIMS 2001]**

**Answer :** (c) tension and surface tension

Explanation : Dimension of tension = Dimension of force

= [ MLT** ^{-2}** ]

Dimension of surface tension = dimension of force / dimension of length

= [ MLT** ^{-2}** ] / [ L ]

= [ ML** ^{2}**T

**]**

^{-2}*Hence dimension of tension and surface tension are not similar.*

**Question 4 : The dimensions of plank’s constant equals to that of **

** (a) energy (b) momentum **

** (c) angular momentum (d) power **

**[CBSE PMT 2001]**

**Answer :** (c) angular momentum

**Explanation :** [ angular momentum ] = [ mass] * [ velocity ] * [ radius ]

= [ ML** ^{2}**T

**]**

^{-1}[ planck’s constant ] = [ energy ] / [ frequency ]

= [ ML** ^{2}**T

**] / [ T**

^{-2}**]**

^{-1} = [ ML** ^{2}**T

**]**

^{-1}**Question 5 : A force is given by F = at + bt ^{2} where t is time. What are the dimensions of a and b? **

** (a) [ MLT ^{-3} ] and [ MLT^{-4}] (b) [ MLT^{-3} ] and [ ML^{2}T^{-4} ]**

** (c) [ MLT ^{-1} ] and [ MLT^{0} ] (d) [ MLT^{-4} ] and [ MLT^{4} ]**

** [ AFMC – pune 2000 ]**

**Answer :** (a) [ MLT** ^{-3}** ] and [ MLT

**]**

^{-4}**Explanation :** F = at + bt2

According to principal of homogeneity of dimension

⇒ [ MLT** ^{-2}** ] = [ a ][ T ] + [ b ][ T

**]**

^{2}Now

[ MLT** ^{-2}** ] = [ a ][ T ]

⇒ [ MLT** ^{-2}** ] / [ T ] = [ a ]

⇒ [ MLT** ^{-1}** ] = [ a ]

and

[ MLT** ^{-2}** ] = [ b ][ T

**]**

^{2}⇒ [ MLT** ^{-2}** ] / [ T

**] = [ b ]**

^{2}⇒ [ MLT** ^{-4 }**] = [ b ]

**Question 6 : Which of the following pair does not have similar dimensions?**

** (a) energy and torque (b) force and impulse **

** (c) angular momentum and plank’s constant (d) elastic modulus and pressure **

** [ CBSE PMT 2000 ]**

**Answers :** (b) force and impulse

**Explanation :** [ force ] = [ MLT** ^{-2}** ]

[ impulse ] = [ MLT** ^{-1}** ]

Hence, force and impulse have not similar dimension.

**Question 7 : Dimensions of gravitational constant are **

** (a) [ ML ^{2}T^{-1} ] (b) [ ML^{2}T^{-2} ]**

** (c) [ M ^{-1}L^{3}T^{-2} ] (d) [ ML^{-1}T^{-1} ]**

** [ AIIMS 2000 ]**

**Answer :** (c) [ M** ^{-1}**L

**T**

^{3}**]**

^{-2}**Explanation :** According to newton’s law of gravitation

F = GMm / r^{2}

⇒ G = Fr** ^{2 }**/ Mm

[ G ] = [ MLT-2 ][ L2 ] / [ M ][ M ]

= [ M** ^{-1}**L

**T**

^{3}**]**

^{-2}**Question 8 : The dimensions of Hubble’s constant are **

** (a) [ T ^{–1} ] (b) [ M**

^{0}**L**

**T ]**^{0}** (c) [ MLT ^{4} ] (d) [ MLT^{-4} ]**

** [ Pb PMT 2000 ]**

**Answer :** (a) [ T** ^{-1 }**]

**Explanation :** [ Hubble’s constant ] = [ velocity ] / [ distance ]

= [ LT** ^{-1}** ] / [ L ]

= [ T** ^{-1}** ]

**Question 9 : The dimensions of RC **

** (a) time (b) inverse time **

** (c) square of time (d) square of inverse of time **

** [ AIIMS 1999 ]**

**Answers :** (a) time

**Explanation :** Χ = 1/2fC

C = 1/2fΧ

Unit of f = sec

Unit of Χ = ohm

Unit of C = 1/( Ωs** ^{-1}** )

= Ω** ^{-1}**s

Unit of RC = -1s

= s

Dimensions of RC = dimension of time

**Question 10 : The dimensions of magnetic flux are **

** (a) [ ML ^{2}T^{-2}A^{-1 }] (b) [ ML^{0}T^{-2}A^{-2} ]**

** (c) [ M ^{0}L^{-2}T^{2}A^{-2} ] (d) [ ML^{2}T^{-3}A^{3} ]**

** [ CBSE PMT 1999 ]**

**Answer **: (a) [ ML** ^{2}**T

**A**

^{-2}**]**

^{-1}**Explanation :** F = iBl

Where f = force, i = current, B = magnetic field intensity, l = length

B = F/il

[ B ] = [ MLT** ^{-2}** ] / [ A ][ L ]

= [ MT** ^{-2}**A

**]**

^{-1 }[ Magnetic flux ] = [ B ][ area ]

= [ MT** ^{-2}**A

**][ L**

^{-1}**]**

^{2} = [ ML** ^{2}**T

**A**

^{-2}**]**

^{-1}**Question 11 : The dimensions of impulse are **

** (a) [ ML ^{-1}T^{-2} ] (b) [ MLT^{-1 }]**

** (c) [ MT ^{-2} ] (d) [ ML^{-1}T^{-3} ]**

** [ AFMC pune 1998 ]**

Answer : (a) [ ML^{-1}T^{-2} ]

Explanation : [ impulse ] = [ force ][ time ]

= [ MLT^{-2} ][ T ]

= [ MLT^{-1} ]

**Question 11 : Which of the following is not a dimensionless constant ? **

** (a) refractive index (b) poisson’s ratio **

** (c) relative density (d) gravitational constant **

** [ AIIMS 1996 ]**

**Answer :** (d) gravitational constant

**Explanation :** refractive index, Poisson’s ratio and relative density are the ratio of similar physical quantities but G (gravitational constant) is the ratio of the product of force and radius ( which have the dimension of length ) and square of mass.

[ G ] = [ M^{-1}L^{3}T^{-2} ]

**Question 13 : A sphere of radius ‘a’ moves with velocity v in a medium and force ‘F’ acting on it is given by F = 6 πηav. The dimensions of will be **

** (a) [ML ^{-1}T^{-1} ] (b) [ MT^{-1} ]**

** (c) [ MLT ^{-1} ] (d) [ ML^{-3} ]**

** [CBSE PMT 1997]**

**Answer :** (a) [ ML** ^{-1}**T

**]**

^{-1}**Explanation :** F = 6πηav

η = F / ( 6πav )

[ η ] = [ F ] / ( [ a ][ v ] )

[ η ] = [ MLT** ^{-2}** ] / ( [ L ][ LT

**] )**

^{-1}[ η ] =[ ML** ^{-1}**T

**]**

^{-1}**Question 14 : The velocity of a particle at time t is given by v=at+b/(t+c). The dimensions of a,b,c are respectively **

** (a) LT ^{-2}, L,T (b) L^{2},T, LT^{2}**

** (c) LT ^{2}, LT, L (d) L, LT, T^{2}**

** **

**Answer :** (a) LT** ^{-2}**, L, T

Explanation: v = at + b / ( t + c )

[ v ] = [ a ] [ T ] + [ b ] ( [ T ] + [ c ] )

According to principal of homogeneity of dimensions

[ v ] = [ a ] [ T ]

[ a ] = [ v ] / [ T ]

=[ LT** ^{-1 }**] / [ T ]

= [ LT** ^{-2 }**]

And

[ v ] = [ b ] / ( [ T ] + [ c ] )

Where ( [ T ] + [ c ] ) should have dimension of time

Therefore, [ c ] = [ T ]

[ b ] = [ v ] * [ T ]

= [ LT** ^{-1 }**] * [ T ]

= [ L ]

**Question 15 : **The time dependence of a physical quantity p is given by p = p_{0} exp (- αt** ^{2 }**) where α is a constant and t is the time . Then constant α

(a) is dimensionless (b) has dimension T-^{2}

(c) has dimensions T** ^{2}** (d) none of the above .

**[CBSE PMT 1993]**

Answer: (d) has dimension of T-^{2}

Explanation: p = p_{0} exp (- αt** ^{2 }**)

The index of e should be dimensionless . So , should have the dimension of

Inverse of square of time i.e

Dimension of = [ T-** ^{2 }**]

- Two quantities A & B have different dimensions . Which mathematical operation given below is physically meaningful ?

(a) A/B (b) A+B

(c) A-B (d) none

**[CPMT 1997]**

Answer: (a) A/B

Explanation : addition and subtraction will be possible if and only if all the quantities

have the similar dimensions but it is not mandatory for multiplication and division .

- Turpentine oil is flowing through a tube of length l, and radius r. The pressure difference between the two ends of the tube is P . the viscosity of the oil is given by =p(r2-x2)/4vl where v is the velocity of oil at a distance x from the axis of the tube . the dimensions of are

(a) [M0L0T0] (b) [MLT-1]

(c) [ML2T-2] (d) [ML-1T-3]

**[CBSE PMT 1993]**

Answer: (d) [ML-1T-3]

Explanation : =p(r2-x2)/4vl

[]=[ML-1T-2][L2]/[LT-1][L]

[]=[ML-1T-3]

- In the standard equation s=u+a/2*(2n-1) what dimension do you view for s ?

(a) [M0LT0] (b) [M0LT-1]

(c) [M0L-1T] (d) [M0L0T]

**[DPMT 1994]**

Answer: (b) [M0LT-1]

Explanation: s=u+a/2*(2n-1)

[s]=[M0LT-1]+[M0LT-2]*[M0L0T]

=[M0LT-1]

- Which physical quantities have the same dimensions ?

(a) moment of couple and work

(b) force and power

(c) latent heat and specific heat

(d) work and power

**[CPMT 1997]**

Answer: (a) moment of couple and work

Explanation : moment of couple = force * distance

Work = force * displacement

[distance]=[displacement]

[work]=[moment of couple]

- The frequency of vibration f of mass m suspended from a spring constant k is given by the relation f=cmxky where c is a dimensionless constant . The values of x and y are

(a) x=½ , y=½ (b) x=-½ , y=-½

(c) x=½ ,y=-½ (d) x=-½ , y=½

**[CBSE PMT 1990]**

Answer: (d) x=-½ , y=½

Explanation: f=cmxky

According to principal of homogeneity of dimensions

[f]=[m]x[k]y

[T-1]=[M]x[MT-2]y

[M0L0T-1]=[Mx+yLyT-2y]

By equating the dimensions , we get

x+y=0———————–(1)

-2y=-1 ⇒ y=½

Putting the value of y in equation (1)

x=-½

- The SI unit of gravitational constant G is

(a) Nmkg-1 (b) Nm2kg-1

(c) Nmkg-2 (d) Nm2kg-2

**[Haryana CET 2001]**

Answer: (c) Nm2kg-2

Explanation : gravitational constant, G=Fr2/Mm

Unit of G =Nm2kg-2

- Which of the following pair does not have the same dimensions ?

(a) angular momentum and planck’s constant

(b) moment of inertia and moment of force

(c) work and torque

(d)impulse and momentum

**[BHU 1997]**

Answer: (b) moment of inertia and moment of force

Explanation: [Moment of inertia] = [ML2]

[moment of force]=[ML2T-2]

- A quantity X is defined by equation X=3CB2 where C is the capacitance in farad and B represents magnetic field in tesla . The dimensions of X are

(a) [ML-2] (b) [ML-2T2A]

(c) [ML-2T-2A] (d) [L-1A-1]

**[AMU 1998]**

Answer: (a) [ML-2]

Explanation: [X]=[C][B]2

= [Q]/[V]*[F/IL]

=[Q]/[(energy/Q)][MT-4A-2]

=([A2T2]/[ML2T-2])*[MT-4A-2]

=[M-1L-2T4A2]*[MT-4A-2]

=[ML-2]

- A quantity X is given by X=L(v/t)where is the permittivity of free space , L is the length , vis the potential difference and t is time interval the dimensions of X are the same as for

(a) resistance (b) charge

(c) voltage (d) current

**[IIT 2001]**

Answer: (d) current

Explanation:[L]=[capacitance]

=[Q/v]

[L(v/t)]=[Q/v]*(v/t)

=[Q/t]

= dimension of current

- The SI unit of pole strength is

(a) Am2 (b) Am

(c) Am-1 (d) Am-2

**[AMU 1998]**

Answer:(b) Am

Explanation: Unit of Pole strength ,m=unit of M/unit of 2l

=Am2/m

=Am

- If speed of light (c) , acceleration due to gravity (g), and pressure (p) are taken as fundamental units the dimensions of the gravitational constant G are

(a) c0gp-3 (b) c2g3p-2

(c) c0g2p-1 (d)c2g2p-2

**[AMU 1999]**

Answer: (c) c0g2p-1

Explanation: let G=cxgypz

[M-1L3T-2]=[LT-1]x[LT-2]y[ML-1T-2]z

=[MzLx+y-zT-x-2y-2z]

By comparing the dimensions on both sides , we get

z=-1; x+y-z=3; -x-2y-2z=-2

Putting z=-1 in x+y-z=3 we get

x+y-(-1)=3

⇒ x+y=2

⇒ x=-y+2

Now putting x=y-1 & z=-1 in -x-2y-2z=-2 we get

-(-y+2)-2y-2*(-1)=-2

⇒ y-2-2y+2=-2

⇒ -y=-2

⇒ y=2

Putting the values of y and z in x+y-z=3 , we get x=0

G=c0g2p-1

- The dimensions of 0E2/2 (0=permittivity of free space , E=electric field ) are

(a) MLT-1 (b) ML2T-2

(c) ML-1T-2 (d) ML2T-1

**[IIT 2000]**

Answer:(c) ML-1T-2

Explanation: F=Qq/4r2

E=F/q

E2/2=(Qq/4Fr2)*(F2/q2)

Therefore unit of E2/2 is equal to F/r2

Unit of E2/2=kg ms-2/m2

=kgm-1s-2

Dimensions of E2/2=[ML-1T-2]

- The dimensional formula for 0is given by

(a) M0LT (b) MLT-2A-2

(c) M0L2T-1A2 (d) none of the above

**[Haryana CET 2000]**

Answer: (b) MLT-2A-2

Explanation: from Biot-savart law

F=0(Mm/r2)/4

⇒ 0=4Fr2/(Mm)

Where M and m are pole strength of the magnet

Dimensional formula of 0=[MLT-2][L2]/[AL]2

=[MLT-2A-2]

- If L and R denote inductance and capacitance respectively , which of the following has the dimensions of frequency ?

(a) R/L (b) L/R

(c) R/L (d) L/R

**[MP CEE 1999]**

Answer: (a) R/L

explanation:[L]=[e][dt/dI]

Where e = e.m.f

[L]=([ML2T-2]/[AT])*([T]/[A])

=[ML2T-2A-2]

[R]=[V]/[I]

=[energy]/[(charge*I)]

=[ML2T-2]/([AT]*[A])

=[ML2T-3A-2]

R/L=[ML2T-3A-2]/[ML2T-2A-2]

=[T-1]

**Question 9 : The ratio of one micron to one nanometre is **

** (a) 103 (b) 10-3**

** (c) 10-6 (d)10-9**

**[MNR 1993]**

Answer: (a)103

Explanation: one micron=10-6

One nanometre=10-9

One micron : one nanometre

⇒ 10-6 : 10-9

⇒ 103 : 1

Hence , required ratio is 103

- Which of the following is equal to joule-ohm/second-volt?

(a) ampere (b) watt

(c) tesla (d) volt

**[Haryana CET 1998]**

Answer: (d) volt

Explanation: (joule/sec)*(ohm/volt)=(joule/charge)*(charge/sec)*((volt/ampere)/volt)

=volt*ampere*(1/ampere)

= volt

- The unit of permittivity of free space is

(a) coulomb2/newton-meter2

(b) coulomb/(newton-meter)2

(c) newton-metre2/coulomb2

(d)coulomb/newton-metre

**[Pb CET 1998]**

Answer: (a) coulomb2/newton-metre2

Explanation:According to coulomb’s law of electrostat

F=Qq/(4r2)

Where Q and Q are electric point charges , is the permittivity and r is the distance between the two charges

=Qq/(4Fr2)

Unit of =coloumb2/(newton-metre2)

- The physical quantity which has the dimensional formula MT-3 is

(a) surface tension (b) solar constant

(c) density (d) compressibility

**[Pb CET 1998]**

Answer: (b) solar constant

Explanation: [Solar constant] = [Energy/(Area x Time)]

= [ML2T-2/(L2 T)]

= [MT-3]

- The dimensions of the quantities in one of the following pairs are the same . Identify the pair

(a) torque and work

(b) angular momentum and work

(c) energy and young’s modulus

(d) light year and frequency

**[IIT 1996]**

Answer: (a) torque and work

Explanation: torque=force *perpendicular distance

And work = force * displacement

[displacement]=[perpendicular distance]

Hence torque and work have the same dimensions .

- The SI unit of inductance ,the henry can be

(a) weber/ampere (b) volt-second /ampere

(c) joule/ampere2 (d) all of the above

**[IIT 1998]**

Answer:

Explanation: e=L(dI/dt)

L=e/(dI/dt)

=volt-sec/ampere

L=N/I

Where n is no of turns ,is magnetic flux and I is the current

Unit of L = weber/ampere

Again we have I=V/X

=V/2fL

L=V/2fI

Unit of L = volt -sec/ampere

=joule/ampere2

- A gas bubble from an explosion under water oscillates with a period T proportional to padbEc where p is static pressure ,d is the density of water and E is the total energy of explosion . The calculated values of a,b,and c on the dimensional analysis will be

(a) a=-⅚,b=-½ ,c=⅓ (b) a=b=c=1

(c) a=5,b=½,c=⅓ (d)a=-⅚,b=½,c=⅓

**[Haryana CET 1994]**

Answer: (a) a=-⅚,b=-½ ,c=⅓

Explanation: Given that

TpadbEc

According to principal of homogeneity of dimensions

[T]=[ML-1T-2]a[ML-3]b[ML2T-2]c

[M0L0T]=[Ma+b+cL-a-3b+2cT-2a-2c]

By equating dimensions on both sides, we get

a+b+c=0—————————(1)

-a-3b+2c=0————————(2)

-2a-2c=1—————————(3)

Multiplying 3 in the equation (1) on both sides and the adding it to equ.(2)

2a+5c=0—————————(4)

Now solving equ.(3)&(4), we get

3c=1

⇒ c=⅓

Therefore , a=(1+⅔)/(-2)

=-⅚

Putting the values of a and c in equ. (1) , we get

b=-⅓

- In the equation (p+a/v2)(v-b)=constant (p=pressure ,v=volume), the units of a are

(a) dyne*cm2 (b) dyne*cm4

(c) dyne/cm2 (d) dyne*cm3

**[MNR 1995]**

Answer: (a) dyne*cm4

Explanation: (p+a/v2)(v-b)=constant

The units of a/v2 must be that of pressure .

Unit of a/v2 =unit of pressure

Unit of a = unit of pressure * unit of v2

=(dyne/cm2)*(cm3)2

=dyne*cm4

- Finding dimensions of resistance R and inductance L , what physical quantity L/R represents?

(a) time (b)frequency

(c) energy (d) power

**[Roorkee 1995]**

Answer: (a) time

Explanation: [R]=[V]/[I]

=[energy]/[(charge*I)]

=[ML2T-2]/([AT]*[A])

=[ML2T-3A-2]

[L]=[e][dt/dI]

Where e = e.m.f

[L]=([ML2T-2]/[AT])*([T]/[A])

=[ML2T-2A-2]

[L]/[R]=[ML2T-3A-2]/[ML2T-2A-2]

=[T]

- L,C and R represent the physical quantities inductance , capacitance and resistance respectively . Identify the quantity that represents frequency .

(a) 1/RC (b) L/R

(c) C/L (d) L/C

**[IIT 1984] **

Answer: (a) 1/RC

Explanation: [R]=[V]/[I]

=[energy]/[(charge*I)]

=[ML2T-2]/([AT]*[A])

=[ML2T-3A-2]

[C]=[Q]/[V]

=[Q]/[(energy/Q)]

=[A2T2]/[ML2T-2]

=[M-1L-2T4A2]

1/RC=1/([M-1L-2T4A2]*[ML2T-3A-2])

=[T-1]

=dimensions of frequency

Hence 1/RC represents frequency

- The unit of young’s modulus is

(a) Nm-1 (b) Nm

(c) Nm-2 (d) Nm2

**[MP PET 1995]**

Answer: Nm-2

Explanation: Young’s modulus=stress/strain

Since strain has no unit

Unit of Y=unit of stress

=unit of force/unit of area

=N/m2

=Nm-2

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