68 + Interview Questions in Dimensions Interview Questions in Dimensions Tutorial

1.	If Force (f), Velocity (v) And Acceleration (a) Are Taken As The Fundamental Units Instead Of Mass, Length And Time, Express Pressure And Impulse In Terms Of F, V And A?
Answer» We know that Force = MASS ✕ acceleration ⇒ mass = FA-1 and length = VELOCITY ✕ time = velocity ✕ velocity ÷ acceleration = V2A-1 and time = VA-1 ∵ Pressure = Force ÷ Area = F ÷ (V2A-1)2 = FV-4A2 IMPULSE = Force ✕ time = FVA-1 We know that Force = mass ✕ acceleration ⇒ mass = FA-1 and length = velocity ✕ time = velocity ✕ velocity ÷ acceleration = V2A-1 and time = VA-1 ∵ Pressure = Force ÷ Area = F ÷ (V2A-1)2 = FV-4A2 Impulse = Force ✕ time = FVA-1

1.

If Force (f), Velocity (v) And Acceleration (a) Are Taken As The Fundamental Units Instead Of Mass, Length And Time, Express Pressure And Impulse In Terms Of F, V And A?

Answer»

We know that Force = MASS ✕ acceleration

⇒ mass = FA-1

and length = VELOCITY ✕ time = velocity ✕ velocity ÷ acceleration = V2A-1

and time = VA-1

∵ Pressure = Force ÷ Area = F ÷ (V2A-1)2 = FV-4A2

IMPULSE = Force ✕ time = FVA-1

We know that Force = mass ✕ acceleration

⇒ mass = FA-1

and length = velocity ✕ time = velocity ✕ velocity ÷ acceleration = V2A-1

and time = VA-1

∵ Pressure = Force ÷ Area = F ÷ (V2A-1)2 = FV-4A2

Impulse = Force ✕ time = FVA-1

Discussion

2.	Express Capacitance In Terms Of Dimensions Of Fundamental Quantities I.e. Mass (m), Length(l), Time(t) And Ampere(a)?
Answer» Capacitance(C) is defined as the ability of a electric body to store electric charge. ∴ Capacitance (C) = Total Charge(Q) / potential difference between two plates (V) = COULOMB/ Volt ∵ Volt = Work done (W)/ Charge(q) = Joule/Coulomb ⇒ Capacitance (C) = Charge(q)2/ Work(W) ∵ Charge (q) = Current (I) × Time(t) DIMENSION of [q] = [AT] ———– (I) Dimension of Work = Force × distance = [MLT-2][L] = [ML2T-2] ——— (II) PUTTING values of I and II, [C] = ([AT])2/ [ML2T-2] = [M-1L-2T2+2A2] = [M-1L-2T4A2] Physical Quantities having the same dimensional formula: a. impulse and momentum. b. force, thrust. c. work, energy, TORQUE, moment of force, energy d. angular momentum, Planck’s constant, rotational impulse e. force constant, surface tension, surface energy. f. stress, pressure, modulus of elasticity. g. angular velocity, frequency, velocity gradient h. latent heat, gravitational potential. i. thermal capacity, entropy, universal gas constant and Boltzmann’s constant. j. power, luminous flux. Capacitance(C) is defined as the ability of a electric body to store electric charge. ∴ Capacitance (C) = Total Charge(q) / potential difference between two plates (V) = Coulomb/ Volt ∵ Volt = Work done (W)/ Charge(q) = Joule/Coulomb ⇒ Capacitance (C) = Charge(q)2/ Work(W) ∵ Charge (q) = Current (I) × Time(t) Dimension of [q] = [AT] ———– (I) Dimension of Work = Force × distance = [MLT-2][L] = [ML2T-2] ——— (II) Putting values of I and II, [C] = ([AT])2/ [ML2T-2] = [M-1L-2T2+2A2] = [M-1L-2T4A2] Physical Quantities having the same dimensional formula: a. impulse and momentum. b. force, thrust. c. work, energy, torque, moment of force, energy d. angular momentum, Planck’s constant, rotational impulse e. force constant, surface tension, surface energy. f. stress, pressure, modulus of elasticity. g. angular velocity, frequency, velocity gradient h. latent heat, gravitational potential. i. thermal capacity, entropy, universal gas constant and Boltzmann’s constant. j. power, luminous flux.

2.

Express Capacitance In Terms Of Dimensions Of Fundamental Quantities I.e. Mass (m), Length(l), Time(t) And Ampere(a)?

Answer»

Capacitance(C) is defined as the ability of a electric body to store electric charge.

∴ Capacitance (C) = Total Charge(Q) / potential difference between two plates (V)

= COULOMB/ Volt

∵ Volt = Work done (W)/ Charge(q) = Joule/Coulomb

⇒ Capacitance (C) = Charge(q)2/ Work(W)

∵ Charge (q) = Current (I) × Time(t)

DIMENSION of [q] = [AT] ———– (I)

Dimension of Work = Force × distance = [MLT-2][L] = [ML2T-2] ——— (II)

PUTTING values of I and II,

[C] = ([AT])2/ [ML2T-2] = [M-1L-2T2+2A2] = [M-1L-2T4A2]

Physical Quantities having the same dimensional formula:

a. impulse and momentum.

b. force, thrust.

c. work, energy, TORQUE, moment of force, energy

d. angular momentum, Planck’s constant, rotational impulse

e. force constant, surface tension, surface energy.

f. stress, pressure, modulus of elasticity.

g. angular velocity, frequency, velocity gradient

h. latent heat, gravitational potential.

i. thermal capacity, entropy, universal gas constant and Boltzmann’s constant.

j. power, luminous flux.

Capacitance(C) is defined as the ability of a electric body to store electric charge.

∴ Capacitance (C) = Total Charge(q) / potential difference between two plates (V)

= Coulomb/ Volt

∵ Volt = Work done (W)/ Charge(q) = Joule/Coulomb

⇒ Capacitance (C) = Charge(q)2/ Work(W)

∵ Charge (q) = Current (I) × Time(t)

Dimension of [q] = [AT] ———– (I)

Dimension of Work = Force × distance = [MLT-2][L] = [ML2T-2] ——— (II)

Putting values of I and II,

[C] = ([AT])2/ [ML2T-2] = [M-1L-2T2+2A2] = [M-1L-2T4A2]

Physical Quantities having the same dimensional formula:

a. impulse and momentum.

b. force, thrust.

c. work, energy, torque, moment of force, energy

d. angular momentum, Planck’s constant, rotational impulse

e. force constant, surface tension, surface energy.

f. stress, pressure, modulus of elasticity.

g. angular velocity, frequency, velocity gradient

h. latent heat, gravitational potential.

i. thermal capacity, entropy, universal gas constant and Boltzmann’s constant.

j. power, luminous flux.

Discussion

3.	A Student While Doing An Experiment Finds That The Velocity Of An Object Varies With Time And It Can Be Expressed As Equation: V = Xt2 + Yt +z . If Units Of V And T Are Expressed In Terms Of Si Units, Determine The Units Of Constants X, Y And Z In The Given Equation?
Answer» Given, v = Xt2 + Yt +Z Dimensions of velocity v = [M0L1T-1] Applying applying principle of homogeneity in dimensions, TERMS MUST have same dimension. [v] = [Xt2] + [Yt] + [Z] ∴ [v] = [Xt2] ⇒ [X] = [v] /[t2] = [M0L1T-1] / [M0L0T2] = [M0L1T-3] ….(i) SIMILARLY, [v] = [Yt] ⇒ [Y] = [v] / [t] = [M0L1T-1]/ [M0L0T-1] = [M0L1T-2] …(ii) Similarly, [v]= [Z] [Z] = [M0L1T-1] …(III) ⇒ Unit of X = m-s-3 ⇒ Unit of Y = m-s-2 ⇒ Unit of Z = m-s-1 Given, v = Xt2 + Yt +Z Dimensions of velocity v = [M0L1T-1] Applying applying principle of homogeneity in dimensions, terms must have same dimension. [v] = [Xt2] + [Yt] + [Z] ∴ [v] = [Xt2] ⇒ [X] = [v] /[t2] = [M0L1T-1] / [M0L0T2] = [M0L1T-3] ….(i) Similarly, [v] = [Yt] ⇒ [Y] = [v] / [t] = [M0L1T-1]/ [M0L0T-1] = [M0L1T-2] …(ii) Similarly, [v]= [Z] [Z] = [M0L1T-1] …(iii) ⇒ Unit of X = m-s-3 ⇒ Unit of Y = m-s-2 ⇒ Unit of Z = m-s-1

3.

A Student While Doing An Experiment Finds That The Velocity Of An Object Varies With Time And It Can Be Expressed As Equation: V = Xt2 + Yt +z . If Units Of V And T Are Expressed In Terms Of Si Units, Determine The Units Of Constants X, Y And Z In The Given Equation?

Answer»

Given, v = Xt2 + Yt +Z

Dimensions of velocity v = [M0L1T-1]

Applying applying principle of homogeneity in dimensions, TERMS MUST have same dimension.

[v] = [Xt2] + [Yt] + [Z]

∴ [v] = [Xt2]

⇒ [X] = [v] /[t2] = [M0L1T-1] / [M0L0T2] = [M0L1T-3] ….(i)

SIMILARLY, [v] = [Yt]

⇒ [Y] = [v] / [t] = [M0L1T-1]/ [M0L0T-1] = [M0L1T-2] …(ii)

Similarly, [v]= [Z]

[Z] = [M0L1T-1] …(III)

⇒ Unit of X = m-s-3

⇒ Unit of Y = m-s-2

⇒ Unit of Z = m-s-1

Given, v = Xt2 + Yt +Z

Dimensions of velocity v = [M0L1T-1]

Applying applying principle of homogeneity in dimensions, terms must have same dimension.

[v] = [Xt2] + [Yt] + [Z]

∴ [v] = [Xt2]

⇒ [X] = [v] /[t2] = [M0L1T-1] / [M0L0T2] = [M0L1T-3] ….(i)

Similarly, [v] = [Yt]

⇒ [Y] = [v] / [t] = [M0L1T-1]/ [M0L0T-1] = [M0L1T-2] …(ii)

Similarly, [v]= [Z]

[Z] = [M0L1T-1] …(iii)

⇒ Unit of X = m-s-3

⇒ Unit of Y = m-s-2

⇒ Unit of Z = m-s-1

Discussion

4.	If The Velocity Of Light C, Gravitational Constant G And Planks Constant H Be Chosen As Fundamental Units, Find The Value Of A Gram, A Cm And A Sec In Term Of New Unit Of Mass, Length And Time Respectively. (take C = 3 X 1010 Cm/sec, G = 6.67 X 108 Dyn Cm2/gram2 And H = 6.6 X 10-27 Erg Sec)?
Answer» Given, c = 3 x 1010 cm/sec G = 6.67 x 108 dyn cm2/gm2 h = 6.6 x 10-27 erg sec Putting respective DIMENSIONS, Dimension formula for c = [M0L1T-1] = 3 x 1010 cm/sec …. (I) Dimensions of G = [M-1L3T-2] = 6.67 x 108dyn cm2/gm2 …(II) Dimensions of h = [M1L2T-1] = 6.6 x 10-27erg sec …(III) (Note: Applying newton’s law of gravitation, you can find dimensions of G i.e. G = Fr2/(mM) Similarly, Planck’s Constant (h) = Energy / frequency) To get M, multiply eqn-I and III and divide by eqn.-II, ⇒ [M0L1T-1].[M1L2T-1].[M1L-3T2] = ( 3 x 1010 cm/sec).( 6.6 x 10-27 erg sec)/ 6.67 x 108 dyn cm2/gm2 ⇒[M2] = 2.968 x 10-9 ⇒[M] = 0.5448 x 10-4 gm or 1gm = [M]/0.5448 x 10-4 = 1.835 x 10-4 unit of MASS To obtain length [L], eqn.-II x eqn-III / cube of eqn.-I i.e. [M-1L3T-2].[M1L2T-1].[M0L-3T3] = (6.67 x 108 dyn cm2/gm2 ).( 6.6 x 10-27erg sec)/(3 x 1010 cm/sec)3 ⇒ [L2] = 1.6304 x 10-65cm2 ⇒ [L] = 0.4038 x 10-32 cm or 1cm = [L]/ 0.4038 x 10-32 = 2.47 x 10-32unit of length In eqn-I, [M0L1T-1] = 3 x 1010cm/sec ⇒ [T] = [L] ÷ 3 x 1010cm/s ⇒ [T] = 0.4038 x 10-32 cm ÷ 3 x 1010cm/s = 0.1345 x 10-42 s or 1S = [T]/0.1345 x 10-42s = 7.42 x 1042unit of time Given, c = 3 x 1010 cm/sec G = 6.67 x 108 dyn cm2/gm2 h = 6.6 x 10-27 erg sec Putting respective dimensions, Dimension formula for c = [M0L1T-1] = 3 x 1010 cm/sec …. (I) Dimensions of G = [M-1L3T-2] = 6.67 x 108dyn cm2/gm2 …(II) Dimensions of h = [M1L2T-1] = 6.6 x 10-27erg sec …(III) (Note: Applying newton’s law of gravitation, you can find dimensions of G i.e. G = Fr2/(mM) Similarly, Planck’s Constant (h) = Energy / frequency) To get M, multiply eqn-I and III and divide by eqn.-II, ⇒ [M0L1T-1].[M1L2T-1].[M1L-3T2] = ( 3 x 1010 cm/sec).( 6.6 x 10-27 erg sec)/ 6.67 x 108 dyn cm2/gm2 ⇒[M2] = 2.968 x 10-9 ⇒[M] = 0.5448 x 10-4 gm or 1gm = [M]/0.5448 x 10-4 = 1.835 x 10-4 unit of mass To obtain length [L], eqn.-II x eqn-III / cube of eqn.-I i.e. [M-1L3T-2].[M1L2T-1].[M0L-3T3] = (6.67 x 108 dyn cm2/gm2 ).( 6.6 x 10-27erg sec)/(3 x 1010 cm/sec)3 ⇒ [L2] = 1.6304 x 10-65cm2 ⇒ [L] = 0.4038 x 10-32 cm or 1cm = [L]/ 0.4038 x 10-32 = 2.47 x 10-32unit of length In eqn-I, [M0L1T-1] = 3 x 1010cm/sec ⇒ [T] = [L] ÷ 3 x 1010cm/s ⇒ [T] = 0.4038 x 10-32 cm ÷ 3 x 1010cm/s = 0.1345 x 10-42 s or 1s = [T]/0.1345 x 10-42s = 7.42 x 1042unit of time

4.

If The Velocity Of Light C, Gravitational Constant G And Planks Constant H Be Chosen As Fundamental Units, Find The Value Of A Gram, A Cm And A Sec In Term Of New Unit Of Mass, Length And Time Respectively. (take C = 3 X 1010 Cm/sec, G = 6.67 X 108 Dyn Cm2/gram2 And H = 6.6 X 10-27 Erg Sec)?

Answer»

Given,

c = 3 x 1010 cm/sec

G = 6.67 x 108 dyn cm2/gm2

h = 6.6 x 10-27 erg sec

Putting respective DIMENSIONS,

Dimension formula for c = [M0L1T-1] = 3 x 1010 cm/sec …. (I)

Dimensions of G = [M-1L3T-2] = 6.67 x 108dyn cm2/gm2 …(II)

Dimensions of h = [M1L2T-1] = 6.6 x 10-27erg sec …(III)

(Note: Applying newton’s law of gravitation, you can find dimensions of G i.e. G = Fr2/(mM)

Similarly, Planck’s Constant (h) = Energy / frequency)

To get M, multiply eqn-I and III and divide by eqn.-II,

⇒ [M0L1T-1].[M1L2T-1].[M1L-3T2]

= ( 3 x 1010 cm/sec).( 6.6 x 10-27 erg sec)/ 6.67 x 108 dyn cm2/gm2

⇒[M2] = 2.968 x 10-9

⇒[M] = 0.5448 x 10-4 gm

or 1gm = [M]/0.5448 x 10-4 = 1.835 x 10-4 unit of MASS

To obtain length [L], eqn.-II x eqn-III / cube of eqn.-I i.e.

[M-1L3T-2].[M1L2T-1].[M0L-3T3]

= (6.67 x 108 dyn cm2/gm2 ).( 6.6 x 10-27erg sec)/(3 x 1010 cm/sec)3

⇒ [L2] = 1.6304 x 10-65cm2

⇒ [L] = 0.4038 x 10-32 cm

or 1cm = [L]/ 0.4038 x 10-32 = 2.47 x 10-32unit of length

In eqn-I, [M0L1T-1] = 3 x 1010cm/sec

⇒ [T] = [L] ÷ 3 x 1010cm/s

⇒ [T] = 0.4038 x 10-32 cm ÷ 3 x 1010cm/s = 0.1345 x 10-42 s

or 1S = [T]/0.1345 x 10-42s = 7.42 x 1042unit of time

Given,

c = 3 x 1010 cm/sec

G = 6.67 x 108 dyn cm2/gm2

h = 6.6 x 10-27 erg sec

Putting respective dimensions,

Dimension formula for c = [M0L1T-1] = 3 x 1010 cm/sec …. (I)

Dimensions of G = [M-1L3T-2] = 6.67 x 108dyn cm2/gm2 …(II)

Dimensions of h = [M1L2T-1] = 6.6 x 10-27erg sec …(III)

(Note: Applying newton’s law of gravitation, you can find dimensions of G i.e. G = Fr2/(mM)

Similarly, Planck’s Constant (h) = Energy / frequency)

To get M, multiply eqn-I and III and divide by eqn.-II,

⇒ [M0L1T-1].[M1L2T-1].[M1L-3T2]

= ( 3 x 1010 cm/sec).( 6.6 x 10-27 erg sec)/ 6.67 x 108 dyn cm2/gm2

⇒[M2] = 2.968 x 10-9

⇒[M] = 0.5448 x 10-4 gm

or 1gm = [M]/0.5448 x 10-4 = 1.835 x 10-4 unit of mass

To obtain length [L], eqn.-II x eqn-III / cube of eqn.-I i.e.

[M-1L3T-2].[M1L2T-1].[M0L-3T3]

= (6.67 x 108 dyn cm2/gm2 ).( 6.6 x 10-27erg sec)/(3 x 1010 cm/sec)3

⇒ [L2] = 1.6304 x 10-65cm2

⇒ [L] = 0.4038 x 10-32 cm

or 1cm = [L]/ 0.4038 x 10-32 = 2.47 x 10-32unit of length

In eqn-I, [M0L1T-1] = 3 x 1010cm/sec

⇒ [T] = [L] ÷ 3 x 1010cm/s

⇒ [T] = 0.4038 x 10-32 cm ÷ 3 x 1010cm/s = 0.1345 x 10-42 s

or 1s = [T]/0.1345 x 10-42s = 7.42 x 1042unit of time

Discussion

5.	The Centripetal Force (f) Acting On A Particle (moving Uniformly In A Circle) Depends On The Mass (m) Of The Particle, Its Velocity (v) And Radius (r) Of The Circle. Derive Dimensionally Formula For Force (f)?
Answer» GIVEN, F ∝ ma.vb.rc ∴ F = kma.vb.rc (where K is CONSTANT) Putting dimensions of each quantity in the equation, [M1L1T-2] = [M1L0T0]a.[M0L1T-1]b. [M0L1T0]C = [MaLb+cT+cT-b] ⇒ a =1, b +c = 1, -b = -2 ⇒ a= 1, b = 2, c = -1 ∴ F = km1.v2.r-1= kmv2/r Given, F ∝ ma.vb.rc ∴ F = kma.vb.rc (where k is constant) Putting dimensions of each quantity in the equation, [M1L1T-2] = [M1L0T0]a.[M0L1T-1]b. [M0L1T0]c = [MaLb+cT+cT-b] ⇒ a =1, b +c = 1, -b = -2 ⇒ a= 1, b = 2, c = -1 ∴ F = km1.v2.r-1= kmv2/r

5.

The Centripetal Force (f) Acting On A Particle (moving Uniformly In A Circle) Depends On The Mass (m) Of The Particle, Its Velocity (v) And Radius (r) Of The Circle. Derive Dimensionally Formula For Force (f)?

Answer»

GIVEN, F ∝ ma.vb.rc

∴ F = kma.vb.rc (where K is CONSTANT)

Putting dimensions of each quantity in the equation,

[M1L1T-2] = [M1L0T0]a.[M0L1T-1]b. [M0L1T0]C = [MaLb+cT+cT-b]

⇒ a =1, b +c = 1, -b = -2

⇒ a= 1, b = 2, c = -1

∴ F = km1.v2.r-1= kmv2/r

Given, F ∝ ma.vb.rc

∴ F = kma.vb.rc (where k is constant)

Putting dimensions of each quantity in the equation,

[M1L1T-2] = [M1L0T0]a.[M0L1T-1]b. [M0L1T0]c = [MaLb+cT+cT-b]

⇒ a =1, b +c = 1, -b = -2

⇒ a= 1, b = 2, c = -1

∴ F = km1.v2.r-1= kmv2/r

Discussion

6.	Convert 1 Newton Into Dyne Using Method Of Dimensions?
Answer» Dimensions of Force = [M1L1T-2] Considering dimensional unit CONVERSION formula i.e. n1[M1aL1bT1c] = N2[M2aL2bT2c] ⇒ a = 1, b = 1 and c = -2 In SI system, M1 = 1kg, L1 = 1m and T1 = 1s In cgs system, M2 = 1g, L2 = 1cm and T2 = 1s PUTTING the values in the conversion formula, n2 = n1(1Kg/1g)1.(1m/1cm)1(1s/1s)-2= 1.(103/1g)(102cm) = 105dyne Dimensions of Force = [M1L1T-2] Considering dimensional unit conversion formula i.e. n1[M1aL1bT1c] = n2[M2aL2bT2c] ⇒ a = 1, b = 1 and c = -2 In SI system, M1 = 1kg, L1 = 1m and T1 = 1s In cgs system, M2 = 1g, L2 = 1cm and T2 = 1s Putting the values in the conversion formula, n2 = n1(1Kg/1g)1.(1m/1cm)1(1s/1s)-2= 1.(103/1g)(102cm) = 105dyne

6.

Convert 1 Newton Into Dyne Using Method Of Dimensions?

Answer»

Dimensions of Force = [M1L1T-2]

Considering dimensional unit CONVERSION formula i.e. n1[M1aL1bT1c] = N2[M2aL2bT2c]

⇒ a = 1, b = 1 and c = -2

In SI system, M1 = 1kg, L1 = 1m and T1 = 1s

In cgs system, M2 = 1g, L2 = 1cm and T2 = 1s

PUTTING the values in the conversion formula,

n2 = n1(1Kg/1g)1.(1m/1cm)1(1s/1s)-2= 1.(103/1g)(102cm) = 105dyne

Dimensions of Force = [M1L1T-2]

Considering dimensional unit conversion formula i.e. n1[M1aL1bT1c] = n2[M2aL2bT2c]

⇒ a = 1, b = 1 and c = -2

In SI system, M1 = 1kg, L1 = 1m and T1 = 1s

In cgs system, M2 = 1g, L2 = 1cm and T2 = 1s

Putting the values in the conversion formula,

n2 = n1(1Kg/1g)1.(1m/1cm)1(1s/1s)-2= 1.(103/1g)(102cm) = 105dyne

Discussion

7.	The Kinetic Energy K Of A Rotating Body Depends On Its Moment Of Inertia I And Its Angular Speed ω. Considering The Relation To Be K = Kiaωb Where K Is Dimensionless Constant. Find A And B. Moment Of Inertia Of A Spehere About Its Diameter Is (2/5)mr2?
Answer» DIMENSIONS of Kinetic ENERGY K = [M1L2T-2] Dimensions of MOMENT of Inertia (I) = [ (2/5)Mr2] = [ML2T0] Dimensions of angular speed ω = [θ/t] = [M0L0T-1] Applying principle of HOMOGENEITY in dimensions in the equation K = kIaωb [M1L2T-2] = k ( [ML2T0])a([M0L0T-1])B [M1L2T-2] = k [MaL2aT-b] ⇒ a = 1 and b = 2 ⇒ K = kIω2 Dimensions of Kinetic energy K = [M1L2T-2] Dimensions of Moment of Inertia (I) = [ (2/5)Mr2] = [ML2T0] Dimensions of angular speed ω = [θ/t] = [M0L0T-1] Applying principle of homogeneity in dimensions in the equation K = kIaωb [M1L2T-2] = k ( [ML2T0])a([M0L0T-1])b [M1L2T-2] = k [MaL2aT-b] ⇒ a = 1 and b = 2 ⇒ K = kIω2

7.

The Kinetic Energy K Of A Rotating Body Depends On Its Moment Of Inertia I And Its Angular Speed ω. Considering The Relation To Be K = Kiaωb Where K Is Dimensionless Constant. Find A And B. Moment Of Inertia Of A Spehere About Its Diameter Is (2/5)mr2?

Answer»

DIMENSIONS of Kinetic ENERGY K = [M1L2T-2]

Dimensions of MOMENT of Inertia (I) = [ (2/5)Mr2] = [ML2T0]

Dimensions of angular speed ω = [θ/t] = [M0L0T-1]

Applying principle of HOMOGENEITY in dimensions in the equation K = kIaωb

[M1L2T-2] = k ( [ML2T0])a([M0L0T-1])B

[M1L2T-2] = k [MaL2aT-b]

⇒ a = 1 and b = 2

⇒ K = kIω2

Dimensions of Kinetic energy K = [M1L2T-2]

Dimensions of Moment of Inertia (I) = [ (2/5)Mr2] = [ML2T0]

Dimensions of angular speed ω = [θ/t] = [M0L0T-1]

Applying principle of homogeneity in dimensions in the equation K = kIaωb

[M1L2T-2] = k ( [ML2T0])a([M0L0T-1])b

[M1L2T-2] = k [MaL2aT-b]

⇒ a = 1 and b = 2

⇒ K = kIω2

Discussion

8.	A Calorie Is A Unit Of Heat Or Energy And It Equals About 4.2 J Where 1j = 1 Kg M2 S–2. Suppose We Employ A System Of Units In Which The Unit Of Mass Equals α Kg, The Unit Of Length Equals β M, The Unit Of Time Is γ S. Show That A Calorie Has A Magnitude 4.2 α–1 β–2 γ2 In Terms Of The New Units?
Answer» Considering the unit CONVERSION formula, n1U1 = n1U2 N1[M1aL1bT1c] = n2[M2aL2bT2c] Given here, 1 Cal = 4.2 J = 4.2 KG m2 s–2. n1 = 4.2, M1 = 1kg, L1 = 1m, T1 = 1 sec and n2 = ?, M2 = α kg, L2 = βm, T2 = γ sec The dimensional formula of ENERGY is = [M1L2T-2] ⇒ a = 1, b =1 and c = -2 Putting these values in above equation, n2= n1[M1/M2]a[L1/L2]b[T1/T2]c = n1[M1/M2]1[L1/L2]2[T1/T2]-2 = 4.2[1Kg/α kg]1[1m/βm]2[1sec/γ sec]-2 = 4.2 α–1 β–2 γ2 Considering the unit conversion formula, n1U1 = n1U2 n1[M1aL1bT1c] = n2[M2aL2bT2c] Given here, 1 Cal = 4.2 J = 4.2 kg m2 s–2. n1 = 4.2, M1 = 1kg, L1 = 1m, T1 = 1 sec and n2 = ?, M2 = α kg, L2 = βm, T2 = γ sec The dimensional formula of energy is = [M1L2T-2] ⇒ a = 1, b =1 and c = -2 Putting these values in above equation, n2= n1[M1/M2]a[L1/L2]b[T1/T2]c = n1[M1/M2]1[L1/L2]2[T1/T2]-2 = 4.2[1Kg/α kg]1[1m/βm]2[1sec/γ sec]-2 = 4.2 α–1 β–2 γ2

8.

A Calorie Is A Unit Of Heat Or Energy And It Equals About 4.2 J Where 1j = 1 Kg M2 S–2. Suppose We Employ A System Of Units In Which The Unit Of Mass Equals α Kg, The Unit Of Length Equals β M, The Unit Of Time Is γ S. Show That A Calorie Has A Magnitude 4.2 α–1 β–2 γ2 In Terms Of The New Units?

Answer»

Considering the unit CONVERSION formula,

n1U1 = n1U2

N1[M1aL1bT1c] = n2[M2aL2bT2c]

Given here, 1 Cal = 4.2 J = 4.2 KG m2 s–2.

n1 = 4.2, M1 = 1kg, L1 = 1m, T1 = 1 sec

and

n2 = ?, M2 = α kg, L2 = βm, T2 = γ sec

The dimensional formula of ENERGY is = [M1L2T-2]

⇒ a = 1, b =1 and c = -2 Putting these values in above equation,

n2= n1[M1/M2]a[L1/L2]b[T1/T2]c

= n1[M1/M2]1[L1/L2]2[T1/T2]-2

= 4.2[1Kg/α kg]1[1m/βm]2[1sec/γ sec]-2 = 4.2 α–1 β–2 γ2

Considering the unit conversion formula,

n1U1 = n1U2

n1[M1aL1bT1c] = n2[M2aL2bT2c]

Given here, 1 Cal = 4.2 J = 4.2 kg m2 s–2.

n1 = 4.2, M1 = 1kg, L1 = 1m, T1 = 1 sec

and

n2 = ?, M2 = α kg, L2 = βm, T2 = γ sec

The dimensional formula of energy is = [M1L2T-2]

⇒ a = 1, b =1 and c = -2 Putting these values in above equation,

n2= n1[M1/M2]a[L1/L2]b[T1/T2]c

= n1[M1/M2]1[L1/L2]2[T1/T2]-2

= 4.2[1Kg/α kg]1[1m/βm]2[1sec/γ sec]-2 = 4.2 α–1 β–2 γ2

Discussion

9.	Compute The Dimensional Formula Of Electrical Resistance (r)?
Answer» ACCORDING to OHM’s law V = IR or R = V/I Since Work DONE (W) = QV where Q is the charge ⇒ R = W/QI = W/I2t (I = Q/t) DIMENSIONS of Work [W] = [M1L2T-2] ∴Dimension of R = [R] = [M1L2T-2][A-2T-1] = [M1L2T-3A-2] According to Ohm’s law V = IR or R = V/I Since Work done (W) = QV where Q is the charge ⇒ R = W/QI = W/I2t (I = Q/t) Dimensions of Work [W] = [M1L2T-2] ∴Dimension of R = [R] = [M1L2T-2][A-2T-1] = [M1L2T-3A-2]

9.

Compute The Dimensional Formula Of Electrical Resistance (r)?

Answer»

ACCORDING to OHM’s law

V = IR or R = V/I

Since Work DONE (W) = QV where Q is the charge

⇒ R = W/QI = W/I2t (I = Q/t)

DIMENSIONS of Work [W] = [M1L2T-2]

∴Dimension of R = [R] = [M1L2T-2][A-2T-1] = [M1L2T-3A-2]

According to Ohm’s law

V = IR or R = V/I

Since Work done (W) = QV where Q is the charge

⇒ R = W/QI = W/I2t (I = Q/t)

Dimensions of Work [W] = [M1L2T-2]

∴Dimension of R = [R] = [M1L2T-2][A-2T-1] = [M1L2T-3A-2]

Discussion

10.	Hooke’s Law States That The Force, F, In A Spring Extended By A Length X Is Given By F = −kx. According To Newton’s Second Law F = Ma, Where M Is The Mass And A Is The Acceleration. Calculate The Dimension Of The Spring Constant K?
Answer» Given, F = -KX ⇒ k = – F/x F = ma, the DIMENSIONS of force is: [F] = ma = [M1L0T0].[M0L1T-2] = [M1L1T-2] Therefore, DIMENSION of spring CONSTANT (k) is: [k] = [F]/[x] = [M1L1T-2].[M0L-1T0] = [M1L0T-2] or [MT-2] ….. Given, F = -kx ⇒ k = – F/x F = ma, the dimensions of force is: [F] = ma = [M1L0T0].[M0L1T-2] = [M1L1T-2] Therefore, dimension of spring constant (k) is: [k] = [F]/[x] = [M1L1T-2].[M0L-1T0] = [M1L0T-2] or [MT-2] …..

10.

Hooke’s Law States That The Force, F, In A Spring Extended By A Length X Is Given By F = −kx. According To Newton’s Second Law F = Ma, Where M Is The Mass And A Is The Acceleration. Calculate The Dimension Of The Spring Constant K?

Answer»

Given, F = -KX

⇒ k = – F/x

F = ma,

the DIMENSIONS of force is:

[F] = ma = [M1L0T0].[M0L1T-2] = [M1L1T-2]

Therefore, DIMENSION of spring CONSTANT (k) is:

[k] = [F]/[x] = [M1L1T-2].[M0L-1T0] = [M1L0T-2] or [MT-2] …..

Given, F = -kx

⇒ k = – F/x

F = ma,

the dimensions of force is:

[F] = ma = [M1L0T0].[M0L1T-2] = [M1L1T-2]

Therefore, dimension of spring constant (k) is:

[k] = [F]/[x] = [M1L1T-2].[M0L-1T0] = [M1L0T-2] or [MT-2] …..

Discussion

11.	Ncert): A Man Walking Briskly In Rain With Speed V Must Slant His Umbrella Forward Making An Angle θ With The Vertical. A Student Derives The Following Relation Between θ And V : Tan θ = V And Checks That The Relation Has A Correct Limit: As V → 0, θ →0, As Expected. (we Are Assuming There Is No Strong Wind And That The Rain Falls Vertically For A Stationary Man). Do You Think This Relation Can Be Correct ? If Not, Guess The Correct Relation?
Answer» Given, V = tanθ Dimensions of LHS = [v] = [M0L1T-1] Dimension of RHS = [tanθ] = [M0L0T0] (trigonometric ratios are dimensionless) Since [LHS] ≠ [RHS]. Equation is dimensionally INCORRECT. To MAKE the equation dimensionally correct, LHS should also be dimension less. It may be possible if consider speed of rainfall (Vr) and the equation will become: TAN θ = v/Vr Given, v = tanθ Dimensions of LHS = [v] = [M0L1T-1] Dimension of RHS = [tanθ] = [M0L0T0] (trigonometric ratios are dimensionless) Since [LHS] ≠ [RHS]. Equation is dimensionally incorrect. To make the equation dimensionally correct, LHS should also be dimension less. It may be possible if consider speed of rainfall (Vr) and the equation will become: tan θ = v/Vr

11.

Ncert): A Man Walking Briskly In Rain With Speed V Must Slant His Umbrella Forward Making An Angle θ With The Vertical. A Student Derives The Following Relation Between θ And V : Tan θ = V And Checks That The Relation Has A Correct Limit: As V → 0, θ →0, As Expected. (we Are Assuming There Is No Strong Wind And That The Rain Falls Vertically For A Stationary Man). Do You Think This Relation Can Be Correct ? If Not, Guess The Correct Relation?

Answer»

Given, V = tanθ

Dimensions of LHS = [v] = [M0L1T-1]

Dimension of RHS = [tanθ] = [M0L0T0] (trigonometric ratios are dimensionless)

Since [LHS] ≠ [RHS]. Equation is dimensionally INCORRECT.

To MAKE the equation dimensionally correct, LHS should also be dimension less. It may be possible if consider speed of rainfall (Vr) and the equation will become:

TAN θ = v/Vr

Given, v = tanθ

Dimensions of LHS = [v] = [M0L1T-1]

Dimension of RHS = [tanθ] = [M0L0T0] (trigonometric ratios are dimensionless)

Since [LHS] ≠ [RHS]. Equation is dimensionally incorrect.

To make the equation dimensionally correct, LHS should also be dimension less. It may be possible if consider speed of rainfall (Vr) and the equation will become:

tan θ = v/Vr

Discussion

12.	Check The Following Equation For Calculating Displacement Is Dimensionally Correct Or Not (a) X = X0 + Ut + (1/2) At2 Where, X Is Displacement At Given Time T Xo Is The Displacement At T = 0 U Is The Velocity At T = 0 A Represents The Acceleration. (b) P = (ρgh)½ Where P Is The Pressure, ρ Is The Density G Is Gravitational Acceleration H Is The Height.
Answer» (a) x = x0 + ut + (1/2) at2 Applying PRINCIPLE of homogeneity, all the sub-expressions of the equation must have the same DIMENSION and be equal to [LHS] Dimension of x = [M0L1T0] Dimensions of sub-expressions of [RHS] must be [M0L1T0] ⇒ Dimension of x0 (displacement) = [M0L1T0] = [LHS] Dimension of ut = VELOCITY x time = [M0L1T-1][M0L0T1] = [M0L1T0] = [LHS] Dimension of at2 = acceleration x (time)2 = [M0L1T-2][M0L0T-2] = [M0L1T0] = [LHS] ∴ The equation is dimensionally correct. (b) P = (ρgh)½ Dimensions of LHS i.e. Pressure [P] = [M1L-1T-2] Dimensions of ρ = mass/volume = [M1L-3T0] Dimensions of G (acceleration) = [M0L1T-2] Dimensions of h (height) = [M0L1T0] Dimensions of RHS = [(ρgh)½] = ([M1L-3T0]. [M0L1T-2].[M0L1T0])½ = ([M1L-1T-2])½ = [M½L-½T-1] ≠ [LHS] (a) x = x0 + ut + (1/2) at2 Applying principle of homogeneity, all the sub-expressions of the equation must have the same dimension and be equal to [LHS] Dimension of x = [M0L1T0] Dimensions of sub-expressions of [RHS] must be [M0L1T0] ⇒ Dimension of x0 (displacement) = [M0L1T0] = [LHS] Dimension of ut = velocity x time = [M0L1T-1][M0L0T1] = [M0L1T0] = [LHS] Dimension of at2 = acceleration x (time)2 = [M0L1T-2][M0L0T-2] = [M0L1T0] = [LHS] ∴ The equation is dimensionally correct. (b) P = (ρgh)½ Dimensions of LHS i.e. Pressure [P] = [M1L-1T-2] Dimensions of ρ = mass/volume = [M1L-3T0] Dimensions of g (acceleration) = [M0L1T-2] Dimensions of h (height) = [M0L1T0] Dimensions of RHS = [(ρgh)½] = ([M1L-3T0]. [M0L1T-2].[M0L1T0])½ = ([M1L-1T-2])½ = [M½L-½T-1] ≠ [LHS]

12.

Check The Following Equation For Calculating Displacement Is Dimensionally Correct Or Not (a) X = X0 + Ut + (1/2) At2 Where, X Is Displacement At Given Time T Xo Is The Displacement At T = 0 U Is The Velocity At T = 0 A Represents The Acceleration. (b) P = (ρgh)½ Where P Is The Pressure, ρ Is The Density G Is Gravitational Acceleration H Is The Height.

Answer»

(a) x = x0 + ut + (1/2) at2

Applying PRINCIPLE of homogeneity, all the sub-expressions of the equation must have the same DIMENSION and be equal to [LHS]

Dimension of x = [M0L1T0]

Dimensions of sub-expressions of [RHS] must be [M0L1T0]

⇒ Dimension of x0 (displacement) = [M0L1T0] = [LHS]

Dimension of ut = VELOCITY x time = [M0L1T-1][M0L0T1] = [M0L1T0] = [LHS]

Dimension of at2 = acceleration x (time)2 = [M0L1T-2][M0L0T-2] = [M0L1T0] = [LHS]

∴ The equation is dimensionally correct.

(b) P = (ρgh)½

Dimensions of LHS i.e. Pressure [P] = [M1L-1T-2]

Dimensions of ρ = mass/volume = [M1L-3T0]

Dimensions of G (acceleration) = [M0L1T-2]

Dimensions of h (height) = [M0L1T0]

Dimensions of RHS = [(ρgh)½] = ([M1L-3T0]. [M0L1T-2].[M0L1T0])½ = ([M1L-1T-2])½

= [M½L-½T-1] ≠ [LHS]

(a) x = x0 + ut + (1/2) at2

Applying principle of homogeneity, all the sub-expressions of the equation must have the same dimension and be equal to [LHS]

Dimension of x = [M0L1T0]

Dimensions of sub-expressions of [RHS] must be [M0L1T0]

⇒ Dimension of x0 (displacement) = [M0L1T0] = [LHS]

Dimension of ut = velocity x time = [M0L1T-1][M0L0T1] = [M0L1T0] = [LHS]

Dimension of at2 = acceleration x (time)2 = [M0L1T-2][M0L0T-2] = [M0L1T0] = [LHS]

∴ The equation is dimensionally correct.

(b) P = (ρgh)½

Dimensions of LHS i.e. Pressure [P] = [M1L-1T-2]

Dimensions of ρ = mass/volume = [M1L-3T0]

Dimensions of g (acceleration) = [M0L1T-2]

Dimensions of h (height) = [M0L1T0]

Dimensions of RHS = [(ρgh)½] = ([M1L-3T0]. [M0L1T-2].[M0L1T0])½ = ([M1L-1T-2])½

= [M½L-½T-1] ≠ [LHS]

Discussion

13.	(ncert): A Famous Relation In Physics Relates ‘moving Mass’ M To The ‘rest Mass’ Mo Of A Particle In Terms Of Its Speed V And The Speed Of Light, C. (this Relation First Arose As A Consequence Of Special Relativity Due To Albert Einstein). A Boy Recalls The Relation Almost Correctly But Forgets Where To Put The Constant C. He Writes?
Answer» Dimension of m (MASS) = [M1L0T0] Dimension of m0 (mass) = [M1L0T0] Dimension of V (velocity) = [M0L1T-1] ∴ Dimension of v2= [M0L2T-2] Dimension of c (velocity) = [M0L1T-1] Applying principle of HOMOGENEITY of dimensions, [LHS] = [RHS] = [M1L0T0] ⇒ The equation (1- v2)½ must be dimension less, which is possible if we have the expressions as: (1 – v2/c2) The equation after placing ‘c’ Dimension of m (mass) = [M1L0T0] Dimension of m0 (mass) = [M1L0T0] Dimension of v (velocity) = [M0L1T-1] ∴ Dimension of v2= [M0L2T-2] Dimension of c (velocity) = [M0L1T-1] Applying principle of homogeneity of dimensions, [LHS] = [RHS] = [M1L0T0] ⇒ The equation (1- v2)½ must be dimension less, which is possible if we have the expressions as: (1 – v2/c2) The equation after placing ‘c’

13.

(ncert): A Famous Relation In Physics Relates ‘moving Mass’ M To The ‘rest Mass’ Mo Of A Particle In Terms Of Its Speed V And The Speed Of Light, C. (this Relation First Arose As A Consequence Of Special Relativity Due To Albert Einstein). A Boy Recalls The Relation Almost Correctly But Forgets Where To Put The Constant C. He Writes?

Answer»

Dimension of m (MASS) = [M1L0T0]

Dimension of m0 (mass) = [M1L0T0]

Dimension of V (velocity) = [M0L1T-1]

∴ Dimension of v2= [M0L2T-2]

Dimension of c (velocity) = [M0L1T-1]

Applying principle of HOMOGENEITY of dimensions, [LHS] = [RHS] = [M1L0T0]

⇒ The equation (1- v2)½ must be dimension less, which is possible if we have the expressions as:

(1 – v2/c2) The equation after placing ‘c’

Dimension of m (mass) = [M1L0T0]

Dimension of m0 (mass) = [M1L0T0]

Dimension of v (velocity) = [M0L1T-1]

∴ Dimension of v2= [M0L2T-2]

Dimension of c (velocity) = [M0L1T-1]

Applying principle of homogeneity of dimensions, [LHS] = [RHS] = [M1L0T0]

⇒ The equation (1- v2)½ must be dimension less, which is possible if we have the expressions as:

(1 – v2/c2) The equation after placing ‘c’

Discussion

14.	(ncert): A Book With Many Printing Errors Contains Four Different Formulas For The Displacement Y Of A Particle Undergoing A Certain Periodic Motion: (a) Y = A Sin 2π T/t (b) Y = A Sin Vt (c) Y = (a/t) Sin T/a (d) Y = (a 2) (sin 2πt / T + Cos 2πt / T ) (a = Maximum Displacement Of The Particle, V = Speed Of The Particle. T = Time-period Of Motion). Rule Out The Wrong Formulas On Dimensional Grounds.
Answer» GIVEN, Dimension of a = displacement = [M0L1T0] Dimension of V (speed) = distance/time = [M0L1T-1] Dimension of t or T (time period) = [M0L0T1] Trigonometric function sine is a ratio, hence it must be dimensionless. (a) y = a sin 2π t/T (correct ✓ ) Dimensions of RHS = [L1] sin([T].[T-1] ) = [M0L1T0] = LHS (eqation is correct). (b) y = a sin vt (wrong ✗) RHS = [L1] sin([LT-1] [T1]) = [L1] sin([L]) = wrong, since trigonometric function must be dimension less. (c) y = (a/T) sin t/a (wrong ✗) RHS = [L1] sin([T].[L-1] ) = [L1] sin([TL-1] ) = wrong, sine function must be dimensionless. (d) y = (a 2) (sin 2πt / T + cos 2πt / T ) (correct ✓ ) RHS = [L1] ( sin([T].[T-1] + cos([T].[T-1] ) = [L1] ( sin(M0L1T0) + cos(M0L1T0) ) = [L1] = RHS = equation is dimensionally correct. Given, Dimension of a = displacement = [M0L1T0] Dimension of v (speed) = distance/time = [M0L1T-1] Dimension of t or T (time period) = [M0L0T1] Trigonometric function sine is a ratio, hence it must be dimensionless. (a) y = a sin 2π t/T (correct ✓ ) Dimensions of RHS = [L1] sin([T].[T-1] ) = [M0L1T0] = LHS (eqation is correct). (b) y = a sin vt (wrong ✗) RHS = [L1] sin([LT-1] [T1]) = [L1] sin([L]) = wrong, since trigonometric function must be dimension less. (c) y = (a/T) sin t/a (wrong ✗) RHS = [L1] sin([T].[L-1] ) = [L1] sin([TL-1] ) = wrong, sine function must be dimensionless. (d) y = (a 2) (sin 2πt / T + cos 2πt / T ) (correct ✓ ) RHS = [L1] ( sin([T].[T-1] + cos([T].[T-1] ) = [L1] ( sin(M0L1T0) + cos(M0L1T0) ) = [L1] = RHS = equation is dimensionally correct.

14.

(ncert): A Book With Many Printing Errors Contains Four Different Formulas For The Displacement Y Of A Particle Undergoing A Certain Periodic Motion: (a) Y = A Sin 2π T/t (b) Y = A Sin Vt (c) Y = (a/t) Sin T/a (d) Y = (a 2) (sin 2πt / T + Cos 2πt / T ) (a = Maximum Displacement Of The Particle, V = Speed Of The Particle. T = Time-period Of Motion). Rule Out The Wrong Formulas On Dimensional Grounds.

Answer»

GIVEN,

Dimension of a = displacement = [M0L1T0]

Dimension of V (speed) = distance/time = [M0L1T-1]

Dimension of t or T (time period) = [M0L0T1]

Trigonometric function sine is a ratio, hence it must be dimensionless.

(a) y = a sin 2π t/T (correct ✓ )

Dimensions of RHS = [L1] sin([T].[T-1] ) = [M0L1T0] = LHS (eqation is correct).

(b) y = a sin vt (wrong ✗)

RHS = [L1] sin([LT-1] [T1]) = [L1] sin([L]) = wrong, since trigonometric function must be dimension less.

(c) y = (a/T) sin t/a (wrong ✗)

RHS = [L1] sin([T].[L-1] ) = [L1] sin([TL-1] ) = wrong, sine function must be dimensionless.

(d) y = (a 2) (sin 2πt / T + cos 2πt / T ) (correct ✓ )

RHS = [L1] ( sin([T].[T-1] + cos([T].[T-1] ) = [L1] ( sin(M0L1T0) + cos(M0L1T0) )

= [L1] = RHS = equation is dimensionally correct.

Given,

Dimension of a = displacement = [M0L1T0]

Dimension of v (speed) = distance/time = [M0L1T-1]

Dimension of t or T (time period) = [M0L0T1]

Trigonometric function sine is a ratio, hence it must be dimensionless.

(a) y = a sin 2π t/T (correct ✓ )

Dimensions of RHS = [L1] sin([T].[T-1] ) = [M0L1T0] = LHS (eqation is correct).

(b) y = a sin vt (wrong ✗)

RHS = [L1] sin([LT-1] [T1]) = [L1] sin([L]) = wrong, since trigonometric function must be dimension less.

(c) y = (a/T) sin t/a (wrong ✗)

RHS = [L1] sin([T].[L-1] ) = [L1] sin([TL-1] ) = wrong, sine function must be dimensionless.

(d) y = (a 2) (sin 2πt / T + cos 2πt / T ) (correct ✓ )

RHS = [L1] ( sin([T].[T-1] + cos([T].[T-1] ) = [L1] ( sin(M0L1T0) + cos(M0L1T0) )

= [L1] = RHS = equation is dimensionally correct.

Discussion

15.	What Are The Main Uses Of Dimensional Equations?
Answer» To test of CORRECTNESS of equations. To derive the equations To CONVERT ONE system of UNITS into another. To RECAPITULATE important formulae.

15.

What Are The Main Uses Of Dimensional Equations?

Answer»

To test of CORRECTNESS of equations.
To derive the equations
To CONVERT ONE system of UNITS into another.
To RECAPITULATE important formulae.

Discussion

16.	What Is Physical Quantity Of Dimesion?
Answer» These are the powers to which the FUNDAMENTAL UNITS must be raised, when the quantity is expressed INTERMS of these units. These are the powers to which the fundamental units must be raised, when the quantity is expressed interms of these units.

16.

What Is Physical Quantity Of Dimesion?

Answer»

These are the powers to which the FUNDAMENTAL UNITS must be raised, when the quantity is expressed INTERMS of these units.

These are the powers to which the fundamental units must be raised, when the quantity is expressed interms of these units.

Discussion

17.	What Is Dimensional Analysis?
Answer» A method used to find a relation between various PHYSICAL QUANTITIES. Also to calculate how a physical quantity will DEPEND in terms of the powers of FUNDAMENTAL units on which it intuitively depends. The method is based on the prinicple that the dimensions of the fudamental quantities ( M, L and T ) must be the same on both sides of an equation. A method used to find a relation between various physical quantities. Also to calculate how a physical quantity will depend in terms of the powers of fundamental units on which it intuitively depends. The method is based on the prinicple that the dimensions of the fudamental quantities ( M, L and T ) must be the same on both sides of an equation.

17.

What Is Dimensional Analysis?

Answer»

A method used to find a relation between various PHYSICAL QUANTITIES. Also to calculate how a physical quantity will DEPEND in terms of the powers of FUNDAMENTAL units on which it intuitively depends.

The method is based on the prinicple that the dimensions of the fudamental quantities ( M, L and T ) must be the same on both sides of an equation.

A method used to find a relation between various physical quantities. Also to calculate how a physical quantity will depend in terms of the powers of fundamental units on which it intuitively depends.

The method is based on the prinicple that the dimensions of the fudamental quantities ( M, L and T ) must be the same on both sides of an equation.

Discussion

18.	What Is Metastable State?
Answer» This is a STATE of a SYSTEM in which it is apparently in a stable EQUILIBRIUM, however if slightly distrubed the system changes to a new state of LOWER energy. This is a state of a system in which it is apparently in a stable equilibrium, however if slightly distrubed the system changes to a new state of lower energy.

18.

What Is Metastable State?

Answer»

This is a STATE of a SYSTEM in which it is apparently in a stable EQUILIBRIUM, however if slightly distrubed the system changes to a new state of LOWER energy.

This is a state of a system in which it is apparently in a stable equilibrium, however if slightly distrubed the system changes to a new state of lower energy.

Discussion

19.	Which Are Called Non - Contact Forces?
Answer» Some forces are only PRODUCED when the one object TOUCHES ANOTHER. These force are called non - contact forces. Some forces are only produced when the one object touches another. These force are called non - contact forces.

19.

Which Are Called Non - Contact Forces?

Answer»

Some forces are only PRODUCED when the one object TOUCHES ANOTHER. These force are called non - contact forces.

Some forces are only produced when the one object touches another. These force are called non - contact forces.

Discussion

20.	Which Scientist Proved The Electro Weak Force They?
Answer» ABDUS Salam became the FIRST person from pakistan who won a NOBEL prize for PROVE this theory. Abdus Salam became the first person from pakistan who won a nobel prize for prove this theory.

20.

Which Scientist Proved The Electro Weak Force They?

Answer»

ABDUS Salam became the FIRST person from pakistan who won a NOBEL prize for PROVE this theory.

Abdus Salam became the first person from pakistan who won a nobel prize for prove this theory.

Discussion

21.	Which Are The Basic Forces?
Answer» The basic FORCES are gravity, ELECTRICITY, magnetism and two kinds of NUCLEAR forces CALLED weak and strong forces. The basic forces are gravity, electricity, magnetism and two kinds of nuclear forces called weak and strong forces.

21.

Which Are The Basic Forces?

Answer»

The basic FORCES are gravity, ELECTRICITY, magnetism and two kinds of NUCLEAR forces CALLED weak and strong forces.

The basic forces are gravity, electricity, magnetism and two kinds of nuclear forces called weak and strong forces.

Discussion

22.	What Is Escape Velocity?
Answer» The MINIMUM speed that a space rocket MUST reach to escape the EARTH’s GRAVITY. The minimum speed that a space rocket must reach to escape the earth’s gravity.

22.

What Is Escape Velocity?

Answer»

The MINIMUM speed that a space rocket MUST reach to escape the EARTH’s GRAVITY.

The minimum speed that a space rocket must reach to escape the earth’s gravity.

Discussion

23.	Which Indian Physcist Worked With Albert Einstein ?
Answer» Satyendranath Bose, who with Einstein DEVELOPED a system of statical QUANTUM mecahnics now known SA Bose Einstein Statistics. Satyendranath Bose, who with Einstein developed a system of statical quantum mecahnics now known sa Bose Einstein Statistics.

23.

Which Indian Physcist Worked With Albert Einstein ?

Answer»

Satyendranath Bose, who with Einstein DEVELOPED a system of statical QUANTUM mecahnics now known SA Bose Einstein Statistics.

Satyendranath Bose, who with Einstein developed a system of statical quantum mecahnics now known sa Bose Einstein Statistics.

Discussion

24.	What Is Velocity?
Answer» Velocity is the RATE of change of the position, equal to SPEED in a particular DIRECTION. Velocity is the rate of change of the position, equal to speed in a particular direction.

24.

What Is Velocity?

Answer»

Velocity is the RATE of change of the position, equal to SPEED in a particular DIRECTION.

Velocity is the rate of change of the position, equal to speed in a particular direction.

Discussion

25.	What Is Horizontal Velocity Of Projectile ?
Answer» The horizontal component of the velocity of the body remains same THROUGHOUT because there is no acceleration ( DUE to GRAVITY ) in the horizontal direction. The vertical component of the veocity initially ( i.e. at t = 0 ) is zero and the vertical component keeps on increasing TILL body touches the GROUND. The horizontal component of the velocity of the body remains same throughout because there is no acceleration ( due to gravity ) in the horizontal direction. The vertical component of the veocity initially ( i.e. at t = 0 ) is zero and the vertical component keeps on increasing till body touches the ground.

25.

What Is Horizontal Velocity Of Projectile ?

Answer»

The horizontal component of the velocity of the body remains same THROUGHOUT because there is no acceleration ( DUE to GRAVITY ) in the horizontal direction.

The vertical component of the veocity initially ( i.e. at t = 0 ) is zero and the vertical component keeps on increasing TILL body touches the GROUND.

The horizontal component of the velocity of the body remains same throughout because there is no acceleration ( due to gravity ) in the horizontal direction.

The vertical component of the veocity initially ( i.e. at t = 0 ) is zero and the vertical component keeps on increasing till body touches the ground.

Discussion

26.	What Is Range Of Projectile?
Answer» The distance between the point of projection and the point where the TRAJECTORY MEETS the HORIZONTAL plane through the point of projection is CALLED its range ( horizontal ). The distance between the point of projection and the point where the trajectory meets the horizontal plane through the point of projection is called its range ( horizontal ).

26.

What Is Range Of Projectile?

Answer»

The distance between the point of projection and the point where the TRAJECTORY MEETS the HORIZONTAL plane through the point of projection is CALLED its range ( horizontal ).

The distance between the point of projection and the point where the trajectory meets the horizontal plane through the point of projection is called its range ( horizontal ).

Discussion

27.	What Is Angle Of Departure Or Angle Of Projection?
Answer» The angle that the of projection MAKES with the HORIZONTAL is CALLED angle of departure or angle of projection. CLEARLY angle of projection for a horizontal projectile is ZERO. The angle that the of projection makes with the horizontal is called angle of departure or angle of projection. Clearly angle of projection for a horizontal projectile is zero.

27.

What Is Angle Of Departure Or Angle Of Projection?

Answer»

The angle that the of projection MAKES with the HORIZONTAL is CALLED angle of departure or angle of projection. CLEARLY angle of projection for a horizontal projectile is ZERO.

The angle that the of projection makes with the horizontal is called angle of departure or angle of projection. Clearly angle of projection for a horizontal projectile is zero.

Discussion

28.	Why We Do Not Ever Consider Rate Of Change Of Acceleration?
Answer» It is FOUND that the BASIC LAWS of motion involve only acceleration and not the rate of CHANGE of acceleration, so we never CONSIDER the rate of change of acceleration. It is found that the basic laws of motion involve only acceleration and not the rate of change of acceleration, so we never consider the rate of change of acceleration.

28.

Why We Do Not Ever Consider Rate Of Change Of Acceleration?

Answer»

It is FOUND that the BASIC LAWS of motion involve only acceleration and not the rate of CHANGE of acceleration, so we never CONSIDER the rate of change of acceleration.

It is found that the basic laws of motion involve only acceleration and not the rate of change of acceleration, so we never consider the rate of change of acceleration.

Discussion

29.	How Is The Position Time Graph Helpful In Studying The Motion Of The Body?
Answer» With the HELP of this GRAPH, we can determine, distance TRAVELLED during any interval of time and ALSO the velocity of the BODY at any instant of time. With the help of this graph, we can determine, distance travelled during any interval of time and also the velocity of the body at any instant of time.

29.

How Is The Position Time Graph Helpful In Studying The Motion Of The Body?

Answer»

With the HELP of this GRAPH, we can determine, distance TRAVELLED during any interval of time and ALSO the velocity of the BODY at any instant of time.

With the help of this graph, we can determine, distance travelled during any interval of time and also the velocity of the body at any instant of time.

Discussion

30.	What Is The Direction Of The Velocity And Acceleration When It Is Thrown Down Wards?
Answer» Velocity is VERTICALLY DOWNWARD and ACCELERATION is ALSO vertically downwards. Velocity is vertically downward and acceleration is also vertically downwards.

30.

What Is The Direction Of The Velocity And Acceleration When It Is Thrown Down Wards?

Answer»

Velocity is VERTICALLY DOWNWARD and ACCELERATION is ALSO vertically downwards.

Velocity is vertically downward and acceleration is also vertically downwards.

Discussion

31.	What Is The Direction Of The Velocity And Acceleration When It Is Thrown Upwards ?
Answer» VELOCITY is VERTICALLY UPWARD and ACCELERATION is vertically DOWNWARDS. Velocity is vertically upward and acceleration is vertically downwards.

31.

What Is The Direction Of The Velocity And Acceleration When It Is Thrown Upwards ?

Answer»

VELOCITY is VERTICALLY UPWARD and ACCELERATION is vertically DOWNWARDS.

Velocity is vertically upward and acceleration is vertically downwards.

Discussion

32.	What Is Retardation?
Answer» If the velocity increases, the acceleration is POSITIVE and if the velocity DECREASES, the acceleration is negative. The negative acceleration is called RETARDATION. If the velocity increases, the acceleration is positive and if the velocity decreases, the acceleration is negative. The negative acceleration is called retardation.

32.

What Is Retardation?

Answer»

If the velocity increases, the acceleration is POSITIVE and if the velocity DECREASES, the acceleration is negative. The negative acceleration is called RETARDATION.

If the velocity increases, the acceleration is positive and if the velocity decreases, the acceleration is negative. The negative acceleration is called retardation.

Discussion

33.	What Is Instantaneous Acceleration?
Answer» If the motion changes of a BODY is such that its VELOCITY changes by unequal values in equal intervals of time, then the value of the accleration at any INSTANT is called instantaneous ACCELERATION. If the motion changes of a body is such that its velocity changes by unequal values in equal intervals of time, then the value of the accleration at any instant is called instantaneous acceleration.

33.

What Is Instantaneous Acceleration?

Answer»

If the motion changes of a BODY is such that its VELOCITY changes by unequal values in equal intervals of time, then the value of the accleration at any INSTANT is called instantaneous ACCELERATION.

If the motion changes of a body is such that its velocity changes by unequal values in equal intervals of time, then the value of the accleration at any instant is called instantaneous acceleration.

Discussion

34.	What Is Uniform Acceleration?
Answer» A BODY is SAID to be moving with uniform acceleration, if its velocity changes by equal VALUES in equal INTERVALS of time. A body is said to be moving with uniform acceleration, if its velocity changes by equal values in equal intervals of time.

34.

What Is Uniform Acceleration?

Answer»

A BODY is SAID to be moving with uniform acceleration, if its velocity changes by equal VALUES in equal INTERVALS of time.

A body is said to be moving with uniform acceleration, if its velocity changes by equal values in equal intervals of time.

Discussion

35.	What Is Acceleration?
Answer» It is the RATE of CHANGE of VELOCITY with TIME. It is the rate of change of velocity with time.

35.

What Is Acceleration?

Answer»

It is the RATE of CHANGE of VELOCITY with TIME.

It is the rate of change of velocity with time.

Discussion

36.	What Is Instantaneous Velocity?
Answer» The velocity of a body in a NON - uniform motion at any instant is CALLED INSTANTANEOUS velocity. It is different from the average velocity over an interval of TIME. The velocity of a body in a non - uniform motion at any instant is called instantaneous velocity. It is different from the average velocity over an interval of time.

36.

What Is Instantaneous Velocity?

Answer»

The velocity of a body in a NON - uniform motion at any instant is CALLED INSTANTANEOUS velocity. It is different from the average velocity over an interval of TIME.

The velocity of a body in a non - uniform motion at any instant is called instantaneous velocity. It is different from the average velocity over an interval of time.

Discussion

37.	What Is Average Velocity?
Answer» It is the RATIO of the TOTAL DISTANCE travelled to the total time TAKEN by the body. It is the ratio of the total distance travelled to the total time taken by the body.

37.

What Is Average Velocity?

Answer»

It is the RATIO of the TOTAL DISTANCE travelled to the total time TAKEN by the body.

It is the ratio of the total distance travelled to the total time taken by the body.

Discussion

38.	What Is Variable Velocity?
Answer» If a body COVERS unequal distances in equal intervals of time along a straight line or if the body CHANGES the direction of motion ( though it MAY be covering equal intervals of time ) , it is said to process variable velocity. If a body covers unequal distances in equal intervals of time along a straight line or if the body changes the direction of motion ( though it may be covering equal intervals of time ) , it is said to process variable velocity.

38.

What Is Variable Velocity?

Answer»

If a body COVERS unequal distances in equal intervals of time along a straight line or if the body CHANGES the direction of motion ( though it MAY be covering equal intervals of time ) , it is said to process variable velocity.

If a body covers unequal distances in equal intervals of time along a straight line or if the body changes the direction of motion ( though it may be covering equal intervals of time ) , it is said to process variable velocity.

Discussion

39.	What Is Non - Uniform Motion?
Answer» When a BODY travels unequal distance in equal intervals of TIME, the motion is SAID to be non - UNIFORM motion. When a body travels unequal distance in equal intervals of time, the motion is said to be non - uniform motion.

39.

What Is Non - Uniform Motion?

Answer»

When a BODY travels unequal distance in equal intervals of TIME, the motion is SAID to be non - UNIFORM motion.

When a body travels unequal distance in equal intervals of time, the motion is said to be non - uniform motion.

Discussion

40.	What Is Uniform Velocity?
Answer» If the BODY MOVES equal distances in equal intervals of TIME and always in the same DIRECTION, then it is said to possess uniform velocity. If the body moves equal distances in equal intervals of time and always in the same direction, then it is said to possess uniform velocity.

40.

What Is Uniform Velocity?

Answer»

If the BODY MOVES equal distances in equal intervals of TIME and always in the same DIRECTION, then it is said to possess uniform velocity.

If the body moves equal distances in equal intervals of time and always in the same direction, then it is said to possess uniform velocity.

Discussion

41.	What Is Uniform Motion?
Answer» A motion is said to be uniform if the BODY moves equal DISTANCES in equal intervals of TIME and always in the same direction. For such a motion, the ACTUAL DISTANCE covered in time t is the magnitude of the displacement. A motion is said to be uniform if the body moves equal distances in equal intervals of time and always in the same direction. For such a motion, the actual distance covered in time t is the magnitude of the displacement.

41.

What Is Uniform Motion?

Answer»

A motion is said to be uniform if the BODY moves equal DISTANCES in equal intervals of TIME and always in the same direction. For such a motion, the ACTUAL DISTANCE covered in time t is the magnitude of the displacement.

A motion is said to be uniform if the body moves equal distances in equal intervals of time and always in the same direction. For such a motion, the actual distance covered in time t is the magnitude of the displacement.

Discussion

42.	What Is Meant By Instanteous Position Of An Object?
Answer» The position CO - ordinate in a moving body describes TE definite and EXACT position of the body at any time. The position of a body at any instant is CALLED instantaneous position. The position co - ordinate in a moving body describes te definite and exact position of the body at any time. The position of a body at any instant is called instantaneous position.

42.

What Is Meant By Instanteous Position Of An Object?

Answer»

The position CO - ordinate in a moving body describes TE definite and EXACT position of the body at any time. The position of a body at any instant is CALLED instantaneous position.

The position co - ordinate in a moving body describes te definite and exact position of the body at any time. The position of a body at any instant is called instantaneous position.

Discussion

43.	What Is Meant By Negative Time And Positive Time?
Answer» If we ASSIGN a negative time to an EVENT, it MEANS that it OCCURED before the event to which positive time was assigned. If we assign a negative time to an event, it means that it occured before the event to which positive time was assigned.

43.

What Is Meant By Negative Time And Positive Time?

Answer»

If we ASSIGN a negative time to an EVENT, it MEANS that it OCCURED before the event to which positive time was assigned.

If we assign a negative time to an event, it means that it occured before the event to which positive time was assigned.

Discussion

44.	Which Type Of Motions Are The Following ?
Answer» Motion ALONG a CIRCLE. Motion along a CURVE Both are MOTIONS in TWO dimensions Both are motions in two dimensions

44.

Which Type Of Motions Are The Following ?

Answer»

Motion ALONG a CIRCLE.
Motion along a CURVE

Both are MOTIONS in TWO dimensions

Both are motions in two dimensions

Discussion

45.	What Is Motion?
Answer» It is the CHANGE of position of an object in the course of time. The BODY in motion is treated as a particle. The motion has been classfied as :- Motion in one dimension - i.e atrain running on a RAILWAY track. Motion in two dimensions - i.e motion of a stone which is thrown in the horizontal direction from the top of a tower. Motion in THREE dimensions - i.e the motion of the molecules of a GAS. It is the change of position of an object in the course of time. The body in motion is treated as a particle. The motion has been classfied as :-

45.

What Is Motion?

Answer»

It is the CHANGE of position of an object in the course of time.

The BODY in motion is treated as a particle.

The motion has been classfied as :-

Motion in one dimension - i.e atrain running on a RAILWAY track.
Motion in two dimensions - i.e motion of a stone which is thrown in the horizontal direction from the top of a tower.
Motion in THREE dimensions - i.e the motion of the molecules of a GAS.

It is the change of position of an object in the course of time.

The body in motion is treated as a particle.

The motion has been classfied as :-

Discussion

46.	What Is Mechanics?
Answer» This DEALS with all the SUBJECTS NAMELY, kinematics, DYNAMICS and statics. This deals with all the subjects namely, kinematics, dynamics and statics.

46.

What Is Mechanics?

Answer»

This DEALS with all the SUBJECTS NAMELY, kinematics, DYNAMICS and statics.

This deals with all the subjects namely, kinematics, dynamics and statics.

Discussion

47.	What Is Statics?
Answer» It is the STUDY of objects at REST i.e. when a large number of forces acting on a BODY are in equilibrium. It is the study of objects at rest i.e. when a large number of forces acting on a body are in equilibrium.

47.

What Is Statics?

Answer»

It is the STUDY of objects at REST i.e. when a large number of forces acting on a BODY are in equilibrium.

It is the study of objects at rest i.e. when a large number of forces acting on a body are in equilibrium.

Discussion

48.	What Is Dynamics?
Answer» It is dervied from the greek WORD “ dynamics ” which means POWER. It deals with the study of motion TAKING into CONSIDERATION the forces which cause motion. It is dervied from the greek word “ dynamics ” which means power. It deals with the study of motion taking into consideration the forces which cause motion.

48.

What Is Dynamics?

Answer»

It is dervied from the greek WORD “ dynamics ” which means POWER. It deals with the study of motion TAKING into CONSIDERATION the forces which cause motion.

It is dervied from the greek word “ dynamics ” which means power. It deals with the study of motion taking into consideration the forces which cause motion.

Discussion

49.	What Is Kinematics?
Answer» The word kinematics is derived from the greek word “ Knemia ” which MEANS MOTION. Thus kinematics is the study of motion. We study the position, velocity, ACCELERATION etc. of a BODY without specifyng the nature of the body and the nature of the forces which cause motion. In this branch we study ways to DESCRIBE motion of object independent of causes of motion and independent of the nature of the body. The word kinematics is derived from the greek word “ Knemia ” which means motion. Thus kinematics is the study of motion. We study the position, velocity, acceleration etc. of a body without specifyng the nature of the body and the nature of the forces which cause motion. In this branch we study ways to describe motion of object independent of causes of motion and independent of the nature of the body.

49.

What Is Kinematics?

Answer»

The word kinematics is derived from the greek word “ Knemia ” which MEANS MOTION. Thus kinematics is the study of motion. We study the position, velocity, ACCELERATION etc. of a BODY without specifyng the nature of the body and the nature of the forces which cause motion. In this branch we study ways to DESCRIBE motion of object independent of causes of motion and independent of the nature of the body.

The word kinematics is derived from the greek word “ Knemia ” which means motion. Thus kinematics is the study of motion. We study the position, velocity, acceleration etc. of a body without specifyng the nature of the body and the nature of the forces which cause motion. In this branch we study ways to describe motion of object independent of causes of motion and independent of the nature of the body.

Discussion

50.	How Many Base Units Are There In The S.i System ?
Answer» There are only SEVEN BASE UNITS and TWO SUPPLEMENTARY units. There are only seven base units and two supplementary units.

50.

How Many Base Units Are There In The S.i System ?

Answer»

There are only SEVEN BASE UNITS and TWO SUPPLEMENTARY units.

There are only seven base units and two supplementary units.

Discussion

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Compute The Dimensional Formula Of Electrical Resistance (r)?

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What Are The Main Uses Of Dimensional Equations?

What Is Physical Quantity Of Dimesion?

What Is Dimensional Analysis?

What Is Metastable State?

Which Are Called Non - Contact Forces?

Which Scientist Proved The Electro Weak Force They?

Which Are The Basic Forces?

What Is Escape Velocity?

Which Indian Physcist Worked With Albert Einstein ?

What Is Velocity?

What Is Horizontal Velocity Of Projectile ?

What Is Range Of Projectile?

What Is Angle Of Departure Or Angle Of Projection?

Why We Do Not Ever Consider Rate Of Change Of Acceleration?

How Is The Position Time Graph Helpful In Studying The Motion Of The Body?

What Is The Direction Of The Velocity And Acceleration When It Is Thrown Down Wards?

What Is The Direction Of The Velocity And Acceleration When It Is Thrown Upwards ?

What Is Retardation?

What Is Instantaneous Acceleration?

What Is Uniform Acceleration?

What Is Acceleration?

What Is Instantaneous Velocity?

What Is Average Velocity?

What Is Variable Velocity?

What Is Non - Uniform Motion?

What Is Uniform Velocity?

What Is Uniform Motion?

What Is Meant By Instanteous Position Of An Object?

What Is Meant By Negative Time And Positive Time?

Which Type Of Motions Are The Following ?

What Is Motion?

What Is Mechanics?

What Is Statics?

What Is Dynamics?

What Is Kinematics?

How Many Base Units Are There In The S.i System ?