Q 11 :

Match the List I with List II

  List I   List II
(A) Gravitational constant (I) [LT–2]
(B) Gravitational potential energy (II) [L2T–2]
(C) Gravitational potential (III) [ML2T–2]
(D) Acceleration due to gravity (IV) [M–1L3T–2]

 

Choose the correct answer from the options given below:          [2025]

  • A-IV, B-III, C-II, D-I

     

  • A-III, B-II, C-I, D-IV

     

  • A-II, B-IV, C-III, D-I

     

  • A-I, B-III, C-IV, D-II

     

(1)

(A) G=Fr2m2

         [G]=[MLT–2][L2][M2]=[M–1L3T–2] ⇒ A→IV

(B) P.E. U=mgh=[MLT–2L]=[ML2T–2] ⇒ B→III

(C) Gravitational Potential, V=GMr

                                                            V=[M–1L3T–2][M][L]=[M0L2T–2]=[L2T–2] ⇒ C→II

(D) Acceleration due to gravity =[g]=[LT–2] ⇒ D→I



Q 12 :

Match the List I with List II

  List I   List II
(A) Boltzmann constant (I) ML2T–1
(B) Coefficient of viscosity (II) MLT–3K–1
(C) Planck's constant (III) ML2T–2K–1
(D) Thermal conductivity (IV) ML–1T–1

 

Choose the correct answer from the options given below:          [2025]

  • A-III, B-IV, C-I, D-II

     

  • A-II, B-III, C-IV, D-I

     

  • A-III, B-II, C-I, D-IV

     

  • A-III, B-IV, C-II, D-I

     

(1)

(A) [k]=PVNT=ML2T–2K=ML2T–2K–1

(B) [η]=F6πrv=MLT–2L2T–1=ML–1T–1

(C) [h]=Ef=ML2T–2T–1=ML2T–1

(D) dQdt=kAdTdx, k=(ML2T–3)LL2·K=MLT–3K–1



Q 13 :

In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of MPLQTRAS, where value of 'Q' and 'R' are          [2025]

  • (3, –5)

     

  • (–2, 2)

     

  • (–2, 1)

     

  • (1, –1)

     

(4)

ϕEϕM=EABA=EB

EB=c ⇒ [EB]=[c]=[LT–1]



Q 14 :

In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of [MPLQTRAS]. The value of P and Q are:        [2025]

  • –1, 0

     

  • –1, 1

     

  • 1, –1

     

  • 0, –1

     

(4)

Electric dipole moment (P→)=q×2l

Magnetic dipole moment =(M→)=IA

               [PM]=[LTAL2A]=L–1T=M0L–1T1A0

After comparing values of P and Q are 0, –1.



Q 15 :

The dimension of μ0ε0 is equal to that of : (μ0 = Vacuum permeability and ε0 = Vacuum permittivity)          [2025]

  • Voltage

     

  • Capacitance

     

  • Inductance

     

  • Resistance

     

(4)

[μ0]=[MLT–2A–2]

[ε0]=[M–1L–3T4A2]

[μ0ε0]=[ML2T–3A–2]≃[R]



Q 16 :

Match the List-I with List-II.

  List-I   List-II
(A) Mass density (I) [ML2T-3]
(B) Impulse (II) [MLT-1]
(C) Power (III) [ML2T0]
(D) Moment of inertia (IV) [ML-3T0]

 

Choose the correct answer from the options given below :          [2025]

  • (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

     

  • (A)-(I), (B)-(III), (C)-(IV), (D)-(II)

     

  • (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

     

  • (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

     

(3)

(A) Mass density =MV=[M1L–3]                      ... (IV)

(B) Impulse =M×u=[M1L1T–1]                     ... (II)

(C) Power =F·v=[M1L2T–3]                          ... (I)

(D) Moment of inertia =Mr2=[M1L2]          ... (III)



Q 17 :

In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of [MaLbTc]. If b = –3, the value of c is __________.          [2025]



(0)

[Modulus of elasticity] =ML–1T–2

[Torque] =ML2T–2

[Modulus of elasticity per unit torque] =ML–1T–2ML2T–2=L–3.



Q 18 :

The frequency (v) of an oscillating liquid drop may depend upon radius (r) of the drop, density (ρ) of liquid and the surface tension (s) of the liquid as : v=raρbsc. The values of a, b and c respectively are          [2023]

  • (–32,12,12)

     

  • (–32,–12,12)

     

  • (32,12,–12)

     

  • (32,–12,12)

     

(2)

[T–1]=[L1]a[M1L–3]b[MLT-2L]c

⇒ T–1=Mb+c·La–3b·T–2c

              c=12, b=–12, a–3b=0

              a+32=0 ⇒ a=–32



Q 19 :

Match the following Column – I with Column – II

  Column –I   Column – II
A. Surface tension I. kg m–1s–1
B. Pressure II. kg ms–1
C. Viscosity III. kg m–1s–2
D. Impulse IV. kg s–2

Choose the correct answer from the options given below:          [2023]

  • A – III; B – IV; C – I; D – II

     

  • A – IV; B – III; C – II; D – I

     

  • A – IV; B – III; C – I; D – II

     

  • A – II; B – I; C – III; D – IV

     

(3)

Surface tension – N/m →kg s–2

Pressure – N/m2→kg m–1s–2

Viscosity – kg m–1s–1

Impulse – kg ms–1

A –IV, B –III, C – I, D – II



Q 20 :

Match List I with List II

  List I   List II
A. Young's Modulus (Y) I. [ML–1T–1]
B. Co-efficient of Viscosity (η) II. [ML2T–1]
C. Planck's Constant (h) III. [ML–1T–2]
D. Work Function (ϕ) IV. [ML2T–2]

Choose the correct answer from the options given below:          [2023]

  • A – I, B – III, C – IV, D – II

     

  • A – I, B – II, C – III, D – IV

     

  • A – II, B – III, C – IV, D – I

     

  • A – III, B – I, C – II, D – IV

     

(4)

Y=StressStrain=F/A△l/l=[MLT–2][L2]=[ML–1T–2]

F=6πηrv ⇒ η=F6πrv

[η]=[MLT–2][L][LT–1]=[ML–1T–1]

E=hv ⇒ h=Ev=[ML2T–2][T–1]=[ML2T–1]

Work function has same dimension as that of energy, so

[ϕ]=[ML2T–2]



Q 21 :

Match List I with List II:

  List I   List II
A. Pressure gradient I. [M0L2T–2]
B. Energy density II. [M1L–1T–2]
C. Electric field III. [M1L–2T–2]
D. Latent heat IV. [M1L1T–3A–1]

Choose the correct answer from the options given below:          [2023]

  • A – III, B – II, C – I, D – IV

     

  • A – II, B – III, C – IV, D – I

     

  • A – III, B – II, C – IV, D – I

     

  • A – II, B – III, C – I, D – IV

     

(3)

Pressure gradient = dpdx=[ML–1T–2][L]=[M1L–2T–2]

Energy density = energyvolume=[ML2T–2][L3]=[M1L–1T–2]

Electric field = Forcecharge=[MLT–2][A·T]=[M1L1T–3A–1]

Latent heat = heatmass=[ML2T–2][M]=[M0L2T–2]



Q 22 :

The equation of a circle is given by x2+y2=a2, where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation : (x–At)2+(y–tB)2=a2

The dimensions of t is given as [T–1].          [2023]

  • A=[L–1T], B=[L–1T]

     

  • A=[LT], B=[L–1T–1]

     

  • A=[L–1T–1], B=[L–1]

     

  • A=[L–1T–1], B=[LT]

     

(2)

(x–At)2+(y–tB)2=a2

    [At]=A×1T=L

∴  [A]=T1L1

      tB is in meters

∴  1T[B]=L ⇒ [B]=T–1L–1



Q 23 :

Match List I with List II.

  List I   List II
A. Torque I. kg m–1s–2
B. Energy density II. kg ms–1
C. Pressure gradient III. kg m–2s–2
D. Impulse IV. kg m2s–2

Choose the correct answer from the options given below:          [2023]

  • A – IV, B – III, C – I, D – II

     

  • A – I, B – IV, C – III, D – II

     

  • A – IV, B – I, C – II, D – III

     

  • A – IV, B – I, C – III, D – II

     

(4)

Torque →kg m2s–2                     (IV)

Energy density →kg m–1s–2                 (I)

Pressure gradient →kg m–2s–2           (III)

Impulse →kg ms–1                  (II)



Q 24 :

Match List I with List II.

  List I   List II
A. Angular momentum I. [ML2T–2]
B. Torque II. [ML–2T–2]
C. Stress III. [ML2T–1]
D. Pressure gradient IV. [ML–1T–2]

Choose the correct answer from the options given below:          [2023]

  • A – I, B – IV, C – III, D – II

     

  • A – III, B – I, C – IV, D – II

     

  • A – II, B – III, C – IV, D – I

     

  • A – IV, B – II, C – I, D – III

     

(2)

L¯=r→×p→ ⇒ [L]=[M0L1T0][M1L1T–1]=[M1L2T–1]

τ¯=r→×F→ ⇒ [τ]=[L1][MLT–2]=[ML2T–2]

Stress ≡ Pressure = FA ⇒ [Stress]=[ML–1T–2]

Pressure Gradient = dPdx

⇒ [Pressure Gradient] = [ML–1T–2]



Q 25 :

(P+aV2)(V–b)=RT represents the equation of state of some gases. Where P is the pressure, V is the volume, T is the temperature and a, b, R are the constants. The physical quantity, which has dimensional formula as that of b2a, will be          [2023]

  • bulk modulus

     

  • modulus of rigidity

     

  • compressibility

     

  • energy density

     

(3)

[b] = [V]

[ab2]=[P]

∴  [b2a]=1[P]=1[B]=[K]



Q 26 :

If the velocity of light c, universal gravitational constant G and planck's constant h are chosen as fundamental quantities. The dimensions of mass in the new system is:          [2023]

  • [h12c–12G1]

     

  • [h1c1G–1]

     

  • [h–12c12G12]

     

  • [h12c12G–12]

     

(4)

Dimensional formale of mass is HxCyGz

M1=(ML2T–1)x(LT–1)(M–1L3T–2)z

M1L0T0=Mx–zL2x+y+3zT–x–y–2z

on comparing both side

x–z=1

2x+y+3z=0

–x–y–2z=0

On solving above equations we get

x=12, y=12, z=–12



Q 27 :

If R, XL and XC represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless:          [2023]

  • RXLXC

     

  • RXLXC

     

  • RXLXC

     

  • RXLXC

     

(2)

All three have same dimension therefore RXLXC is dimensionless.



Q 28 :

Match List I with List II

  List I   List II
A. Planck's constant (h) I, [M1L2T–2]
B. Stopping potential (VS) II. [M1L1T–1]
C. Work function (ϕ) III. [M1L2T–1]
D. Momentum (p) IV. [M1L2T–3A–1]

Choose the correct answer from the options given below:           [2023]

  • A – III, B – IV, C – II, D – I

     

  • A – III, B – IV, C – I, D – II

     

  • A – I, B – III, C – IV, D – I

     

  • A – III, B – I, C – II, D – IV

     

(2)

(A) Planck's constant

     hv = E

     h=Ev=M1L2T–2T–2=M1L2T–1

(B) E = qv

     V=Eq=M1L2T–2A1T1=M1L2T–3A–1

(C) ϕ (work function) = energy = M1L2T–2

(D) Momentum (p) = F.t = M1L1T–2T1=M1L1T–1



Q 29 :

Dimension of 1μ0ε0 should be equal to           [2023]

  • LT

     

  • L2T2

     

  • TL

     

  • T2L2

     

(2)

1μ0ε0=c2 ⇒ [1μ0ε0]=[c2]=[L2T–2]



Q 30 :

Match List I with List II

  List I   List II
A. Torque I. ML–2T–2
B. Stress II. ML2T–2
C, Pressure gradient III. ML–1T–1
D. Coefficient of viscosity IV. ML–1T–2

Choose the correct answer from the options given below:         [2023]

  • A – II, B – I, C – IV, D – III

     

  • A – II, B – IV, C – I, D – III

     

  • A – III, B – IV, C – I, D – II

     

  • A – IV, B – II, C – III, D – I

     

(2)

A.  Torque ⇒τ→=r→×F→

       [τ]=[L][MLT–2] ⇒ ML2T–2

B.  Stress = FA ⇒ MLT–2L2

       [Stress] = ML–1T–2

C.  Pressure gradient = △P△X

       ⇒ [F/A][L]=MLT–2L3 ⇒ ML–2T–2

D.  Coefficient of viscosity ⇒ F=6πηrv

       MLT–2=[η]L2T–1

       [η]=ML–1T–1



Q 31 :

If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be          [2023]

  • FV–2T2

     

  • FV–4T2

     

  • FV4T–6

     

  • F2V–2T6

     

(2)

[ML–3]=[MLT–2]a[LT–1]b[T]c=[MaLa+bT–2a+b+c]

             a = 1

             a + b = –3 ⇒ b = –4

Also, – 2a – b + c = 0

              c = 2



Q 32 :

Match List-I with List-II

  List I   List II
A. Spring constant I. (T–1)
B. Angular speed II. (MT–2)
C. Angular momentum III. (ML2)
D. Moment of Inertia IV. (ML2T–1)

Choose the correct answer from the options given below:          [2023]

  • A – II, B – I, C – IV, D – III

     

  • A – IV, B – I, C – III, D – II

     

  • A – II, B – III, C – I, D – IV

     

  • A – I, B – III, C – II, D – IV

     

(1)

Spring Constant

[K]=[F][x]=MLT–2L=MT–2

[ω]=[θ][t]=1T=T–1



Q 33 :

In the equation [X+aY2][Y–b]=RT, X is pressure, Y is volume, R is universal gas constant and T is temperature. The physical quantity a equivalent to the ratio ab is          [2023]

  • Energy

     

  • Impulse

     

  • Pressure gradient

     

  • Coefficient of viscosity

     

(1)

X and aY2 have same dimensions

Y and b have same dimensions

∴  [a]=[ML5T–2]

       [b]=[L3]

      [a][b]=[ML2T–2] has dimensions of energy.



Q 34 :

The speed of a wave produced in water is given by v=λagbρc. When λ, g and ρ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of a, b and c respectively, are          [2023]

  • 12,12,0

     

  • 1, 1, 0

     

  • 1, –1, 0

     

  • 12,0,12

     

(1)

v=λagbρc

using dimension formula

⇒ [M0L1T–1]=[L1]a[L1T–2]b[M1L–3]c

⇒ [M0L1T–1]=[McLa+b–3cT–2b]

∴  c=0, a+b–3c=1, –2b=–1 ⇒ b=12

Now, a + b –3c = 1

⇒ a+12–0=1

⇒ a=12

∴  a=12, b=12, c=0



Q 35 :

Consider a modified Bernoulli equation.

(P+ABt2)+ρg(h+Bt)+12ρv2=constant.

If t has the dimension of time then the dimensions of A and B are ________, ________ respectively.        [2026]

  • [ML0T-1] and [M0LT-1]

     

  • [ML0T-2] and [M0LT-2]

     

  • [ML0T-1] and [M0LT]

     

  • [ML0T-2] and [M0LT-1]

     

(1)

⇒  [P]=[ABt2]     .....(1)

⇒  [h]=[Bt]     .....(2)

⇒  [B]=[ht]=[LT]=[LT-1]

Putting B in equation (1)

[ML-1T-2]=[ALT-1×T2]

[A]=[ML0T-1]



Q 36 :

Match the LIST-I with LIST-II                     [2026]

  List - I   List - II
A. Magnetic induction I. MLT-2A-2
B. Magnetic flux II. ML2T-2A-2
C. Magnetic permeability III. ML0T-2A-1
D. Self inductance IV. ML2T-2A-1

 

Choose the correct answer from the options given below:

  • A-III, B-IV, C-II, D-I

     

  • A-I, B-III, C-IV, D-II

     

  • A-IV, B-III, C-I, D-II

     

  • A-III, B-IV, C-I, D-II

     

(4)

Magnetic induction

F=qvB

[B]=[Fqv]

[B]=[MT-2A-1]

Magnetic Flux (ϕ)

ϕ=(B)·(Area)

[ϕ]=[ML2T-2A-1]

Magnetic Permeability

[μ]=[MLT-2A-2]

Self inductance

Using U=12LI2

[Self inductance]=[ML2T-2A-1]

A-III, B-IV, C-I, D-II



Q 37 :

Match the LIST-I with LIST-II:                [2026]

  List - I   List - II
A. Spring constant I. ML2T-2K-1
B. Thermal conductivity II. ML0T-2
C. Boltzmann constant III. ML2T-3A-2
D. Inductive reactance IV. MLT-3K-1

 

Choose the correct answer from the options given below:

  • A-II, B-IV, C-I, D-III

     

  • A-II, B-I, C-IV, D-III

     

  • A-I, B-IV, C-II, D-III

     

  • A-III, B-II, C-IV, D-I

     

(1)

(A)  F=Kx

[MLT-2]=[K][L]

[K]=ML0T-2

(B) Thermal conductivity

dQdt=kAℓ ΔT

ML2T-3=[k]L2KL

[k]=MLT-3K-1

(C) Boltzmann constant

[K]=ML2T-2K-1

(D) Inductive reactance

[V][I]=ML2T-3A-1A

=ML2T-3A-2



Q 38 :

Match List – I with List – II      [2026]

List – I

List – II

A. Coefficient of viscosity

I. [ML⁻¹T⁻²]

B. Surface tension

II. [ML²T⁻²]

C. Pressure

III. [ML⁰T⁻²]

D. Surface energy

IV. [ML⁻¹T⁻¹]

 

Choose the correct answer from the options given below :

  • A–I, B–II, C–IV, D–III

     

  • A–I, B–III, C–II, D–IV

     

  • A–IV, B–III, C–I, D–II

     

  • A–IV, B–I, C–II, D–III

     

(3)

(A)  η=FdrAdv=[MLT-2][L][L2][LT-1]=[ML-1T-1]

(B)  S=FL=[MLT-2][L]=[MT-2]

(C)  P=FA=[MLT-2][L2]=[ML-1T-2]

(D)  E=S×A=[MT-2][L2]=[ML2T-2]



Q 39 :

The force of interaction between two atoms is given by F=αβexp(-x2αkT); where 'x' is the distance, 'k' is the Boltzmann constant and T is temperature and α and β are two constants. The dimensions of β is:

  • M0L2T-4

     

  • M2LT-4

     

  • MLT-2

     

  • M2L2T-2

     

(2)

Force of interaction between two atoms, F=αβe(-x2αkT)

Since exponential terms are dimensionless

∴   αβe(-x2αkT)

[F]=[α][β]

MLT-2=M-1T2[β]

⇒ [β]=M2LT-4



Q 40 :

Match the List-I with List-II and select the correct answer using the codes given below the list.

  List-I   List-II
A. Boltzmann constant I. [ML2T-1]
B. Coefficient of viscosity II. [ML-1T-1]
C. Planck constant III. [MLT-3K-1]
D. Thermal conductivity IV. [ML2T-2K-1]

Choose the correct answer from the options given below:

  • A-III, B-I, C-II, D-IV

     

  • A-III, B-II, C-I, D-IV

     

  • A-IV, B-II, C-I, D-III

     

  • A-IV, B-I, C-II, D-III

     

(3)

conceptual