Q 21 :

If force (F), velocity (V) and time (T) are considered as fundamental physical quantities, then dimensional formula of density will be:

  • FV-2T2

     

  • FV-4T-2

     

  • FV4T2

     

  • F2V-2T6

     

(2)

M1L-3T0=(M1LT-2)a(LT-1)bTc    

a=1, b=-4, c=-2



Q 22 :

The electric resistance of medium depends upon permeability (μ) and permittivity (ε) as given below R∝(μ)a(ε)b. If a-b=N. The value of N is ...........



(1)

ML2T-3A-2=[MLT-2A-2]a[M-1L-3T4A2]b

a-b=1

a-3b=2

a=12,  b=-12



Q 23 :

An expression of energy density is given by u=αβsin(αxkt). Where α,β are constants, x is displacement, k is Boltzmann constant and t is the absolute temperature. The dimensions of β will be:

  • [ML2T-2θ-1]

     

  • [M0L2T-2]

     

  • [M0L0T0]

     

  • [M0L2T0]

     

(4)

sin(αxkt)=Dimensionless

∴ α[L][ML2T-2]=[M0L0T0]⇒α=[MLT-2]

αβ=EnergyVolume=[ML2T-2][L3]

⇒β=[MLT-2][L3][ML2T-2]

=[M0L2T0]



Q 24 :

If F is force, V is velocity, H is latent heat, and s is specific heat, then match the quantity given in Column I with the dimensions given in Column II.

  Column I   Column II
i. F a. L2T-2
ii. V b. L2T-2K-1
iii. H c. MLT-2
iv. S d. LT-1

 

  • i → c, ii → d, iii → a, iv → b

     

  • i → a, ii → b, iii → c, iv → d

     

  • i → d, ii → a, iii → b, iv → c

     

  • i → c, ii → a, iii → b, iv → d

     

(1)

iii.   [H]=[HeatMass]=ML2T-2M=L2T-2

iv.   [s]=[HeatMass×Temperature]=ML2T-2MK=L2T-2K-1



Q 25 :

Match list I with list II:

  List I   List II
a) Torque (i) MLT-1
b) Impulse (ii) MT-2
c) Tension (iii) ML2T-2
d) Surface tension (iv) MLT-2

 

Choose the most appropriate answer from the options given below:

  • a)-(iii), b)-(iv), c)-(i), d)-(ii)

     

  • a)-(iii), b)-(i), c)-(iv), d)-(ii)

     

  • a)-(i), b)-(iii), c)-(iv), d)-(ii)

     

  • a)-(ii), b)-(i), c)-(iv), d)-(iii)

     

(2)

Torque-ML2T-2

Impulse-MLT-1

Tension-MLT-2

Surface tension-MT-2



Q 26 :

The dimensional formula of 12ε0E2 (ε0 = permittivity of vacuum and E = electric field) is MaLbTc. The value of 2a-b+c = ______.       [2026]

  • 0

     

  • 1

     

  • -1

     

  • 2

     

(2)

12ε0E2=dudv=EnergyVolume

∴ [12ε0E2]=ML2T-2L3=ML-1T-2

∴ a=1, b=-1, c=-2

∴ 2a-b+c=2+1-2=1



Q 27 :

Dimensions of universal gravitational constant (G) in terms of Planck's constant (h), distance (L), mass (M) and time (T) are ______.              [2026]

  • [hTLM-2]

     

  • [hT-1LM-2]

     

  • [hTL2M-2]

     

  • [h-1T-1LM-2]

     

(2)

G=M-1L3T-2                        ...(i)

h=ML2T-1                              ...(ii)

Dividing eq. (i) to eq. (ii)

Gh=M-1L3T-2ML2T-1

G=M-2LT-1h1



Q 28 :

Match List-I with List-II.  

  List-I   List-II
A. Planck’s constant I. ML2T-2
B. Stopping potential II. T-1
C. Work function III. ML2T-1
D. Threshold frequency IV. ML2T-3A-1

 

Choose the correct answer from the options given below:                      [2026]

  • A-III, B-IV, C-I, D-II

     

  • A-I, B-II, C-III, D-IV

     

  • A-IV, B-III, C-I, D-II

     

  • A-I, B-IV, C-III, D-II

     

(1)

(A)  [h]=[E][ν]=ML2T-2T-1=ML2T-1

(B)  Stopping potential: E=q·V0

       [V0]=[E][q]=ML2T-2AT=ML2T-3A-1

(C)  Work function (ϕ)→Energy form [ϕ]=ML2T-2

(D)  Threshold freq. [ν0]=T-1



Q 29 :

L, C and R represents physical quantities inductance capacitance and resistance respectively. The dimensional formula ML2T-4A-2 corresponds to ______.      [2026]

  • RLC

     

  • RLC

     

  • CLR

     

  • 1RLC

     

(1)

R=VI=WqI=[ML2T-2][A2T]=[ML2T-3A-2]

U=12LI2⇒[ML2T-2]=[L][A2]

[L]=[ML2T-2A-2]

Q=CV⇒[AT]=[C][ML2T-3A-1]

[C]=[M-1L-2T4A2]

[LC]=[ML2T-2A-2]×[M-1L-2T4A2]=[T2]

RLC=[ML2T-3A-2][T]=[ML2T-4A-2]



Q 30 :

Match List-I with List-II.

  List-I   List-II
A. Meter (L) I. hcG
B. Second (S) II. Ghc5
C. Kilogram (M) III. K2L2c3Gh
D. Kelvin (K) IV. Ghc3

 

where h (Planck’s constant), G (gravitational constant) and c (speed of light in vacuum) as fundamental units.

Choose the correct answer from the options given below:                              [2026]

  • A-II, B-IV, C-I, D-III

     

  • A-IV, B-II, C-I, D-III

     

  • A-IV, B-I, C-II, D-III

     

  • A-III, B-I, C-II, D-IV

     

(2)

[h]=[ML2T-1]

[c]=[LT-1] and [G]=[M-1L3T-2]

then [hcG]=[ML2T-1 LT-1M-1L3T-2]1/2=[M]

[Ghc5]=[M-1L3T-2 ML2T-1L5T-5]1/2=[T]

[K2L2c3Gh]=[K2L2L3T-3M-1L3T-2 ML2T-1]1/2=[K]

[Ghc3]=[M-1L3T-2 ML2T-1L3T-3]1/2=[L]

A→IV, B→II, C→I, D→III



Q 31 :

The potential energy of a particle changes with distance x from a fixed origin as V=Axx+B, where A and B are constant with appropriate dimensions. The dimension of AB are ______.                    [2026]

  • [M1L5/2T-2]

     

  • [M3/2L5/2T-2]

     

  • [M1L2T-2]

     

  • [M1L7/2T-2]

     

(4)

[B]=[x]=[L]

[A]=[PE]×[length]length=ML2T-2×LL1/2

[A]=[ML5/2T-2]

[A·B]=[ML5/2T-2]×[L]=[ML7/2T-2]



Q 32 :

Match List-I with List-II.                            [2026]

  List-I   List-II
A. Boltzmann constant I. [M-1L3T-2]
B. Stefan’s constant II. [ML2T-1]
C. Planck’s constant III. [ML2T-2K-1]
D. Gravitational constant IV. [ML0T-3K-4]

 

  • A-I, B-II, C-III, D-IV

     

  • A-IV, B-III, C-II, D-I

     

  • A-III, B-IV, C-II, D-I

     

  • A-II, B-I, C-IV, D-III

     

(3)

Boltzmann constant=EnergyTemp.

⇒[k]=[ML2T-2K-1]

σ⇒PAT4⇒[σ]=[M1T-3K-4]

h=Eν=E·t⇒[h]=[M1L2T-1]

G=Fr2m2⇒[G]=[M-1L3T-2]

Option (3) is correct.



Q 33 :

Consider the equation H=xpεqErts

Where H = magnetising intensity; E = electric field; ε = permittivity, x = distance, t = time. The values of p, q, r and s respectively are            [2026]

  • 1, 1, 1, 1

     

  • -1, 1, 2, 1

     

  • 1, -1, -2, 1

     

  • -1, -2, -2, 1

     

(1)

As per problem if we take H as magnetic field then no option matching but if we take H as magnetising intensity instead of magnetic field then only we get answer

H=Bμ0-Ni⇒current per unit length

⇒[H]=[AL-1]

      [E]=[MLT-2][IT]=[MLA-1T-3]

      [ε]=[I2T2][MLT-2L2]=[M-1L-3I2T4]

       H=xpεqErts

        AL-1=Lp(M-1L-3T4A2)q(MLT-3A-1)rTs

        -q+r=0                   ...(i)

         p-3q+r=-1        ...(ii)

         4q-3r-s=0          ...(iii)

          2q-r=1                   ...(iv)

Solving (i), (ii), (iii) and (iv)

p=1, q=1, r=1, s=1



Q 34 :

A ball of radius r and density ρ dropped through a viscous liquid of density σ and viscosity η attains its terminal velocity at time t, given by t=Aρarbηcσd, where A is a constant and a,b,c and d are integers. The value of b+ca+d is ______.            [2026]



(1)

T1=[ML-3]a[L]b[ML-1T-1]c[ML-3]d

M0L0T1=Ma+c+dL-3a+b-c-3dT-c

On comparing

a+c+d=0                            ...(i)

-3a+b-c-3d=0            ...(ii)

-c=1                                      ...(iii)

From (i) and (iii): a+d=1

From (ii): a+d=b-c3

b+13=1⇒b=2

b+c=2+(-1)=1

⇒b+ca+d=11=1