Q21JEE Main 2025MCQ

If $\mu _0$ and $\epsilon _0$ are the permeability and permittivity of free space, respectively, then the dimension of $(\;\frac{1}{\mu _0 \epsilon _0})$ is[2 Apr 2025 Shift 2]

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📖 Explanation

$\;C=\;{1}/{{\sqrt{\mu _0 \epsilon _0}}}$ $\;\mu _0 \epsilon _0 \equiv \;\frac{1}{C^2} \equiv [L^{-2} T^2]$ $\; \therefore [\;\frac{1}{\mu _0 \epsilon _0}] =\;\frac{L^2}{T^2}$

Q22JEE Main 2025MCQ

Match the LIST-I with LIST-II. LIST-ILIST-IIA.Boltzmann constantI. ${\text{ML}}^2 {\text{T}}^{-1}$ B.Coefficient of viscosityII. ${\text{MLT}}^{-3} {\text{K}}^{-1}$ C.Planck's constantIII. ${\text{ML}}^2 {\text{T}}^{-2} {\text{K}}^{-1}$ D.Thermal conductivityIV. ${\text{ML}}^{-1} {\text{T}}^{-1}$ Choose the correct answer from the options given below. [3 Apr 2025 Shift 2]

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📖 Explanation

$\;k t=\;\text{ Energy }\; \Rightarrow k= {\text{ML}}^2 {\text{T}}^{-2} {\text{K}}^{-1}$ $\; {\text{A}} \longrightarrow {\text{III}}$ $\;F=6 \pi η r u \Rightarrow η=\;{ {\text{MLT}}^{-2}}/{ {\text{LLT}}^{-1}}$ $\;η= {\text{ML}}^{-1} {\text{T}}^{-1}$ $\; {\text{B}} \longrightarrow {\text{IV}}$ $\;E=h v=h= {\text{ML}}^2 {\text{T}}^{-1}$ $\; {\text{C}} \longrightarrow {\text{I}}$ $\;H=\;\frac{k A \Delta T}{I}=k=\;{ {\text{ML}}^2 {\text{T}}^{-3}}/{ {\text{LK}}}= {\text{MLT}}^{-3} {\text{K}}^{-1}$ $\; {\text{D}} \longrightarrow {\text{II}}$

Q23JEE Main 2025MCQ

The dimension of $\sqrt{\;\frac{\mu _0}{\epsilon _0}}$ is equal to that of : ( $\mu _0=$ Vacuum permeability and $\epsilon _0=$ Vacuum permittivity)[7 Apr 2025 Shift 2]

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📖 Explanation

$\;{ [\mu _0] = \$ [M L T^{-2} A^{-2}]$ };{ [\epsilon _0] = $[M^{-1} L^{-3} T^4 A^2]$ };{ [\sqrt{;\frac{\mu _0}{\epsilon _0}}] = $[M L^2 T^{-3} A^{-2}]$ =[R]}$

Q24JEE Main 2025MCQ

The maximum percentage error in the measurement of density of a wire is[Given, mass of wire $=(0.60 ± 0.003) {\text{g}}$ radius of wire $=(0.50 ± 0.01) {\text{cm}}$ length of wire $=(10.00 ± 0.05) {\text{cm}}]$ [22 Jan 2025 Shift 2]

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📖 Explanation

$\rho =\;\frac{m}{v}=\;\frac{m}{\pi r^2 l}$ $\;\;\frac{\Delta \rho }{\rho }=\;\frac{\Delta m}{m}+\;\frac{\Delta l}{l}+\;\frac{2 \Delta r}{r}$ $\;=\;\frac{1}{2} \%+\;\frac{1}{2} \%+2 \times 2 \%$ $\;=5 \%$

Q25JEE Main 2025MCQ

Given a charge $q$ , current $I$ and permeability of vacuum $\mu _0$ . Which of the following quantity has the dimension of momentum?[2 Apr 2025 Shift 2]

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📖 Explanation

$\;\;\text{ Momentum }\;=m v$ $\;F \equiv B v q$ $\;\;\frac{m v}{t} \equiv \;\frac{\mu _0 I v q}{r}$ $\;m v \equiv \mu _0 I q$

Q26JEE Main 2025MCQ

A person measures mass of 3 different particles as $435.42 {\text{g}}, 226.3 {\text{g}}$ and 0.125 g . According to the rules for arithmetic operations with significant figures, the addition of the masses of 3 particles will be[3 Apr 2025 Shift 1]

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📖 Explanation

$\;435.42$ $\;226.3$ $\;\;\;\;\;0.125$ $\;───$ $661.845$ For addition, minimum decimal number is considered for reporting the measured value $\Rightarrow 661.8$

Q27JEE Main 2025MCQ

Match List- 1 with List-II List - I List - II (A) Permeability of free space (I) $[ {\text{M}}^2 {\text{T}}^{-2}]$ (B) Magnetic field (II) $[ {\text{MT}}^{-2} {\text{A}}^{-1}]$ (C) Magnetic moment (III) $[ {\text{MLT}}^{-2} {\text{A}}^{-2}]$ (D) Torsional constant (IV) $[ {\text{L}}^2 {\text{A}}]$ Choose the correct answer from the options given below[23 Jan 2025 Shift 2]

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📖 Explanation

(A) $B=\;\frac{\mu _0 i}{2 \pi r} & q v B=F$ $\mu _0 \equiv \;\frac{B r}{i} \equiv \;\frac{F r}{q v i} \equiv \;\frac{MLT^{-2} \times L}{A^2 TLT^{-1}} \equiv MLT^{-2} A^{-2} \;\;(\text{III})$ (B) $B \equiv \;\frac{F}{q v} \equiv \;\frac{ MLT ^{-2}}{ ATLT ^{-1}}= MT ^{-2} A^{-1} \;\;(\text{II})$ (C) $M=i A \equiv AL ^2 \;\;(\text{IV})$ (D) $\tau =C \theta \Rightarrow C \equiv ML ^2 T^{-2} \;\;(\text{I})$

Q28JEE Main 2025MCQ

Match List - I with List - II. List - I List - II (A) Young's Modulus (I) ${\text{ML}}^{-1} {\text{T}}^{-1}$ (B) Torque (II) ${\text{ML}}^{-1} {\text{T}}^{-2}$ (C) Coefficient of Viscosity (III) ${\text{M}}^{-1} {\text{L}}^3 {\text{T}}^{-2}$ (D) Gravitational Constant (IV) ${\text{ML}}^2 {\text{T}}^{-2}$ Choose the correct answer from the options given below : [29 Jan 2025 Shift 2]

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📖 Explanation

(A) Young's Modulus $(Y)=Y=\;\frac{F}{A (\;\frac{\Delta I}{I}) }$ $[Y]= [ML^{-1} T^{-2}]$ (B) Torque $(ξ^{-1}) =\vec{r} \times \vec{F}$ $[ξ]= [ML^2 T^{-2}]$ (C) Coefficient of Viscosity $(η)$ $F=η A \;\frac{d V}{d t}$ $[η]= [M L^{-1} T^{-1}]$ (D) Gravitational Constant $[G]=\;\frac{F \cdot d^2}{m_1 m_2}$ $[G]= [M^{-1} L^3 T^{-2}]$

Q29JEE Main 2025MCQ

The expression given below shows the variation of velocity $(v)$ with lime $(t), \;\; v=A t^{2+}\;\frac{B t}{C+t}$ . The dimension of $A B C$ is : [29 Jan 2025 Shift 1]

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📖 Explanation

$\;v=A t^{2+}\;\frac{B t}{C+t}$ $\;{[v]=[A] t^2= [\;\frac{B t}{C+t}] }$ $\;{[A]= LT ^{-3}}$ $\;{[C]= T }$ $\;{[B]= LT ^{-1}}$ $\;{[A B C]= L ^2 T^{-3}}$

Q30JEE Main 2025NAT

A physical quantity $Q$ is related to four observables $a, b, c, d$ as follows : $Q=\;\frac{a b^4}{c d}$ where, $a=(60 ± 3) {\text{Pa}} ; b=(20 ± 0.1) {\text{m}} ; c=(40 ±$ $0.2) {\text{Nsm}}^{-2}$ and $d=(50 ± 0.1) {\text{m}}$ , then the percentage error in $Q$ is $\;\frac{x}{1000}$ , where $x=$ ______ .[29 Jan 2025 Shift 2]

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📖 Explanation

$\;\;\text{ Here, }\; \;\;\frac{\Delta Q}{Q} \times 100= [\;\frac{\Delta a}{a}+4 \;\frac{\Delta b}{b}+\;\frac{\Delta c}{c}+\;\frac{\Delta d}{d}] \times 100$ $\;\; = [\;\frac{3}{60}+\;\frac{0.4}{20}+\;\frac{0.2}{40}+\;\frac{0.1}{50}]$ $\;\;\frac{x}{1000}= [\;\frac{3}{60}+\;\frac{0.4}{20}+\;\frac{0.2}{40}+\;\frac{0.1}{50}] \times 100$ $\;x=7700$

Q31JEE Main 2025MCQ

The position of a particle moving on $x$ -axis is given by $x(t)=A sin \;t+B \cos ^2 t+C t^{2+}D$ , where $t$ is time. The dimension of $\;\frac{A B C}{D}$ is[23 Jan 2025 Shift 1]

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📖 Explanation

$\;A \equiv {\text{L}}$ $\;B \equiv {\text{L}}$ $\;C \equiv {\text{LT}}^{-2}$ $\;D \equiv {\text{L}}$ $\;\Rightarrow \;\frac{A B C}{D}=\;{ {\text{L}} \times {\text{L}} \times {\text{LT}}^2}/{ {\text{L}}}= {\text{L}}^2 {\text{T}}^{-2}$

Q32JEE Main 2025MCQ

Match List-I with List-II. LIST-ILIST-II(A)Coefficient of viscosity(I) $[ {\text{ML}}^0 {\text{T}}^{-3}]$ (B)Intensity of wave(II) $[ {\text{ML}}^{-2} {\text{T}}^{-2}]$ (C)Pressure gradient(III)[{\text{M}}^{-1} {\text{LT}}^2 $](D)Compressibility(IV)$ [ {\text{ML}}^{-1} {\text{T}}^{-1}]$ Choose the correct answer from the options given below. [2 Apr 2025 Shift 1]

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📖 Explanation

$η \equiv \;\frac{F}{r V} \equiv \$ [ {\text{ML}}^{-1} {\text{T}}^{-1}]$ ; ; ; ; ; {\text{A}} \longrightarrow {\text{IV}}I \equiv ;\frac{P}{A} \equiv $[ {\text{MT}}^{-3}]$ ;; ; ; ; ; ; ; ; ; ; ; {\text{B}} \longrightarrow {\text{I}};\frac{d P}{d x} \equiv $[ {\text{ML}}^{-2} {\text{T}}^{-2}]$ ; ; ; ; ; ; ; ; ; ; ; {\text{C}} \longrightarrow {\text{II}}K \equiv ;\frac{1}{D} \equiv $[ {\text{M}}^{-1} {\text{LT}}^2]$ ; ; ; ; ; ; ; ; {\text{D}} \longrightarrow {\text{III}}$

Q33JEE Main 2025MCQ

Given below are two statements : Statement-I: In a vernier callipers, one vernier scale division is always smaller than one main scale division.Statement-II: The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions.In the light of the above statements, choose the correct answer from the options given below. [22 Jan 2025 Shift 1]

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Q34JEE Main 2025MCQ

Match List-I with List-II. List-I List-II (A) Angular Impulse (I) $[ {\text{M}}^0 {\text{L}}^2 {\text{T}}^{-2}]$ (B) Latent Heat (II) $[ {\text{M}} {\text{L}}^2 {\text{T}}^{-3} {\text{A}}^{-1}]$ (C) Electrical resistivity (III) $[ {\text{M}} {\text{L}}^2 {\text{T}}^{-1}]$ (D) Electromotive force (IV) $[ {\text{M}} {\text{L}}^3 {\text{T}}^{-3} {\text{A}}^{-2}]$ Choose the correct answer from the options given below. [28 Jan 2025 Shift 2]

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📖 Explanation

[Angular Impulse] $= M L ^2 T^{-1}$ [Latent Heat] $= M ^0 L^2 T^{-2}$ [Electrical resistivity] $= M L ^3 T^{-3} A^{-2}$ [Electromotive force] $= M L ^2 T^{-3} A^{-1}$

Q35JEE Main 2025NAT

A physical quantity $C$ is related to four other quantities $p, q, r$ and $s$ as follows $C=\;{p q^2}/{r^3 {\sqrt{s}}}$ . The percentage errors in the measurement of $p, q, r$ and $s$ are $1 \%, 2 \%, 3 \%$ and $2 \%$ , respectively. The percentage error in the measurement of $C$ will be $\_\_\_\_$ $\%$ .[3 Apr 2025 Shift 2]

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📖 Explanation

$\;\;\frac{\Delta C}{C}=\;\frac{\Delta p}{p}+2 \;\frac{\Delta q}{q}+\;\frac{3 \Delta r}{r}+\;\frac{1}{2} \;\frac{\Delta s}{s}$ $\;\% \;\text{ error }\;=1+2 \times 2+3 \times 3+\;\frac{1}{2} \times 2$ $\; \;\; =1+4+9+1$ $\; \;\; =15$

Q36JEE Main 2025MCQ

In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of $M^P L^Q T^R A^S$ , where value of ' $Q$ ' and ' $R$ ' is[4 Apr 2025 Shift 1]

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📖 Explanation

Electric flux $\Phi_E$ has dimensions $[E][Area] = \$ [MLT^{-3}A^{-1}]$$[L^2]$ = [ML^3T^{-3}A^{-1}] $. Magnetic flux$ \Phi_B $has dimensions$ [B][Area] = $[MT^{-2}A^{-1}]$$[L^2]$ = [ML^2T^{-2}A^{-1}] $. Taking the ratio,$ \frac{\Phi_E}{\Phi_B} = \frac{[ML^3T^{-3}A^{-1}]$}{[ML^2T^{-2}A^{-1}]$} = $[M^{1-1} L^{3-2} T^{-3-(-2)} A^{-1-(-1)}]$ = [M^0 L^1 T^{-1} A^0] $. Comparing this result to$ M^P L^Q T^R A^S $, we identify$ Q = 1 $and$ R = -1 $. Therefore, the values are$ (1, -1)$.

Q37JEE Main 2025NAT

A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm , respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be $\;\frac{x}{100}$ where $x$ is ______[28 Jan 2025 Shift 1]

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📖 Explanation

Least count of screw gauge = $\; \;\;\;\; \;\frac{\;\text{ Pitch }\;}{\;\text{ Number of circular scale divisions }\;}$ $\;\Rightarrow LC=\;\frac{0.75 mm}{15}=0.05 mm$ $\;\;\text{ Length, }\; ℓ=5 mm$ $\;\;\text{ Breadth, }\; b=2.5 mm$ $\;\;\text{ Area }\;=ℓ b$ $\;\Rightarrow \% \;\text{ error \in area }\;=(\% \;\text{ error \in length }\;)+(\% \;\text{ error }\;$ $\;\;\text{ \in breadth }\;)=\;\frac{\Delta ℓ}{ℓ} \times 100+\;\frac{\Delta b}{b} \times 100$ $\;=\;\frac{0.05}{5} \times 100+\;\frac{0.05}{2.5} \times 100$ $\;=3 \%$

Q38JEE Main 2025MCQ

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm . The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm)[4 Apr 2025 Shift 2]

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📖 Explanation

The value of 1 Main Scale Division (MSD) is $1 \text{ MSD} = \frac{15 \text{ cm}}{300} = 0.05 \text{ cm}$ .
Given that 25 Vernier Scale Divisions (VSD) equal 24 MSD, we have $1 \text{ VSD} = \frac{24}{25} \text{ MSD}$ .
The formula for the Least Count (LC) is $LC = 1 \text{ MSD} - 1 \text{ VSD}$ .
Substituting the values, $LC = 1 \text{ MSD} - \frac{24}{25} \text{ MSD} = \frac{1}{25} \text{ MSD}$ .
Finally, calculating with $1 \text{ MSD} = 0.05 \text{ cm}$ , $LC = \frac{0.05 \text{ cm}}{25} = 0.002 \text{ cm}$ .

Q39JEE Main 2024MCQ

Choose the correct answer from the options given below : Match List - I with List - II.List - I (Number)List - II (Significant figure)(A) 1001(I) 3(B) 010.1(II) 4(C) 100.100(III) 5(D) 0.0010010(IV) 6 [1-Feb-2024 Shift 2]

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Q40JEE Main 2024MCQ

There are 100 divisions on the circular scale of a screw gauge of pitch $1 {\text{mm}}$ . With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found the 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :[8 Apr 2024 Shift 2]

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📖 Explanation

$\;\;\text{ Least count }\;=\;\frac{1}{100} {\text{mm}}=0.01 {\text{mm}}$ $\;\;\text{ zero error }\;=+0.05 {\text{mm}}$ $\;\;\text{ Reading }\;=4 \times 1 {\text{mm}}+60 \times 0.01 {\text{mm}}-0.05 {\text{mm}}$ $\;=4.55 {\text{mm}}$