A tiny metallic rectangular sheet has length and breadth of 5 ~mm and 2.5 ~mm , respectively. Using a specially…

Year	2026	2025	2024	Total
Questions	14	22	14	50

Q26 jee_main_2026_22_january_evening Dimensional Analysis

If

in

, E and t represent the free space permittivity, electric field and time respectively, then the unit of

(in E)/(t)

will be :

A. $Am$ $\mathrm{Am}$
B. $Am²$ $\mathrm{Am}^2$
C. $A/m²$ $\mathrm{A/m}^2$
D. $A/m$ $\mathrm{A/m}$

Solution

Related Formula

E = (1)/(4πin) \cdot (q)/(r²)

[(in E)/(t)] = [(q)/(t \cdot r²)]

Core Logic

Substituting electric field equation into the expression gives:

(in E)/(t) = (in)/(t) \cdot (1)/(4πin) (q)/(r²) = (q)/(4π t r²)

Substituting dimensional formulas for current ( $I = q/t arrow A$ $I = q/t \rightarrow \mathrm{A}$) and area ( $r² arrow m²$ $r^2 \rightarrow \mathrm{m}^2$):

[(in E)/(t)] = A \cdot TT \cdot L² = A L⁻² = A/m²

Step 1: Final Conclusion

Hence, the unit of $(in E)/(t)$ $\frac{\in E}{t}$ is $A/m²$ $\mathrm{A/m}^2$.

Pattern Recognition

Sees: $in E / t$ $\in E / t$ product. Shortcut: Permittivity times Electric Field is Displacement Field $D = in E$ $D = \in E$, which has units of Charge per unit Area ( $C/m²$ $C/m^2$). Dividing by time yields $C/(s \cdot m²) = A/m²$ $C/(s \cdot m^2) = A/m^2$ directly.

Chapter Mix

Class 11 Physics: Units and Measurements Class 12 Physics: Electrostatics

Q29 jee_main_2026_23_january_morning Errors in Measurement

Four persons measure the length of a rod as 20.00 cm, 19.75 cm, 17.01 cm and 18.25 cm. The relative error in the measurement of average length of the rod is :

A. $0.24$ $0.24$
B. $0.18$ $0.18$
C. $0.06$ $0.06$
D. $0.08$ $0.08$

Solution

Related Formula

lmean = Σ lᵢn

Δ lmean = Σ |Δ lᵢ|n

Relative Error = Δ lmeanlmean

Step 1: Calculate Mean Value

lmean = (20.00 + 19.75 + 17.01 + 18.25)/(4)

lmean = (75.01)/(4) = 18.7525 \approx 18.75 cm

Step 2: Calculate Mean Absolute Error

Deviations from the mean: $|Δ l₁| = |20.00 - 18.75| = 1.25$ $|Δ l₂| = |19.75 - 18.75| = 1.00$ $|Δ l₃| = |17.01 - 18.75| = 1.74$ $|Δ l₄| = |18.25 - 18.75| = 0.50$

Δ lmean = (1.25 + 1.00 + 1.74 + 0.50)/(4)

Δ lmean = (4.49)/(4) = 1.1225 \approx 1.12 cm

Step 3: Calculate Relative Error

Relative Error = Δ lmeanlmean

Relative Error = (1.12)/(18.75) = 0.05973 \approx 0.06

Pattern Recognition

Sees: "relative error" + "multiple readings" → first find mean, then find absolute differences from mean, average those differences, and finally divide by the mean.

Chapter Mix

Class 11 Physics: Units and Measurements

Q44 jee_main_2026_23_january_morning Screw Gauge

In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument thickness of a sheet is measured. If pitch scale reading is 1 mm and the circular scale reading is 51 then the correct thickness of the sheet is ____ mm. [Assume least count is 0.01 mm]

A. $1.50$ $1.50$
B. $1.48$ $1.48$
C. $1.54$ $1.54$
D. $1.51$ $1.51$

Solution

Related Formula

Zero Error = Division \times Least Count

True Reading = Measured Reading - Zero Error

Core Logic

Since the zero of the circular scale lies above the horizontal reference line when the studs are in contact, the screw gauge has a negative zero error. This means the instrument fundamentally "reads" a value less than the actual value, so the error must be added to the raw reading.

Step 1: Evaluate Zero Error

Zero error e = -3 \times LC = -3 \times 0.01 mm = -0.03 mm

Step 2: Calculate Reading

Measured Reading = Pitch Scale Reading + (Circular Scale Reading \times LC)

Measured Reading = 1 mm + (51 \times 0.01 mm)

Measured Reading = 1.51 mm

Step 3: Apply Zero Correction

Correct Thickness = Measured Reading - e

Correct Thickness = 1.51 - (-0.03) = 1.54 mm

Pattern Recognition

Sees: "zero lies above reference line" → Negative zero error. True value = Measured + |Error|. If it lies below, positive error.

Chapter Mix

Class 11 Physics: Units and Measurements

Q48 jee_main_2026_23_january_evening Dimensional Analysis

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 = A ρa rb ηc σd

, where A is a constant and a, b c and d are integers. The value of

(b + c)/(a + d)

is ____.

Numerical Answer. Answer: 1 to 1

Solution

Related Formula

Time dimension: $[T] = T$ $[T] = \mathrm{T}$ Density dimension: $[ρ] = [σ] = ML⁻³$ $[\rho] = [\sigma] = \mathrm{ML^{-3}}$ Radius dimension: $[r] = L$ $[r] = \mathrm{L}$ Viscosity dimension: $[η] = ML⁻¹T⁻¹$ $[\eta] = \mathrm{ML^{-1}T^{-1}}$

Core Logic

Given dimensional equation:

T = [ρ]^a [r]^b [η]^c [σ]^d

T = ( ML⁻³)^a (L)^b ( ML⁻¹T⁻¹)^c ( ML⁻³)^d

Expand the bases:

T¹ = Ma+c+d L-3a+b-c-3d T-c

Step 1: Compare Exponents

For Time ( $T$ $T$):

-c = 1 c = -1

For Mass ( $M$ $M$):

a + c + d = 0 a - 1 + d = 0 a + d = 1

For Length ( $L$ $L$):

-3a + b - c - 3d = 0

b - c - 3(a + d) = 0

Substitute $c = -1$ $c = -1$ and $a + d = 1$ $a + d = 1$:

b - (-1) - 3(1) = 0

b + 1 - 3 = 0 b = 2

Step 2: Final Calculation

We need the value of $(b+c)/(a+d)$ $\frac{b+c}{a+d}$: Numerator $b+c = 2 + (-1) = 1$ $b+c = 2 + (-1) = 1$ Denominator $a+d = 1$ $a+d = 1$

(b+c)/(a+d) = (1)/(1) = 1

Pattern Recognition

Whenever variables are lumped in a product string $X = y^a z^b$ $X = y^a z^b$, equating dimensions on both sides produces a solvable linear system. The sum groups $(a+d)$ $(a+d)$ can sometimes be substituted directly without fully isolating $a$ $a$ or $d$ $d$ individually.

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids

Q27 jee_main_2026_24_january_evening Vernier Callipers

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division = 0.05 mm, then the least count of the vernier callipers is ____ mm.

A. $0.002$ $0.002$
B. $0.05$ $0.05$
C. $0.02$ $0.02$
D. $0.005$ $0.005$

Solution

Related Formula

Least Count (LC) = 1 MSD - 1 VSD

Core Logic

Given $50 VSD = 48 MSD$ $50 \text{ VSD} = 48 \text{ MSD}$, we have:

1 VSD = (48)/(50) MSD

LC = 1 MSD - (48)/(50) MSD = (2)/(50) MSD

Step 1: Calculation

Substitute $1 MSD = 0.05 mm$ $1 \text{ MSD} = 0.05 \text{ mm}$:

LC = (2)/(50) \times 0.05 mm = 0.002 mm

Pattern Recognition

For non-standard vernier calipers where $N VSD = (N-x) MSD$ $N \text{ VSD} = (N-x) \text{ MSD}$, the least count is $(x/N) \times MSD$ $(x/N) \times \text{MSD}$.

Chapter Mix

Class 11 Physics: Units and Measurements

Solution & Explanation

Core Logic

Step 1: Final Value Match

Pattern Recognition

Chapter Mix

Reference Study Guides

More Units and Measurements Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Calculate Mean Value

Step 2: Calculate Mean Absolute Error

Step 3: Calculate Relative Error

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Zero Error

Step 2: Calculate Reading

Step 3: Apply Zero Correction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compare Exponents

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculation

Pattern Recognition

Chapter Mix

More Units and Measurements Questions — jee_main_2025_28_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back