A person measures mass of 3 different particles as 435.42~g, 226.3~g and 0.125~g. According to the rules for arithmetic operations with significant figures, the additions of the masses of 3 particles will be.

Significant Figures Rule for Addition/Subtraction: The final result must be rounded off to keep only as many decimal places as there are in the measurement with the **least** number of decimal places. Let's look at the decimal places of each measurement: - 435.42~g has **2 decimal places**. - 226.3~g has **1 decimal place**. - 0.125~g has **3 decimal places**. The minimum number of decimal places is **1 decimal place** (from 226.3~g). First, perform the standard mathematical addition: Now, round this raw sum off to **1 decimal place**: - The digit after tenths place is 4 (4 < 5), so we round down. - Net rounded sum = 661.8~g. Bust the common myth: Addition depends on the least number of decimal places, whereas multiplication/division depends on the least number of significant figures. Always distinguish between these two rules during exams! Class 11 Physics: Units and Measurements: Error Analysis and Significant Figures

A person measures mass of 3 different particles as 435.42~g, 226.3~g and 0.125~g. According to the rules for arithmetic operations with significant figures, the additions of the masses of 3 particles

Significant Figures Rule for Addition/Subtraction: The final result must be rounded off to keep only as many decimal places as there are in the measurement with the **least** number of decimal places. Let's look at the decimal places of each measurement: - 435.42~g has **2 decimal places**. - 226.3~g has **1 decimal place**. - 0.125~g has **3 decimal places**. The minimum number of decimal places is **1 decimal place** (from 226.3~g). First, perform the standard mathematical addition: Now, round

A person measures mass of 3 different particles as 435.42~g, 226.3~g and 0.125~g. According to the rules for arithmetic operations with significant figures, the additions of the masses of 3 particles will be.

Significant Figures Rule for Addition/Subtraction: The final result must be rounded off to keep only as many decimal places as there are in the measurement with the **least** number of decimal places. Let's look at the decimal places of each measurement: - 435.42~g has **2 decimal places**. - 226.3~g has **1 decimal place**. - 0.125~g has **3 decimal places**. The minimum number of decimal places is **1 decimal place** (from 226.3~g). First, perform the standard mathematical addition: Now, round this raw sum off to **1 decimal place**: - The digit after tenths place is 4 (4 < 5), so we round down. - Net rounded sum = 661.8~g. Bust the common myth: Addition depends on the least number of decimal places, whereas multiplication/division depends on the least number of significant figures. Always distinguish between these two rules during exams! Class 11 Physics: Units and Measurements: Error Analysis and Significant Figures

A person measures mass of 3 different particles as 435.42~g, 226.3~g and 0.125~g. According to the rules for arithmetic operations with significant figures, the additions of the masses of 3 particles

Significant Figures Rule for Addition/Subtraction: The final result must be rounded off to keep only as many decimal places as there are in the measurement with the **least** number of decimal places. Let's look at the decimal places of each measurement: - 435.42~g has **2 decimal places**. - 226.3~g has **1 decimal place**. - 0.125~g has **3 decimal places**. The minimum number of decimal places is **1 decimal place** (from 226.3~g). First, perform the standard mathematical addition: Now, round

A person measures mass of 3 different particles as 435.42~g, 226.3~g and 0.125~g. According to the rules for arithmetic…

Year	2026	2025	2024	Total
Questions	14	22	14	50

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Significant Figures Rule for Addition/Subtraction: The final result must be rounded off to keep only as many decimal places as there are in the measurement with the least number of decimal places.

Core Logic

Let's look at the decimal places of each measurement:

$435.42~g$ $435.42\mathrm{~g}$ has 2 decimal places.
$226.3~g$ $226.3\mathrm{~g}$ has 1 decimal place.
$0.125~g$ $0.125\mathrm{~g}$ has 3 decimal places.

The minimum number of decimal places is 1 decimal place (from $226.3~g$ $226.3\mathrm{~g}$).

Step 1: Addition and Rounding

First, perform the standard mathematical addition:

Sum = 435.42 + 226.3 + 0.125 = 661.845~g

Now, round this raw sum off to 1 decimal place:

The digit after tenths place is 4 ( $4 < 5$ $4 < 5$), so we round down.
Net rounded sum = $661.8~g$ $661.8\mathrm{~g}$.

Pattern Recognition

Bust the common myth: Addition depends on the least number of decimal places, whereas multiplication/division depends on the least number of significant figures. Always distinguish between these two rules during exams!

Chapter Mix

Class 11 Physics: Units and Measurements: Error Analysis and Significant Figures

More Units and Measurements Previous-Year Questions — Page 2

Q26 jee_main_2026_22_january_evening Dimensional Analysis

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

A person measures mass of 3 different particles as 435.42~g$435.42\mathrm{~g}$, 226.3~g$226.3\mathrm{~g}$ and 0.125~g$0.125\mathrm{~g}$. According to the rules for arithmetic operations with significant figures, the additions of the masses of 3 particles will be.

Solution & Explanation

Related Formula

Core Logic

Step 1: Addition and Rounding

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_03_april_morning

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A person measures mass of 3 different particles as $435.42~g$ $435.42\mathrm{~g}$, $226.3~g$ $226.3\mathrm{~g}$ and $0.125~g$ $0.125\mathrm{~g}$. According to the rules for arithmetic operations with significant figures, the additions of the masses of 3 particles will be.