When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main…

Year	2026	2025	2024	Total
Questions	14	22	14	50

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Zero Error = + (VSR \times LC)

Reading = MSR + (VSR \times LC) - Zero Error

Core Logic

Since the zero of the vernier scale is to the right of the main scale zero, there is a positive zero error. We must subtract this zero error from the observed reading.

Step 1: Calculate Zero Error

Zero error $= +(4 \times 0.1 ~mm) = +0.4 ~mm$

Step 2: Calculate Final Reading

Observed Reading $= 15 ~mm + (5 \times 0.1 ~mm) = 15.5 ~mm$

True Reading $= Observed Reading - Zero Error$ $= 15.5 ~mm - 0.4 ~mm$ $= 15.1 ~mm$ $= 15.1 \mathrm{~mm}$

Pattern Recognition

Vernier zero right of main zero $arrow$ $\rightarrow$ positive error $arrow$ $\rightarrow$ subtract it. Vernier zero left of main zero $arrow$ $\rightarrow$ negative error $arrow$ $\rightarrow$ add it.

Chapter Mix

Class 11 Physics: Units and Measurements

More Units and Measurements Previous-Year Questions — Page 9

Q38 jee_main_2024_27_jan_morning Measuring Instruments

Identify the physical quantity that cannot be measured using a spherometer:

A. $Radius of curvature of concave surface$
B. $Specific rotation of liquids$ $\text{Specific rotation of liquids}$
C. $Thickness of thin plates$ $\text{Thickness of thin plates}$
D. $Radius of curvature of convex surface$

Solution

Core Logic

A spherometer is a mechanical instrument designed to measure small vertical displacements to calculate the thickness of thin plates or the radius of curvature of spherical (convex/concave) surfaces.

Specific rotation of liquids is an optical property measured via a polarimeter, completely outside the scope of a spherometer.

Pattern Recognition

Spherometers operate purely on linear micrometer screw scale geometry, hence restricted strictly to spatial dimensions.

Chapter Mix

Class 11 Physics: Units and Measurements

Q47 jee_main_2024_27_jan_morning Dimensional Analysis

Given below are two statements: Statement (I): Planck's constant and angular momentum have same dimensions. Statement (II): Linear momentum and moment of force have same dimensions. In the light of the above statements, choose the correct answer from the options given below:

A. $Statement I is true but Statement II is false$
B. $Both Statement I and Statement II are false$
C. $Both Statement I and Statement II are true$
D. $Statement I is false but Statement II is true$

Solution

Core Logic

Evaluate dimensions step-by-step:

Planck's constant ( $h$ $h$):

$E = h u [h] = ([E])/([ u]) = ML²T⁻²T⁻¹ = ML²T⁻¹$

Angular momentum ( $L$ $L$):

L = mvr [L] = M \cdot LT⁻¹ \cdot L = ML²T⁻¹

Since $[h] = [L]$ $[h] = [L]$, Statement I is true.

Linear momentum ( $P$ $P$):

P = mv [P] = MLT⁻¹

Moment of force (Torque $τ$ $\tau$):

τ = F r [τ] = MLT⁻² \cdot L = ML²T⁻²

Since $[P] \neq [τ]$ $[P] \neq [\tau]$, Statement II is false.

Pattern Recognition

Planck's constant can always be paired with angular momentum units (Joule-seconds), while moment of force matches work/energy footprints, not translational momentum.

Chapter Mix

Class 11 Physics: Units and Measurements

Q38 jee_main_2024_29_jan_morning Error Analysis

The resistance

R = (V)/(I)

where

V = (200 \pm 5) ~V

and

I = (20 \pm 0.2) ~A

, the percentage error in the measurement of

R

is:

A. 3.5%
B. 7%
C. 3%
D. 5.5%

Solution

Related Formula

By propagation of maximum relative error in division:

R = (V)/(I) (Δ R)/(R) = (Δ V)/(V) + (Δ I)/(I)

Percentage error in $R$ $R$ is given by:

% error in R = ( (Δ V)/(V) + (Δ I)/(I) ) \times 100

Core Logic

Given values:

V = 200 ~V, Δ V = 5 ~V

I = 20 ~A, Δ I = 0.2 ~A

Step 1: Evaluate Relative Error

(Δ R)/(R) = (5)/(200) + (0.2)/(20)

(Δ R)/(R) = (5)/(200) + (2)/(200)

(Δ R)/(R) = (7)/(200)

Step 2: Calculate Percentage Error

% error in R = (Δ R)/(R) \times 100 = (7)/(200) \times 100 = 3.5%

Thus, the percentage error is $3.5%$ $3.5\%$.

Pattern Recognition

Whenever independent physical quantities are multiplied or divided, their fractional/relative errors always add up. Make sure to keep the base denominator values aligned to make mental calculations quick.

Chapter Mix

Class 11 Physics: Units and Measurements

Q31 jee_main_2024_30_january_evening Vernier Callipers

50

Vernier divisions are equal to

49

main scale divisions of a travelling microscope and one smallest reading of main scale is

0.5 ~mm

, the Vernier constant of travelling microscope is:

A. $0.1 ~mm$ $0.1 \mathrm{~mm}$
B. $0.1 ~cm$ $0.1 \mathrm{~cm}$
C. $0.01 ~cm$ $0.01 \mathrm{~cm}$
D. $0.01 ~mm$ $0.01 \mathrm{~mm}$

Solution

Related Formula

Vernier Constant (Least Count) = 1 ~MSD - 1 ~VSD

Core Logic

Given that $50$ $50$ Vernier Scale Divisions (VSD) equal $49$ $49$ Main Scale Divisions (MSD).

50 ~VSD = 49 ~MSD

1 ~VSD = (49)/(50) ~MSD

Also, the smallest reading of the main scale ( $1 ~MSD$ $1 \mathrm{~MSD}$) is $0.5 ~mm$ $0.5 \mathrm{~mm}$.

Step 1: Calculate Vernier Constant

Vernier Constant = 1 ~MSD - 1 ~VSD

= 1 ~MSD - (49)/(50) ~MSD

= (1)/(50) ~MSD

Substitute the value of $1 ~MSD$ $1 \mathrm{~MSD}$:

= (1)/(50) \times 0.5 ~mm = (0.5)/(50) ~mm = (1)/(100) ~mm

= 0.01 ~mm

Pattern Recognition

In Vernier calipers problems where $N VSD = (N-1) MSD$ $N \text{ VSD} = (N-1) \text{ MSD}$, the Least Count is always exactly $(1)/(N) MSD$ $\frac{1}{N} \text{ MSD}$.

Chapter Mix

Class 11 Physics: Units and Measurements

Q44 jee_main_2024_30_january_evening Dimensional Analysis

If mass is written as

m = k c^p G-1/2 h1/2

then the value of

P

will be: (Constants have their usual meaning with

k

a dimensionless constant)

A. $1/2$ $1/2$
B. $1 / 3$ $1 / 3$
C. $2$ $2$
D. $-1 / 3$ $-1 / 3$

Solution

Related Formula

[m] = [M]¹ [L]⁰ [T]⁰

[c] = [L T⁻¹]

[G] = [M⁻¹ L³ T⁻²]

[h] = [M L² T⁻¹]

Core Logic

By applying the principle of dimensional homogeneity, the dimensions on both sides of the equation must be identical.

[M]¹ [L]⁰ [T]⁰ = [L T⁻¹]^p [M⁻¹ L³ T⁻²]-1/2 [M L² T⁻¹]1/2

Step 1: Substitute Dimensions

[M] = L^p T-p \cdot M1/2 L-3/2 T¹ \cdot M1/2 L¹ T-1/2

Step 2: Collect Powers of L

Equating the powers of $[L]$ $[L]$ on both sides:

0 = p - (3)/(2) + 1

0 = p - (1)/(2)

p = (1)/(2)

Pattern Recognition

The expression $m \propto \sqrt(hc/G)$ $m \propto \sqrt{hc/G}$ is a known fundamental relation representing the Planck mass. The exponent on $c$ $c$ inside the square root gives $p = 1/2$ $p = 1/2$.

Chapter Mix

Class 11 Physics: Units and Measurements

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Zero Error

Step 2: Calculate Final Reading

Pattern Recognition

Chapter Mix

Reference Study Guides

More Units and Measurements Previous-Year Questions — Page 9

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Relative Error

Step 2: Calculate Percentage Error

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Vernier Constant

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute Dimensions

Step 2: Collect Powers of L

Pattern Recognition

Chapter Mix

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