A spherical body of radius r and density sigma falls freely through a viscous liquid having density and viscosity and…

Year	2026	2025	2024	Total
Questions	14	22	14	50

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

v₀ = (2)/(9)(r² g)/(η) (σ - ρ)

Core Logic

Rearranging the formula for viscosity $η$ $\eta$:

η = (2)/(9) (r² g)/(v₀) (σ - ρ)

Since $(2)/(9)$ $\frac{2}{9}$, $g$ $g$, $σ$ $\sigma$, and $ρ$ $\rho$ are constants or have negligible error, the relative error depends only on $r$ $r$ and $v₀$ $v_0$.

Step 1: Error Propagation

Taking natural logarithm on both sides:

ln(η) = ln(constant) + 2ln(r) - ln(v₀)

Differentiating for maximum possible error (adding absolute errors):

(Δ η)/(η) = 2(Δ r)/(r) + (Δ v₀)/(v₀)

Step 2: Final Conclusion

The maximum fractional error is $(2Δ r)/(r) + (Δ v₀)/(v₀)$ .

Pattern Recognition

For maximum fractional error of a derived quantity $Z = X^a Y^b$ $Z = X^a Y^b$, the formula is $Δ Z/Z = |a|Δ X/X + |b|Δ Y/Y$ . Always sum the absolute relative errors.

Chapter Mix

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

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: Error Propagation

Step 2: Final Conclusion

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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