The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:

By taking logarithms and differentiating to find relative error (accuracy): Given values: = 20 cm = 200 mm Delta = 2 mm Ttotal = 40 s for 50 oscillations Delta Ttotal = 1 s Note: The relative error in time period T is equal to the relative error in total time t: (Delta T/T) = (Delta t/t). Percentage change = (Delta g/g) x 100% = (6/100) x 100% = 6%. Thus, N = 6. For pendulum gravity error, always use %g = % + 2(%T). Remember that measuring 50 oscillations reduces absolute error on a single swing, but the relative error Delta t / t remains unchanged whether you use total time or single period. Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the acc

By taking logarithms and differentiating to find relative error (accuracy): Given values: = 20 cm = 200 mm Delta = 2 mm Ttotal = 40 s for 50 oscillations Delta Ttotal = 1 s Note: The relative error in time period T is equal to the relative error in total time t: (Delta T/T) = (Delta t/t). Percentage change = (Delta g/g) x 100% = (6/100) x 100% = 6%. Thus, N = 6. For pendulum gravity error, always use %g = % + 2(%T). Remember that measuring 50 oscillations reduces absolute error on a single swing

The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:

By taking logarithms and differentiating to find relative error (accuracy): Given values: = 20 cm = 200 mm Delta = 2 mm Ttotal = 40 s for 50 oscillations Delta Ttotal = 1 s Note: The relative error in time period T is equal to the relative error in total time t: (Delta T/T) = (Delta t/t). Percentage change = (Delta g/g) x 100% = (6/100) x 100% = 6%. Thus, N = 6. For pendulum gravity error, always use %g = % + 2(%T). Remember that measuring 50 oscillations reduces absolute error on a single swing, but the relative error Delta t / t remains unchanged whether you use total time or single period. Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the acc

By taking logarithms and differentiating to find relative error (accuracy): Given values: = 20 cm = 200 mm Delta = 2 mm Ttotal = 40 s for 50 oscillations Delta Ttotal = 1 s Note: The relative error in time period T is equal to the relative error in total time t: (Delta T/T) = (Delta t/t). Percentage change = (Delta g/g) x 100% = (6/100) x 100% = 6%. Thus, N = 6. For pendulum gravity error, always use %g = % + 2(%T). Remember that measuring 50 oscillations reduces absolute error on a single swing

The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was…

Year	2026	2025	2024	Total
Questions	14	22	14	50

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

T = 2π \sqrt(( )/(g)) g = (4π² )/(T²)

Core Logic

By taking logarithms and differentiating to find relative error (accuracy):

(Δ g)/(g) = (Δ )/( ) + 2(Δ T)/(T)

Step 1: Extrapolating Errors

Given values: $= 20 cm = 200 mm$ $\ell = 20 \text{ cm} = 200 \text{ mm}$ $Δ = 2 mm$ $\Delta \ell = 2 \text{ mm}$ $Ttotal = 40 s$ $T_{\text{total}} = 40 \text{ s}$ for 50 oscillations $Δ Ttotal = 1 s$ $\Delta T_{\text{total}} = 1 \text{ s}$ Note: The relative error in time period $T$ $T$ is equal to the relative error in total time $t$ $t$: $(Δ T)/(T) = (Δ t)/(t)$ .

Step 2: Substitution

(Δ g)/(g) = 0.2 cm20 cm + 2 ( 1 s40 s)

(Δ g)/(g) = (2)/(200) + (2)/(40)

(Δ g)/(g) = (1)/(100) + (5)/(100) = (6)/(100)

Step 3: Percentage Conversion

Percentage change $= (Δ g)/(g) \times 100% = (6)/(100) \times 100% = 6%$ . Thus, $N = 6$ $N = 6$.

Pattern Recognition

For pendulum gravity error, always use $%g = % + 2(%T)$ $\%g = \%\ell + 2(\%T)$. Remember that measuring 50 oscillations reduces absolute error on a single swing, but the relative error $Δ t / t$ $\Delta t / t$ remains unchanged whether you use total time or single period.

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

More Units and Measurements Previous-Year Questions — Page 8

Q jee_main_2025_29_jan_morning Dimensional Homogeneity

The expression given below shows the variation of velocity (v) with time (t),

v = At² + (Bt)/(C + t)

. The dimension of ABC is:

A. $[M⁰ ~L² T⁻³]$
B. $[M⁰ ~L¹ T⁻³]$
C. $[M⁰L¹T⁻²]$
D. $[M⁰ ~L² T⁻²]$

Solution

Related Formula

[v] = [At²] = [(Bt)/(C+t)]

Core Logic

By the principle of dimensional homogeneity:

$[C] = [t] = [T]$ $[C] = [t] = [T]$
$[At²] = [v] [A][T²] = [LT⁻¹] [A] = [LT⁻³]$
$[(Bt)/(T)] = [v] [B] = [LT⁻¹]$

Step 1: Calculate Dimensions of ABC

[ABC] = [LT⁻³] \cdot [LT⁻¹] \cdot [T] = [L² T⁻³]

Pattern Recognition

Denominator terms matched first give $C$ $C$, tracking linear velocity units sets the balance for $A$ $A$ and $B$ $B$

Chapter Mix

Class 11 Physics: Units and Measurements

Q42 jee_main_2024_01_february_morning Vernier Calliper

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is:

A. $(1)/(2)$ $\frac{1}{2}$
B. $(10)/(11)$ $\frac{10}{11}$
C. $(50)/(11)$ $\frac{50}{11}$
D. $(5)/(11)$ $\frac{5}{11}$

Solution

Related Formula

Least Count (LC) definition:

LC = 1 MSD - 1 VSD

Core Logic

From the problem:

10 MSD = 11 VSD 1 VSD = (10)/(11) MSD

Substitute into the Least Count formula:

LC = 1 MSD - (10)/(11) MSD = (1)/(11) MSD

Step 1: Calculate with Main Scale Units

Given that $1 MSD = 5 units$ $1\text{ MSD} = 5\text{ units}$:

LC = (1)/(11) \times 5 = (5)/(11) units

Pattern Recognition

Watch out for unconventional scale arrangements where $VSD > MSD$ $\text{VSD} > \text{MSD}$. The fundamental difference formula $1 MSD - 1 VSD$ $1\text{ MSD} - 1\text{ VSD}$ safely determines magnitudes without needing sign corrections.

Chapter Mix

Class 11 Physics: Units and Measurements

Q44 jee_main_2024_01_february_morning Error Analysis

The radius (

r

), length (

l

) and resistance (

R

) of a metal wire was measured in the laboratory as:

r = (0.35 \pm 0.05)~cm

R = (100 \pm 10)~Ω

l = (15 \pm 0.2)~cm

The percentage error in resistivity of the material of the wire is:

A. 25.6%
B. 39.9%
C. 37.3%
D. 35.6%

Solution

Related Formula

Resistivity formula:

ρ = R(A)/(l) = R(π r²)/(l)

Maximum relative error equation:

(Δρ)/(ρ) = (Δ R)/(R) + 2(Δ r)/(r) + (Δ l)/(l)

Core Logic

Substitute the measured fractions:

(Δρ)/(ρ) = (10)/(100) + 2((0.05)/(0.35)) + (0.2)/(15)

(Δρ)/(ρ) = (1)/(10) + 2((1)/(7)) + (1)/(75)

(Δρ)/(ρ) = 0.1 + 0.2857 + 0.0133 = 0.399

Step 1: Convert to Percentage

% severance = 0.399 \times 100% = 39.9%

Pattern Recognition

The power of 2 in $r²$ $r^2$ doubles its fractional error impact, making it the most dominant source of error in this setup.

Chapter Mix

Class 11 Physics: Units and Measurements Class 12 Physics: Current Electricity

Q45 jee_main_2024_01_february_morning Dimensional Analysis

The dimensional formula of angular impulse is:

A. $[M L⁻²T⁻¹]$
B. $[M L²T⁻²]$
C. $[M L T⁻¹]$
D. $[M L²T⁻¹]$

Solution

Related Formula

Angular Impulse equation:

Angular Impulse = \int τ dt = Δ L

Angular Momentum (L) = mvr

Core Logic

By the impulse-momentum theorem for rotation, angular impulse equals the net change in angular momentum.

Extract dimensions from $[mvr]$ $[mvr]$:

[m] = [M]

[v] = [LT⁻¹]

[r] = [L]

Step 1: Combine Dimensions

[Angular Impulse] = [M] \times [LT⁻¹] \times [L] = [ML²T⁻¹]

Pattern Recognition

Angular impulse shares the exact same dimensional signature as Planck's constant ( $h$ $h$) and angular momentum ( $L$ $L$).

Chapter Mix

Class 11 Physics: Units and Measurements Class 11 Physics: System of Particles and Rotational Motion

Q33 jee_main_2024_29_january_evening Error Analysis

A physical quantity

Q

is found to depend on quantities

a, b, c

by the relation

Q = (a⁴ b³)/(c²)

. The percentage error in

a, b

and

c

are

3%

4%

and

5%

respectively. Then, the percentage error in

Q

is:

A. $66%$ $66\%$
B. $43%$ $43\%$
C. $34%$ $34\%$
D. $14%$ $14\%$

Solution

Related Formula

For a quantity $Q = (a^x b^y)/(c^z)$ $Q = \frac{a^x b^y}{c^z}$, the maximum relative error is:

(Δ Q)/(Q) = x (Δ a)/(a) + y (Δ b)/(b) + z (Δ c)/(c)

Multiplying by $100$ $100$ gives the percentage error equation:

% error in Q = x(% error in a) + y(% error in b) + z(% error in c)

Core Logic

Given formula:

Q = (a⁴ b³)/(c²)

Applying the error propagation rule:

(Δ Q)/(Q) \times 100 = 4 ((Δ a)/(a) \times 100) + 3 ((Δ b)/(b) \times 100) + 2 ((Δ c)/(c) \times 100)

Step 1: Substitute the Percentage Errors

Substitute the given values:

$% error in a = 3%$ $\%\text{ error in } a = 3\%$
$% error in b = 4%$ $\%\text{ error in } b = 4\%$
$% error in c = 5%$ $\%\text{ error in } c = 5\%$

% error in Q = 4(3%) + 3(4%) + 2(5%)

% error in Q = 12% + 12% + 10% = 34%

Pattern Recognition

Error percentages always add up, weighted by their exponents in the mathematical expression, regardless of whether the variable is in the numerator or denominator.

Chapter Mix

Class 11 Physics: Units and Measurements

Solution & Explanation

Related Formula

Core Logic

Step 1: Extrapolating Errors

Step 2: Substitution

Step 3: Percentage Conversion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Units and Measurements Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Calculate Dimensions of ABC

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate with Main Scale Units

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Convert to Percentage

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Combine Dimensions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute the Percentage Errors

Pattern Recognition

Chapter Mix

More Units and Measurements Questions — jee_main_2024_31_jan_evening

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