The measured value of the length of a simple pendulum is $20 text cm$ $20 \text{ cm}$ with $2 text mm$ $2 \text{ mm}$ 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 $N\%$ $N\%$. The value of $N$ $N$ is:

A. $4$ $4$
B. $8$ $8$
C. $6$ $6$
D. $5$ $5$

Solution & Explanation

### Related Formula

T = 2pi sqrtfracellg implies g = frac4pi^2 ellT^2

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

fracDelta gg = fracDelta ellell + 2fracDelta TT

### Step 1: Extrapolating Errors Given values:

ell = 20 text cm = 200 text mm

Delta ell = 2 text mm

T_texttotal = 40 text s

for 50 oscillations

Delta T_texttotal = 1 text s

Note: The relative error in time period

T

is equal to the relative error in total time

t

fracDelta TT = fracDelta tt

. ### Step 2: Substitution

fracDelta gg = frac0.2 text cm20 text cm + 2 left(frac1 text s40 text sright)

fracDelta gg = frac2200 + frac240

fracDelta gg = frac1100 + frac5100 = frac6100

### Step 3: Percentage Conversion Percentage change

= fracDelta gg times 100\% = frac6100 times 100\% = 6\%

. Thus,

N = 6

. ### Pattern Recognition For pendulum gravity error, always use

\%g = \%ell + 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

Reference Study Guides

Units And Measurements

More Units and Measurements Previous-Year Questions — Page 2

Q10 jee_main_2025_08_april_evening Error Analysis

A quantity

Q

is formulated as

X^-2Y^frac32Z^-frac25

X, Y

and

Z

are independent parameters which have fractional errors of

0.1, 0.2

and

0.5

, respectively in measurement. The maximum fractional error of

Q

is:

A. $0.1$ $0.1$
B. $0.8$ $0.8$
C. $0.7$ $0.7$
D. $0.6$ $0.6$

Solution

### Related Formula For a quantity

Q = X^a Y^b Z^c

, the maximum fractional error is:

fracDelta QQ = |a| fracDelta XX + |b| fracDelta YY + |c| fracDelta ZZ

where,

fracDelta XX, fracDelta YY, fracDelta ZZ

are fractional errors of individual variables ### Core Logic Given formula:

Q = X^-2 Y^3/2 Z^-2/5

. Identify the absolute exponents: -

|a| = |-2| = 2

|b| = left|frac32right| = frac32

|c| = left|-frac25right| = frac25

Now write the error expression:

fracDelta QQ = 2 fracDelta XX + frac32 fracDelta YY + frac25 fracDelta ZZ

Substitute the given values: -

fracDelta XX = 0.1

fracDelta YY = 0.2

fracDelta ZZ = 0.5

### Step 1: Compute Maximum Fractional Error Calculate term by term:

fracDelta QQ = 2 (0.1) + frac32 (0.2) + frac25 (0.5)

fracDelta QQ = 0.2 + 0.3 + 0.2 = 0.7

### Pattern Recognition Sees: Exponential algebraic relation for errors. Trap: Exponents are negative, but maximum error is cumulative. Always take the *absolute* value of exponents when summing errors! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

Q16 jee_main_2025_29_jan_evening Dimensional Analysis

Match List-I with List-II.

beginarray|l|l|l|l| hline textbfList-I & & textbfList-II & \\ hline text(A) & textYoung's Modulus & text(I) & mathrmML^-1T^-1 \\ text(B) & textTorque & text(II) & mathrmML^-1T^-2 \\ text(C) & textCoefficient of Viscosity & text(III) & mathrmM^-1L^3T^-2 \\ text(D) & textGravitational Constant & text(IV) & mathrmML^2T^-2 \\ hline endarray

Choose the correct answer from the options given below:

A. $text(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$
B. $text(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$
C. $text(A)-(IV), (B)-(II), (C)-(III), (D)-(I)$
D. $text(A)-(II), (B)-(IV), (C)-(I), (D)-(III)$

Solution

### Related Formula

textYoung's Modulus: Y = fracF/ADelta ell / ell

textTorque: tau = F cdot r

textViscosity Force: F = eta A fracdvdx

textGravitational Force: F = fracG m_1 m_2r^2

### Core Logic Evaluating dimensions component-by-component: - **(A) Young's Modulus**:

[Y] = frac[F][A] = fracmathrmMLT^-2mathrmL^2 = mathrmML^-1T^-2 quad rightarrow text(II)

- **(B) Torque**:

[tau] = [F][r] = (mathrmMLT^-2)(mathrmL) = mathrmML^2T^-2 quad rightarrow text(IV)

- **(C) Coefficient of Viscosity**:

[eta] = frac[F][A][dv/dx] = fracmathrmMLT^-2(mathrmL^2)(mathrmT^-1) = mathrmML^-1T^-1 quad rightarrow text(I)

- **(D) Gravitational Constant**:

[G] = frac[F][r^2][m_1][m_2] = frac(mathrmMLT^-2)(mathrmL^2)mathrmM^2 = mathrmM^-1L^3T^-2 quad rightarrow text(III)

Matching path yields: (A)-(II), (B)-(IV), (C)-(I), (D)-(III). ### Pattern Recognition Torque and energy share the identical dimensional formula

mathrmML^2T^-2

. Modulus and pressure share

mathrmML^-1T^-2

. Spotting these matching associations cuts solving time significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

Q23 jee_main_2025_29_jan_evening Combination of Errors

A physical quantity

Q

is related to four observables

a, b, c, d

as follows:

Q = fracab^4cd

where,

a = (60 pm 3)mathrm~Pa

;

b = (20 pm 0.1)mathrm~m

;

c = (40 pm 0.2)mathrm~Nsm^-2

and

d = (50 pm 0.1)mathrm~m

, then the percentage error in

Q

fracx1000

, where

x =

______.

Numerical Answer. Answer: 7700 to 7700

Solution

### Related Formula

fracDelta QQ = fracDelta aa + 4fracDelta bb + fracDelta cc + fracDelta dd

### Core Logic Write down fractional errors from the raw text configurations: -

fracDelta aa = frac360 = 0.05

fracDelta bb = frac0.120 = 0.005

fracDelta cc = frac0.240 = 0.005

fracDelta dd = frac0.150 = 0.002

Compute the total fractional error expression:

fracDelta QQ = [0.05 + 4(0.005) + 0.005 + 0.002]

fracDelta QQ = 0.05 + 0.02 + 0.005 + 0.002 = 0.077

Percentage error expression configuration:

\% text Error = fracDelta QQ times 100 = 7.7 \%

Given that percentage error equals

fracx1000

fracx1000 = 7.7 implies x = 7700

### Pattern Recognition Powers scale up error contributions via direct multiplication multipliers. The term

b^4

contributes exactly 4 times its basic fraction error component to the compilation step. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

Q21 jee_main_2025_28_jan_morning Errors in Measurement

A tiny metallic rectangular sheet has length and breadth of

5 mathrm~mm

and

2.5 mathrm~mm

, respectively. Using a specially designed screw gauge which has pitch of

0.75 mathrm~mm

and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be

fracmathrmx100

where

mathrmx

is ________.

Numerical Answer. Answer: 3 to 3

Solution

### Core Logic First, find the least count of the measurement tool:

textLeast Count = fractextPitchtextNumber of circular scale divisions = frac0.75 mathrm~mm15 = 0.05 mathrm~mm

Least count calculation tracking diagram for Q21

The area of the rectangular metallic sheet is calculated as:

mathrmA = mathrmL cdot mathrmW

Expressing the absolute error via fractional configuration parts:

fracmathrmdAmathrmA = fracmathrmdLmathrmL + fracmathrmdWmathrmW

Substituting the instrument limits (

mathrmdL = mathrmdW = 0.05 mathrm~mm

fracmathrmdAmathrmA = frac0.055 + frac0.052.5 = frac1100 + frac2100 = frac3100

### Step 1: Final Value Match Comparing this to the target format

fracmathrmx100

gives:

mathrmx = 3

### Pattern Recognition The absolute measurement uncertainty matches the instrument's least count value directly. Sum up individual fractional errors to compute the total area uncertainty parameter. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

Q23 jee_main_2025_28_jan_morning Dimensional Analysis

In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of

left[mathrmM^mathrmamathrmL^mathrmbmathrmT^mathrmcright]

. If

b = 3

, the value of

c

Numerical Answer. Answer: 0 to 0

Solution

### Core Logic Let's find the dimensional formula for the ratio of Modulus of Elasticity to Torque:

textTarget Dimensions = frac[textModulus of Elasticity][textTorque]

textTarget Dimensions = frac[mathrmM L^-1 mathrmT^-2][mathrmM L^2 mathrmT^-2] = [mathrmM^0 mathrmL^-3 mathrmT^0]

### Step 1: Exponent Matching Comparing this output to the target layout formula

[mathrmM^mathrma mathrmL^mathrmb mathrmT^mathrmc]

mathrmc = 0

### Pattern Recognition Both dimensions share identical time dependence factors (

mathrmT^-2

), meaning they cancel out completely. This leaves the time exponent value as exactly zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

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