Solution & Explanation

### Related Formula

K_texteq = K_1 + K_2 quad text(Parallel Combination)

Delta K_texteq = Delta K_1 + Delta K_2

textPercentage Error = fracDelta K_texteqK_texteq times 100

### Core Logic For a parallel combination of springs, the equivalent spring constant is simply the sum of individual constants.

K_texteq = K_1 + K_2 = 10 + 20 = 30text N/m

When quantities are added, their absolute errors are also added:

Delta K_texteq = Delta K_1 + Delta K_2 = 0.2 + 0.3 = 0.5text N/m

### Step 1: Calculating Percentage Error The percentage error in

K

is:

text% Error in K = frac0.530 times 100 = frac53\% approx 1.67\%

### Pattern Recognition Addition operation = add absolute errors. For parallel springs, it's just plain addition. Then to get percentage, divide the absolute error sum by the nominal value sum and multiply by 100. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

More Units and Measurements Previous-Year Questions — Page 2

Q jee_main_2025_03_april_evening Dimensional Analysis and Constants

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. Boltzmann constant	I. $ML^2T^-1$ $ML^2T^{-1}$
B. Coefficient of viscosity	II. $MLT^-3K^-1$ $MLT^{-3}K^{-1}$
C. Planck's constant	III. $ML^2T^-2K^-1$ $ML^2T^{-2}K^{-1}$
D. Thermal conductivity	IV. $ML^-1T^-1$ $ML^{-1}T^{-1}$

Choose the correct answer from the options given below :

A. A-III, B-IV, C-I, D-II
B. A-II, B-III, C-IV, D-I
C. A-III, B-II, C-I, D-IV
D. A-III, B-IV, C-II, D-I

Solution

### Related Formula Formulas to find dimensional formulas: - Boltzmann constant:

k_B = fractextEnergytextTemperature

- Coefficient of viscosity:

eta = fracFA fracdvdx

- Planck's constant:

h = fracEnu

- Thermal conductivity:

fracdQdt = K A fracdTdx Rightarrow K = fractextHeat flow cdot textthicknesstextArea cdot textTemperature difference

### Core Logic Evaluate each constant individually: ### Step 1: Dimensions of Boltzmann constant (

k_B

)

[k_B] = frac[ML^2T^-2][K] = [ML^2T^-2K^-1] quad Rightarrow textMatches III

### Step 2: Dimensions of Coefficient of viscosity (

eta

)

[eta] = frac[MLT^-2][L^2] [T^-1] = [ML^-1T^-1] quad Rightarrow textMatches IV

### Step 3: Dimensions of Planck's constant (

h

)

[h] = frac[ML^2T^-2][T^-1] = [ML^2T^-1] quad Rightarrow textMatches I

### Step 4: Dimensions of Thermal conductivity (

K

)

[K] = frac[ML^2T^-3] [L][L^2] [K] = [MLT^-3K^-1] quad Rightarrow textMatches II

This sequence yields A-III, B-IV, C-I, D-II, matching Option (1). ### Pattern Recognition To solve matching sets efficiently, search for the most recognizable dimensions first. Planck's constant

h

(

ML^2T^-1

) and viscosity coefficient

eta

(

ML^-1T^-1

) are highly unique and usually resolve the options instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

Q25 jee_main_2025_03_april_evening Error Analysis

A physical quantity C is related to four other quantities p, q, r and s as follows

C = fracpq^2r^3sqrts

The percentage errors in the measurement of p, q, r and s are

1\% , 2\% , 3\%

and

2\%

respectively. The percentage error in the measurement of C will be ________ \%.

Numerical Answer. Answer: 15 to 15

Solution

### Related Formula For a physical quantity defined by algebraic powers

C = fracp^a q^br^c s^d

, the maximum fractional error is calculated by summing absolute scaled fractional errors:

fracDelta CC = a fracDelta pp + b fracDelta qq + c fracDelta rr + d fracDelta ss

Expressed as percentages:

\% text error in C = a(\% text error in p) + b(\% text error in q) + c(\% text error in r) + d(\% text error in s)

### Core Logic Given expression:

C = p^1 q^2 r^-3 s^-1/2

Max fractional error equation:

fracDelta CC = 1 left(fracDelta ppright) + 2 left(fracDelta qqright) + 3 left(fracDelta rrright) + frac12 left(fracDelta ssright)

### Step 1: Calculate the total percentage error Substitute the individual percentage errors: - Error in

p = 1\%

- Error in

q = 2\%

- Error in

r = 3\%

- Error in

s = 2\%

\% text error in C = 1(1\%) + 2(2\%) + 3(3\%) + frac12(2\%)

\% text error in C = 1\% + 4\% + 9\% + 1\% = 15\%

The total percentage error in

C

15\%

. ### Pattern Recognition In error propagation, individual errors always combine constructively to produce the maximum possible uncertainty limit. Hence, negative powers (like division by

r^3

s^1/2

) are integrated using positive coefficients during maximum absolute error summation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

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

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open

PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open

Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open

Latest from the Blog

View all →

Loading articles...

Review & Tools

Mistakes To Revise

No conceptual mistakes detected.

🔥 Practice Session Live

Showing All Questions

Exam Questions

Test your concepts live. Choose options or enter numerical values, then verify your answer to reveal the double-box solution matrix.

In an experiment the values of two spring constants were measured as $k_1=(10pm0.2)text N/m$ $k_{1}=(10\pm0.2)\text{ N/m}$ and $k_2=(20pm0.3)text N/m$ $k_{2}=(20\pm0.3)\text{ N/m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :

Solution & Explanation

Reference Study Guides

More Units and Measurements Previous-Year Questions — Page 2

Solution

Solution

Solution

Solution

Solution

More Units and Measurements Questions — jee_main_2026_21_jan_morning

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

In an experiment the values of two spring constants were measured as k_1=(10pm0.2)text N/m$k_{1}=(10\pm0.2)\text{ N/m}$ and k_2=(20pm0.3)text N/m$k_{2}=(20\pm0.3)\text{ N/m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :

Solution & Explanation

Reference Study Guides

More Units and Measurements Previous-Year Questions — Page 2

Solution

Solution

Solution

Solution

Solution

More Units and Measurements Questions — jee_main_2026_21_jan_morning

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

In an experiment the values of two spring constants were measured as $k_1=(10pm0.2)text N/m$ $k_{1}=(10\pm0.2)\text{ N/m}$ and $k_2=(20pm0.3)text N/m$ $k_{2}=(20\pm0.3)\text{ N/m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :