Consider a modified Bernoulli equation. left(mathrmP+fracmathrmAmathrmBt^2right)+rhomathrmg(mathrmh+mathrmBt)+frac12rhomathrmV^2=textconstant If t has the dimension of time then the dimensions of A and B are ____, ____ respectively.

Solution & Explanation

### Related Formula textPrinciple of Homogeneity: Only quantities with same dimensions can be added/subtracted. ### Core Logic From the term (h + Bt), h (height/length) and Bt must have the same dimension. [h] = [Bt] [B] = frac[h][t] = frac[L][T] = [L T^-1] So, B = [M^0 L T^-1]. ### Step 1: Finding Dimension of A From the term left(P + fracABt^2right), pressure P and fracABt^2 must have the same dimension. Dimension of Pressure P = fractextForcetextArea = [M L^-1 T^-2]. [P] = left[fracABt^2right] [M L^-1 T^-2] = frac[A][L T^-1] times [T^2] [M L^-1 T^-2] = frac[A][L T] [A] = [M L^-1 T^-2] times [L T] = [M L^0 T^-1] So, A = [M L^0 T^-1]. ### Pattern Recognition Apply the Principle of Homogeneity across any (X + Y) terms inside brackets. Extract one unknown, plug it into the next additive bracket group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids

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Units And Measurements

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-ILIST-II
A. Boltzmann constantI. ML^2T^-1
B. Coefficient of viscosityII. MLT^-3K^-1
C. Planck's constantIII. ML^2T^-2K^-1
D. Thermal conductivityIV. ML^-1T^-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 is 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 or 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
  • B. 0.8
  • C. 0.7
  • D. 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 is 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

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