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 8

Q50 jee_main_2024_31_jan_evening Dimensional Analysis
Consider two physical quantities A and B related to each other as E = fracB - x^2At where E, x and t have dimensions of energy, length and time respectively. The dimension of AB is
  • A. textL^-2textM^1textT^0
  • B. textL^2textM^-1textT^1
  • C. textL^-2textM^-1textT^1
  • D. textL^0textM^-1textT^1

Solution

### Related Formula By the Principle of Homogeneity, terms added or subtracted must have the same dimensions: [B] = [x^2] ### Core Logic Known dimensional formulas: Length x to [L] Energy E to [ML^2T^-2] Time t to [T] ### Step 1: Dimension of B Since x^2 is subtracted from B: [B] = [x^2] = [L^2] ### Step 2: Dimension of A From the equation E = fracB - x^2At: [A] = frac[B - x^2][E][t] [A] = frac[L^2][ML^2T^-2][T] = frac[L^2][ML^2T^-1] [A] = [M^-1T^1] ### Step 3: Dimension of AB [AB] = [A] times [B] [AB] = [M^-1T^1] times [L^2] [AB] = [L^2 M^-1 T^1] ### Pattern Recognition Identify sums/differences first to instantly isolate B. Once [B] is fixed, the entire numerator is just L^2. Swap out variables to isolate [A]. Combining is just standard exponent addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q39 jee_main_2024_31_jan_morning Errors In Measurement
If the percentage errors in measuring the length and the diameter of a wire are 0.1\% each. The percentage error in measuring its resistance will be:
  • A. 0.2\%
  • B. 0.3\%
  • C. 0.1\%
  • D. 0.144\%

Solution

### Related Formula R = fracrho LA = fracrho Lpi left(fracd2right)^2 = frac4rho Lpi d^2 ### Core Logic To find the maximum percentage error in resistance, apply logarithmic differentiation: fracDelta RR = fracDelta LL + 2fracDelta dd Given percentage errors: fracDelta LL times 100\% = 0.1\% fracDelta dd times 100\% = 0.1\% ### Step 2: Substitution Substituting the values: fracDelta RR times 100\% = 0.1\% + 2(0.1\%)\, = 0.1\% + 0.2\% = 0.3\% ### Pattern Recognition Resistance scales inversely with the square of the diameter. The error multiplier for diameter is 2. Just sum linear components directly: Error = L_error + 2 * d_error. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units And Measurements Class 12 Physics: Current Electricity
Q41 jee_main_2024_31_jan_morning Dimensional Analysis
A force is represented by F = ax^2 + bt^1/2 Where x = distance and t = time. The dimensions of b^2 / a are:
  • A. [ML^3T^-3]
  • B. [MLT^-2]
  • C. [ML^-1T^-1]
  • D. [ML^2T^-3]

Solution

### Related Formula textPrinciple of Homogeneity: [F] = [ax^2] = [bt^1/2] ### Core Logic By the principle of dimensional homogeneity, each additive term must have the same dimension as the left hand side. Dimension of force F = [M L T^-2]. For the term ax^2: [a] = frac[F][x^2] = frac[M L T^-2][L^2] = [M L^-1 T^-2] For the term bt^1/2: [b] = frac[F][t^1/2] = frac[M L T^-2][T^1/2] = [M L T^-5/2] ### Step 2: Computing Required Ratio We need the dimension of fracb^2a: left[ fracb^2a right] = frac[M L T^-5/2]^2[M L^-1 T^-2] left[ fracb^2a right] = frac[M^2 L^2 T^-5][M L^-1 T^-2] left[ fracb^2a right] = [M^2-1 L^2 - (-1) T^-5 - (-2)] left[ fracb^2a right] = [M L^3 T^-3] ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units And Measurements

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