If in, E and t represent the free space permittivity, electric field and time respectively, then the unit of fracin Et will be :

Solution & Explanation

### Related Formula E = frac14piin cdot fracqr^2 left[fracin Etright] = left[fracqt cdot r^2right] ### Core Logic Substituting electric field equation into the expression gives: fracin Et = fracint cdot frac14piin fracqr^2 = fracq4pi t r^2 Substituting dimensional formulas for current (I = q/t rightarrow mathrmA) and area (r^2 rightarrow mathrmm^2): left[fracin Etright] = fracmathrmA cdot mathrmTmathrmT cdot mathrmL^2 = mathrmA mathrmL^-2 = mathrmA/m^2 ### Step 1: Final Conclusion Hence, the unit of fracin Et is mathrmA/m^2. ### Pattern Recognition Sees: in E / t product. Shortcut: Permittivity times Electric Field is Displacement Field D = in E, which has units of Charge per unit Area (C/m^2). Dividing by time yields C/(s cdot m^2) = A/m^2 directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Electrostatics

Reference Study Guides

More Units and Measurements Previous-Year Questions

Q30 jee_main_2026_21_jan_morning Errors in Measurement
In an experiment the values of two spring constants were measured as k_1=(10pm0.2)text N/m and k_2=(20pm0.3)text N/m. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
  • A. 2.67%
  • B. 2.33%
  • C. 1.33%
  • D. 1.67%

Solution

### 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
Q32 jee_main_2026_21_jan_morning Dimensional Analysis
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.
  • A. [ML^0T^-1]text and [M^0LT]
  • B. [ML^0T^-1]text and [M^0LT^-1]
  • C. [ML^0T^-2]text and [M^0LT^-2]
  • D. [ML^0T^-2]text and [M^0LT^-1]

Solution

### 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
Q36 jee_main_2026_21_jan_evening Significant Figures
Keeping the significant figures in view, the sum of the physical quantities 52.01 text m, 153.2 text m and 0.123 text m is :
  • A. 205 text m
  • B. 205.333 text m
  • C. 205.33 text m
  • D. 205.3 text m

Solution

### Related Formula Addition Rule for Significant Figures: The result cannot have more digits to the right of the decimal point than any of the original numbers. ### Core Logic Let's align the numbers and perform exact addition: 52.01 (2 decimal places) 153.2 (1 decimal place - least precise) + 0.123 (3 decimal places) -------- 205.333 ### Step 1: Final Conclusion According to the rules of significant figures for addition, the sum must be rounded to the least number of decimal places present in the given measurements. Here, 153.2 has only 1 decimal place. Rounding 205.333 to one decimal place yields 205.3 text m. ### Pattern Recognition Addition/Subtraction is bounded by decimal places. Multiplication/Division is bounded by total significant figures. In this case, 1 decimal place rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q37 jee_main_2026_21_jan_evening Errors in Measurement
A spherical body of radius r and density sigma falls freely through a viscous liquid having density rho and viscosity eta and attains a terminal velocity v_0. Estimated maximum error in the quantity eta is : (Ignore errors associated with sigma, rho and g, gravitational acceleration)
  • A. 2fracDelta rr-fracDelta v_0v_0
  • B. frac2Delta rr+fracDelta v_0v_0
  • C. 2left[fracDelta rr+fracDelta v_0v_0right]
  • D. 2left[fracDelta rr-fracDelta v_0v_0right]

Solution

### Related Formula v_0 = frac29fracr^2 geta (sigma - rho) ### Core Logic Rearranging the formula for viscosity eta: eta = frac29 fracr^2 gv_0 (sigma - rho) Since frac29, g, sigma, and rho are constants or have negligible error, the relative error depends only on r and v_0. ### Step 1: Error Propagation Taking natural logarithm on both sides: ln(eta) = ln(textconstant) + 2ln(r) - ln(v_0) Differentiating for maximum possible error (adding absolute errors): fracDelta etaeta = 2fracDelta rr + fracDelta v_0v_0 ### Step 2: Final Conclusion The maximum fractional error is frac2Delta rr + fracDelta v_0v_0. ### Pattern Recognition For maximum fractional error of a derived quantity Z = X^a Y^b, the formula is Delta Z/Z = |a|Delta X/X + |b|Delta Y/Y. Always sum the absolute relative errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids
Q44 jee_main_2026_22_january_morning Dimensional Analysis
Match the LIST-I with LIST-II
List-IList-II
(A) Spring constant(I) ML^2T^-2K^-1
(B) Thermal conductivity(II) ML^0T^-2
(C) Boltzmann constant(III) ML^2T^-3A^-2
(D) Inductive reactance(IV) MLT^-3K^-1
Choose the correct answer from the options given below:
  • A. \text{A-II, B-I, C-IV, D-III}
  • B. \text{A-I, B-IV, C-II, D-III}
  • C. \text{A-III, B-II, C-IV, D-I}
  • D. \text{A-II, B-IV, C-I, D-III}

Solution

### Related Formula F = Kx, quad fracdQdt = -KAfracdTdx, quad E = k_B T, quad X_L = omega L ### Core Logic (A) Spring constant K: [ML^0 T^-2] -> (II) (B) Thermal conductivity K: [MLT^-3 K^-1] -> (IV) (C) Boltzmann constant k_B: [ML^2 T^-2 K^-1] -> (I) (D) Inductive reactance X_L: [ML^2 T^-3 A^-2] -> (III) Therefore, matching is A-II, B-IV, C-I, D-III. ### Pattern Recognition Sees: Match List dimensional formulas. Shortcut: Derive or recall key physical constants dimensions (Spring constant, Thermal conductivity, Boltzmann constant, Inductive reactance). Check: Matches option (4). ✓ ### Chapter Mix Class 11 Physics: Units and Measurements

More Units and Measurements Questions — jee_main_2026_22_january_evening

Practice all Units and Measurements previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)