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 4

Q13 jee_main_2025_04_april_evening Least Count and Vernier Instruments
For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm):
  • A. 0.001
  • B. 0.002
  • C. 0.0005
  • D. 0.0025

Solution

### Related Formula 1text MSD = fractextTotal LengthtextTotal Divisions textLeast Count (LC) = 1text MSD - 1text VSD = 1text MSD cdot left(1 - frac2425right) ### Core Logic Calculate the value of one main scale division: 1text MSD = frac15text cm300 = 0.05text cm Given that 25text VSD = 24text MSD, we find: 1text VSD = frac2425text MSD ### Step 1: Compute Least Count textLC = 1text MSD cdot left(frac125right) = frac0.05text cm25 = 0.002text cm ### Pattern Recognition Least count calculation is conventionally textLC = frac1text MSDN where N represents the total count of divisions on the vernier scale whenever (N-1)text MSD = Ntext VSD. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q20 jee_main_2025_04_april_evening Dimensional Analysis
In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of [M^pL^QT^RA^S]. The value of P and Q are :
  • A. -1, 0
  • B. -1, 1
  • C. 1, -1
  • D. 0, -1

Solution

### Related Formula Electric Dipole Moment: P_e = q cdot d implies [P_e] = textAcdottextTcdottextL Magnetic Dipole Moment: M_m = I cdot A implies [M_m] = textAcdottextL^2 ### Core Logic Take the dimensional ratio: left[fracP_eM_mright] = fractextLtextTtextAtextL^2textA = textL^-1textT = textM^0textL^-1textT^1textA^0 Comparing powers with [textM^PtextL^QtextT^RtextA^S], we find: P = 0 Q = -1 ### Pattern Recognition Dipole units match up with fundamental currents and charge definitions. Notice that current A cancels out completely, leaving a simple geometric spatial ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Moving Charges and Magnetism
Q8 jee_main_2025_04_april_morning Dimensional Analysis
In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of mathrmM^PmathrmL^QmathrmT^RmathrmA^S, where value of 'Q' and 'R' are
  • A. (3, -5)
  • B. (-2, 2)
  • C. (-2, 1)
  • D. (1, -1)

Solution

### Related Formula Ratio formulation: fracphi_Ephi_M = fracE cdot AB cdot A = fracEB From Maxwell's electromagnetic wave equations: E = c cdot B implies fracEB = c where c is the speed of light. ### Core Logic Since the ratio reduces to the dimension of speed (c): left[fracphi_Ephi_M ight] = [c] = mathrmM^0 mathrmL^1 mathrmT^-1 mathrmA^0 ### Step 1: Identify Exponent Values Matching indices with mathrmM^PmathrmL^QmathrmT^RmathrmA^S: * P = 0 * Q = 1 * R = -1 * S = 0 Hence, (Q, R) = (1, -1). ### Pattern Recognition Flux areas cancel out immediately. The ratio fracEB always carries the dimension of velocity (LT^-1). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Electromagnetic Waves
Q12 jee_main_2025_07_april_evening Dimensional Formula
Match List-I with List-II.
List-IList-II
(A) Mass density(I) [ML^2T^-3]
(B) Impulse(II) [MLT^-1]
(C) Power(III) [ML^2T^0]
(D) Moment of inertia(IV) [ML^-3T^0]
Choose the correct answer from the options given below: [cite: 107]
  • A. (A)-(IV), (B)-(II), (C)-(III), (D)-(I) [cite: 108]
  • B. (A)-(I), (B)-(III), (C)-(IV), (D)-(II) [cite: 109]
  • C. (A)-(IV), (B)-(II), (C)-(I), (D)-(III) [cite: 110]
  • D. (A)-(II), (B)-(III), (C)-(IV), (D)-(I) [cite: 111]

Solution

### Core Logic Let's derive the dimensional formulas systematically: * **(A) Mass density:** rho = fractextMasstextVolume = fracML^3 = [M^1 L^-3 T^0] implies text(IV) [cite: 765]. * **(B) Impulse:** I = F cdot Delta t = [M^1 L^1 T^-2] cdot [T] = [M^1 L^1 T^-1] implies text(II) [cite: 767]. * **(C) Power:** P = fractextWorktextTime = frac[M^1 L^2 T^-2][T] = [M^1 L^2 T^-3] implies text(I) [cite: 770]. * **(D) Moment of inertia:** I = M r^2 = [M^1 L^2 T^0] implies text(III) [cite: 772]. Matching all four pairings establishes the layout: (A)-(IV), (B)-(II), (C)-(I), (D)-(III)[cite: 110, 758]. ### Pattern Recognition Isolating the unique matching option for a straightforward parameter like Mass Density (A-IV) or Moment of Inertia (D-III) easily allows exclusion of multiple invalid answer branches instantly[cite: 765, 772]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q4 jee_main_2025_24_jan_morning Significant Figures
For an experimental expression y=frac32.3times112527.4 , where all the digits are significant. Then to report the value of y we should write :-
  • A. y=1326.2
  • B. y=1326.19
  • C. y=1326.186
  • D. y=1330

Solution

### Related Formula In multiplication and division arithmetic rules, the final product or quotient must be rounded off to retain as many significant figures as are present in the least precise operand. ### Core Logic Let us check the significant digit count of the operands in the expression : * 32.3 has 3 significant figures. * 1125 has 4 significant figures. * 27.4 has 3 significant figures. The minimum number of significant figures among the numbers is 3. ### Step 1: Rounding Off Direct calculation yield : y = 1326.186... Rounding this value off to contain exactly 3 significant figures means changing it to 1330, since the digit after 2 is 6 (which is greater than 5), updating the hundreds spot upwards. ### Pattern Recognition Never keep unearned precision from automated calculation. The output is bounded strictly by your least precise entry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

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