Given a charge q, current I and permeability of vacuum mu_0 . Which of the following quantity has the dimension of momentum?

Solution & Explanation

### Related Formula Let momentum be P ([P] = [M L T^-1]). We assume: P = q^x mu_0^y I^z ### Core Logic Let's find the dimensional formulas of the individual variables: 1. **Charge (q):** [q] = [A T] 2. **Current (I):** [I] = [A] 3. **Permeability of vacuum (mu_0):** From Biot-Savart law or force between parallel wires: F = fracmu_0 I^2 L2pi d: [mu_0] = frac[F][I]^2 = frac[M L T^-2][A]^2 = [M L T^-2 A^-2] ### Step 1: Apply Dimensional Homogeneity Substitute these into our assumed dimensional equation: [M L T^-1] = [A T]^x [M L T^-2 A^-2]^y [A]^z [M L T^-1] = [M]^y [L]^y [T]^x - 2y [A]^x - 2y + z Comparing exponents on both sides: - For [M]: y = 1 - For [L]: y = 1 quad text(consistent) - For [T]: x - 2y = -1 implies x - 2(1) = -1 implies x = 1 - For [A]: x - 2y + z = 0 implies 1 - 2(1) + z = 0 implies z = 1 Thus, x = 1, y = 1, z = 1. Therefore, the required quantity is: q^1 mu_0^1 I^1 = q mu_0 I ### Pattern Recognition Sees: Permeability, charge, and current linked to momentum. Trap: Deriving the dimensions of mu_0 using complex magnetic formulas. Remember [mu_0] = [textForce]/[textCurrent]^2 is the quickest way to get its dimensions. Shortcut: Since [q] = AT and [I] = A, [q mu_0 I] = [A T] [M L T^-2 A^-2] [A] = [M L T^-1], which is exactly the dimensions of momentum. Hence, option (2) is correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 12 Physics: Moving Charges and Magnetism

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Q38 jee_main_2024_29_jan_morning Error Analysis
The resistance R = fracVI where V = (200 pm 5) mathrm~V and I = (20 pm 0.2) mathrm~A, the percentage error in the measurement of R is:
  • A. 3.5%
  • B. 7%
  • C. 3%
  • D. 5.5%

Solution

### Related Formula By propagation of maximum relative error in division: R = fracVI implies fracDelta RR = fracDelta VV + fracDelta II Percentage error in R is given by: \% text error in R = left( fracDelta VV + fracDelta II right) times 100 ### Core Logic Given values: V = 200 mathrm~V, quad Delta V = 5 mathrm~V I = 20 mathrm~A, quad Delta I = 0.2 mathrm~A ### Step 1: Evaluate Relative Error fracDelta RR = frac5200 + frac0.220 fracDelta RR = frac5200 + frac2200 fracDelta RR = frac7200 ### Step 2: Calculate Percentage Error \% text error in R = fracDelta RR times 100 = frac7200 times 100 = 3.5\% Thus, the percentage error is 3.5\%. ### Pattern Recognition Whenever independent physical quantities are multiplied or divided, their fractional/relative errors always add up. Make sure to keep the base denominator values aligned to make mental calculations quick. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q31 jee_main_2024_30_january_evening Vernier Callipers
If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0.5 mathrm~mm, the Vernier constant of travelling microscope is:
  • A. 0.1 mathrm~mm
  • B. 0.1 mathrm~cm
  • C. 0.01 mathrm~cm
  • D. 0.01 mathrm~mm

Solution

### Related Formula textVernier Constant (Least Count) = 1 mathrm~MSD - 1 mathrm~VSD ### Core Logic Given that 50 Vernier Scale Divisions (VSD) equal 49 Main Scale Divisions (MSD). 50 mathrm~VSD = 49 mathrm~MSD 1 mathrm~VSD = frac4950 mathrm~MSD Also, the smallest reading of the main scale (1 mathrm~MSD) is 0.5 mathrm~mm. ### Step 1: Calculate Vernier Constant textVernier Constant = 1 mathrm~MSD - 1 mathrm~VSD = 1 mathrm~MSD - frac4950 mathrm~MSD = frac150 mathrm~MSD Substitute the value of 1 mathrm~MSD: = frac150 times 0.5 mathrm~mm = frac0.550 mathrm~mm = frac1100 mathrm~mm = 0.01 mathrm~mm ### Pattern Recognition In Vernier calipers problems where N text VSD = (N-1) text MSD, the Least Count is always exactly frac1N text MSD. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q44 jee_main_2024_30_january_evening Dimensional Analysis
If mass is written as m = k c^p G^-1/2 h^1/2 then the value of P will be: (Constants have their usual meaning with k a dimensionless constant)
  • A. 1/2
  • B. 1 / 3
  • C. 2
  • D. -1 / 3

Solution

### Related Formula [m] = [M]^1 [L]^0 [T]^0 [c] = [L T^-1] [G] = [M^-1 L^3 T^-2] [h] = [M L^2 T^-1] ### Core Logic By applying the principle of dimensional homogeneity, the dimensions on both sides of the equation must be identical. [M]^1 [L]^0 [T]^0 = [L T^-1]^p [M^-1 L^3 T^-2]^-1/2 [M L^2 T^-1]^1/2 ### Step 1: Substitute Dimensions [M] = L^p T^-p cdot M^1/2 L^-3/2 T^1 cdot M^1/2 L^1 T^-1/2 ### Step 2: Collect Powers of L Equating the powers of [L] on both sides: 0 = p - frac32 + 1 0 = p - frac12 p = frac12 ### Pattern Recognition The expression m propto sqrthc/G is a known fundamental relation representing the Planck mass. The exponent on c inside the square root gives p = 1/2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q31 jee_main_2024_30_jan_morning Dimensional Analysis
Match List-I with List-II.
List-IList-II
A. Coefficient of viscosityI. [M L^2T^-2]
B. Surface TensionII. [M L^2T^-1]
C. Angular momentumIII. [M L^-1T^-1]
D. Rotational kinetic energyIV. [M L^0T^-2]
  • A. textA-II, B-I, C-IV, D-III
  • B. textA-I, B-II, C-III, D-IV
  • C. textA-III, B-IV, C-II, D-I
  • D. textA-IV, B-III, C-II, D-I

Solution

### Related Formula F = eta A fracdvdy textSurface Tension = fracFl L = mvr K.E = frac12 I omega^2 ### Core Logic Let us determine the dimensional formula for each quantity sequentially: **A. Coefficient of viscosity (eta):** Using F = eta A fracdvdy, we have: [M L T^-2] = eta [L^2] [T^-1] eta = [M L^-1 T^-1] Rightarrow text(III) **B. Surface Tension (S.T.):** textS.T = fracFell = frac[M L T^-2][L] = [M L^0 T^-2] Rightarrow text(IV) **C. Angular momentum (L):** L = mvr = [M] [L T^-1] [L] = [M L^2 T^-1] Rightarrow text(II) **D. Rotational kinetic energy (K.E.):** textK.E = frac12 I omega^2 = [M L^2 T^-2] Rightarrow text(I) ### Step 1: Final Matching Matching the derived dimensional formulas: A rightarrow III B rightarrow IV C rightarrow II D rightarrow I ### Pattern Recognition Kinetic energy (whether translational or rotational) always carries the dimension of Work: [M L^2 T^-2]. Surface tension is force per unit length, dropping the L term. Viscosity commonly includes L^-1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q33 jee_main_2024_31_jan_evening Errors in Measurement
The measured value of the length of a simple pendulum is 20 text cm with 2 text mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N\%. The value of N is:
  • A. 4
  • B. 8
  • C. 6
  • D. 5

Solution

### Related Formula T = 2pi sqrtfracellg implies g = frac4pi^2 ellT^2 ### Core Logic By taking logarithms and differentiating to find relative error (accuracy): fracDelta gg = fracDelta ellell + 2fracDelta TT ### Step 1: Extrapolating Errors Given values: ell = 20 text cm = 200 text mm Delta ell = 2 text mm T_texttotal = 40 text s for 50 oscillations Delta T_texttotal = 1 text s Note: The relative error in time period T is equal to the relative error in total time t: fracDelta TT = fracDelta tt. ### Step 2: Substitution fracDelta gg = frac0.2 text cm20 text cm + 2 left(frac1 text s40 text sright) fracDelta gg = frac2200 + frac240 fracDelta gg = frac1100 + frac5100 = frac6100 ### Step 3: Percentage Conversion Percentage change = fracDelta gg times 100\% = frac6100 times 100\% = 6\%. Thus, N = 6. ### Pattern Recognition For pendulum gravity error, always use \%g = \%ell + 2(\%T). Remember that measuring 50 oscillations reduces absolute error on a single swing, but the relative error Delta t / t remains unchanged whether you use total time or single period. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Oscillations

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