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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Q22 jee_main_2025_24_jan_morning Screw Gauge and Least Count
The least count of a screw guage is 0.01 mathrm~mm . If the pitch is increased by 75 \% and number of divisions on the circular scale is reduced by 50 \% , the new least count will be \_ times 10^-3 mathrm~mm .
Numerical Answer. Answer: 35 to 35

Solution

### Related Formula The Least Count (LC) of a screw gauge tool is defined by: textL.C. = fractextPitchtextTotal Number of Circular Divisions (N) ### Core Logic The initial least count is given as [cite: 167, 788]: textL.C.textinitial = fracPN = 0.01text mm Now, calculate the modified parameters from the text details : * New Pitch: P' = P(1 + 0.75) = 1.75P * New Divisions: N' = N(1 - 0.50) = 0.5N ### Step 1: Calculating the New Least Count Set up the updated least count expression ratio : textL.C.textnew = fracP'N' = frac1.75P0.5N = 3.5 times left(fracPN ight) Substitute the initial least count value : textL.C.textnew = 3.5 times 0.01text mm = 0.035text mm Converting into the requested scientific prefix units (10^-3text mm) : textL.C.textnew = 35 times 10^-3text mm Therefore, the requested value is 35. ### Pattern Recognition Least count scales proportionally with pitch increases, and inversely with reductions in circular divisions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q3 jee_main_2025_28_jan_evening Dimensional Analysis
Match List-I with List-II:
List-IList-II
(A) Angular Impulse(I) [M^0L^2T^-2]
(B) Latent Heat(II) [M L^2T^-3A^-1]
(C) Electrical resistivity(III) [M L^2T^-1]
(D) Electromotive force(IV) [M L^3T^-3"A^-2]
Choose the correct answer from the options given below:
  • A. text(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  • B. text(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  • C. text(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  • D. text(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Solution

### Related Formula * **Angular Impulse** = Change in Angular Momentum = tau cdot Delta t = [M L^2 T^-2] cdot [T] = [M L^2 T^-1] [cite: 670, 672] * **Latent Heat** (L) = fracQm = frac[M L^2 T^-2][M] = [M^0 L^2 T^-2] * **Electrical resistivity** (\rho) = fracR cdot Al = frac[M L^2 T^-3 A^-2] cdot [L^2][L] = [M L^3 T^-3 A^-2] * **Electromotive force** (V) = fracWq = frac[M L^2 T^-2][A T] = [M L^2 T^-3 A^-1] ### Core Logic By comparing the formulas derived for each physical quantity with the options given in List-II [cite: 670, 673, 674]: * (A) matches with (III) [cite: 670, 672] * (B) matches with (I) * (C) matches with (IV) * (D) matches with (II) Hence, the correct matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II). ### Pattern Recognition In match-the-column dimensional analysis questions, identifying even one or two straightforward quantities like Latent Heat (L = Q/m) often immediately eliminates three incorrect options, securing a quick correct answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements
Q jee_main_2025_29_jan_morning Dimensional Analysis
The pair of physical quantities not having same dimensions is :
  • A. textTorque and energy
  • B. textSurface tension and impulse
  • C. textAngular momentum and Planck\'s constant
  • D. textPressure and Young\'s modulus

Solution

### Core Logic Let\'s check the dimensions of each pair : * textTorque = [textEnergy] = [ML^2T^-2] * textSurface Tension = [MT^-2] vs textImpulse = [MLT^-1] * [textAngular Momentum] = [textPlanck\'s Constant] = [ML^2T^-1] * [textPressure] = [textYoung\'s Modulus] = [ML^-1T^-2] ### Step 1: Identify Non-Matching Pair Surface tension and impulse do not share matching dimension frameworks. ### Pattern Recognition Surface tension is force per unit length ([MT^-2]), while impulse is force times time ([MLT^-1]) ### Chapter Mix Class 11 Physics: Units and Measurements
Q jee_main_2025_29_jan_morning Dimensional Homogeneity
The expression given below shows the variation of velocity (v) with time (t), v = At^2 + fracBtC + t . The dimension of ABC is:
  • A. left[mathrmM^0 mathrm~L^2 mathrmT^-3right]
  • B. left[mathrmM^0 mathrm~L^1 mathrmT^-3right]
  • C. [mathrmM^0mathrmL^1mathrmT^-2]
  • D. left[mathrmM^0 mathrm~L^2 mathrmT^-2right]

Solution

### Related Formula [v] = [At^2] = left[fracBtC+tright] ### Core Logic By the principle of dimensional homogeneity: 1. [C] = [t] = [T] 2. [At^2] = [v] implies [A][T^2] = [LT^-1] implies [A] = [LT^-3] 3. left[fracBtTright] = [v] implies [B] = [LT^-1] ### Step 1: Calculate Dimensions of ABC [ABC] = [LT^-3] cdot [LT^-1] cdot [T] = [L^2 T^-3] ### Pattern Recognition Denominator terms matched first give C, tracking linear velocity units sets the balance for A and B ### Chapter Mix Class 11 Physics: Units and Measurements
Q42 jee_main_2024_01_february_morning Vernier Calliper
10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is:
  • A. frac12
  • B. frac1011
  • C. frac5011
  • D. frac511

Solution

### Related Formula Least Count (LC) definition: textLC = 1text MSD - 1text VSD ### Core Logic From the problem: 10text MSD = 11text VSD implies 1text VSD = frac1011text MSD Substitute into the Least Count formula: textLC = 1text MSD - frac1011text MSD = frac111text MSD ### Step 1: Calculate with Main Scale Units Given that 1text MSD = 5text units: textLC = frac111 times 5 = frac511text units ### Pattern Recognition Watch out for unconventional scale arrangements where textVSD > textMSD. The fundamental difference formula 1text MSD - 1text VSD safely determines magnitudes without needing sign corrections. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

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