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 :

Solution & Explanation

### 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

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Units And Measurements

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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
Q7 jee_main_2025_02_april_evening Dimensional Analysis
Given a charge q, current I and permeability of vacuum mu_0 . Which of the following quantity has the dimension of momentum?
  • A. q I / mu_0
  • B. q mu_0 I
  • C. q^2 mu_0 I
  • D. q mu_0 / I

Solution

### 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
Q jee_main_2025_02_april_morning Dimensional Analysis
Match List-I with List-II. beginarray|l|l| hline textbfList-I & textbfList-II \\ hline text(A) Coefficient of viscosity & text(I) [mathrmML^0mathrmT^-3] \\ hline text(B) Intensity of wave & text(II) [mathrmML^-2mathrmT^-2] \\ hline text(C) Pressure gradient & text(III) [mathrmM^-1mathrmLT^2] \\ hline text(D) Compressibility & text(IV) [mathrmML^-1mathrmT^-1] \\ hline endarray Choose the correct answer from the options given below:
  • A. (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  • B. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  • C. (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  • D. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Solution

### Related Formula Definitions of physical quantities: 1. Viscosity: F = -eta A fracdvdx 2. Intensity: I = fractextPowertextArea 3. Pressure gradient: fracdPdx 4. Compressibility: K = frac1B = fractextStraintextStress ### Core Logic Let's calculate each dimensional formula: 1. **Coefficient of viscosity (eta):** [eta] = frac[F][A]left[fracdvdxright] = fractextM L T^-2textL^2 left(fractextL T^-1textLright) = textM L^-1textT^-1 Matches **(IV)**. 2. **Intensity of wave (I):** [I] = frac[textPower][textArea] = fractextM L^2textT^-3textL^2 = textM T^-3 = textM L^0textT^-3 Matches **(I)**. 3. **Pressure gradient (fracdPdx):** left[fracdPdxright] = frac[textPressure][textLength] = fractextM L^-1textT^-2textL = textM L^-2textT^-2 Matches **(II)**. 4. **Compressibility (K):** Compressibility is the inverse of Bulk Modulus: [K] = frac1[textPressure] = frac1textM L^-1textT^-2 = textM^-1textLtextT^2 Matches **(III)**. Thus: (A)-(IV), (B)-(I), (C)-(II), (D)-(III). ### Step 1: Final Conclusion The correct option is (2). ### Pattern Recognition Target basic matching terms first: Compressibility is the reciprocal of pressure, giving [M^-1LT^2]. Wave intensity has units of power per unit area, giving [MT^-3]. This immediately isolates option (2). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids Class 11 Physics: Mechanical Properties of Solids
Q2 jee_main_2025_02_april_morning Dimensional Analysis
The equation for real gas is given by left(P + fracaV^2right)(V - b) = RT, where P, V, T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of ab^-2 is equivalent to that of:
  • A. Planck's constant
  • B. Compressibility
  • C. Strain
  • D. Energy density

Solution

### Related Formula By the principle of dimensional homogeneity, terms added or subtracted must have the same dimensions: [P] = left[fracaV^2right] implies [a] = [P][V]^2 [V] = [b] ### Core Logic Let's find the dimensional formula of the quantities: - Pressure P: [P] = textM L^-1textT^-2 - Volume V: [V] = textL^3 Substituting these to find [a] and [b]: [a] = (textM L^-1textT^-2)(textL^6) = textM L^5textT^-2 [b] = textL^3 implies [b^-2] = textL^-6 Now, compute the dimensions of ab^-2: [ab^-2] = (textM L^5textT^-2)(textL^-6) = textM L^-1textT^-2 This matches the dimensions of pressure. Let's evaluate the options: 1. Planck's constant: [h] = textM L^2textT^-1 2. Compressibility: [beta] = textM^-1textLtextT^2 3. Strain: dimensionless 4. Energy density (energy per unit volume): left[fracEVright] = fractextM L^2textT^-2textL^3 = textM L^-1textT^-2 ### Step 1: Final Conclusion Therefore, the dimension of ab^-2 is equivalent to that of Energy density. ### Pattern Recognition By writing the relation directly as [ab^-2] = frac[a][b]^2, and noting [a] = [P][V]^2 and [b] = [V], we get [ab^-2] = frac[P][V]^2[V]^2 = [P] (Pressure). Since pressure and energy density have identical dimensions, the answer is immediately Energy density. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Kinetic Theory of Gases

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