A physical quantity Q$Q$ is related to four observables a, b, c, d$a, b, c, d$ as follows: Q = fracab^4cd$Q = \frac{ab^4}{cd}$ where, a = (60 pm 3)mathrm~Pa$a = (60 \pm 3)\mathrm{~Pa}$ ; b = (20 pm 0.1)mathrm~m$b = (20 \pm 0.1)\mathrm{~m}$ ; c = (40 pm 0.2)mathrm~Nsm^-2$c = (40 \pm 0.2)\mathrm{~Nsm}^{-2}$ and d = (50 pm 0.1)mathrm~m$d = (50 \pm 0.1)\mathrm{~m}$ , then the percentage error in Q$Q$ is fracx1000$\frac{x}{1000}$ , where x = $x = $ ______.

### Related Formula By the Principle of Homogeneity, terms added or subtracted must have the same dimensions: [B] = [x^2]$[B] = [x^2]$ ### Core Logic Known dimensional formulas: Length x to [L]$x \to [L]$ Energy E to [ML^2T^-2]$E \to [ML^2T^{-2}]$ Time t to [T]$t \to [T]$ ### Step 1: Dimension of B Since x^2$x^2$ is subtracted from B$B$: [B] = [x^2] = [L^2]$$[B] = [x^2] = [L^2]$$ ### Step 2: Dimension of A From the equation E = fracB - x^2At$E = \frac{B - x^2}{At}$: [A] = frac[B - x^2][E][t]$$[A] = \frac{[B - x^2]}{[E][t]}$$ [A] = frac[L^2][ML^2T^-2][T] = frac[L^2][ML^2T^-1]$$[A] = \frac{[L^2]}{[ML^2T^{-2}][T]} = \frac{[L^2]}{[ML^2T^{-1}]}$$ [A] = [M^-1T^1]$$[A] = [M^{-1}T^1]$$ ### Step 3: Dimension of AB [AB] = [A] times [B]$$[AB] = [A] \times [B]$$ [AB] = [M^-1T^1] times [L^2]$$[AB] = [M^{-1}T^1] \times [L^2]$$ [AB] = [L^2 M^-1 T^1]$$[AB] = [L^2 M^{-1} T^1]$$ ### Pattern Recognition Identify sums/differences first to instantly isolate B$B$. Once [B]$[B]$ is fixed, the entire numerator is just L^2$L^2$. Swap out variables to isolate [A]$[A]$. Combining is just standard exponent addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements

### Related Formula R = fracrho LA = fracrho Lpi left(fracd2right)^2 = frac4rho Lpi d^2$$R = \frac{\rho L}{A} = \frac{\rho L}{\pi \left(\frac{d}{2}\right)^2} = \frac{4\rho L}{\pi d^2}$$ ### Core Logic To find the maximum percentage error in resistance, apply logarithmic differentiation: fracDelta RR = fracDelta LL + 2fracDelta dd$$\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d}$$ Given percentage errors: fracDelta LL times 100\% = 0.1\%$\frac{\Delta L}{L} \times 100\% = 0.1\%$ fracDelta dd times 100\% = 0.1\%$\frac{\Delta d}{d} \times 100\% = 0.1\%$ ### Step 2: Substitution Substituting the values: fracDelta RR times 100\% = 0.1\% + 2(0.1\%)\, = 0.1\% + 0.2\% = 0.3\%$$\frac{\Delta R}{R} \times 100\% = 0.1\% + 2(0.1\%)\, = 0.1\% + 0.2\% = 0.3\%$$ ### Pattern Recognition Resistance scales inversely with the square of the diameter. The error multiplier for diameter is 2. Just sum linear components directly: Error = L_error + 2 * d_error. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units And Measurements Class 12 Physics: Current Electricity

### Related Formula textPrinciple of Homogeneity: [F] = [ax^2] = [bt^1/2]$$\text{Principle of Homogeneity: } [F] = [ax^2] = [bt^{1/2}]$$ ### Core Logic By the principle of dimensional homogeneity, each additive term must have the same dimension as the left hand side. Dimension of force F = [M L T^-2]$F = [M L T^{-2}]$. For the term ax^2$ax^2$: [a] = frac[F][x^2] = frac[M L T^-2][L^2] = [M L^-1 T^-2]$$[a] = \frac{[F]}{[x^2]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]$$ For the term bt^1/2$bt^{1/2}$: [b] = frac[F][t^1/2] = frac[M L T^-2][T^1/2] = [M L T^-5/2]$$[b] = \frac{[F]}{[t^{1/2}]} = \frac{[M L T^{-2}]}{[T^{1/2}]} = [M L T^{-5/2}]$$ ### Step 2: Computing Required Ratio We need the dimension of fracb^2a$\frac{b^2}{a}$: left[ fracb^2a right] = frac[M L T^-5/2]^2[M L^-1 T^-2]$$\left[ \frac{b^2}{a} \right] = \frac{[M L T^{-5/2}]^2}{[M L^{-1} T^{-2}]}$$ left[ fracb^2a right] = frac[M^2 L^2 T^-5][M L^-1 T^-2]$$\left[ \frac{b^2}{a} \right] = \frac{[M^2 L^2 T^{-5}]}{[M L^{-1} T^{-2}]}$$ left[ fracb^2a right] = [M^2-1 L^2 - (-1) T^-5 - (-2)]$$\left[ \frac{b^2}{a} \right] = [M^{2-1} L^{2 - (-1)} T^{-5 - (-2)}]$$ left[ fracb^2a right] = [M L^3 T^-3]$$\left[ \frac{b^2}{a} \right] = [M L^3 T^{-3}]$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units And Measurements