A physical quantity P is given by P= A³B⁰.⁵/(C⁻⁴D¹.⁵). The quantity which brings in the maximum percentage error in P is

The correct option is C.

Step-by-Step Explanation:

To find which quantity brings in the maximum percentage error in P, we first write down the given relation for the physical quantity P:

$P=\frac{A^3{B}^{0.5}}{C^{-4}{D}^{1.5}}$

We can rewrite the expression by moving the term in the denominator to the numerator using laws of exponents (specifically, $\frac{1}{C^{-4}}=C^4$):

$P=A^3{B}^{0.5}{C}^4{D}^{-1.5}$

When calculating the maximum possible relative error in a product or quotient of quantities, we sum the absolute value of the exponents multiplied by their respective fractional errors. Therefore, the maximum relative error in P is given by:

$\frac{\Delta P}{P}=3\left(\frac{\Delta A}{A}\right)+0.5\left(\frac{\Delta B}{B}\right)+4\left(\frac{\Delta C}{C}\right)+1.5\left(\frac{\Delta D}{D}\right)$

Multiplying by 100 on both sides gives the expression for the maximum percentage error:

$\left(\frac{\Delta P}{P}\times 100\right)=3\left(\frac{\Delta A}{A}\times 100\right)+0.5\left(\frac{\Delta B}{B}\times 100\right)+4\left(\frac{\Delta C}{C},x\right)\times 100+1.5\left(\frac{\Delta D}{D},\times ,100\right)$

From the expression above, we can see the coefficients associated with the measurement errors of each quantity:
- For A, the multiplier is 3.
- For B, the multiplier is 0.5.
- For C, the multiplier is 4.
- For D, the multiplier is 1.5.

Since the coefficient of the relative error of C is the largest (which is 4), any error in the measurement of C is scaled up by a factor of 4. Consequently, the quantity C contributes the most to the overall percentage error in P.