Question

A physical quantity Q is related to four observables a, b, c, d as follows : $Q = \frac{ab^4}{cd}$

where, a = (60 ±3)Pa; b = (20 ± 0.1)m; c = (40 ± 0.2)Nsm⁻² and d = (50 ± 0.1)m, then the percentage error in Q is $\frac{x}{1000}$, where x = ________.

Answer 1

7700

Answer 2

To find the percentage error in Q, we use the rules of error propagation for multiplication and division.

Given the relation: $Q = \frac{ab^4}{cd}$

The formula for the maximum fractional error in Q is given by: $\frac{\Delta Q}{Q} = \left|1 \cdot \frac{\Delta a}{a}\right| + \left|4 \cdot \frac{\Delta b}{b}\right| + \left|-1 \cdot \frac{\Delta c}{c}\right| + \left|-1 \cdot \frac{\Delta d}{d}\right|$

Since errors always add up in magnitude, we consider the absolute values of the powers: $\frac{\Delta Q}{Q} = 1 \cdot \frac{\Delta a}{a} + 4 \cdot \frac{\Delta b}{b} + 1 \cdot \frac{\Delta c}{c} + 1 \cdot \frac{\Delta d}{d}$

Now, let's calculate the fractional errors for each observable:

1. For 'a': $a = (60 \pm 3)$ Pa $\frac{\Delta a}{a} = \frac{3}{60} = 0.05$

2. For 'b': $b = (20 \pm 0.1)$ m $\frac{\Delta b}{b} = \frac{0.1}{20} = 0.005$

3. For 'c': $c = (40 \pm 0.2)$ Nsm⁻² $\frac{\Delta c}{c} = \frac{0.2}{40} = 0.005$

4. For 'd': $d = (50 \pm 0.1)$ m $\frac{\Delta d}{d} = \frac{0.1}{50} = 0.002$

Substitute these values into the error propagation formula: $\frac{\Delta Q}{Q} = (0.05) + 4 \times (0.005) + (0.005) + (0.002)$ $\frac{\Delta Q}{Q} = 0.05 + 0.020 + 0.005 + 0.002$ $\frac{\Delta Q}{Q} = 0.077$

The percentage error in Q is given by: Percentage Error = $\frac{\Delta Q}{Q} \times 100\%$ Percentage Error = $0.077 \times 100\%$ Percentage Error = $7.7\%$

The problem states that the percentage error in Q is $\frac{x}{1000}$. So, we have: $7.7 = \frac{x}{1000}$ To find x, multiply 7.7 by 1000: $x = 7.7 \times 1000$ $x = 7700$

The final answer is 7700.

Explanation of the solution:

The percentage error in a quantity derived from products and quotients of other quantities is the sum of the percentage errors of the individual quantities, with each percentage error multiplied by the power to which that quantity is raised.

Given $Q = \frac{ab^4}{cd}$, the fractional error is $\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$.

Calculate individual fractional errors: $\frac{\Delta a}{a} = \frac{3}{60} = 0.05$ $\frac{\Delta b}{b} = \frac{0.1}{20} = 0.005$ $\frac{\Delta c}{c} = \frac{0.2}{40} = 0.005$ $\frac{\Delta d}{d} = \frac{0.1}{50} = 0.002$

Summing them up: $\frac{\Delta Q}{Q} = 0.05 + 4(0.005) + 0.005 + 0.002 = 0.05 + 0.02 + 0.005 + 0.002 = 0.077$.

Percentage error = $0.077 \times 100\% = 7.7\%$.

Given that percentage error is $\frac{x}{1000}$, we have $7.7 = \frac{x}{1000}$, which gives $x = 7700$.