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

The problem requires us to calculate the percentage error in a physical quantity Q, which is defined in terms of four other observables a, b, c, and d. The relationship is given by $Q = \frac{ab^4}{cd}$. We are also provided with the values and their absolute errors for a, b, c, and d.

The rule for propagation of errors for a quantity $Q = \frac{a^p b^q}{c^r d^s}$ is given by:

$\frac{\Delta Q}{Q} = p \frac{\Delta a}{a} + q \frac{\Delta b}{b} + r \frac{\Delta c}{c} + s \frac{\Delta d}{d}$

For the given relation $Q = \frac{ab^4}{cd}$, the powers are $p=1$, $q=4$, $r=1$, $s=1$. So, the fractional error in Q is:

$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4 \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$

Now, let's calculate the fractional error for each observable using the given values:

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 fractional errors into the equation for $\frac{\Delta Q}{Q}$:

$\frac{\Delta Q}{Q} = 0.05 + 4(0.005) + 0.005 + 0.002$

$\frac{\Delta Q}{Q} = 0.05 + 0.020 + 0.005 + 0.002$

$\frac{\Delta Q}{Q} = 0.077$

To find the percentage error in Q, we multiply the fractional error by 100:

$\text{Percentage error in Q} = \left( \frac{\Delta Q}{Q} \right) \times 100\%$

$\text{Percentage error in Q} = 0.077 \times 100\% = 7.7\%$

The question states that the percentage error in Q is $\frac{x}{1000}$. Let the numerical value of the percentage error be $P$. So, $P = 7.7$. According to the problem statement, $P = \frac{x}{1000}$. Therefore, we can write:

$7.7 = \frac{x}{1000}$

Solving for x:

$x = 7.7 \times 1000$

$x = 7700$

The final answer is $\boxed{7700}$.

Explanation of the solution:

1. Identify the formula for Q and the given values with their absolute errors.
2. Apply the error propagation rule for products and quotients: $\frac{\Delta Q}{Q} = \sum (\text{power}) \times \frac{\Delta \text{observable}}{\text{observable}}$.
3. Calculate the fractional error for each observable: $\frac{\Delta a}{a}$, $\frac{\Delta b}{b}$, $\frac{\Delta c}{c}$, $\frac{\Delta d}{d}$.
4. Substitute these values into the error propagation formula to find the total fractional error $\frac{\Delta Q}{Q}$.
5. Convert the fractional error to percentage error by multiplying by 100.
6. Equate the calculated percentage error to $\frac{x}{1000}$ and solve for $x$.

Answer: The value of x is 7700.