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

For $$Q = \frac{ab^4}{cd}$$, the percentage error is given by

$$ \frac{\Delta Q}{Q} \times 100 = \left(\frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}\right) \times 100 $$

Using the provided data, we calculate each relative error:

$$ \frac{\Delta a}{a} = \frac{3}{60} = 0.05 $$

$$ \frac{\Delta b}{b} = \frac{0.1}{20} = 0.005 $$

$$ \frac{\Delta c}{c} = \frac{0.2}{40} = 0.005 $$

$$ \frac{\Delta d}{d} = \frac{0.1}{50} = 0.002 $$

Substituting these results into the percentage error formula yields

$$ = (0.05 + 4 \times 0.005 + 0.005 + 0.002) \times 100 $$

$$ = (0.05 + 0.02 + 0.005 + 0.002) \times 100 $$

$$ = 0.077 \times 100 = 7.7\% $$

Since the percentage error can be expressed in the form $$\frac{x}{1000}$$, we set

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

$$ x = 7700 $$

The answer is 7700.