Question

20. A quantity z is given by the relation, z = 2√ab/c^3
-. Find the percentage error in z if the percentage error in a,
b and c are respectively 2%, 1% and 4%.​ help me solve this que. quickly .
I will mark the best ans. brainliast
# plz help

Answer 1

Heya!!

______________________________⭐⭐⭐

→ given,

z= 2√ab/c^3

(here, 2 is constant)

→ ∆z/z ×100 = a^1/2 +b^1/2 + c^3/2

→∆z/z ×100 = 1/2×2 + 1/2×1 + 3/2×4

→ ∆z/z ×100= 7.5%

Note:- error is always added

hope it helps!!

#PhoeniX

Answer 2

Answer:

The percentage error in z is 14 %.

Explanation:

Given,

z is related as:

$z = 2\times \frac{\sqrt{a}\times b}{c^3}$

The percentage error in a is (a% = 2%).

The percentage error in b is (b% = 1%).

The percentage error in c is (c% = 4%).

To find,

The percentage error in z.

Calculation,

$z = 2 \times \frac{\sqrt{a}\times b}{c^3}$...(1)

Then, equation (1) can be written as:

$\log z =\log(2 \times \frac{\sqrt{a}\ b }{c^3} )$

$\implies \log z = \log 2 + \log{a^{1/2}} + \log{b} + \log{c^3}$

$\implies \log{z} = \log{2} + \frac{1}{2} \log{a} + \log b + 3\log c$

Now we differentiate the above equation

$\frac{\Delta z}{z} =0 + \frac{1}{2} \frac{\Delta a}{a} + \frac{\Delta b}{b}+ 3\frac{\Delta c}{c}$$\implies \frac{\Delta z}{z} \times 100 =0 + \frac{1}{2} \frac{\Delta a}{a}\times 100 + \frac{\Delta b}{b}\times 100+ 3\frac{\Delta c}{c}\times 100$

⇒ z % = 1/2(a %) + b % + 3(c %)

Now substituting a% = 2%, b% = 1%, and c% = 4% in the above equation

z % = 1/2(2 %) + 1 % + 3(4 %)

z % = 1 % + 1 % + 12 %

z % = 14 %

Therefore, the percentage error in z is 14 %.

#SPJ2