Question

two measured quantities are given as A = 2.0+-0.1m and B = 2.0 +- 0.1m. the correct value of (AB)^1/2 will be

​

Answer 1

Given : Two measured quantities are given as

A = (2.0 ± 0.1) m and B = (2.0 ± 0.1) m

To find : The correct value of AB½

solution : let y = AB½

squaring both sides we get,

y² = AB

taking log both sides we get,

logy² = log(AB)

⇒2logy = logA + logB

now differentiating both sides we get,

2 dy/y = dA/A + dB/A

⇒dy = y(dB/2B + dA/2A)

therefore, ∆y = y(∆B/2B + ∆A/2A)

here A = 2 , B = 2

so, y = AB½ = (2 × 2)½ = 2

again, ∆A = 0.1 , ∆B = 0.1

then, ∆y = 2[0.1/(2 × 2) + 0.1/(2 × 2)]

= 2[0.1/4 + 0.1/4]

= 2 × 0.1/2 = 0.1

Therefore the correct value of (AB)^1/2 will be (2.0 ± 0.1)