Question

if the measure values of two quantities are a plus minus Delta A and B plus minus Delta B Delta a and Delta B being the mean absolute error what is the maximum possible error in a plus minus b show that Z is equal to a upon B Delta Z upon Z is equal to 2 upon 3 + Delta B upon B​

Answer 1

Answer:

Putting the question in a right way:

If the measured values of two quantities are A±ΔA and B±ΔB, ΔA and ΔB being the mean absolute errors. What is the maximum possible error in A+B? Show that if Z=A/B then ΔZ/Z= ΔA/A + ΔB/B

Solution:

Error in A = ±ΔA

Error in B = ±ΔB

Maximum possible error in A + B = ΔA + ΔB        (Ans)

If Z = A/B

If error in Z is ±ΔZ

Then

$Z\±\Delta Z =  (A\±\Delta A)/(B\±\Delta B)$

$\implies Z\±\Delta Z =  \frac{A\±\Delta A}{B\±\Delta B}$

$\implies Z\±\Delta Z =  \frac{A}{B}(\frac{1\±\Delta A/A}{1\±\Delta B/B})$

$\implies Z\±\Delta Z =  Z(\frac{1\±\Delta A/A}{1\±\Delta B/B})$

$\implies \frac{Z\±\Delta Z}{Z}  =  \frac{1\±\Delta A/A}{1\±\Delta B/B}$

$\implies 1\±\frac{\Delta Z}{Z}   =  \frac{1\±\Delta A/A}{1\±\Delta B/B}$

$\implies (1\±\frac{\Delta Z}{Z})(1\±\frac{\Delta B}{B})   =  1\±\frac{\Delta A}{A}$

$\implies 1\±\frac{\Delta Z}{Z} \±\frac{\Delta B}{B} \±\frac{\Delta Z}{Z}\frac{\Delta B}{B} =  1\±\frac{\Delta A}{A}$

ΔZ.ΔB will be very small quantity

Thereofre neglecting it we get

$\±\frac{\Delta Z}{Z} \±\frac{\Delta B}{B} =  \±\frac{\Delta A}{A}$

or, $\±\frac{\Delta Z}{Z}=  \±\frac{\Delta A}{A}\±\frac{\Delta B}{B}$

or, $\boxed{\frac{\Delta Z}{Z}=  \frac{\Delta A}{A}+\frac{\Delta B}{B}}$

Answer 2

Answer:ΔZ/Z= ΔA/A + ΔB/B

Explanation:See the photo You will love it