Question

The maximum percentage error in the measurement of

mass, length and density of a thin cylindrical rod are 1%, 2% and 7% respectively. Find the maximum

percentage error in the measurement of its diameter.​

Answer 1

Given : The maximum percentage error in the measurement of  mass, length and density of a thin cylindrical rod are 1%, 2% and 7%  respectively

To Find :  the maximum percentage error in the measurement of its diameter.​

Solution:

Volume of cylinder =  πR²H

R = Radius = Diameter /2  = D/2

H = Length

Volume of cylinder =  πD²H/4

Density = Mass / Volume

=> Density = Mass / ( πD²H/4)

=> Density = 4 Mass / ( πD²H )

=> D² =   4 Mass / π Density H

=> 2 ΔD/D  =  Δmass/Mass  +  Δdensity/density +   Δlength/length

=>  100 * 2 ΔD/D   =100 *  Δmass/Mass  +  100 *Δdensity/Density +   100 *Δlength/length

=>  (100 * ΔD/D) * 2 = 1 % +  2 %   + 7   %

=>  (100 * ΔD/D) * 2 =  10 %

=>  (100 * ΔD/D)  =  5 %

percentage error in the measurement of its diameter.   =  5 %

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Answer 2

Given:

The maximum percentage error in the measurement of mass, length and density of a thin cylindrical rod are 1%, 2% and 7% respectively.

To find:

% error in measurement of diameter?

Calculation:

$\therefore \: mass = volume \times density$

$\implies \: m = \rho \times V$

$\implies \: m = \rho \times (\pi {r}^{2} h)$

$\implies \: m = \rho \times (\pi \times \dfrac{ {d}^{2} }{4} \times h)$

$\implies \: {d}^{2} = \dfrac{4m}{\pi \rho h}$

$\implies \:d = \sqrt{ \dfrac{4m}{\pi \rho h} }$

So, for small changes in m, $\rho$ and h, we can say :

$\implies \: \dfrac{\Delta d}{d} = \dfrac{1}{2} \dfrac{\Delta m}{m} + \dfrac{1}{2} \dfrac{\Delta \rho}{ \rho} + \dfrac{1}{2} \dfrac{\Delta h}{ h}$

$\implies \: \dfrac{\Delta d}{d} \% = \dfrac{1}{2} \bigg( \dfrac{\Delta m}{m} \% + \dfrac{\Delta \rho}{ \rho} \% + \dfrac{\Delta h}{ h} \% \bigg)$

$\implies \: \dfrac{\Delta d}{d} \% = \dfrac{1}{2} \bigg( 1 + 2 + 7 \bigg)\%$

$\implies \: \dfrac{\Delta d}{d} \% = 5\%$

So, max error in diameter is 5%.