**Solve the numerical example.**

If the length of a cylinder is l = (4.00 ± 0.001) cm, radius r = (0.0250 ± 0.001) cm and mass m = (6.25 ± 0.01) g. Calculate the percentage error in the determination of density.

#### Solution

**Given:**

l = (4.00 ± 0.001) cm,

In order to have same precision, we use,

(4.000 ± 0.001), r = (0.0250 ± 0.001) cm,

In order to have same precision, we use, (0.025 ± 0.001) m = (6.25 ± 0.01) g

**To find:** percentage error in density

**Formulae: **

1. Relative error in volume, `(Delta"V")/"V" = (2Delta"r")/"r" + (Deltal)/l` ....(∵ Volume of cylinder, V = πr^{2}l)

2. Relative error `(Deltarho)/rho = (Delta"m")/"m" + (Delta"V")/"V"` ....[∵ Density (ρ) = `("mass"("m"))/("volume"("v"))`]

3. Percentage error = Relative error × 100 %

**Calculation:**

From formulae (i) and (ii),

∴ `(Deltarho)/rho = (Delta"m")/"m" + (2Delta"r")/"r" + (Deltal)/l`

`= 0.01/6.25 + (2(0.001))/0.025 + 0.001/4.000`

= 0.0016 + 0.08 + 0.00025

= 0.08185

From formula (iii),

% error in density = `(Deltarho)/rho xx 100`

= 0.08185 × 100

= 8.185%

Percentage error in density is 8.185%.