Question

The length of a wire is measured with a metre scale having least count 1mm. Its diameter is measued with a vernier calipers of least count 0.1mm. Given that length and diameter of the wire are measured as 5cm and 4mm, the percentage error in the calculated value of volume of the wire will be

Answer 1

$\underline{\boxed{ \huge\purple{ \rm{Answer}}}}$

$\rm \: volume \: of \: wire = \pi {r}^{2} h \\ \\ \therefore \rm \: \red{v \propto {r}^{2} h}$

Given :

length of wire = 5cm = 50mm

least count of meter scale = 1mm

radius of wire = 2mm

least count of vernier scale = 0.1mm

To Find :

Percentage error in measurement

Formula :

$\underline{ \boxed{ \bold{ \rm{ \pink{ \frac{ \triangle{v}}{v} = \frac{2 \triangle{r}}{r} + \frac{ \triangle{h}}{h}}}}}} \: \dag$

Calculation :

$\rm \frac{ \triangle{v}}{v} = \frac{2(0.1)}{2} + \frac{1}{50} \\ \\ \rm \frac{ \triangle{v}}{v} = 0.1 + 0.02 = 0.12 \\ \\ \therefore \: \underline{ \boxed{ \rm{ \orange{\% \frac{ \triangle{v}}{v} = 12\%}}}}$

Answer 2

The percentage error in the calculated volume of wire will be 12%.

Given,

The least count of the meter scale = 1mm.

The radius of the wire = 4/2 = 2mm

The least count of the vernier scale = 0.1mm

The length of the wire = 5 cm = 50 mm

To Find,

The percentage error in the calculated value of volume.

Solution,

The volume of wire is V = πr²π

So,

V ∝ r²h

Now, percentage error will be

Δv/v = 2Δr/r + Δh/h

Δv/v = 2(0.1)/2+1/50

Δv/v = 0.1+0.02

Δv/v = 0.12

Percentage error = Δv/v*100 = 12%

Hence, the percentage error in the calculated volume of wire will be 12%.

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