Question

A rectangular metal slab of mass 33.333g has its length 8.0 cm, breadth 5.0 cm and
thickness 1 mm. The mass is measured with accuracy upto 1 mg with a sensitive balance. The length and breadth are
measured with a vernier calipers having a least count of 0.01 cm. The thickness is measured with a screw gauge of least count 0.01 mm. Calculate the percentage accuracy in density calculated from above measurements​

Answer 1

AnSwer :-

The percentage error itself gives us the percentage accuracy.

$\sf d = \dfrac{mass (M)}{lenght (\ell) \times breadth(b) \times height(h) } = \dfrac{M}{ \ell bh}$

$\therefore$The relative error is given by

$\sf \frac{ \Delta d}{d} = \bigg( \dfrac{ \Delta M}{M} \bigg) + \bigg( \dfrac{ \Delta \ell}{ \ell} \bigg) + \bigg( \dfrac{ \Delta b}{ b} \bigg) + \bigg( \dfrac{ \Delta h}{ h} \bigg)$

Given that,

$\ell$= 8.0 cm and $\Delta \ell$ = 0.01cm

b = 5.0 cm and $\Delta$b = 0.01 cm

h = 1 mm and $\Delta$h = 0.01 mm

M = 33.333 g and $\Delta$M = 1 mg=0.001g

hence,

$\sf \dfrac{ \Delta M}{M} = \dfrac{0.001}{33.333}$

$\dfrac{ \Delta \ell }{ \ell} = \dfrac{0.01}{8.0}$

$\sf \dfrac{ \Delta b}{b} = \dfrac{0.01}{5}$

$\sf\dfrac{ \Delta h}{h} = \dfrac{0.01}{1}$

The percentage error is given by

$\sf \bigg( \dfrac{ \Delta M}{M} + \dfrac{ \Delta \ell}{ \ell} + \dfrac{ \Delta b}{ b} + \dfrac{ \Delta h}{ h} \bigg) \times 100$

$\sf = \bigg( \dfrac{0.001}{33.333} + \dfrac{0.01}{8.0} + \dfrac{0.01}{5} + \dfrac{0.01}{1} \bigg) \times 100$

$\sf = 0.003 + 0.125 + 0.2 + 1.0 = 1.328 \cong 1.3$

The percentage error is 1.3%