Question

The density of the material of a cube is measured by measuring its mass and length of its side. If the maximum error in the measurement of mass and the length are $3\%$ and $2\%$ respectively, the maximum error in the measurement of density is

• A. $1\%$
• B. $5\%$
• C. $7\%$
• D. $9\%$

Answer 1

Answer: (D)

Step 1: Given data:

Error in the measurement of mass of the cube = $3\%$

Error in the measurement of length of the cube = $2\%$

Step 2: Formula used:

The density of a body is given by;

$Density=\frac{Massofthebody}{Volume}$

For a cube;

The volume of the cube is,

$Volume\left(V\right)ofacube={\left\{length\left(l\right)\right\}}^3$

Thus,

$Density\left(d\right)=\frac{Mass\left(m\right)}{{\left(l\right)}^3}$

For maximum error in the measurement of the density of the cube, we have

$d=\frac{m}{l^3}$

$Takinglogonbothsides,weget$

$logd=logm-3logl$

$Differentiatingbothsides,weget\frac{\Delta d}{d}=\frac{\Delta m}{m}-3\frac{\Delta l}{l}$

$Formaximumerror,$

$\frac{\Delta d}{d}=\frac{\Delta m}{m}+3\frac{\Delta l}{l}$

For percentage error;

$\frac{\Delta d}{d}\times 100\%=\frac{\Delta m}{m}\times 100\%+3\frac{\Delta l}{l}\times 100\%$

Step 3: Calculation of maximum error in the measurement of density of the cube:

$\frac{\Delta d}{d}\times 100\%=3\%+3\times 2\%$

$\frac{\Delta d}{d}\times 100\%=3\%+6\%$

$\frac{\Delta d}{d}\times 100\%=9\%$

Thus,

Option D) 9% is the correct option.