Question

The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is \[\dfrac{x}{{100}}\% \] . If the relative errors in measuring the mass and the diameter are $6.0\% $ and $1.5\% $ respectively, then what is the value of x?

Answer 1

Hint: We are given the percentage errors of density ( $\rho $ ), mass (m), and diameter (d) of a solid metal sphere. So first we write down the formula of the density of the sphere as $\rho = \dfrac{m}{V}$ where V is the volume of the sphere. Now write the percentage error formula for density and substitute the given values in it and find the value of x.

Formula Used:
$density = \dfrac{{mass}}{{volume}}$

Complete step by step Solution:
Given: Maximum error in the density ($\rho $ ) of solid metal sphere $ = \dfrac{x}{{100}}\% $
Error in measuring mass (m) of the solid metal sphere $ = 6.0\% $
Error in measuring diameter (d) of the solid metal sphere $ = 1.5\% $
We know that, $density = \dfrac{{mass}}{{volume}}$ i.e., $\rho = \dfrac{m}{V}$ where V is the volume of the solid metal sphere.
                                                                                                                                                                        ...(1)
 $V = \dfrac{4}{3}\pi {r^3}$ where r is the radius of the solid metal sphere. ...(2)
Substitute $r = \dfrac{d}{2}$ in volume formula in equation (2), we get
 $V = \dfrac{4}{3}\pi {\left( {\dfrac{d}{2}} \right)^3}$ ...(3)
Substitute value V from equation (3) in equation (1), and we get
 $\rho = \dfrac{m}{{\dfrac{4}{3}\pi {{\left( {\dfrac{d}{2}} \right)}^3}}}$ ...(4)
Now we write the percentage error of equation (4), and we get
 $\left( {\dfrac{{\vartriangle \rho }}{\rho } \times 100} \right)\% = \left( {\dfrac{{\vartriangle m}}{m} \times 100} \right)\% + 3 \times \left( {\dfrac{{\vartriangle d}}{d} \times 100} \right)\% $ ...(5)
Given: $\left( {\dfrac{{\vartriangle \rho }}{\rho } \times 100} \right)\% = \dfrac{x}{{100}}\% $
 $\left( {\dfrac{{\vartriangle m}}{m} \times 100} \right)\% = 6.0\% $
$\left( {\dfrac{{\vartriangle d}}{d} \times 100} \right)\% = 1.5\% $
By substituting these values in equation (5), we get
 $\left( {\dfrac{x}{{100}}} \right) = 6.0 + 3 \times 1.5$
 $\dfrac{x}{{100}} = 10.5$
Solving this further
 $x = 10.5 \times 100$
 $x = 1050$
Hence, the value of $x = 1050$ .

Note: One of the important points to be noted here is that the errors get added both in the case of multiplication as well as division. Another point to be noted here is that the constants vanish when we write the percentage error formula of density as there can never be any error in a constant.