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 ( x \over 100 ) %. If the relative errors in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of x is \underline\hspace2cm .

Sign in

Units and Measurements question

2020 · 6 Sep · Shift 1 · Q60

Units and Measurements question

2020 · 6 Sep · Shift 1 · Q60

View written solutionFree

Correct answer: 1050

The density of a sphere is

$\rho = \frac{m}{V}$

For a sphere,

$V = \frac{\pi d^3}{6}$

So,

$\rho = \frac{6m}{\pi d^3}$

Hence, density depends on mass and diameter as

$\rho \propto m d^{-3}$

For maximum relative error, we add absolute percentage errors:

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

Given:

$\frac{\Delta m}{m} = 6.0\%$

$\frac{\Delta d}{d} = 1.5\%$

Thus,

$\frac{\Delta \rho}{\rho} = 6.0 + 3(1.5) = 6.0 + 4.5 = 10.5\%$

The question says the maximum error is

$\left(\frac{x}{100}\right)\%$

So,

$\frac{x}{100} = 10.5$

$x = 10.5 \times 100 = 1050$

Therefore, the required integer is

$\boxed{1050}$

More from Units and Measurements

Sign in

Units and Measurements question

2020 · 6 Sep · Shift 1 · Q60

Units and Measurements question

2020 · 6 Sep · Shift 1 · Q60

View written solutionFree

Correct answer: 1050

The density of a sphere is

$\rho = \frac{m}{V}$

For a sphere,

$V = \frac{\pi d^3}{6}$

So,

$\rho = \frac{6m}{\pi d^3}$

Hence, density depends on mass and diameter as

$\rho \propto m d^{-3}$

For maximum relative error, we add absolute percentage errors:

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

Given:

$\frac{\Delta m}{m} = 6.0\%$

$\frac{\Delta d}{d} = 1.5\%$

Thus,

$\frac{\Delta \rho}{\rho} = 6.0 + 3(1.5) = 6.0 + 4.5 = 10.5\%$

The question says the maximum error is

$\left(\frac{x}{100}\right)\%$

So,

$\frac{x}{100} = 10.5$

$x = 10.5 \times 100 = 1050$

Therefore, the required integer is

$\boxed{1050}$