Question

if the error in measuring the radius of the sphere is 2%and that in measuring its mass is 3%,then the error in measuring the density of material of the sphere is​

Answer 1

$volume = 4 \div 3 \times \pi \: r ^{3}$

Error in volume =3 times the error in radius

dV/V=3×dr/r

so error in volume =3×2%=6%

Error in mass=3%

Density =mass/volume

errors in physics are always additive

so error in density is 3+6=9%

Answer 2

Given that,

The error in measuring the radius of the sphere is 2% and that in measuring its mass is 3%.

To find,

The error in measuring the density of material of the sphere.

Solution,

The formula of density is given by :

$d=\dfrac{m}{V}$

V is the volume of sphere

$d=\dfrac{m}{(4/3)\pi r^3}\\\\d=\dfrac{3}{4}\dfrac{m}{\pi r^3}$

The percentage error in density is given by :

$\dfrac{\Delta d}{d}=(\dfrac{\Delta m}{m}+3\dfrac{\Delta r}{r})\\\\\dfrac{\Delta d}{d}=(3+3\times 2)\\\\\dfrac{\Delta d}{d}=9\%$

So, the error in measuring the density of material of the sphere is​ 9 percent.

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