Question

if 2% is the error in the measurement of the diameter of a sphere then error in the measurement of its volume is

Answer 1

Answer:

If 2% is the error in the measurement of the diameter of a sphere then the error in the measurement of its volume is $6$%

Explanation:

• An error gives the difference between the true value and measured value
• The percentage error is the difference between the true value and calculated value divided by true value multiplied by $100$

                     percentage error= $\frac{x_{calcu}-x_{true}}{x_{true} }$

• For the equation given $A=nx^{2}$ the percentage error in the quantity A is

                     $\frac{\Delta A}{A}\times 100=2\frac{\Delta x}{x}\times 100$

• The volume of a sphere with radius R is given by the formula  $V=\frac{4}{3} \pi R^{3}$

From the question, it is given that

the percentage error in diameter is $\Delta R} /{R}\times 100=2$%

Now the error in the volume of the sphere is

$\frac{\Delta V}{V}\times 100=3\frac{\Delta R}{R}\times 100$

                 $=3*2=6$%