Question

if the error in the measurements of length is2% then error in the measurements of volume is

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Answer 1

Answer:

If the error in measuring the radius of a sphere is 2%, then the error in the measurement of volume is:

ANSWER

Percentage error in radius is given as 2% i.e.

r

Δr

×100=2 %

Volume of sphere V=

3

4π

r

3

Percentage error in volume

V

ΔV

×100=3×

r

Δr

×100=3×2=6 %

Answer 2

Answer:

Volume error =6%

Explanation:

Percentage error in radius is given as 2%

i.e.

∆r/r ×100=2 %

Volume of sphere V= 4π/3 × r³

Percentage error in volume

ΔV/V ×100= 3×∆r/r × 100

=3×2=6 %