the %age error in the measurements of radius of sphere is 2%.the % error in the measurements of its surface area will be?
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Given that
Δr/r * 100 = 2%
A = 4 π r^2
ΔA/A * 100 = 2 * Δr/r * 100
= 2 * 2%
= 4%
Percentage error in its surface area is 4%
Δr/r * 100 = 2%
A = 4 π r^2
ΔA/A * 100 = 2 * Δr/r * 100
= 2 * 2%
= 4%
Percentage error in its surface area is 4%
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