6. The diameter of a sphere is measured to be 40cm. If an error of 0.02 cm is made in it
approximate errors in volume and surface area of the sphere.
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Step-by-step explanation:
radius of the sphere = 20 cm
Δr= 0.01 cm
volume of the sphere
we need to find error in calculating volume ΔV
=
=4/3π(20)^2×0.01
= 16.75 cm^2
Also surface area
S= 4πr^2
now we need to find ΔS
ΔS = 8πrΔr
= 8π20×0.01 = 1.6π= 5.0265 cm^2
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