Question

(3) (c) and (e) (4) (a) and (d)
5. If the error in the measurement of radius of a
sphere is 2%, then the error in the determination
of volume of the sphere will be
(AIPMT (Prelims)-2008]
(1) 2%
(3) 6%
(4) 8%​

Answer 1

Answer:

6%

Explanation:

% error (in radius) = 2%

∴ Δr/r (relative error in measuring radius) = 0.02 ----------------(i)

We know that

Volume of sphere (v) = 4/3(πr³)

∴ Δv/v = 3(Δr/r) [∵ 4/3π is a constant]

=> Δv/v = 3 x 0.02 -------------[from (i)]

=> Δv/v = 0.06

∴ % error (in volume) = 0.06 x 100 = 6%

Hope it is helpful to you!

Answer 2

Answer:

Answer =6%

Explanation:

It is a correct.