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The diameter of a sphere is measured to be 40 cm. If an error of 0.02 cm is made in it, then find the approximate errors in volume and surface area of the sphere.

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50.24 {cm}^{3} and 5.024 {cm}^{2}

25.54 {cm}^{3} and 7.04 {cm}^{2}

10.54 {cm}^{3} and 3.034 {cm}^{2}

30.28 {cm}^{3} and 5.025 {cm}^{2}

answer is A.

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Detailed Solution

Given,
Diameter of a sphere d= 40 cm
So, its radius,

\begin{matrix} r = \frac{(d)}{2} \\ r = \frac{40}{2} \\ r = 20 cm \end{matrix}

Error in measuring the diameter = 0.02 cm.
Therefore, the error in radius will half of it.
(

Δr)

= 0.01 cm
Now, the volume of sphere,

(Δv)

= \frac{4}{3} π r^{3}

Δv = (\frac{d}{dr} v) \times Δr

\begin{matrix} Δv = \frac{4}{3} π (3 r^{2}) \times (0.01) \\ = 50.24 {cm}^{3} \end{matrix}

Now, surface area of sphere

s = 4 π r^{2}

For approximate error differentiate S with respect to r. Also, multiply with

Δ r

Δs = (\frac{ds}{dr}) Δ r

\begin{matrix} \Rightarrow Δs = (4 \times 2 \times π r) Δr \\ \Rightarrow Δs = (8 π \times 20) (0.01) \\ \Rightarrow Δs = (1.6 π) \\ = 5.024 {cm}^{2} \end{matrix}

Option 1 is correct.

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