The approximate value of square root of 25.2 is

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Question

The approximate value of square root of 25.2 is

Options:

A . 5.01

B . 5.02

C . 5.03

D . 5.04

Answer: Option B
:
B
Let f (x) =

\sqrt{x}

Now,

f (x + δ x) - f (x) = f^{'} (x) . δ x = \frac{δ x}{2 \sqrt{x}}

We may write, 25.2 = 25 + 0.2
Taking x = 25 and

δ x = 0.2

We have

f (25.2) - f (25) = \frac{0.2}{2 \sqrt{25}} = 0.02 ∴ f (25.2) = f (25) + 0.02 = \sqrt{25} + 0.02 = 5.02 \Rightarrow \sqrt{(25.2)} = 5.02

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