JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 23)
The radius of a sphere is measured to be (7.50 $$\pm$$ 0.85) cm. Suppose the percentage error in its volume is x.
The value of x, to the nearest x, is __________.
The value of x, to the nearest x, is __________.
Answer
34
Explanation
Given, radius of sphere = (7.50 $$\pm$$ 0.85) cm
$$ \therefore $$ r = 7.50 and dr = 0.85
We know, volume of a sphere $$v = {4 \over 3}\pi {r^3}$$
taking log both sides, we get
$$\ln v = \ln {{4\pi } \over 3} + 3\ln r$$
Differentiating both sides,
$${{dv} \over v} = 0 + 3{{dr} \over r}$$
$$ \therefore $$ Fractional error in volume $${{dv} \over v} = 3{{dr} \over r}$$
$$ \therefore $$ % error in volume,
$${{dv} \over v} \times 100 = 3{{dr} \over r} \times 100$$
$$ = 3 \times {{0.85} \over {7.50}} \times 100$$
= 34%
$$ \therefore $$ r = 7.50 and dr = 0.85
We know, volume of a sphere $$v = {4 \over 3}\pi {r^3}$$
taking log both sides, we get
$$\ln v = \ln {{4\pi } \over 3} + 3\ln r$$
Differentiating both sides,
$${{dv} \over v} = 0 + 3{{dr} \over r}$$
$$ \therefore $$ Fractional error in volume $${{dv} \over v} = 3{{dr} \over r}$$
$$ \therefore $$ % error in volume,
$${{dv} \over v} \times 100 = 3{{dr} \over r} \times 100$$
$$ = 3 \times {{0.85} \over {7.50}} \times 100$$
= 34%
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