If the error in the measurement of radius of a sphere is 2%, then the ...
Error in the measurement of radius of a sphere = 2%
Volume of the sphere =

∴ Error in the volume =

= 3 × 2% = 6%
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If the error in the measurement of radius of a sphere is 2%, then the ...
To understand why the error in the measurement of the radius of a sphere affects the determination of its volume, we need to understand the relationship between the radius and the volume of a sphere.
The volume of a sphere is given by the formula V = (4/3)πr^3, where V is the volume and r is the radius of the sphere.
Let's consider the error in the measurement of the radius as a percentage. The error in the radius is given as 2% or 0.02.
Now, let's calculate the error in the determination of the volume of the sphere.
**Calculating the Error in the Volume:**
To find the error in the volume, we need to consider how the error in the radius affects the volume formula.
The formula for the volume of a sphere is V = (4/3)πr^3. Taking the derivative of this formula with respect to r, we get dV/dr = 4πr^2.
Now, let's calculate the relative error in the volume (ΔV/V), which is the ratio of the error in the volume (ΔV) to the actual volume (V).
(ΔV/V) = (dV/dr * Δr) / V
(ΔV/V) = (4πr^2 * Δr) / [(4/3)πr^3]
(ΔV/V) = (3 * Δr) / r
Substituting the given error in the radius (Δr = 0.02) into the equation, we get:
(ΔV/V) = (3 * 0.02) / r
(ΔV/V) = 0.06 / r
From the above equation, we can see that the relative error in the volume is directly proportional to the error in the radius (Δr).
**Conclusion:**
Since the relative error in the volume is 0.06/r, the error in the determination of the volume of the sphere will be 6% when the error in the measurement of the radius is 2%.
Therefore, the correct answer is option B) 6%.
If the error in the measurement of radius of a sphere is 2%, then the ...
Volume of sphere = 4/3 π r³
Error = 3 ×∆r / r ×100
= 3×2
= 6%