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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: [2008]
  • a)
    4%
  • b)
    6%
  • c)
    8%
  • d)
    2%
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
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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Most Upvoted Answer
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%.
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Community Answer
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%
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