If errors of 1% each are made in te base radius and height of a cylinder, then the percentage error in its volume, is

Options

A. `1%`
B. `2%`
C. `3%`
D. none f these

Explanation

To find the percentage error in the volume of a cylinder when there are 1% errors in both the base radius and height, we can follow these steps:

### Step 1: Understand the formula for the volume of a cylinder
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( r \) is the base radius and \( h \) is the height.

### Step 2: Determine the percentage errors in radius and height
We are given that the percentage error in the base radius \( r \) is 1%. This can be expressed mathematically as:
\[ \frac{dr}{r} \times 100 = 1 \implies \frac{dr}{r} = 0.01 \]
Similarly, the percentage error in height \( h \) is also 1%:
\[ \frac{dh}{h} \times 100 = 1 \implies \frac{dh}{h} = 0.01 \]

### Step 3: Differentiate the volume formula
To find the error in volume, we differentiate the volume \( V \) with respect to its variables \( r \) and \( h \):
\[ dV = \pi (2rh \, dr + r^2 \, dh) \]

### Step 4: Express the relative error in volume
The relative error in volume can be expressed as:
\[ \frac{dV}{V} = \frac{dV}{\pi r^2 h} \]
Substituting the expression for \( dV \):
\[ \frac{dV}{V} = \frac{\pi (2rh \, dr + r^2 \, dh)}{\pi r^2 h} = \frac{2rh \, dr + r^2 \, dh}{r^2 h} \]
This simplifies to:
\[ \frac{dV}{V} = \frac{2rh \, dr}{r^2 h} + \frac{r^2 \, dh}{r^2 h} = \frac{2 \, dr}{r} + \frac{dh}{h} \]

### Step 5: Substitute the percentage errors
Now substituting the values of \( \frac{dr}{r} \) and \( \frac{dh}{h} \):
\[ \frac{dV}{V} = 2 \times 0.01 + 0.01 = 0.02 + 0.01 = 0.03 \]

### Step 6: Convert to percentage
To find the percentage error in volume:
\[ \text{Percentage error in volume} = \frac{dV}{V} \times 100 = 0.03 \times 100 = 3\% \]

### Final Answer
The percentage error in the volume of the cylinder is **3%**.

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