The radius of metal sphere at room temperature T is R and the coefficient of linear expansion of the metal is `alpha`. The sphere is heated a little by a temperature T, so that new temperature is `T+DeltaT`. The increase in volume of sphere is approximately

Options

A. `2piRalphaDeltaT`
B. `piR^(2)alphaDeltaT`
C. `4piR^(3) alphaDeltaT//3`
D. `4piR^(3) alpha DeltaT`

Explanation

To find the increase in volume of a metal sphere when it is heated, we can follow these steps:

### Step 1: Understand the relationship between volume expansion and temperature change
The volume expansion of a material is related to its coefficient of volume expansion (γ), which is approximately three times the coefficient of linear expansion (α) for isotropic materials. Thus, we can express this relationship as:
\[ \gamma \approx 3\alpha \]

### Step 2: Write the formula for volume expansion
The change in volume (ΔV) can be expressed in terms of the original volume (V) and the change in temperature (ΔT):
\[ \gamma = \frac{\Delta V}{V \Delta T} \]
From this, we can rearrange to find ΔV:
\[ \Delta V = \gamma V \Delta T \]

### Step 3: Calculate the original volume of the sphere
The volume (V) of a sphere with radius R is given by:
\[ V = \frac{4}{3} \pi R^3 \]

### Step 4: Substitute the volume into the volume expansion formula
Now, substituting the expression for V into the equation for ΔV, we have:
\[ \Delta V = \gamma \left(\frac{4}{3} \pi R^3\right) \Delta T \]

### Step 5: Substitute the value of γ
Substituting γ with 3α, we get:
\[ \Delta V = 3\alpha \left(\frac{4}{3} \pi R^3\right) \Delta T \]

### Step 6: Simplify the expression
The 3s will cancel out:
\[ \Delta V = 4\pi R^3 \alpha \Delta T \]

### Final Answer
Thus, the increase in volume of the sphere when heated is:
\[ \Delta V \approx 4\pi R^3 \alpha \Delta T \]