Sunrays are allowed to fall on a lens of diameter 20 cm. They are then brought to focus on a calorimeter containing 20 g of ice. If the absorption by the lens is neglible, the time required to melt all the ice is
(solar constant =`1.9 cal min^(-1) cm^(-2) and L = 80 calg^(-1)`)

Options

A. 6.4 min
B. 3.2 min
C. 7.2 min
D. 2.7 min

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Calculate the heat required to melt the ice

The heat required (Q) to melt the ice can be calculated using the formula:
\[ Q = m \times L \]
where:
- \( m \) = mass of ice = 20 g
- \( L \) = latent heat of fusion of ice = 80 cal/g

Substituting the values:
\[ Q = 20 \, \text{g} \times 80 \, \text{cal/g} = 1600 \, \text{cal} \]

### Step 2: Calculate the area of the lens

The area (A) of the lens can be calculated using the formula for the area of a circle:
\[ A = \pi \left(\frac{D}{2}\right)^2 \]
where:
- \( D \) = diameter of the lens = 20 cm

Substituting the value:
\[ A = \pi \left(\frac{20}{2}\right)^2 = \pi \times 10^2 = 100\pi \, \text{cm}^2 \]

### Step 3: Calculate the power incident on the lens

The power (P) incident on the lens can be calculated using the solar constant and the area:
\[ P = \text{solar constant} \times A \]
Given the solar constant = 1.9 cal/min/cm², we have:
\[ P = 1.9 \, \text{cal/min/cm}^2 \times 100\pi \, \text{cm}^2 \]
Using \( \pi \approx 3.14 \):
\[ P \approx 1.9 \times 100 \times 3.14 \approx 597 \, \text{cal/min} \]

### Step 4: Calculate the time required to melt the ice

The time (T) required to melt the ice can be calculated using the formula:
\[ T = \frac{Q}{P} \]
Substituting the values:
\[ T = \frac{1600 \, \text{cal}}{597 \, \text{cal/min}} \approx 2.68 \, \text{min} \]

### Step 5: Final result

Rounding off the result, the time required to melt all the ice is approximately:
\[ T \approx 2.7 \, \text{min} \]

### Summary

The time required to melt all the ice is approximately **2.7 minutes**.

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