Let's break down the problem: **Question:** A cubical box of length 3 cm is solidified in 5 min in a sand casting. If the box is melted and casted into a spherical object, then calculate the time required to solidify the spherical object in the same atmospheric condition. ### Step-by-Step Solution #### 1. **Volume Conservation** The volume of the cube and the sphere will be the same. - Volume of cube, \( V_{cube} = a^3 \) - \( a = 3 \) cm - \( V_{cube} = 3^3 = 27 \) cm³ - Volume of sphere, \( V_{sphere} = \frac{4}{3} \pi r^3 \) Equate the volumes: \[ 27 = \frac{4}{3} \pi r^3 \] \[ r^3 = \frac{27 \times 3}{4\pi} = \frac{81}{4\pi} \] \[ r = \left(\frac{81}{4\pi}\right)^{1/3} \] #### 2. **Chvorinov's Rule** Solidification time (\( t \)) is proportional to the square of the volume-to-surface-area ratio: \[ t \propto \left(\frac{V}{A}\right)^2 \] Let: - \( t_1 = 5 \) min (for cube) - \( t_2 = ? \) (for sphere) \[ \frac{t_2}{t_1} = \left(\frac{(V/A)_{sphere}}{(V/A)_{cube}}\right)^2 \] #### 3. **Surface Areas** - Cube: \( A_{cube} = 6a^2 = 6 \times 9 = 54 \) cm² - Sphere: \( A_{sphere} = 4\pi r^2 \) #### 4. **V/A Ratios** - Cube: \( (V/A)_{cube} = \frac{27}{54} = 0.5 \) - Sphere: \( (V/A)_{sphere} = \frac{\frac{4}{3}\pi r^3}{4\pi r^2} = \frac{r}{3} \) Using \( r = \left(\frac{81}{4\pi}\right)^{1/3} \): Calculate \( r \): \[ r = \left(\frac{81}{4\pi}\right)^{1/3} \approx \left(\frac{81}{12.566}\right)^{1/3} \approx (6.45)^{1/3} \approx 1.87 \text{ cm} \] So, \[ (V/A)_{sphere} = \frac{1.87}{3} \approx 0.623 \] #### 5. **Ratio and Time Calculation** \[ \frac{t_2}{5} = \left(\frac{0.623}{0.5}\right)^2 = (1.246)^2 \approx 1.55 \] \[ t_2 = 5 \times 1.55 = 7.75 \text{ min} \approx 7 \text{ min } 45 \text{ sec} \] #### **Final Answer** The closest answer choice is **7 min 42 s**. **So, the answer is:** > **7 min 42 s** (Option 2)