Read the statements carefully and select the correct option .
Statement - I : If a hemisphere of lead of radius 7 cm is melted and recast into a right circular cone of height 49 cm, then the radius of the base is 7 cm.
Statement - II : Lead spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged . If the diameter of the beaker is 18 cm and water rises by 40 cm, then the number of lead spheres dropped in the water is 40.

Options

A. Both Statement - I and Statement - II are false.
B. Both Statement-I and Statement-II are true.
C. Statement-I is true but Statement-II is false.
D. Statement-I is false but Statement-II is true.

Answer

To solve the given problem, we need to analyze both statements and verify their correctness step by step.

### Statement I:
**If a hemisphere of lead of radius 7 cm is melted and recast into a right circular cone of height 49 cm, then the radius of the base is 7 cm.**

1. **Calculate the volume of the hemisphere:**
The formula for the volume \( V \) of a hemisphere is given by:
\[ V = \frac{2}{3} \pi r^3 \]
Substituting \( r = 7 \) cm:
\[ V = \frac{2}{3} \pi (7)^3 = \frac{2}{3} \pi (343) = \frac{686}{3} \pi \text{ cm}^3 \]

2. **Calculate the volume of the cone:**
The formula for the volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi R^2 h \]
where \( R \) is the radius of the base and \( h \) is the height. Here, \( h = 49 \) cm.
\[ V = \frac{1}{3} \pi R^2 (49) \]

3. **Set the volumes equal:**
Since the volume of the melted hemisphere equals the volume of the cone:
\[ \frac{686}{3} \pi = \frac{1}{3} \pi R^2 (49) \]
Canceling \( \pi \) and \( \frac{1}{3} \):
\[ 686 = R^2 (49) \]

4. **Solve for \( R^2 \):**
\[ R^2 = \frac{686}{49} = 14 \]
Therefore, \( R = \sqrt{14} \approx 3.74 \) cm.

5. **Conclusion for Statement I:**
The radius of the base of the cone is approximately 3.74 cm, not 7 cm. Thus, Statement I is **false**.

### Statement II:
**Lead spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18 cm and water rises by 40 cm, then the number of lead spheres dropped in the water is 40.**

1. **Calculate the radius of the lead sphere:**
The diameter of the sphere is 6 cm, so the radius \( r \) is:
\[ r = \frac{6}{2} = 3 \text{ cm} \]

2. **Calculate the volume of one lead sphere:**
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting \( r = 3 \) cm:
\[ V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36 \pi \text{ cm}^3 \]

3. **Calculate the volume of water displaced in the beaker:**
The radius of the beaker is:
\[ R = \frac{18}{2} = 9 \text{ cm} \]
The volume of water displaced (which equals the volume of the submerged spheres) can be calculated using the formula for the volume of a cylinder:
\[ V = \pi R^2 h \]
Substituting \( R = 9 \) cm and \( h = 40 \) cm:
\[ V = \pi (9)^2 (40) = \pi (81)(40) = 3240 \pi \text{ cm}^3 \]

4. **Set the volumes equal to find the number of spheres:**
Let \( n \) be the number of spheres:
\[ n \cdot 36 \pi = 3240 \pi \]
Canceling \( \pi \):
\[ n \cdot 36 = 3240 \]
Solving for \( n \):
\[ n = \frac{3240}{36} = 90 \]

5. **Conclusion for Statement II:**
The number of lead spheres is 90, not 40. Thus, Statement II is also **false**.

### Final Conclusion:
Both Statement I and Statement II are false.