CBSE Class 10 Maths Chapter 13: Surface Areas and Volumes NCERT Solutions

Given the volume of each cube is 64 cm³.

The formula for the volume of a cube is $V\; =\; (\backslash text\{Edge\})^3$.

So, $(\backslash text\{Edge\})^3\; =\; 64\; \backslash text\{\; cm\}^3$.

Taking the cube root of both sides, we find the edge length:

$\backslash text\{Edge\}\; =\; \backslash sqrt[3]\{64\}\; =\; 4\; \backslash text\{\; cm\}$.

When two such cubes are joined end to end, they form a cuboid. The dimensions of this cuboid will be:

• Length (l) = 4 cm + 4 cm = 8 cm
• Breadth (b) = 4 cm
• Height (h) = 4 cm

The formula for the surface area of a cuboid is $A\; =\; 2(lb\; +\; bh\; +\; lh)$.

Substituting the dimensions:

$$A\; =\; 2((8\; \backslash times\; 4)\; +\; (4\; \backslash times\; 4)\; +\; (8\; \backslash times\; 4))$$

$A\; =\; 2(32\; +\; 16\; +\; 32)$

$A\; =\; 2(80)$

$A\; =\; 160\; \backslash text\{\; cm\}^2$.

Therefore, the surface area of the resulting cuboid is 160 cm².

The vessel consists of a hollow cylinder mounted on a hollow hemisphere.

Given the diameter of the hemisphere is 14 cm. Therefore, the radius (r) of the hemisphere is half of the diameter:

$r\; =\; \backslash frac\{14\}\{2\}\; =\; 7\; \backslash text\{\; cm\}$.

Since the cylinder is mounted on the hemisphere, the radius of the cylinder is also the same as the radius of the hemisphere, so $r\; =\; 7\; \backslash text\{\; cm\}$.

The total height of the vessel is given as 13 cm.

The height of the hemispherical part is equal to its radius, which is 7 cm.

The height of the cylindrical part (h) is the total height minus the height of the hemisphere:

$h\; =\; \backslash text\{Total\; height\}\; -\; \backslash text\{Radius\; of\; hemisphere\}\; =\; 13\; \backslash text\{\; cm\}\; -\; 7\; \backslash text\{\; cm\}\; =\; 6\; \backslash text\{\; cm\}$.

The inner surface area of the vessel is the sum of the curved surface area (CSA) of the cylindrical part and the CSA of the hemispherical part.

CSA of cylinder = $2\backslash pi\; rh$

CSA of hemisphere = $2\backslash pi\; r^2$

Inner surface area = $2\backslash pi\; rh\; +\; 2\backslash pi\; r^2$

Substitute the values $\backslash pi\; =\; \backslash frac\{22\}\{7\}$, $r\; =\; 7\; \backslash text\{\; cm\}$, and $h\; =\; 6\; \backslash text\{\; cm\}$:

Inner surface area = $\backslash left(2\; \backslash times\; \backslash frac\{22\}\{7\}\; \backslash times\; 7\; \backslash times\; 6\backslash right)\; +\; \backslash left(2\; \backslash times\; \backslash frac\{22\}\{7\}\; \backslash times\; 7\; \backslash times\; 7\backslash right)$

Inner surface area = $(2\; \backslash times\; 22\; \backslash times\; 6)\; +\; (2\; \backslash times\; 22\; \backslash times\; 7)$

Inner surface area = $264\; +\; 308$

Inner surface area = $572\; \backslash text\{\; cm\}^2$.

Thus, the inner surface area of the vessel is 572 cm².