CBSE Class 9 Maths Exemplar Chapter 13: Surface Areas and Volumes NCERT Solutions

We are given that the radius of the sphere is \(R = 2r\).

The formula for the volume of a sphere is \(V = \frac{4}{3}\pi R^3\).

Substituting the given radius into the formula:

$$V\; =\; \backslash frac\{4\}\{3\}\backslash pi\; (2r)^3$$

First, calculate \((2r)^3\):

$$(2r)^3\; =\; 2^3\; \backslash times\; r^3\; =\; 8r^3$$

Now, substitute this back into the volume formula:

$$V\; =\; \backslash frac\{4\}\{3\}\backslash pi\; \backslash times\; 8r^3$$

$$V\; =\; \backslash frac\{32\}\{3\}\backslash pi\; r^3$$

Therefore, the correct option is (D).

The formula for the total surface area of a cube with edge length 'a' is \(A = 6a^2\).

We are given that the total surface area is 96 cm².

$$6a^2\; =\; 96$$

To find the edge length, we first solve for \(a^2\):

$$a^2\; =\; \backslash frac\{96\}\{6\}$$

$$a^2\; =\; 16$$

Now, take the square root to find the edge length 'a':

$$a\; =\; \backslash sqrt\{16\}$$

$$a\; =\; 4\; \backslash text\{\; cm\}$$

The formula for the volume of a cube with edge length 'a' is \(V = a^3\).

Substitute the value of 'a' we found:

$$V\; =\; (4\; \backslash text\{\; cm\})^3$$

$$V\; =\; 64\; \backslash text\{\; cm\}^3$$

Hence, the correct option is (C).

First, we calculate the volume of the cone. The formula for the volume of a cone is \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\), where 'r' is the base radius and 'h' is the height.

Given: \(h = 8.4\) cm and \(r = 2.1\) cm.

Volume of the cone:

$$V\_\{\backslash text\{cone\}\}\; =\; \backslash frac\{1\}\{3\}\backslash pi\; (2.1\; \backslash text\{\; cm\})^2\; \backslash times\; 8.4\; \backslash text\{\; cm\}$$

When the cone is melted and recast into a sphere, the volume remains the same. Let the radius of the sphere be \(r_1\). The formula for the volume of a sphere is \(V_{\text{sphere}} = \frac{4}{3}\pi r_1^3\).

According to the problem, the volume of the cone is equal to the volume of the sphere:

$$V\_\{\backslash text\{sphere\}\}\; =\; V\_\{\backslash text\{cone\}\}$$

$$\backslash frac\{4\}\{3\}\backslash pi\; r\_1^3\; =\; \backslash frac\{1\}\{3\}\backslash pi\; (2.1)^2\; \backslash times\; 8.4$$

We can cancel \(\frac{1}{3}\pi\) from both sides:

$$4\; r\_1^3\; =\; (2.1)^2\; \backslash times\; 8.4$$

Now, solve for \(r_1^3\):

$$r\_1^3\; =\; \backslash frac\{(2.1)^2\; \backslash times\; 8.4\}\{4\}$$

Simplify the expression:

$$r\_1^3\; =\; (2.1)^2\; \backslash times\; \backslash frac\{8.4\}\{4\}$$

$$r\_1^3\; =\; (2.1)^2\; \backslash times\; 2.1$$

$$r\_1^3\; =\; (2.1)^3$$

Taking the cube root of both sides to find \(r_1\):

$$r\_1\; =\; 2.1\; \backslash text\{\; cm\}$$

Hence, the correct option is (B).

Let the original radius of the cylinder be 'r' and the original height be 'h'.

The formula for the curved surface area (CSA) of a cylinder is \(CSA = 2\pi rh\).

According to the problem, the radius is doubled, so the new radius \(r' = 2r\).

The height is halved, so the new height \(h' = \frac{h}{2}\).

Now, let's calculate the new curved surface area (CSA') using the new dimensions:

$$CSA\text{'}\; =\; 2\backslash pi\; r\text{'}\; h\text{'}$$

Substitute the new radius and height:

$$CSA\text{'}\; =\; 2\backslash pi\; (2r)\; \backslash left(\backslash frac\{h\}\{2\}\backslash right)$$

Simplify the expression:

$$CSA\text{'}\; =\; 2\backslash pi\; \backslash times\; 2\; \backslash times\; r\; \backslash times\; \backslash frac\{h\}\{2\}$$

$$CSA\text{'}\; =\; 2\backslash pi\; r\; h$$

Comparing the new curved surface area (CSA') with the original curved surface area (CSA), we see that \(CSA' = CSA\).

Therefore, the curved surface area remains the same. The correct option is (C).