The curved surface area of a cylinder of radius 7cm and height 10cm is:

$$\text{CSA} = 2\pi rh$$

where $r$ is the radius and $h$ is the height.

Given $r = 7$ cm and $h = 10$ cm.

$$\text{CSA} = 2 \times \pi \times 7 \text{ cm} \times 10 \text{ cm}$$

$$\text{CSA} = 140\pi \text{ cm}^2$$

The correct option is A.
The curved surface area is $\boxed*{140\pi \text{ cm}^2*}$.

10. The total surface area of a closed cylinder of radius 5cm and height 5cm is:

$$\text{TSA} = 2\pi r(r+h)$$

where $r$ is the radius and $h$ is the height.

Given $r = 5$ cm and $h = 5$ cm.

$$\text{TSA} = 2 \times \pi \times 5 \text{ cm} \times (5 \text{ cm} + 5 \text{ cm})$$ $$\text{TSA} = 10\pi \text{ cm} \times (10 \text{ cm})$$

$$\text{TSA} = 100\pi \text{ cm}^2$$

The correct option is B.
The total surface area is $\boxed*{100\pi \text{ cm}^2*}$.

11. The curved surface area of a right circular cone of radius 6cm and slant height 9cm is:

$$\text{CSA} = \pi rl$$

where $r$ is the radius and $l$ is the slant height.

Given $r = 6$ cm and $l = 9$ cm.

$$\text{CSA} = \pi \times 6 \text{ cm} \times 9 \text{ cm}$$

$$\text{CSA} = 54\pi \text{ cm}^2$$

The correct option is C.
The curved surface area is $\boxed*{54\pi \text{ cm}^2*}$.

12. The Volume of a cone is 154cm$^3$. If its height is 6cm and $\pi = \frac{22}{7}$, find its radius:

$$V = \frac{1}{3}\pi r^2 h$$

where $V$ is the volume, $r$ is the radius, and $h$ is the height.

Given $V = 154$ cm$^3$, $h = 6$ cm, and $\pi = \frac{22}{7}$.

$$154 \text{ cm}^3 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 6 \text{ cm}$$

$$154 = \frac{22 \times 6}{3 \times 7} r^2$$ $$154 = \frac{132}{21} r^2$$ $$154 = \frac{44}{7} r^2$$

$$r^2 = 154 \times \frac{7}{44}$$ $$r^2 = \frac{1078}{44}$$ $$r^2 = 24.5$$

Wait, let me recheck the calculation.

$$r^2 = 154 \times \frac{7}{44} = (11 \times 14) \times \frac{7}{(11 \times 4)} = \frac{14 \times 7}{4} = \frac{98}{4} = 24.5$$

This value does not lead to a simple integer or half-integer radius from the options. Let me re-evaluate the simplification.

$$154 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 6$$ $$154 = \frac{22 \times 6}{21} r^2$$ $$154 = \frac{132}{21} r^2$$ $$154 = \frac{44}{7} r^2$$ $$r^2 = 154 \times \frac{7}{44}$$

Since $154 = 44 \times 3.5$,

$$r^2 = (44 \times 3.5) \times \frac{7}{44}$$ $$r^2 = 3.5 \times 7$$ $$r^2 = 24.5$$

This is still 24.5. Let me check the options again.
A. 3.5cm B. 5cm C. 9cm D. 14cm
If $r = 3.5$, then $r^2 = (3.5)^2 = 12.25$.
If $r = 5$, then $r^2 = 25$.
It seems there might be a slight discrepancy in the problem or options, as 24.5 is not a perfect square that matches the options. However, 24.5 is very close to 25. Let's assume the closest option is intended.
If $r^2 = 25$, then $r = 5$ cm.
Let's check if $r=5$ cm gives $V=154$ cm$^3$.
$V = \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 6 = \frac{1}{3} \times \frac{22}{7} \times 25 \times 6 = \frac{22 \times 25 \times 2}{7} = \frac{1100}{7} \approx 157.14$ cm$^3$.
This is not exactly 154 cm$^3$.

Let's re-examine the calculation $154 \times \frac{7}{44}$.
$154 / 44 = 3.5$.
$3.5 \times 7 = 24.5$.
So $r^2 = 24.5$.
If the volume was $154 \times \frac{25}{24.5} \approx 157.14$, then $r$ would be 5.
If the volume was 154, then $r = \sqrt{24.5} \approx 4.95$.
The closest option to 4.95 is 5 cm. It's possible the problem intends for $r=5$ and the volume is an approximation, or there's a typo in the volume or height. Given the multiple-choice format, we should select the closest reasonable answer.

Let's assume the question intended for $r=3.5$ cm.
If $r = 3.5$ cm, then $r^2 = (3.5)^2 = 12.25$.
$V = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 6 = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 6$
$V = \frac{22 \times 12.25 \times 2}{7} = \frac{22 \times 24.5}{7} = 22 \times 3.5 = 77$ cm$^3$.
This is not 154 cm$^3$.

Let's assume the question intended for $r=7$ cm.
If $r = 7$ cm, then $r^2 = 49$.
$V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 6 = \frac{1}{3} \times \frac{22}{7} \times 49 \times 6 = 22 \times 7 \times 2 = 308$ cm$^3$.
This is not 154 cm$^3$.

Let's re-check the calculation for $r^2 = 24.5$.
If $r^2 = 24.5$, then $r = \sqrt{24.5} \approx 4.9497$.
The closest option is 5 cm.
Let's consider if the height was different.
If $r=5$ and $V=154$, then $154 = \frac{1}{3} \times \frac{22}{7} \times 5^2 \times h$.
$154 = \frac{22 \times 25}{21} h$.
$h = 154 \times \frac{21}{550} = \frac{3234}{550} \approx 5.88$ cm. This is close to 6 cm.

Given the options, and the common practice in such problems, it's highly probable that $r=5$ cm is the intended answer, with a slight rounding in the volume or height.
Let's proceed with $r=5$ cm as the most plausible answer among the choices.
The value of $r$ is approximately 4.95 cm, which is closest to 5 cm.
The correct option is B.
The radius is $\boxed*{5 \text{ cm}*}$.

13. A closed cylindrical tank has radius 7cm and height 12cm. Calculate:

i) Its total surface area

$$\text{TSA} = 2\pi r(r+h)$$

where $r$ is the radius and $h$ is the height.

Given $r = 7$ cm, $h = 12$ cm, and $\pi = \frac{22}{7}$.

$$\text{TSA} = 2 \times \frac{22}{7} \times 7 \text{ cm} \times (7 \text{ cm} + 12 \text{ cm})$$ $$\text{TSA} = 2 \times 22 \times (19) \text{ cm}^2$$

$$\text{TSA} = 44 \times 19 \text{ cm}^2$$ $$\text{TSA} = 836 \text{ cm}^2$$

The total surface area is $\boxed*{836 \text{ cm}^2*}$.

ii) Its volume (Take $\pi = \frac{22}{7}$)

$$V = \pi r^2 h$$

where $r$ is the radius and $h$ is the height.

Given $r = 7$ cm, $h = 12$ cm, and $\pi = \frac{22}{7}$.

$$V = \frac{22}{7} \times (7 \text{ cm})^2 \times 12 \text{ cm}$$ $$V = \frac{22}{7} \times 49 \text{ cm}^2 \times 12 \text{ cm}$$

$$V = 22 \times 7 \times 12 \text{ cm}^3$$ $$V = 154 \times 12 \text{ cm}^3$$ $$V = 1848 \text{ cm}^3$$

The volume is $\boxed*{1848 \text{ cm}^3*}$.

14. A Cone has radius 5cm and slant height 13cm.

i) Calculate its Curved surface area.

$$\text{CSA} = \pi rl$$

where $r$ is the radius and $l$ is the slant height.

Given $r = 5$ cm and $l = 13$ cm.

$$\text{CSA} = \pi \times 5 \text{ cm} \times 13 \text{ cm}$$

$$\text{CSA} = 65\pi \text{ cm}^2$$

The curved surface area is $\boxed*{65\pi \text{ cm}^2*}$.

ii) If the vertical height of the Cone is 12cm, Calculate its volume (Take $\pi = \frac{22}{7}$)

$$V = \frac{1}{3}\pi r^2 h$$

where $r$ is the radius and $h$ is the vertical height.

Given $r = 5$ cm, $h = 12$ cm, and $\pi = \frac{22}{7}$.

$$V = \frac{1}{3} \times \frac{22}{7} \times (5 \text{ cm})^2 \times 12 \text{ cm}$$ $$V = \frac{1}{3} \times \frac{22}{7} \times 25 \text{ cm}^2 \times 12 \text{ cm}$$

$$V = \frac{22 \times 25 \times 12}{3 \times 7} \text{ cm}^3$$ $$V = \frac{22 \times 25 \times 4}{7} \text{ cm}^3$$ $$V = \frac{2200}{7} \text{ cm}^3$$ $$V \approx 314.29 \text{ cm}^3$$

The volume is $\boxed*{\frac{2200}{7} \text{ cm}^3*}$.