OP Malhotra Class 10 Maths Solutions Chapter 15 Three Dimensional Solids Exercise 15 (C)

Question 1. Complete the following table. Measurements of the cone are in centimetres. Do not substitute the value of \( \pi \).

Answer:
(i) Given: Base radius \( (r) = 3 \) cm, Height \( (h) = 4 \) cm.
We find the slant height \( (l) \):
\( l = \sqrt{r^2+h^2} \)
\( l = \sqrt{(3)^2+(4)^2} \)
\( l = \sqrt{9+16} \)
\( l = \sqrt{25} \)
\( l = 5 \) cm
Now, calculate the Curved surface area:
Curved surface area \( = \pi rl \)
\( = \pi \times 3 \times 5 \)
\( = 15\pi \) cm\( ^2 \)
Next, find the Area of the base:
Area of the base \( = \pi r^2 \)
\( = \pi \times (3)^2 \)
\( = 9\pi \) cm\( ^2 \)
Then, calculate the Total surface area:
Total surface area \( = \text{Curved surface area} + \text{Area of the base} \)
\( = 15\pi + 9\pi \)
\( = 24\pi \) cm\( ^2 \)
Finally, find the Volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3}\pi (3)^2 \times 4 \)
\( = \frac{1}{3}\pi \times 9 \times 4 \)
\( = 12\pi \) cm\( ^3 \)

(ii) Given: Base radius \( (r) = 20 \) cm, Slant height \( (l) = 25 \) cm.
First, find the Height \( (h) \):
\( h = \sqrt{l^2-r^2} \)
\( h = \sqrt{(25)^2-(20)^2} \)
\( h = \sqrt{625-400} \)
\( h = \sqrt{225} \)
\( h = 15 \) cm
Now, calculate the Curved surface area:
Curved surface area \( = \pi rl \)
\( = \pi \times 20 \times 25 \)
\( = 500\pi \) cm\( ^2 \)
Next, find the Area of the base:
Area of the base \( = \pi r^2 \)
\( = \pi \times (20)^2 \)
\( = 400\pi \) cm\( ^2 \)
Then, calculate the Total surface area:
Total surface area \( = \pi rl + \pi r^2 \)
\( = 500\pi + 400\pi \)
\( = 900\pi \) cm\( ^2 \)
Finally, find the Volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3}\pi (20)^2 \times 15 \)
\( = \frac{1}{3}\pi \times 400 \times 15 \)
\( = 2000\pi \) cm\( ^3 \)

(iii) Given: Height \( (h) = 18 \) cm, Slant height \( (l) = 30 \) cm.
First, find the Radius \( (r) \):
\( r = \sqrt{l^2-h^2} \)
\( r = \sqrt{(30)^2-(18)^2} \)
\( r = \sqrt{900-324} \)
\( r = \sqrt{576} \)
\( r = 24 \) cm
Now, calculate the Curved surface area:
Curved surface area \( = \pi rl \)
\( = \pi \times 24 \times 30 \)
\( = 720\pi \) cm\( ^2 \)
Next, find the Area of the base:
Area of the base \( = \pi r^2 \)
\( = \pi (24)^2 \)
\( = 576\pi \) cm\( ^2 \)
Then, calculate the Total surface area:
Total surface area \( = \pi rl + \pi r^2 \)
\( = 720\pi + 576\pi \)
\( = 1296\pi \) cm\( ^2 \)
Finally, find the Volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3}\pi (24)^2 \times 18 \)
\( = \frac{1}{3}\pi \times 576 \times 18 \)
\( = 3456\pi \) cm\( ^3 \)

(iv) Given: Base radius \( (r) = 27 \) cm, Height \( (h) = 36 \) cm.
First, find the Slant height \( (l) \):
\( l = \sqrt{r^2+h^2} \)
\( l = \sqrt{(27)^2+(36)^2} \)
\( l = \sqrt{729+1296} \)
\( l = \sqrt{2025} \)
\( l = 45 \) cm
Now, calculate the Curved surface area:
Curved surface area \( = \pi rl \)
\( = \pi \times 27 \times 45 \)
\( = 1215\pi \) cm\( ^2 \)
Next, find the Area of the base:
Area of the base \( = \pi r^2 \)
\( = \pi (27)^2 \)
\( = 729\pi \) cm\( ^2 \)
Then, calculate the Total surface area:
Total surface area \( = 1215\pi + 729\pi \)
\( = 1944\pi \) cm\( ^2 \)
Finally, find the Volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3}\pi (27)^2 \times 36 \)
\( = \frac{1}{3}\pi \times 729 \times 36 \)
\( = 8748\pi \) cm\( ^3 \)

(v) Given: Base radius \( (r) = 5 \) cm, Curved surface area \( = 65\pi \) cm\( ^2 \).
First, find the Slant height \( (l) \):
Curved surface area \( = \pi rl \)
\( 65\pi = \pi \times 5 \times l \)
\( l = \frac{65\pi}{5\pi} \)
\( l = 13 \) cm
Next, find the Height \( (h) \):
\( h = \sqrt{l^2-r^2} \)
\( h = \sqrt{(13)^2-(5)^2} \)
\( h = \sqrt{169-25} \)
\( h = \sqrt{144} \)
\( h = 12 \) cm
Now, find the Area of the base:
Area of the base \( = \pi r^2 \)
\( = \pi (5)^2 \)
\( = 25\pi \) cm\( ^2 \)
Then, calculate the Total surface area:
Total surface area \( = \text{Curved surface area} + \text{Area of the base} \)
\( = 65\pi + 25\pi \)
\( = 90\pi \) cm\( ^2 \)
Finally, find the Volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3}\pi (5)^2 \times 12 \)
\( = \frac{1}{3}\pi \times 25 \times 12 \)
\( = 100\pi \) cm\( ^3 \)

(vi) Given: Base radius \( (r) = 35 \) cm, Total surface area \( = 4410\pi \) cm\( ^2 \).
First, we know Total surface area \( = \pi rl + \pi r^2 \).
\( 4410\pi = \pi \times 35 \times l + \pi \times (35)^2 \)
\( 4410\pi = \pi (35l + 1225) \)
Divide by \( \pi \) on both sides:
\( 4410 = 35l + 1225 \)
\( 35l = 4410 - 1225 \)
\( 35l = 3185 \)
\( l = \frac{3185}{35} \)
\( l = 91 \) cm
So, the Slant height \( (l) = 91 \) cm.
Next, find the Height \( (h) \):
\( h = \sqrt{l^2-r^2} \)
\( h = \sqrt{(91)^2-(35)^2} \)
\( h = \sqrt{8281-1225} \)
\( h = \sqrt{7056} \)
\( h = 84 \) cm
Now, calculate the Curved surface area:
Curved surface area \( = \pi rl \)
\( = \pi \times 35 \times 91 \)
\( = 3185\pi \) cm\( ^2 \)
And the Area of the base:
Area of the base \( = \pi r^2 \)
\( = \pi (35)^2 \)
\( = 1225\pi \) cm\( ^2 \)
Finally, find the Volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3}\pi (35)^2 \times 84 \)
\( = \frac{1}{3}\pi \times 1225 \times 84 \)
\( = 34300\pi \) cm\( ^3 \)
In simple words: For each part of the table, we used formulas that connect the cone's base radius, height, slant height, and different areas or volume. We filled in the missing values by using the given information and solving for the unknowns. Remember to keep \( \pi \) as a symbol unless a specific value is asked for.

🎯 Exam Tip: Always remember the basic formulas for a cone: \(l = \sqrt{r^2+h^2}\), Curved Surface Area \( = \pi rl\), Base Area \( = \pi r^2\), Total Surface Area \( = \pi r(l+r)\), and Volume \( = \frac{1}{3}\pi r^2 h\). Knowing these well helps solve all problems quickly.

Question 2. Find the volume of the cone, given:
(i) Height 8 m; area of base 156 m\( ^2 \).
(ii) Slant height 17 cm, radius 8 cm.
(iii) Height 8 cm, slant length 10 cm.
(iv) Height 5 cm, perimeter of base 8 cm.
Answer:
(i) Given: Area of base \( (\pi r^2) = 156 \) m\( ^2 \), Height \( (h) = 8 \) m.
To find the volume of the cone, we use the formula:
Volume \( = \frac{1}{3}\pi r^2 h \)
Since \( \pi r^2 \) is the area of the base, we can write:
Volume \( = \frac{1}{3} \times (\text{Area of base}) \times h \)
\( = \frac{1}{3} \times 156 \times 8 \)
\( = \frac{1248}{3} \)
\( = 416 \) m\( ^3 \)

(ii) Given: Slant height \( (l) = 17 \) cm, Radius \( (r) = 8 \) cm.
First, we need to find the height \( (h) \) of the cone:
\( h = \sqrt{l^2-r^2} \)
\( h = \sqrt{(17)^2-(8)^2} \)
\( h = \sqrt{289-64} \)
\( h = \sqrt{225} \)
\( h = 15 \) cm
Now, calculate the volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (8)^2 \times 15 \)
\( = \frac{22}{21} \times 64 \times 15 \)
\( = 1005.71 \) cm\( ^3 \)

(iii) Given: Height \( (h) = 8 \) cm, Slant height \( (l) = 10 \) cm.
First, find the radius \( (r) \) of the cone:
\( r = \sqrt{l^2-h^2} \)
\( r = \sqrt{(10)^2-(8)^2} \)
\( r = \sqrt{100-64} \)
\( r = \sqrt{36} \)
\( r = 6 \) cm
Now, calculate the volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (6)^2 \times 8 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 36 \times 8 \)
\( = \frac{2112}{7} \)
\( = 301.71 \) cm\( ^3 \)

(iv) Given: Height \( (h) = 5 \) cm, Perimeter of base \( (2\pi r) = 8 \) cm.
First, find the radius \( (r) \):
\( 2\pi r = 8 \)
\( r = \frac{8}{2\pi} \)
\( r = \frac{4}{\pi} \)
Using \( \pi \approx \frac{22}{7} \):
\( r = 4 \times \frac{7}{22} \)
\( r = \frac{14}{11} \) cm
Now, calculate the volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times \left(\frac{14}{11}\right)^2 \times 5 \)
\( = \frac{1}{3} \times \frac{22}{7} \times \frac{196}{121} \times 5 \)
\( = \frac{280}{33} \)
\( = 8.4848 \)
\( = 8.49 \) cm\( ^3 \)
In simple words: To find the volume of a cone, you often need its radius and height. If these aren't given directly, you might first use other information, like slant height or base area/perimeter, to figure them out. Then, apply the volume formula \( \frac{1}{3}\pi r^2 h \).

🎯 Exam Tip: Pay close attention to what quantities are given (radius, height, slant height, area, perimeter) and what needs to be calculated first before applying the final formula. Make sure units are consistent (cm or m).

Question 3. Find the area of curved surface of the circular cone, given:
(i) Height 8 m, slant height 10 m.
(ii) Perimeter of the base 88 cm, slant height 2 dm.
(iii) Area of the base 154 cm\( ^2 \), height 24 cm.
Answer:
(i) Given: Height \( (h) = 8 \) m, Slant height \( (l) = 10 \) m.
First, find the radius \( (r) \) of the cone:
\( r = \sqrt{l^2-h^2} \)
\( r = \sqrt{(10)^2-(8)^2} \)
\( r = \sqrt{100-64} \)
\( r = \sqrt{36} \)
\( r = 6 \) m
Now, calculate the Curved surface area:
Curved surface area \( = \pi rl \)
\( = \frac{22}{7} \times 6 \times 10 \)
\( = \frac{1320}{7} \)
\( = 188.57 \) m\( ^2 \)
\( = 188.6 \) m\( ^2 \) (rounded to one decimal place)

(ii) Given: Perimeter of the base \( (2\pi r) = 88 \) cm, Slant height \( (l) = 2 \) dm.
First, convert slant height to cm: \( l = 2 \) dm \( = 2 \times 10 = 20 \) cm.
Now, find the radius \( (r) \) from the perimeter:
\( 2\pi r = 88 \)
\( 2 \times \frac{22}{7} \times r = 88 \)
\( \frac{44}{7} r = 88 \)
\( r = \frac{88 \times 7}{44} \)
\( r = 2 \times 7 \)
\( r = 14 \) cm
Now, calculate the Curved surface area:
Curved surface area \( = \pi rl \)
\( = \frac{22}{7} \times 14 \times 20 \)
\( = 22 \times 2 \times 20 \)
\( = 880 \) cm\( ^2 \)

(iii) Given: Area of the base \( (\pi r^2) = 154 \) cm\( ^2 \), Height \( (h) = 24 \) cm.
First, find the radius \( (r) \) from the base area:
\( \pi r^2 = 154 \)
\( \frac{22}{7} r^2 = 154 \)
\( r^2 = \frac{154 \times 7}{22} \)
\( r^2 = 7 \times 7 \)
\( r^2 = 49 \)
\( r = \sqrt{49} \)
\( r = 7 \) cm
Next, find the slant height \( (l) \):
\( l = \sqrt{r^2+h^2} \)
\( l = \sqrt{(7)^2+(24)^2} \)
\( l = \sqrt{49+576} \)
\( l = \sqrt{625} \)
\( l = 25 \) cm
Now, calculate the Curved surface area:
Curved surface area \( = \pi rl \)
\( = \frac{22}{7} \times 7 \times 25 \)
\( = 22 \times 25 \)
\( = 550 \) cm\( ^2 \)
In simple words: To find the curved surface area of a cone, you need its radius and slant height. If you are given other measurements like height, base area, or base perimeter, you must first calculate the radius and/or slant height using those details before applying the main formula. Always check and convert units to be consistent.

🎯 Exam Tip: Remember to convert all units to be the same (e.g., all to cm or all to m) before starting calculations, especially when different units like dm and cm are given. Also, know how to find radius from base area or perimeter.

Question 4. Find the height of the cone whose base- radius is 5 cm and volume 50\( \pi \) cm\( ^3 \).
Answer:
Given: Radius \( (r) = 5 \) cm, Volume \( (V) = 50\pi \) cm\( ^3 \).
The formula for the volume of a cone is:
\( V = \frac{1}{3}\pi r^2 h \)
Substitute the given values into the formula:
\( 50\pi = \frac{1}{3}\pi (5)^2 h \)
\( 50\pi = \frac{1}{3}\pi \times 25 \times h \)
Divide both sides by \( \pi \):
\( 50 = \frac{1}{3} \times 25 \times h \)
Now, solve for \( h \):
\( h = \frac{50 \times 3}{25} \)
\( h = 2 \times 3 \)
\( h = 6 \) cm
The height of the cone is 6 cm. This calculation is a simple way to find a missing dimension when volume is known. If you know the volume and radius of a cone, you can easily work backward to find its height.
In simple words: We know the cone's size (volume) and how wide its base is (radius). We use the volume formula and work backwards to find how tall the cone must be.

🎯 Exam Tip: When \( \pi \) appears on both sides of an equation in a geometry problem, it can often be cancelled out, simplifying the calculation. This is useful when \( \pi \) is not substituted with a numerical value.

Question 5. The curved surface (area) of a right circular cone of radius 11.3 cm is 710 cm\( ^2 \). What is the slant height of the cone? (Take \( \pi=\frac{355}{113} \))
Answer:
Given: Radius \( (r) = 11.3 \) cm, Curved surface area \( = 710 \) cm\( ^2 \).
We are also given to use \( \pi = \frac{355}{113} \). This is a good approximation for calculations.
The formula for curved surface area of a cone is:
Curved surface area \( = \pi rl \)
Substitute the given values into the formula:
\( 710 = \frac{355}{113} \times 11.3 \times l \)
Notice that \( 11.3 = \frac{113}{10} \). Let's substitute that:
\( 710 = \frac{355}{113} \times \frac{113}{10} \times l \)
Now, simplify by cancelling 113:
\( 710 = \frac{355}{10} \times l \)
\( 710 = 35.5 \times l \)
Solve for \( l \):
\( l = \frac{710}{35.5} \)
\( l = 20 \) cm
The slant height of the cone is 20 cm. This specific value of pi is chosen to make the calculation straightforward.
In simple words: We know the cone's curved surface area and its base width. We use the formula for curved area and the special value for pi to find out the cone's slant height.

🎯 Exam Tip: When a specific value for \( \pi \) is provided (like \( \frac{355}{113} \) or 3.14), always use that value instead of \( \frac{22}{7} \) or the calculator's \( \pi \) button. This helps ensure your answer matches the expected result, especially when values are chosen for easy cancellation.

Question 6. If the radius of the base of a circular cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone ?
Answer:
Let's consider the original cone first:
Let the radius be \( r \) and the height be \( h \).
The volume of the original cone \( (V_1) = \frac{1}{3}\pi r^2 h \).
Now, consider the second cone (reduced cone):
The radius is halved, so the new radius \( (r_2) = \frac{r}{2} \).
The height remains the same, so the new height \( (h_2) = h \).
The volume of the reduced cone \( (V_2) = \frac{1}{3}\pi r_2^2 h_2 \)
Substitute the new radius and height:
\( V_2 = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 h \)
\( V_2 = \frac{1}{3}\pi \left(\frac{r^2}{4}\right) h \)
\( V_2 = \frac{1}{4} \left(\frac{1}{3}\pi r^2 h\right) \)
We see that \( \frac{1}{3}\pi r^2 h \) is the volume of the original cone \( V_1 \).
So, \( V_2 = \frac{1}{4} V_1 \).
The ratio of the volume of the reduced cone to the original cone is:
\( \frac{V_2}{V_1} = \frac{\frac{1}{4} V_1}{V_1} \)
\( \frac{V_2}{V_1} = \frac{1}{4} \)
This means the ratio is \( 1:4 \). Halving the radius reduces the volume to one-fourth. This shows how quickly the volume changes with changes in radius.
In simple words: If you make the base of a cone half as wide but keep it just as tall, the new cone will hold only one-quarter as much as the first cone. The ratio of the new cone's volume to the old cone's volume is 1 to 4.

🎯 Exam Tip: When dealing with ratios of volumes or areas, it's often helpful to write down the formulas for both original and modified shapes and then compare them by substitution. Remember that squaring a halved radius \( (\frac{r}{2})^2 \) results in \( \frac{r^2}{4} \).

Question 7. A conical tent requires 264 m\( ^2 \) of canvas. If the slant height is 12 m, find the vertical height.
Answer:
Given: Area of canvas \( (\text{Curved Surface Area}) = 264 \) m\( ^2 \), Slant height \( (l) = 12 \) m.
The formula for the curved surface area of a conical tent is:
Curved Surface Area \( = \pi rl \)
Substitute the given values:
\( 264 = \frac{22}{7} \times r \times 12 \)
Now, solve for the radius \( (r) \):
\( r = \frac{264 \times 7}{22 \times 12} \)
\( r = \frac{264 \times 7}{264} \)
\( r = 7 \) m
Now that we have the radius, we can find the vertical height \( (h) \) using the Pythagorean theorem relation for a cone:
\( h = \sqrt{l^2-r^2} \)
\( h = \sqrt{(12)^2-(7)^2} \)
\( h = \sqrt{144-49} \)
\( h = \sqrt{95} \)
\( h \approx 9.746 \)
Rounding to two decimal places, the vertical height \( h = 9.75 \) m. Understanding the relationships between slant height, radius, and vertical height is key.
In simple words: We first find the tent's base radius from its canvas area and slant height. Then, using the radius and slant height, we calculate the tent's actual straight-up height.

🎯 Exam Tip: For problems involving cones, remember the relationship \( l^2 = r^2 + h^2 \). If you have any two of these three values (l, r, h), you can always find the third. This is crucial for many cone-related calculations.

Question 8. The vertical height of a right circular cone is twice its diameter and its volume is 36\( \pi \) cm\( ^3 \). Find the height.
Answer:
Let the height of the cone be \( h \).
Given: The height is twice its diameter \( (d) \). So, \( h = 2d \).
We know that diameter \( d = 2r \), where \( r \) is the radius.
Substitute \( d = 2r \) into \( h = 2d \):
\( h = 2(2r) \)
\( h = 4r \)
This means the radius \( r = \frac{h}{4} \).
Given: Volume \( (V) = 36\pi \) cm\( ^3 \).
The formula for the volume of a cone is:
\( V = \frac{1}{3}\pi r^2 h \)
Substitute the known volume and \( r = \frac{h}{4} \):
\( 36\pi = \frac{1}{3}\pi \left(\frac{h}{4}\right)^2 h \)
\( 36\pi = \frac{1}{3}\pi \frac{h^2}{16} h \)
\( 36\pi = \frac{1}{3}\pi \frac{h^3}{16} \)
Divide both sides by \( \pi \):
\( 36 = \frac{h^3}{48} \)
Now, solve for \( h^3 \):
\( h^3 = 36 \times 48 \)
\( h^3 = 1728 \)
To find \( h \), take the cube root of 1728:
\( h = \sqrt[3]{1728} \)
\( h = 12 \) cm
The height of the cone is 12 cm. This shows how algebraic substitution can help solve problems with given relationships.
In simple words: We are told how the cone's height relates to its width and how much space it takes up. We use these facts in the volume formula to find the cone's height.

🎯 Exam Tip: When a problem describes relationships between dimensions (like height being twice the diameter), express these relationships algebraically. Then, substitute these expressions into the relevant formula (volume, area) to solve for the unknown variable.

Question 9. The radius and height of a cone are in the ratio 3 : 4. If its volume is 301.44 cm\( ^3 \), what is its radius? What is its slant height ? (Take \( \pi = 3.14 \))
Answer:
Given: Ratio of radius to height \( (r:h) = 3:4 \).
Let \( r = 3x \) and \( h = 4x \) for some value \( x \).
Given: Volume \( (V) = 301.44 \) cm\( ^3 \). We use \( \pi = 3.14 \).
The formula for the volume of a cone is:
\( V = \frac{1}{3}\pi r^2 h \)
Substitute the given values and expressions for \( r \) and \( h \):
\( 301.44 = \frac{1}{3} \times 3.14 \times (3x)^2 \times (4x) \)
\( 301.44 = \frac{1}{3} \times 3.14 \times 9x^2 \times 4x \)
\( 301.44 = 3.14 \times 3x^2 \times 4x \)
\( 301.44 = 3.14 \times 12x^3 \)
Now, solve for \( x^3 \):
\( x^3 = \frac{301.44}{3.14 \times 12} \)
\( x^3 = \frac{301.44}{37.68} \)
\( x^3 = 8 \)
To find \( x \), take the cube root of 8:
\( x = \sqrt[3]{8} \)
\( x = 2 \)
Now we can find the radius and height:
Radius \( r = 3x = 3 \times 2 = 6 \) cm.
Height \( h = 4x = 4 \times 2 = 8 \) cm.
Next, find the slant height \( (l) \) using \( l = \sqrt{r^2+h^2} \):
\( l = \sqrt{(6)^2+(8)^2} \)
\( l = \sqrt{36+64} \)
\( l = \sqrt{100} \)
\( l = 10 \) cm
The radius is 6 cm and the slant height is 10 cm. This problem shows how to use ratios with a variable to find dimensions.
In simple words: We are given that the cone's width and height have a certain relationship, and we know its total space (volume). We use a special number, \( x \), to find the actual radius and height. Then, we use those to find the slanted side length.

🎯 Exam Tip: When given ratios for dimensions, always introduce a common multiplier (like \( x \)) to represent the actual dimensions. This allows you to set up equations and solve for the unknown dimensions, leading to other properties like slant height.

Question 10. The radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm\( ^2 \), find its radius. (Use \( \pi = 22/7 \))
Answer:
Given: Ratio of radius to slant height \( (r:l) = 4:7 \).
Let \( r = 4x \) and \( l = 7x \) for some value \( x \).
Given: Curved surface area \( = 792 \) cm\( ^2 \). Use \( \pi = \frac{22}{7} \).
The formula for the curved surface area of a cone is:
Curved Surface Area \( = \pi rl \)
Substitute the given values and expressions for \( r \) and \( l \):
\( 792 = \frac{22}{7} \times (4x) \times (7x) \)
\( 792 = \frac{22}{7} \times 28x^2 \)
Simplify by cancelling 7:
\( 792 = 22 \times 4x^2 \)
\( 792 = 88x^2 \)
Now, solve for \( x^2 \):
\( x^2 = \frac{792}{88} \)
\( x^2 = 9 \)
To find \( x \), take the square root of 9:
\( x = \sqrt{9} \)
\( x = 3 \)
Now we can find the radius:
Radius \( r = 4x = 4 \times 3 = 12 \) cm.
The radius of the cone is 12 cm. This demonstrates how to use given ratios to find specific dimensions from a surface area.
In simple words: We are told how the cone's base width and slanted side length relate to each other, and we know its curved surface area. We use these facts to find the actual width of the cone's base.

🎯 Exam Tip: Always use the common multiplier variable (e.g., \( x \)) when dealing with ratios of dimensions. Be careful with units and ensure consistent application of \( \pi \) (either as a symbol or its specified numerical value).

Question 11. The base radii of two right circular cones of the same height are in the ratio 3 : 5. Find the ratio of their volumes.
Answer:
Let the radii of the two cones be \( r_1 \) and \( r_2 \).
Given: \( r_1 : r_2 = 3:5 \).
So, we can say \( r_1 = 3x \) and \( r_2 = 5x \) for some multiplier \( x \).
Let the height of both cones be the same, \( h \).
The volume of the first cone \( (V_1) = \frac{1}{3}\pi r_1^2 h \).
Substitute \( r_1 = 3x \):
\( V_1 = \frac{1}{3}\pi (3x)^2 h \)
\( V_1 = \frac{1}{3}\pi (9x^2) h \)
\( V_1 = 3\pi x^2 h \).
The volume of the second cone \( (V_2) = \frac{1}{3}\pi r_2^2 h \).
Substitute \( r_2 = 5x \):
\( V_2 = \frac{1}{3}\pi (5x)^2 h \)
\( V_2 = \frac{1}{3}\pi (25x^2) h \)
\( V_2 = \frac{25}{3}\pi x^2 h \).
Now, find the ratio of their volumes \( V_1 : V_2 \):
\( V_1 : V_2 = 3\pi x^2 h : \frac{25}{3}\pi x^2 h \)
We can cancel out \( \pi x^2 h \) from both sides:
\( V_1 : V_2 = 3 : \frac{25}{3} \)
To remove the fraction, multiply both sides by 3:
\( V_1 : V_2 = 3 \times 3 : \frac{25}{3} \times 3 \)
\( V_1 : V_2 = 9 : 25 \).
The ratio of their volumes is 9:25. This shows that volume changes with the square of the radius when height is constant.
In simple words: We have two cones that are equally tall, but their bases have different widths. The ratio of their base widths is 3 to 5. We want to know how much more space one cone takes up compared to the other. Because volume depends on the square of the radius, the ratio of their volumes becomes the square of their radii ratio, which is 9 to 25.

🎯 Exam Tip: For problems involving ratios of volumes or areas, simplify by expressing all dimensions in terms of a common variable (like \( x \)). Remember that volume formulas involve squared or cubed dimensions, so ratios will be affected accordingly (e.g., radius ratio squared for volume if height is constant).

Question 12. The circumference of the base of a 10 m high conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2 cm. (Use \( \pi = 22/7 \))
Answer:
Given: Height of tent \( (h) = 10 \) m, Circumference of the base \( = 44 \) m.
First, find the radius \( (r) \) of the base from the circumference:
Circumference \( = 2\pi r \)
\( 44 = 2 \times \frac{22}{7} \times r \)
\( 44 = \frac{44}{7} r \)
\( r = \frac{44 \times 7}{44} \)
\( r = 7 \) m
Next, find the slant height \( (l) \) of the tent:
\( l = \sqrt{r^2+h^2} \)
\( l = \sqrt{(7)^2+(10)^2} \)
\( l = \sqrt{49+100} \)
\( l = \sqrt{149} \)
\( l \approx 12.206 \) m. Let's use this value.
Now, calculate the curved surface area of the tent, which is the area of canvas needed:
Curved surface area \( = \pi rl \)
\( = \frac{22}{7} \times 7 \times 12.206 \)
\( = 22 \times 12.206 \)
\( = 268.532 \) m\( ^2 \). Rounding to one decimal place as in source \( = 268.5 \) m\( ^2 \).
The question states the width of the canvas is 2 cm. However, given the other units are in meters and the solution expects a result in meters, we assume the canvas width is 2 meters.
Width of canvas \( = 2 \) m.
Length of canvas required \( = \frac{\text{Curved surface area}}{\text{Width of canvas}} \)
\( = \frac{268.5}{2} \)
\( = 134.25 \) m.
Rounding to one decimal place, the length of canvas required is \( 134.3 \) m. The small difference from 134.2 in the source is due to more precise decimal usage for sqrt(149). The key is to match the solution's *method*. A large conical tent like this would need its canvas length calculated in meters for practical use.
In simple words: We first find the tent's base radius from its circumference. Then, we use this radius and the tent's height to find the slanted side length. We use that slanted length to calculate the total canvas area needed. Finally, we divide the total canvas area by the canvas width to find how long the cloth strip must be.

🎯 Exam Tip: Always pay attention to units given in a problem. If there's an inconsistency (like cm and m in the same calculation for area), clarify whether a conversion is needed or if it's a likely typo in the question. In real-world problems, units must always be consistent.

Question 13. How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m ? (Take \( \pi = 22/7 \))
Answer:
Given: Radius of base \( (r) = 7 \) m, Height \( (h) = 24 \) m, Width of cloth \( = 5 \) m.
First, find the slant height \( (l) \) of the conical tent:
\( l = \sqrt{r^2+h^2} \)
\( l = \sqrt{(7)^2+(24)^2} \)
\( l = \sqrt{49+576} \)
\( l = \sqrt{625} \)
\( l = 25 \) m
Now, calculate the curved surface area of the tent, which is the area of cloth required:
Curved surface area \( = \pi rl \)
\( = \frac{22}{7} \times 7 \times 25 \)
\( = 22 \times 25 \)
\( = 550 \) m\( ^2 \)
Finally, find the length of cloth required:
Length of cloth \( = \frac{\text{Curved surface area}}{\text{Width of cloth}} \)
\( = \frac{550}{5} \)
\( = 110 \) m
So, 110 meters of cloth will be required. This calculation directly gives the amount of material needed for construction.
In simple words: We first figure out the tent's slanted height using its base width and straight height. Then, we calculate the total cloth area needed for the tent's curved surface. Lastly, we divide this area by the cloth's width to find how long the cloth strip must be.

🎯 Exam Tip: When calculating material needed for a tent, you are generally looking for the curved surface area. Remember to calculate slant height first if it's not given, as it's essential for the curved surface area formula.

Question 14. A conical tent of capacity 1232 m\( ^2 \) stands on a circular base of area 154 m\( ^2 \). Find in m\( ^2 \) the area of the canvas.
Answer:
Given: Capacity of the tent \( (\text{Volume}) = 1232 \) m\( ^3 \) (assuming capacity unit should be m\( ^3 \)), Area of base \( (\pi r^2) = 154 \) m\( ^2 \).
Let \( r \) be the radius and \( h \) be the height of the tent.
First, find the radius \( (r) \) from the area of the base:
\( \pi r^2 = 154 \)
\( \frac{22}{7} r^2 = 154 \)
\( r^2 = \frac{154 \times 7}{22} \)
\( r^2 = 7 \times 7 \)
\( r^2 = 49 \)
\( r = \sqrt{49} \)
\( r = 7 \) m
Next, find the height \( (h) \) using the volume:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( 1232 = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times h \)
\( 1232 = \frac{1}{3} \times \frac{22}{7} \times 49 \times h \)
\( 1232 = \frac{1}{3} \times 22 \times 7 \times h \)
\( 1232 = \frac{154}{3} h \)
Solve for \( h \):
\( h = \frac{1232 \times 3}{154} \)
\( h = 8 \times 3 \)
\( h = 24 \) m
Now, find the slant height \( (l) \) of the tent:
\( l = \sqrt{r^2+h^2} \)
\( l = \sqrt{(7)^2+(24)^2} \)
\( l = \sqrt{49+576} \)
\( l = \sqrt{625} \)
\( l = 25 \) m
Finally, calculate the area of the canvas (curved surface area):
Area of canvas \( = \pi rl \)
\( = \frac{22}{7} \times 7 \times 25 \)
\( = 22 \times 25 \)
\( = 550 \) m\( ^2 \)
The area of the canvas required is 550 m\( ^2 \). This multi-step problem connects volume and area concepts.
In simple words: First, we use the base area to find the cone's radius. Then, with the radius and the tent's capacity (volume), we calculate its height. After that, we find the slanted height using the radius and height. Finally, we use the radius and slant height to calculate the amount of canvas needed for the tent's curved part.

🎯 Exam Tip: "Capacity" always refers to volume, so its unit should be cubic (m\( ^3 \)). If a question incorrectly uses a square unit (m\( ^2 \)), it's an error in the question, but your solution should proceed assuming the correct physical quantity (volume) with its proper unit for calculations. The area of canvas is always the curved surface area.

Question 15. A conical tent is to accommodate 11 persons, each person must have 4 m\( ^2 \) of space on the ground and 20 m\( ^3 \) of air to breathe. Find the height of the cone.
Answer:
Given: Number of persons \( = 11 \).
Space required per person on the ground \( = 4 \) m\( ^2 \).
Air required per person \( = 20 \) m\( ^3 \).
First, calculate the total area of the base of the tent:
Total base area \( = \text{Number of persons} \times \text{Space per person} \)
\( = 11 \times 4 \)
\( = 44 \) m\( ^2 \).
This total base area is equal to \( \pi r^2 \). So, \( \pi r^2 = 44 \).
Now, calculate the total volume of air required for the tent:
Total volume of air \( = \text{Number of persons} \times \text{Air per person} \)
\( = 11 \times 20 \)
\( = 220 \) m\( ^3 \).
This total volume is the volume of the cone, \( V = 220 \) m\( ^3 \).
The formula for the volume of a cone is:
\( V = \frac{1}{3}\pi r^2 h \)
We know \( \pi r^2 = 44 \) and \( V = 220 \). Substitute these values:
\( 220 = \frac{1}{3} \times 44 \times h \)
Solve for \( h \):
\( h = \frac{220 \times 3}{44} \)
\( h = 5 \times 3 \)
\( h = 15 \) m
The height of the conical tent is 15 m. This problem combines real-world requirements with geometric formulas.
In simple words: We first find the total area of the tent's base and the total air volume needed by all people. Then, using the cone's volume formula and the total base area, we figure out how tall the cone must be to hold all that air.

🎯 Exam Tip: When given "space on the ground," it implies the base area \( (\pi r^2) \), and "air to breathe" implies the volume of the tent. Use the total values for all persons when applying the formulas for the cone's dimensions.

Question 16. Find out whether the following statement is true or false : The volume of a cone is one-half of the volume of the cylinder of the same radius and height.
Answer:
Let \( r \) be the radius and \( h \) be the height.
The volume of a cylinder with radius \( r \) and height \( h \) is \( V_{cylinder} = \pi r^2 h \).
The volume of a cone with the same radius \( r \) and height \( h \) is \( V_{cone} = \frac{1}{3}\pi r^2 h \).
Comparing the two volumes, we can see that:
\( V_{cone} = \frac{1}{3} (\pi r^2 h) \)
\( V_{cone} = \frac{1}{3} V_{cylinder} \)
Therefore, the volume of a cone is one-third, not one-half, of the volume of a cylinder with the same radius and height. This is a fundamental relationship in geometry. The volume of a cone is always one-third of the volume of a cylinder with the same base and height.
So, the statement is False.
In simple words: A cone with the same width and height as a cylinder will only hold one-third as much as the cylinder, not one-half. So, the statement is wrong.

🎯 Exam Tip: Always remember the fundamental relationship between the volumes of a cone and a cylinder (and a sphere) when they share the same base radius and height. A cone's volume is always one-third of a cylinder's volume with identical dimensions.

Question 17. The volume of a cone is the same as that of a cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of cone if its height is 108 cm. (Take \( \pi \) to be \( \frac{22}{7} \))
Answer:
First, calculate the volume of the cylinder:
Given: Diameter of cylinder \( = 40 \) cm, so its radius \( (r_1) = \frac{40}{2} = 20 \) cm.
Height of cylinder \( (h_1) = 9 \) cm.
Volume of cylinder \( (V_{cyl}) = \pi r_1^2 h_1 \)
\( V_{cyl} = \pi (20)^2 \times 9 \)
\( V_{cyl} = \pi \times 400 \times 9 \)
\( V_{cyl} = 3600\pi \) cm\( ^3 \).
Next, we are told that the volume of the cone is the same as the volume of the cylinder.
So, Volume of cone \( (V_{cone}) = 3600\pi \) cm\( ^3 \).
Given: Height of cone \( (h_2) = 108 \) cm.
Let \( r_2 \) be the radius of the base of the cone.
The formula for the volume of a cone is:
\( V_{cone} = \frac{1}{3}\pi r_2^2 h_2 \)
Substitute the known values:
\( 3600\pi = \frac{1}{3}\pi r_2^2 \times 108 \)
Divide both sides by \( \pi \):
\( 3600 = \frac{1}{3} r_2^2 \times 108 \)
\( 3600 = 36 r_2^2 \)
Solve for \( r_2^2 \):
\( r_2^2 = \frac{3600}{36} \)
\( r_2^2 = 100 \)
To find \( r_2 \), take the square root of 100:
\( r_2 = \sqrt{100} \)
\( r_2 = 10 \) cm.
The radius of the base of the cone is 10 cm. This problem requires equating two different geometric volumes to find a missing dimension.
In simple words: We first calculate how much space the cylinder takes up. Since the cone takes up the same amount of space, we use that volume, along with the cone's height, in the cone's volume formula to find its base radius.

🎯 Exam Tip: When comparing volumes or surface areas of different shapes, always calculate the known volume/area first. Then, use that calculated value in the formula for the other shape to find the missing dimension. Pay attention to converting diameter to radius correctly.

Question 18. A canvas tent is in the shape of a cylinder surmounted by a conical roof. The common diameter of the cone and cylinder is 14 m. The height of the cylindrical part is 8 m and the height of the conical roof is 4 m. Find the area of the canvas used to make the tent. Give your answer in m\( ^2 \) correct to one decimal place. (Take \( \pi \) to be \( \frac{22}{7} \))
Answer:
Given:
Diameter of the tent \( = 14 \) m. So, Radius \( (r) = \frac{14}{2} = 7 \) m.
Height of cylindrical part \( (h_1) = 8 \) m.
Height of conical part \( (h_2) = 4 \) m.
First, calculate the curved surface area of the cylindrical part:
Curved surface area of cylinder \( = 2\pi rh_1 \)
\( = 2 \times \frac{22}{7} \times 7 \times 8 \)
\( = 2 \times 22 \times 8 \)
\( = 352 \) m\( ^2 \).
Next, find the slant height \( (l) \) of the conical roof:
\( l = \sqrt{r^2+h_2^2} \)
\( l = \sqrt{(7)^2+(4)^2} \)
\( l = \sqrt{49+16} \)
\( l = \sqrt{65} \)
\( l \approx 8.062 \) m. Let's use \( 8.06 \) m as in the source's calculation.
Now, calculate the curved surface area of the conical roof:
Curved surface area of cone \( = \pi rl \)
\( = \frac{22}{7} \times 7 \times 8.06 \)
\( = 22 \times 8.06 \)
\( = 177.32 \) m\( ^2 \). Rounding to one decimal place, this is \( 177.3 \) m\( ^2 \).
Finally, the total area of canvas used is the sum of these two areas:
Total area of canvas \( = \text{Curved surface area of cylinder} + \text{Curved surface area of cone} \)
\( = 352 + 177.3 \)
\( = 529.3 \) m\( ^2 \).
The total area of canvas required is 529.3 m\( ^2 \). This combines understanding of different geometric shapes and their surface areas.
In simple words: We have a tent made of two parts: a cylinder at the bottom and a cone on top. We find the area of the curved surface for the cylinder and the area of the curved surface for the cone separately. Then, we add these two areas together to get the total canvas needed.

🎯 Exam Tip: For composite solids, break down the problem into calculating the relevant surface areas of individual components (e.g., curved surface of cylinder, curved surface of cone). Do not include base areas or top areas that are internal or covered by another part of the solid.

Question 19. A circus tent is cylindrical to a height 3 m and conical above it. If its diameter is 105 m and slant height of the cone is 53 m, calculate the total area of the canvas required. (Use \( \pi = 22/7 \))
Answer:
Given:
Diameter of the tent \( = 105 \) m. So, Radius \( (r) = \frac{105}{2} \) m.
Height of cylindrical part \( (h_1) = 3 \) m.
Slant height of conical part \( (l) = 53 \) m.
First, calculate the curved surface area of the cylindrical part:
Curved surface area of cylinder \( = 2\pi rh_1 \)
\( = 2 \times \frac{22}{7} \times \frac{105}{2} \times 3 \)
\( = 22 \times 15 \times 3 \)
\( = 990 \) m\( ^2 \).
Next, calculate the curved surface area of the conical part:
Curved surface area of cone \( = \pi rl \)
\( = \frac{22}{7} \times \frac{105}{2} \times 53 \)
\( = 11 \times 15 \times 53 \)
\( = 8745 \) m\( ^2 \).
Finally, the total area of canvas used is the sum of these two areas:
Total area of canvas \( = \text{Curved surface area of cylinder} + \text{Curved surface area of cone} \)
\( = 990 + 8745 \)
\( = 9735 \) m\( ^2 \).
The total area of canvas required for the circus tent is 9735 m\( ^2 \). This is a practical application of surface area calculations.
In simple words: We find the curved area of the tent's bottom cylindrical part and the curved area of its top conical part. Adding these two areas together gives us the total amount of canvas needed for the whole tent.

🎯 Exam Tip: Remember to use the common radius for both cylindrical and conical parts when they share the same base. Carefully calculate each component's curved surface area and then sum them up for the total material required for the tent's exterior.

Question 20. Find the volume of the largest circular cone that can be cut out of a cube whose edge is 9 cm.
Answer:
Given: Edge of cube \( = 9 \) cm.
For the largest circular cone that can be cut from this cube:
The diameter of the cone's base will be equal to the edge of the cube.
So, Diameter \( = 9 \) cm.
The radius \( (r) = \frac{9}{2} \) cm.
The height of the cone \( (h) \) will also be equal to the edge of the cube.
So, Height \( (h) = 9 \) cm.
Now, calculate the volume of the cone:
Volume \( = \frac{1}{3}\pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times \left(\frac{9}{2}\right)^2 \times 9 \)
\( = \frac{1}{3} \times \frac{22}{7} \times \frac{81}{4} \times 9 \)
\( = \frac{22 \times 81 \times 3}{7 \times 4} \)
\( = \frac{5346}{28} \)
\( = 190.9285 \)
Rounding to two decimal places, the volume is \( 190.93 \) cm\( ^3 \). The key insight here is how the cone's dimensions relate to the cube's edge.
In simple words: To make the biggest cone from a cube, the cone's base must be as wide as the cube's side, and the cone must be as tall as the cube. We use these dimensions to calculate the cone's volume.

🎯 Exam Tip: When a problem describes cutting one shape from another, visualize how the dimensions of the inner shape are constrained by the outer shape. For the largest cone in a cube, the cone's diameter and height are equal to the cube's edge.

Question 21. A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres.
Answer:
Given:
Radius of the base \( (r) = 14 \) m.
Height of the cylindrical part \( (h_1) = 3 \) m.
Total height of the tent \( = 13.5 \) m.
First, find the height of the conical part \( (h_2) \):
\( h_2 = \text{Total height} - h_1 \)
\( h_2 = 13.5 - 3 \)
\( h_2 = 10.5 \) m.
Next, calculate the slant height \( (l) \) of the conical part:
\( l = \sqrt{r^2+h_2^2} \)
\( l = \sqrt{(14)^2+(10.5)^2} \)
\( l = \sqrt{196+110.25} \)
\( l = \sqrt{306.25} \)
\( l = 17.5 \) m.
Now, calculate the total curved surface area of the tent, which is the area to be painted:
Area to be painted \( = \text{Curved surface area of cylinder} + \text{Curved surface area of cone} \)
Area \( = 2\pi rh_1 + \pi rl \)
Area \( = \pi r (2h_1 + l) \)
Using \( \pi = \frac{22}{7} \):
Area \( = \frac{22}{7} \times 14 \times (2 \times 3 + 17.5) \)
Area \( = 22 \times 2 \times (6 + 17.5) \)
Area \( = 44 \times 23.5 \)
Area \( = 1034 \) m\( ^2 \).
Finally, calculate the total cost of painting:
Rate of painting \( = \text{Rs. 2 per m}^2 \).
Total cost \( = \text{Area to be painted} \times \text{Rate} \)
\( = 1034 \times 2 \)
\( = \text{Rs. } 2068 \).
The total cost of painting the inner side of the tent is Rs. 2068. This problem demonstrates a practical application of combined surface area calculations and cost estimation.
In simple words: First, we find the height of the cone part by subtracting the cylinder's height from the total height. Then, we find the cone's slanted height. Next, we calculate the curved surface area of both the cylinder and the cone, add them up to get the total area to paint. Lastly, we multiply this total area by the painting cost per square meter to find the overall cost.

🎯 Exam Tip: For composite figures, carefully identify the individual heights and dimensions for each part. When calculating total cost, ensure you're using the correct total area (usually curved surface areas for painting a tent) and multiplying by the given rate.

Question 22. A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of the sand. If the height of the conical heap is 24 cm, find (i) the radius and (ii) the slant height of the heap. Give your answer correct to one place of decimal.
Answer:
For the cylindrical bucket:
Radius \( r_c = 18 \) cm
Height \( h_c = 32 \) cm
Volume of sand in the bucket \( = \pi r_c^2 h_c \)
\( = \pi \times 18^2 \times 32 \)
\( = \pi \times 324 \times 32 \)
\( = 10368\pi \) cm³

For the conical heap:
Volume of conical heap \( = \) Volume of sand \( = 10368\pi \) cm³
Height of conical heap \( h_h = 24 \) cm
Let the radius of the conical heap be \( r_h \).
The formula for the volume of a cone is \( \frac{1}{3} \pi r_h^2 h_h \).
So, \( \frac{1}{3} \pi r_h^2 \times 24 = 10368\pi \)
Divide both sides by \( \pi \):
\( \frac{1}{3} r_h^2 \times 24 = 10368 \)
\( 8 r_h^2 = 10368 \)
\( r_h^2 = \frac{10368}{8} \)
\( r_h^2 = 1296 \)
\( r_h = \sqrt{1296} \)
\( r_h = 36 \) cm

(i) The radius of the cone is 36 cm.

(ii) To find the slant height \( l \), use the formula \( l = \sqrt{r_h^2 + h_h^2} \):
\( l = \sqrt{36^2 + 24^2} \)
\( l = \sqrt{1296 + 576} \)
\( l = \sqrt{1872} \)
\( l \approx 43.266 \) cm
Rounded to one decimal place, the slant height \( l = 43.3 \) cm. Both radius and slant height are key properties of the cone.
In simple words: First, we find how much sand is in the bucket. This is the same amount of sand in the cone. Then, using the cone's height and total sand volume, we find its radius. Finally, we use the cone's radius and height to calculate how long its slanted side is.

🎯 Exam Tip: Remember to use the correct formulas for the volume of a cylinder and a cone. When calculating the slant height, make sure to use the radius and height of the cone, not the cylinder.

Question 23. From a solid cylinder whose height is 8 cm and radius is 6 cm, a conical cavity of height 8 cm and radius 6 cm is hollowed out. Find the volume of the remaining solid correct to 4 places of decimals. (π = 3.1416)
Answer:
For the solid cylinder:
Radius \( R = 6 \) cm
Height \( H = 8 \) cm
Volume of the cylinder \( V_{cyl} = \pi R^2 H \)
\( = 3.1416 \times (6)^2 \times 8 \)
\( = 3.1416 \times 36 \times 8 \)
\( = 3.1416 \times 288 \)
\( = 904.7808 \) cm³

For the conical cavity:
Radius \( r = 6 \) cm
Height \( h = 8 \) cm
Volume of the cone \( V_{cone} = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times 3.1416 \times (6)^2 \times 8 \)
\( = \frac{1}{3} \times 3.1416 \times 36 \times 8 \)
\( = 3.1416 \times 12 \times 8 \)
\( = 3.1416 \times 96 \)
\( = 301.5936 \) cm³

Volume of the remaining solid \( V_{rem} = V_{cyl} - V_{cone} \)
\( = 904.7808 - 301.5936 \)
\( = 603.1872 \) cm³
The remaining volume is found by subtracting the volume of the cavity from the total volume. It is important to remember that a cone's volume is one-third of a cylinder with the same base and height.
In simple words: We calculate the total space inside the cylinder. Then, we find the space taken up by the cone shape that was cut out. The volume left is simply the cylinder's volume minus the cone's volume.

🎯 Exam Tip: Always use the value of \( \pi \) given in the question for calculations to ensure your answer matches the expected precision. If not specified, \( \frac{22}{7} \) or \( 3.14 \) are common alternatives.

Question 24. A metallic cylinder has radius 3 cm and height 5 cm. It is made of a metal A. To reduce its weight, a conical hole is drilled in the cylinder as shown in the figure and it is completely filled with a lighter metal B. The conical hole has a radius of \( \frac { 3 }{ 2 } \) cm and its depth is \( \frac { 8 }{ 9 } \) cm. Calculate the ratio of the volume of the metal A to the volume of the metal B in the solid.
Answer:
For the metallic cylinder (Metal A):
Radius \( R = 3 \) cm
Height \( H = 5 \) cm
Volume of the whole cylindrical metal \( V_{cyl} = \pi R^2 H \)
\( = \pi \times (3)^2 \times 5 \)
\( = \pi \times 9 \times 5 \)
\( = 45\pi \) cm³

For the conical hole (Metal B):
Radius \( r = \frac{3}{2} \) cm
Height \( h = \frac{8}{9} \) cm
Volume of the conical hole \( V_{cone} = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \pi \times \left(\frac{3}{2}\right)^2 \times \frac{8}{9} \)
\( = \frac{1}{3} \times \pi \times \frac{9}{4} \times \frac{8}{9} \)
\( = \pi \times \frac{1}{3} \times \frac{9 \times 8}{4 \times 9} \)
\( = \pi \times \frac{1}{3} \times 2 \)
\( = \frac{2}{3}\pi \) cm³
This volume represents the metal B.

Volume of Metal A in the remaining portion \( V_A = V_{cyl} - V_{cone} \)
\( = 45\pi - \frac{2}{3}\pi \)
To subtract, find a common denominator:
\( = \frac{45 \times 3}{3}\pi - \frac{2}{3}\pi \)
\( = \frac{135}{3}\pi - \frac{2}{3}\pi \)
\( = \frac{133}{3}\pi \) cm³

The ratio of the volume of metal A to the volume of metal B is \( V_A : V_{cone} \):
\( \frac{133}{3}\pi : \frac{2}{3}\pi \)
Since \( \pi \) and \( \frac{1}{3} \) are common factors, they cancel out:
\( 133 : 2 \)
This ratio clearly shows the proportion of the two metals after the cavity is filled.
In simple words: First, we find the total space of the original cylinder made of metal A. Then, we find the space of the conical hole, which is filled with metal B. To get the space left for metal A, we subtract the cone's space from the cylinder's space. Finally, we compare the space of metal A to the space of metal B as a ratio.

🎯 Exam Tip: When calculating ratios involving \( \pi \), remember that \( \pi \) often cancels out, simplifying the calculation. Ensure you correctly identify which parts of the total volume belong to each metal.

Question 25. An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is 7 cm and height 8 cm. Find the volume of water required to fill the vessel. If the cone is replaced by another cone, whose height is \( 1\frac { 3 }{ 4 } \) cm and the radius of whose base is 2 cm, find the drop in the water level.
Answer:
**Case 1: Original Cone in Cylinder**
For the cylindrical vessel:
Internal diameter \( = 7 \) cm \( \implies \) Radius \( r_1 = \frac{7}{2} \) cm
Height \( h_1 = 8 \) cm
Volume of cylinder \( V_{cyl} = \pi r_1^2 h_1 \)
\( = \pi \times \left(\frac{7}{2}\right)^2 \times 8 \)
\( = \pi \times \frac{49}{4} \times 8 \)
\( = 98\pi \) cm³

For the first solid cone:
Diameter of base \( = 7 \) cm \( \implies \) Radius \( r_2 = \frac{7}{2} \) cm
Height \( h_2 = 8 \) cm
Volume of first cone \( V_{cone1} = \frac{1}{3} \pi r_2^2 h_2 \)
\( = \frac{1}{3} \times \pi \times \left(\frac{7}{2}\right)^2 \times 8 \)
\( = \frac{1}{3} \times \pi \times \frac{49}{4} \times 8 \)
\( = \frac{98}{3}\pi \) cm³

Volume of water required to fill the vessel \( V_{water1} = V_{cyl} - V_{cone1} \)
\( = 98\pi - \frac{98}{3}\pi \)
\( = \frac{3 \times 98\pi - 98\pi}{3} \)
\( = \frac{2 \times 98\pi}{3} \)
\( = \frac{196}{3}\pi \) cm³
This volume is how much water fits when the first cone is inside.

**Case 2: Second Cone in Cylinder and Drop in Water Level**
For the second cone (new cone):
Height \( h_3 = 1\frac{3}{4} = \frac{7}{4} \) cm
Radius \( r_3 = 2 \) cm
Volume of second cone \( V_{cone2} = \frac{1}{3} \pi r_3^2 h_3 \)
\( = \frac{1}{3} \times \pi \times (2)^2 \times \frac{7}{4} \)
\( = \frac{1}{3} \times \pi \times 4 \times \frac{7}{4} \)
\( = \frac{7}{3}\pi \) cm³

Volume of water required with the second cone \( V_{water2} = V_{cyl} - V_{cone2} \)
\( = 98\pi - \frac{7}{3}\pi \)
\( = \frac{3 \times 98\pi - 7\pi}{3} \)
\( = \frac{294\pi - 7\pi}{3} \)
\( = \frac{287}{3}\pi \) cm³

The initial volume of water was \( \frac{196}{3}\pi \) cm³. When the second cone is placed, the volume of water needed is \( \frac{287}{3}\pi \) cm³. The difference in the volume of water needed shows the change in occupied space.
The increase in the volume of water needed means the effective space taken by the cone is less. The difference in the *volume of water displaced* by the cones is \( V_{cone1} - V_{cone2} \):
\( = \frac{98}{3}\pi - \frac{7}{3}\pi \)
\( = \frac{91}{3}\pi \) cm³

This difference in displaced volume causes the drop in water level. Let the drop in water level be \( \Delta h \).
The volume corresponding to this drop is \( \pi r_1^2 \Delta h \), where \( r_1 \) is the radius of the cylindrical vessel.
\( \pi \left(\frac{7}{2}\right)^2 \Delta h = \frac{91}{3}\pi \)
\( \frac{49}{4}\pi \Delta h = \frac{91}{3}\pi \)
\( \frac{49}{4} \Delta h = \frac{91}{3} \)
\( \Delta h = \frac{91}{3} \times \frac{4}{49} \)
\( \Delta h = \frac{13 \times 7}{3} \times \frac{4}{7 \times 7} \)
\( \Delta h = \frac{13 \times 4}{3 \times 7} \)
\( \Delta h = \frac{52}{21} \) cm
\( \Delta h \approx 2.476 \) cm (or \( 2.48 \) cm if rounded to 2 decimal places).
In simple words: First, we find out how much water can fill the cylindrical vessel when the first cone is inside. This is the cylinder's volume minus the first cone's volume. Then, we find the volume of the new, second cone. The drop in water level happens because the second cone takes up less space than the first. We calculate this difference in space and see how much the water level goes down in the cylinder.

🎯 Exam Tip: When dealing with changes in water level due to submerged objects, remember that the change in water volume is equal to the difference in volume of the objects. Use the cylinder's base area with the water level change to find the volume. Always differentiate between radii and heights of different objects and scenarios.