Exercise : Surface Area & Volume - General Questions
✔ Surface Area & Volume -
General Questions
11.
If the radius of a cylinder is increased by 10% and the height is decreased by 10%, find the percentage change in volume.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: Let the original radius be \(r\) and height be \(h\). Original volume $$V_1 = \pi r^2 h$$.
Step 2: The new radius is \(1.1r\) and the new height is \(0.9h\).
Step 3: Calculate the new volume: $$V_2 = \pi (1.1r)^2 (0.9h) = \pi (1.21r^2) (0.9h)$$.
Step 4: $$V_2 = 1.089 \pi r^2 h = 1.089 V_1$$.
Step 5: The increase is \(1.089 - 1 = 0.089\), which is 8.9%.
Step 2: The new radius is \(1.1r\) and the new height is \(0.9h\).
Step 3: Calculate the new volume: $$V_2 = \pi (1.1r)^2 (0.9h) = \pi (1.21r^2) (0.9h)$$.
Step 4: $$V_2 = 1.089 \pi r^2 h = 1.089 V_1$$.
Step 5: The increase is \(1.089 - 1 = 0.089\), which is 8.9%.
12.
A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: Equal bases implies they have the same radius \(r\).
Step 2: Volume of cone = $$\frac{1}{3}\pi r^2 h$$. Volume of hemisphere = $$\frac{2}{3}\pi r^3$$.
Step 3: Set them equal: $$\frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^3$$.
Step 4: Cancel common terms ($$\frac{1}{3}, \pi, r^2$$): \(h = 2r\).
Step 5: The ratio of the height of the cone to the radius (height of the hemisphere) is 2 : 1.
Step 2: Volume of cone = $$\frac{1}{3}\pi r^2 h$$. Volume of hemisphere = $$\frac{2}{3}\pi r^3$$.
Step 3: Set them equal: $$\frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^3$$.
Step 4: Cancel common terms ($$\frac{1}{3}, \pi, r^2$$): \(h = 2r\).
Step 5: The ratio of the height of the cone to the radius (height of the hemisphere) is 2 : 1.
13.
A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. Find the radius of the sphere.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: Convert dimensions to cm: length \(h = 3600\) cm, diameter = 0.2 cm, so radius \(r = 0.1\) cm.
Step 2: Volume of wire (cylinder) = $$\pi r^2 h = \pi (0.1)^2 \times 3600 = \pi \times 0.01 \times 3600 = 36\pi$$ cm³.
Step 3: Set the volume of the sphere equal to the volume of the wire: $$\frac{4}{3}\pi R^3 = 36\pi$$.
Step 4: Solve for \(R^3\): $$R^3 = \frac{36 \times 3}{4} = 9 \times 3 = 27$$.
Step 5: Take the cube root: \(R = 3\) cm.
Step 2: Volume of wire (cylinder) = $$\pi r^2 h = \pi (0.1)^2 \times 3600 = \pi \times 0.01 \times 3600 = 36\pi$$ cm³.
Step 3: Set the volume of the sphere equal to the volume of the wire: $$\frac{4}{3}\pi R^3 = 36\pi$$.
Step 4: Solve for \(R^3\): $$R^3 = \frac{36 \times 3}{4} = 9 \times 3 = 27$$.
Step 5: Take the cube root: \(R = 3\) cm.
14.
Find the area of a canvas required to make a conical tent of height 24 m and base radius 7 m.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: The area of canvas required is the Curved Surface Area (CSA) of the cone.
Step 2: Calculate the slant height \(l\) using the Pythagorean theorem: $$l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2}$$.
Step 3: $$l = \sqrt{576 + 49} = \sqrt{625} = 25$$ m.
Step 4: Use the formula $$CSA = \pi rl = \left(\frac{22}{7}\right) \times 7 \times 25$$.
Step 5: Calculate the final area: $$22 \times 25 = 550$$ m².
Step 2: Calculate the slant height \(l\) using the Pythagorean theorem: $$l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2}$$.
Step 3: $$l = \sqrt{576 + 49} = \sqrt{625} = 25$$ m.
Step 4: Use the formula $$CSA = \pi rl = \left(\frac{22}{7}\right) \times 7 \times 25$$.
Step 5: Calculate the final area: $$22 \times 25 = 550$$ m².
15.
A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 12 cm. If the sphere is submerged, find the rise in water level.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: The volume of water displaced is equal to the volume of the submerged sphere.
Step 2: Volume of sphere = $$\frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 6^3 = \frac{4}{3}\pi \times 216 = 288\pi$$ cm³.
Step 3: The volume of displaced water in the cylinder is $$\pi R^2 h_{rise}$$, where \(R = 12\) cm.
Step 4: Set the volumes equal: $$\pi \times 12^2 \times h_{rise} = 288\pi$$.
Step 5: Solve for \(h_{rise}\): $$144 \times h_{rise} = 288 \implies h_{rise} = 2$$ cm.
Step 2: Volume of sphere = $$\frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 6^3 = \frac{4}{3}\pi \times 216 = 288\pi$$ cm³.
Step 3: The volume of displaced water in the cylinder is $$\pi R^2 h_{rise}$$, where \(R = 12\) cm.
Step 4: Set the volumes equal: $$\pi \times 12^2 \times h_{rise} = 288\pi$$.
Step 5: Solve for \(h_{rise}\): $$144 \times h_{rise} = 288 \implies h_{rise} = 2$$ cm.
16.
A cone of height 20 cm is cut by a plane parallel to the base at 10 cm from the base. Find the ratio of the volume of the smaller cone to the frustum.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: The height of the large cone is \(H = 20\) cm. The cut is at 10 cm from the base, so the height of the small cone is \(h = 20 - 10 = 10\) cm.
Step 2: The ratio of heights is \(h:H = 10:20 = 1:2\).
Step 3: The ratio of volumes is the cube of the ratio of heights: $$V_{small} : V_{large} = 1^3 : 2^3 = 1 : 8$$.
Step 4: The volume of the frustum is the difference: $$V_{frustum} = V_{large} - V_{small} = 8 - 1 = 7$$ parts.
Step 5: The ratio of the smaller cone to the frustum is 1 : 7.
Step 2: The ratio of heights is \(h:H = 10:20 = 1:2\).
Step 3: The ratio of volumes is the cube of the ratio of heights: $$V_{small} : V_{large} = 1^3 : 2^3 = 1 : 8$$.
Step 4: The volume of the frustum is the difference: $$V_{frustum} = V_{large} - V_{small} = 8 - 1 = 7$$ parts.
Step 5: The ratio of the smaller cone to the frustum is 1 : 7.
17.
What is the maximum volume of a cone that can be carved out of a solid hemisphere of radius 'r'?
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: To maximize the cone's volume, its base radius must be the radius of the hemisphere (\(r\)) and its height must be the radius of the hemisphere (\(r\)).
Step 2: Use the cone volume formula: $$V = \frac{1}{3}\pi r^2 h$$.
Step 3: Substitute \(h = r\): $$V = \frac{1}{3}\pi r^2 (r) = \frac{1}{3}\pi r^3$$.
Step 4: This is the maximum possible volume.
Step 2: Use the cone volume formula: $$V = \frac{1}{3}\pi r^2 h$$.
Step 3: Substitute \(h = r\): $$V = \frac{1}{3}\pi r^2 (r) = \frac{1}{3}\pi r^3$$.
Step 4: This is the maximum possible volume.
18.
A cylindrical pipe has an inner diameter of 7 cm and water flows through it at 192.5 liters/minute. Find the speed of flow in km/h.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: Convert flow rate to cm³/min: 192.5 liters = 192,500 cm³.
Step 2: Area of pipe cross-section = $$\pi r^2 = \frac{22}{7} \times (3.5)^2 = 38.5$$ cm².
Step 3: Find speed in cm/min: $$Speed = \text{Volume per minute} / \text{Area} = 192,500 / 38.5 = 5,000$$ cm/min.
Step 4: Convert to m/min: $$5,000 / 100 = 50$$ m/min.
Step 5: Convert to km/h: $$50 \times 60 / 1,000 = 3$$ km/h.
Step 2: Area of pipe cross-section = $$\pi r^2 = \frac{22}{7} \times (3.5)^2 = 38.5$$ cm².
Step 3: Find speed in cm/min: $$Speed = \text{Volume per minute} / \text{Area} = 192,500 / 38.5 = 5,000$$ cm/min.
Step 4: Convert to m/min: $$5,000 / 100 = 50$$ m/min.
Step 5: Convert to km/h: $$50 \times 60 / 1,000 = 3$$ km/h.
19.
The total surface area of a solid hemisphere is 1848 cm². Find its volume.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: The formula for TSA of a solid hemisphere is $$3\pi r^2$$.
Step 2: Set $$3 \times \left(\frac{22}{7}\right) \times r^2 = 1848$$.
Step 3: Solve for \(r^2\): $$r^2 = \frac{1848 \times 7}{66} = 196 \implies r = 14$$ cm.
Step 4: Use the volume formula for a hemisphere: $$V = \frac{2}{3}\pi r^3$$.
Step 5: Calculate: $$V = \frac{2}{3} \times \frac{22}{7} \times 14^3 \approx 5,749.33$$ cm³.
Step 2: Set $$3 \times \left(\frac{22}{7}\right) \times r^2 = 1848$$.
Step 3: Solve for \(r^2\): $$r^2 = \frac{1848 \times 7}{66} = 196 \implies r = 14$$ cm.
Step 4: Use the volume formula for a hemisphere: $$V = \frac{2}{3}\pi r^3$$.
Step 5: Calculate: $$V = \frac{2}{3} \times \frac{22}{7} \times 14^3 \approx 5,749.33$$ cm³.
20.
A hollow sphere of internal and external diameters 4 cm and 8 cm is melted into a cone of base diameter 8 cm. Find the height of the cone.
View Answer
Answer: Option A
Explanation:
Explanation:
Step 1: Radii of hollow sphere are \(r = 2\) cm and \(R = 4\) cm. Volume = $$\frac{4}{3}\pi (R^3 - r^3) = \frac{4}{3}\pi (64 - 8) = \frac{4}{3}\pi (56) = \frac{224\pi}{3}$$.
Step 2: Radius of the cone is \(R_{cone} = 4\) cm. Volume = $$\frac{1}{3}\pi R_{cone}^2 h = \frac{1}{3}\pi (16) h = \frac{16\pi h}{3}$$.
Step 3: Equate volumes: $$\frac{224\pi}{3} = \frac{16\pi h}{3}$$.
Step 4: Solve for \(h\): \(16h = 224 \implies h = 14\) cm.
Step 2: Radius of the cone is \(R_{cone} = 4\) cm. Volume = $$\frac{1}{3}\pi R_{cone}^2 h = \frac{1}{3}\pi (16) h = \frac{16\pi h}{3}$$.
Step 3: Equate volumes: $$\frac{224\pi}{3} = \frac{16\pi h}{3}$$.
Step 4: Solve for \(h\): \(16h = 224 \implies h = 14\) cm.