GSEB Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.6

Question 1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³ = 1L)
Answer: Let the base radius of the cylindrical container be \(r\) cm. So, the base circumference of the cylindrical vessel equals \(2\pi r\) cm. As per the problem,
\(2\pi r = 132\)
\(2 \times \frac{22}{7} \times r = 132\)
\(r = \frac{132 \times 7}{2 \times 22}\)
\( \implies r = 21\) cm
The height \(h = 25\) cm.
Therefore, the capacity of the cylindrical vessel is \( \pi r^2 h \).
\( = \frac{22}{7} (21)^2 (25) \) cm³
\( = 34650 \) cm³
To convert to litres, divide by 1000 (since \(1000\) cm³ \( = 1\) L).
\( = \frac{34650}{1000} \) L
\( = 34.65 \) L
Hence, the cylindrical container can hold \(34.65\) L of water.
In simple words: First, find the radius using the circumference. Then, use the radius and height to calculate the volume in cubic centimeters. Finally, convert this volume to litres.

Exam Tip: Remember to convert the volume from cubic centimeters to litres by dividing by 1000 when required in the question.

Question 2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Answer: The inner diameter is \(24\) cm.
So, the inner radius \((r) = \frac{24}{2}\) cm \( = 12\) cm.
The outer diameter is \(28\) cm.
So, the outer radius \((R) = \frac{28}{2}\) cm \( = 14\) cm.
The length of the pipe \((h) = 35\) cm.
The inner volume \( = \pi r^2 h \).
\( = \frac{22}{7} \times (12)^2 \times 35 \)
\( = 15840 \) cm³
The outer volume \( = \pi R^2 h \).
\( = \frac{22}{7} \times (14)^2 \times 35 \)
\( = 21560 \) cm³
The volume of wood used \( = \) Outer volume \( - \) Inner volume.
\( = 21560 \) cm³ \( - 15840 \) cm³ \( = 5720 \) cm³
The mass of the pipe \( = 5720 \times 0.6 \) g
\( = 3432 \) g
\( = 3.432 \) kg
In simple words: First, calculate the inner and outer volumes of the pipe. Subtract the inner volume from the outer volume to find the actual volume of wood. Finally, multiply this wood volume by its density to get the mass.

Exam Tip: When dealing with hollow objects, always calculate the volume of the material used by finding the difference between the outer and inner volumes. Remember to convert units carefully if needed.

Question 3. A soft drink is available in two packs (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with a circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Answer: (i) For the tin can:
Length \(l = 5\) cm
Width \(b = 4\) cm
Height \(h = 15\) cm
Capacity \( = l \times b \times h \)
\( = 5 \times 4 \times 15 \) cm³
\( = 300 \) cm³
(ii) For the plastic cylinder:
Diameter \( = 7\) cm
Radius \((r) = \frac{7}{2}\) cm
Height \((h) = 10\) cm
Capacity \( = \pi r^2 h \)
\( = \frac{22}{7} \times (\frac{7}{2})^2 \times 10 \)
\( = \frac{22}{7} \times \frac{49}{4} \times 10 \)
\( = 11 \times 7 \times 5 \)
\( = 385 \) cm³
It is clear that the second container, which is a plastic cylinder, offers more capacity than the first container, the tin can, by \(385 - 300 = 85\) cm³.
In simple words: Calculate the volume for the rectangular tin can by multiplying its length, width, and height. Then, calculate the volume for the cylindrical plastic container using its radius and height. Compare the two volumes to find which one holds more and by how much.

Exam Tip: Always state the formula used for calculating capacity for each shape clearly. Pay attention to units and ensure all calculations are accurate, especially when comparing values.

Question 4. If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find (i) radius of its base (ii) its volume. (Use \(\pi = 3.14\))
Answer: (i) Let the radius of the base of the cylinder be \(r\) cm.
Height \(h = 5\) cm
Lateral surface area \( = 94.2\) cm²
The formula for lateral surface area is \(2\pi rh\).
\( \implies 2\pi rh = 94.2 \)
\( 2 \times 3.14 \times r \times 5 = 94.2 \)
\( r = \frac{94.2}{2 \times 3.14 \times 5} \)
\( r = \frac{94.2}{31.4} \)
\( \implies r = 3\) cm
Hence, the radius of the base is \(3\) cm.
(ii) Now, we find the volume of the cylinder using the radius \(r = 3\) cm and height \(h = 5\) cm.
Volume of the cylinder \( = \pi r^2 h \)
\( = 3.14 \times (3)^2 \times 5 \)
\( = 3.14 \times 9 \times 5 \)
\( = 141.3 \) cm³
In simple words: First, use the given lateral surface area and height to calculate the radius of the cylinder. Then, use this radius along with the height to determine the cylinder's volume.

Exam Tip: Be careful to use the correct formula for lateral surface area (curved surface area) and volume. Ensure you use the given value of \(\pi\) (like 3.14) if specified in the problem.

Question 5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per m², find
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.
Answer: Height (depth) of the cylindrical vessel \(h = 10\) m.
Total cost of painting \( = \) Rs \(2200\).
Rate of painting \( = \) Rs \(20\) per m².
(i) Inner curved surface area of the vessel \( = \frac{\text{Total cost}}{\text{Rate per m}^2} \)
\( = \frac{2200}{20} \)
\( = 110 \) m²
(ii) Let the radius of the base be \(r\) m.
Inner curved surface area \( = 2\pi rh \)
\( \implies 2\pi rh = 110 \)
\( 2 \times \frac{22}{7} \times r \times 10 = 110 \)
\( r = \frac{110 \times 7}{2 \times 22 \times 10} \)
\( r = \frac{770}{440} \)
\( r = \frac{7}{4} \)
\( r = 1.75 \) m
Hence, the radius of the base is \(1.75\) m.
(iii) Now, we find the capacity of the vessel.
Radius \(r = 1.75\) m
Height \(h = 10\) m
Capacity of the vessel \( = \pi r^2 h \)
\( = \frac{22}{7} \times (1.75)^2 \times 10 \)
\( = \frac{22}{7} \times 3.0625 \times 10 \)
\( = 22 \times 0.4375 \times 10 \)
\( = 96.25 \) m³
Hence, the capacity of the vessel is \(96.25\) m³ (or \(96.25\) kL).
In simple words: First, divide the total painting cost by the cost per square meter to get the curved surface area. Next, use this area and the given depth to calculate the base radius. Finally, with the radius and depth, find the vessel's total capacity.

Exam Tip: Break down complex problems into smaller parts. Ensure correct formulas are applied for surface area and volume, and remember that \(1\) m³ is equivalent to \(1\) kL.

Question 6. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of the metal sheet would be needed to make it?
Answer: Height of the cylindrical vessel \(h = 1\) m.
Capacity \( = 15.4 \) litres.
Convert capacity to m³: \(15.4\) litres \( = \frac{15.4}{1000} \) m³ \( = 0.0154 \) m³.
Let the radius of the base be \(r\) m.
Capacity of a cylinder \( = \pi r^2 h \).
\( \implies \pi r^2 h = 0.0154 \)
\( \frac{22}{7} \times r^2 \times 1 = 0.0154 \)
\( \implies r^2 = \frac{0.0154 \times 7}{22} \)
\( \implies r^2 = 0.0049 \)
\( \implies r = \sqrt{0.0049} \)
\( \implies r = 0.07 \) m
Now, we need to find the total surface area of the closed cylindrical vessel to determine the amount of metal sheet required.
Total surface area \( = 2\pi rh + 2\pi r^2 \)
\( = 2 \times \frac{22}{7} \times 0.07 \times 1 + 2 \times \frac{22}{7} \times (0.07)^2 \)
\( = 2 \times 22 \times 0.01 \times 1 + 2 \times \frac{22}{7} \times 0.0049 \)
\( = 0.44 + 2 \times 22 \times 0.0007 \)
\( = 0.44 + 0.0308 \)
\( = 0.4708 \) m²
Hence, \(0.4708\) m² of the metal sheet should be needed.
In simple words: Convert the given capacity from litres to cubic meters. Use this capacity along with the height to find the cylinder's radius. Finally, calculate the total surface area of the closed cylinder using the radius and height to determine the metal sheet required.

Exam Tip: Ensure consistent units throughout the calculation; convert litres to m³ before proceeding. Remember that the total surface area of a closed cylinder includes both curved surface area and the areas of the two circular bases.

Question 7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Answer: Given length of the pencil \(h = 14\) cm. Convert to mm for consistency with diameters: \(14\) cm \( = 140\) mm.
For the solid cylinder of graphite:
Diameter \( = 1\) mm
Radius \((r) = \frac{1}{2}\) mm \( = 0.5\) mm
Volume of the graphite \( = \pi r^2 h \)
\( = \frac{22}{7} \times (\frac{1}{2})^2 \times 140 \)
\( = \frac{22}{7} \times \frac{1}{4} \times 140 \)
\( = 22 \times 5 \times \frac{1}{2} \)
\( = 11 \times 5 = 55 \) mm³
To convert to cm³: \(55 \) mm³ \( = \frac{55}{10 \times 10 \times 10} \) cm³ \( = \frac{55}{1000} \) cm³ \( = 0.055 \) cm³.
For the cylinder of wood (outer pencil cylinder):
Diameter \( = 7\) mm
Radius \((R) = \frac{7}{2}\) mm \( = 3.5\) mm
Length of the pencil \(h = 140\) mm
Volume of the outer pencil cylinder \( = \pi R^2 h \)
\( = \frac{22}{7} \times (\frac{7}{2})^2 \times 140 \)
\( = \frac{22}{7} \times \frac{49}{4} \times 140 \)
\( = 22 \times \frac{7}{4} \times 140 \)
\( = 11 \times 7 \times 70 \)
\( = 5390 \) mm³
To convert to cm³: \(5390 \) mm³ \( = \frac{5390}{1000} \) cm³ \( = 5.39 \) cm³.
Volume of the wood \( = \) Volume of outer pencil cylinder \( - \) Volume of graphite
\( = 5390 \) mm³ \( - 55 \) mm³ \( = 5335 \) mm³
To convert to cm³: \(5335 \) mm³ \( = \frac{5335}{1000} \) cm³ \( = 5.335 \) cm³.
Therefore, the volume of the wood is \(5.335\) cm³ and the volume of the graphite is \(0.055\) cm³.
In simple words: First, make sure all measurements are in the same unit. Calculate the volume of the graphite cylinder. Then, calculate the total volume of the pencil as if it were solid wood. Subtract the graphite volume from the total pencil volume to find the volume of the wood itself.

Exam Tip: Pay close attention to unit conversions (mm to cm or vice-versa) to avoid errors. Remember that the volume of the wood is the difference between the outer cylinder's volume and the inner graphite's volume.

Question 8. A patient in a hospital is given soup daily in a cylindrical bowl of a diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Answer: Diameter of the cylindrical bowl \( = 7\) cm.
Radius \((r) = \frac{7}{2}\) cm \( = 3.5\) cm.
Height of soup in the bowl \(h = 4\) cm.
Volume of soup in one cylindrical bowl \( = \pi r^2 h \).
\( = \frac{22}{7} \times (\frac{7}{2})^2 \times 4 \) cm³
\( = \frac{22}{7} \times \frac{49}{4} \times 4 \) cm³
\( = 22 \times 7 \) cm³
\( = 154 \) cm³
The hospital needs to serve \(250\) patients daily.
Total volume of soup to be prepared daily to serve \(250\) patients \( = 154 \times 250 \) cm³
\( = 38500 \) cm³
To convert to litres: \(38500 \) cm³ \( = \frac{38500}{1000} \) L \( = 38.5 \) L.
Hence, the hospital has to prepare \(38500\) cm³ (or \(38.5\) L) of soup daily to serve \(250\) patients.
In simple words: Calculate the amount of soup in one bowl using the cylinder's volume formula. Then, multiply this single bowl's volume by the total number of patients to find the total soup needed, converting to litres if necessary.

Exam Tip: First, calculate the volume for a single serving accurately. Then, multiply by the total number of servings. Remember to convert the final volume to litres if asked, as this is a common unit for liquids.