OP Malhotra Class 9 Maths Solutions Chapter 18 Surface Area and Volume of 3D Solids Exercise 18 (B)

Question 1. Find the volume of a rail of uniform cross-section, given area of cross-section 12.8 cm², and length 1.26 cm.
Answer: We are given the area of the cross-section of the rail as \( 12.8 \text{ cm}^2 \) and its length as \( 1.26 \text{ cm} \). To find the volume of the rail, we multiply the cross-sectional area by its length.
Volume of rail \( = \text{Area of cross-section} \times \text{Length} \)
\( = 12.8 \text{ cm}^2 \times 1.26 \text{ cm} \)
\( = 16.128 \text{ cm}^3 \)
Thus, the volume of the rail is \( 16.128 \text{ cm}^3 \). This is a simple application of the formula for the volume of a prism or cylinder.
In simple words: To find how much space the rail takes up, multiply the area of its front face by how long it is.

🎯 Exam Tip: Always ensure the units are consistent (e.g., all in cm or all in m) before performing calculations. For volume, the unit will be cubic (e.g., cm³).

Question 2. Find the area of the cross-section, assuming it to be uniform, of a solid, given that its volume is 92.8 cm³, and length is 6.4 m.
Answer: We are given the volume of the solid as \( 92.8 \text{ cm}^3 \) and its length as \( 6.4 \text{ m} \). First, we need to make sure all units are the same. We will convert the length from meters to centimeters.
Length \( = 6.4 \text{ m} = 6.4 \times 100 \text{ cm} = 640 \text{ cm} \)
Now, we can find the area of the cross-section using the formula: Volume \( = \text{Area of cross-section} \times \text{Length} \).
So, Area of cross-section \( = \frac{\text{Volume}}{\text{Length}} \)
\( = \frac{92.8 \text{ cm}^3}{640 \text{ cm}} \)
\( = 0.145 \text{ cm}^2 \)
Therefore, the area of the cross-section is \( 0.145 \text{ cm}^2 \). It's important to convert units at the start for accuracy.
In simple words: First, change the length to centimeters. Then, divide the total volume by the length to find the area of the front face.

🎯 Exam Tip: Unit consistency is crucial. Always convert all dimensions to the same unit (e.g., all cm or all m) before starting calculations to avoid errors.

Question 3. The dimensions of the L-shaped cross-section of bar, 2 m long, are shown in cm in figure. Find
(i) the volume of the bar.
(ii) the weight, if the material weighs 0.3 kg per cm³.
Answer:
First, let's convert the length of the bar to centimeters: \( 2 \text{ m} = 2 \times 100 \text{ cm} = 200 \text{ cm} \).
The L-shaped cross-section can be divided into two rectangles. Let's refer to the diagram for dimensions (assuming G is at (0,0)):
Area of the left vertical rectangle (similar to ABCG in solution notation, assuming \( AG = 4 - 1/2 = 3.5 \text{ cm} \) and width \( 3/4 \text{ cm} \)):
Area\(_{1} = 3.5 \text{ cm} \times 0.75 \text{ cm} = \frac{7}{2} \times \frac{3}{4} = \frac{21}{8} \text{ cm}^2 \)
Area of the bottom horizontal rectangle (similar to GDEF in solution notation, with length \( 3 \text{ cm} \) and height \( 1/2 \text{ cm} \)):
Area\(_{2} = 3 \text{ cm} \times 0.5 \text{ cm} = \frac{3}{2} \text{ cm}^2 \)
Total area of the cross-section \( = \text{Area}_1 + \text{Area}_2 \)
\( = \frac{21}{8} + \frac{3}{2} = \frac{21 + 12}{8} = \frac{33}{8} \text{ cm}^2 \)
(i) To find the volume of the bar, we multiply the cross-sectional area by its length:
Volume \( = \text{Area of cross-section} \times \text{Length} \)
\( = \frac{33}{8} \text{ cm}^2 \times 200 \text{ cm} \)
\( = 33 \times 25 \text{ cm}^3 \)
\( = 825 \text{ cm}^3 \)
(ii) To find the weight of the bar, we use the given density of the material:
Weight of \( 1 \text{ cm}^3 = 0.3 \text{ kg} \)
Total weight \( = \text{Volume} \times \text{Weight per unit volume} \)
\( = 825 \text{ cm}^3 \times 0.3 \text{ kg/cm}^3 \)
\( = 247.5 \text{ kg} \)
The total weight of the L-shaped bar is \( 247.5 \text{ kg} \). Knowing the density allows us to easily find the mass from the volume.
In simple words: First, find the area of the L-shape by splitting it into two rectangles and adding their areas. Then, multiply this total area by the bar's length to get its volume. Finally, multiply the volume by how much each cubic centimeter weighs to get the total weight.

🎯 Exam Tip: When dealing with composite shapes, break them down into simpler, familiar shapes (like rectangles or triangles) to calculate their areas or volumes accurately. Always check for consistent units.

Question 4. The dimensions of the cross-section of a girder, 2.5 m long, are shown in cm in the diagram. Find
(i) the volume of the girder,
(ii) the weight, if the material weighs 7.8 gm per cm².
Answer:
First, convert the length of the girder from meters to centimeters: \( 2.5 \text{ m} = 2.5 \times 100 \text{ cm} = 250 \text{ cm} \).
The cross-section of the girder is an I-beam shape, which can be divided into three rectangles (top flange, web, and bottom flange) as per the solution's decomposition. Note that the dimensions used in the calculation (especially for widths) are internal to the solution's logic and may differ from literal image labels if the image is only illustrative. We will use the dimensions from the solution's calculation:

Area of rectangle I (bottom flange): \( = 4 \text{ cm} \times 2.5 \text{ cm} = 10 \text{ cm}^2 \)
Area of rectangle II (web): The solution calculates its width as \( (8 - 3 - 3) = 2 \text{ cm} \) and height \( 3.5 \text{ cm} \).
Area\(_{II} = 3.5 \text{ cm} \times 2 \text{ cm} = 7 \text{ cm}^2 \)
Area of rectangle III (top flange): The solution indicates a width of \( 8 \text{ cm} \) and height \( 1.5 \text{ cm} \).
Area\(_{III} = 8 \text{ cm} \times 1.5 \text{ cm} = 12 \text{ cm}^2 \)
Total area of the cross-section \( = \text{Area}_I + \text{Area}_{II} + \text{Area}_{III} \)
\( = 10 \text{ cm}^2 + 7 \text{ cm}^2 + 12 \text{ cm}^2 = 29 \text{ cm}^2 \)
(i) To find the volume of the girder:
Volume \( = \text{Area of cross-section} \times \text{Length} \)
\( = 29 \text{ cm}^2 \times 250 \text{ cm} = 7250 \text{ cm}^3 \)
(ii) To find the weight of the girder, given the material weighs 7.8 gm per cm³:
Weight of \( 1 \text{ cm}^3 = 7.8 \text{ gm} \)
Total weight \( = \text{Volume} \times \text{Weight per unit volume} \)
\( = 7250 \text{ cm}^3 \times 7.8 \text{ gm/cm}^3 \)
\( = 56550 \text{ gm} \)
To convert grams to kilograms, divide by 1000:
Total weight \( = \frac{56550}{1000} \text{ kg} = 56.55 \text{ kg} \)
Approximately, this is \( 57 \text{ kg} \). The total weight depends on both the volume and the material's density.
In simple words: First, change the girder's length to centimeters. Then, find the area of its I-shape by adding the areas of three rectangles. Multiply this total area by the girder's length to get its volume. Finally, multiply the volume by how much the material weighs per cubic centimeter to get the total weight in grams, then convert to kilograms.

🎯 Exam Tip: Pay close attention to how complex shapes are decomposed in the solution and use the dimensions implied by the calculation steps, even if they seem to deviate slightly from visual labels in diagrams. Ensure all unit conversions are handled correctly.

Question 9. The cross-area of a pipe is 42 cm², and water is pouring out of it at the rate of 1.25 m per sec. If the pipe remains full, find the number of litres discharged per minute.
Answer:
Given the cross-sectional area of the pipe \( = 42 \text{ cm}^2 \).
Rate of water pouring \( = 1.25 \text{ m/sec} \).
We need to find the volume of water discharged per minute in litres. First, let's make the units consistent.
Convert cross-sectional area to square meters:
Area \( = 42 \text{ cm}^2 = \frac{42}{(100 \times 100)} \text{ m}^2 = \frac{42}{10000} \text{ m}^2 = 0.0042 \text{ m}^2 \)
Convert the rate of water pouring from meters per second to meters per minute:
Rate \( = 1.25 \text{ m/sec} = 1.25 \times 60 \text{ m/minute} = 75 \text{ m/minute} \)
Now, calculate the volume of water discharged per minute:
Volume \( = \text{Area} \times \text{Rate} \)
\( = 0.0042 \text{ m}^2 \times 75 \text{ m/minute} \)
\( = 0.315 \text{ m}^3/\text{minute} \)
Finally, convert the volume from cubic meters to litres, knowing that \( 1 \text{ m}^3 = 1000 \text{ litres} \):
Volume in litres \( = 0.315 \times 1000 \text{ litres} \)
\( = 315 \text{ litres} \)
So, the pipe discharges 315 litres of water per minute. This problem highlights the importance of unit conversions in practical applications.
In simple words: First, change the pipe's area to square meters and the water speed to meters per minute. Then, multiply the area by the speed to find the volume of water flowing out each minute in cubic meters. Finally, turn that volume into litres.

🎯 Exam Tip: When converting units, remember that area involves squaring the conversion factor (e.g., cm² to m² requires dividing by \(100^2\)), and volume involves cubing it. Always set up your conversions clearly.

Question 6. The figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimetres. Assume that all angles in the figure are right angles.
Answer:
The figure shows a solid with an L-shaped cross-section. We can divide this L-shape into two rectangles to find its area. Let's look at the given dimensions:

We can split the L-shape into:
1. A vertical rectangle: width \( 2 \text{ cm} \), height \( 6 \text{ cm} \). Area \( = 2 \times 6 = 12 \text{ cm}^2 \).
2. A horizontal rectangle (the top-right part): width \( 4 \text{ cm} \) (total width 6 minus 2 of the vertical part), height \( 2 \text{ cm} \). Area \( = 4 \times 2 = 8 \text{ cm}^2 \).
Total area of the cross-section \( = 12 \text{ cm}^2 + 8 \text{ cm}^2 = 20 \text{ cm}^2 \).
The length of the solid (along the 'c' axis in the solution, implied as 4 cm) is \( 4 \text{ cm} \).
Now, calculate the volume of the solid:
Volume \( = \text{Area of cross-section} \times \text{Length} \)
\( = 20 \text{ cm}^2 \times 4 \text{ cm} \)
\( = 80 \text{ cm}^3 \)
Thus, the volume of the solid is \( 80 \text{ cm}^3 \). Breaking down complex shapes makes their calculations straightforward.
In simple words: Imagine the L-shape is made of two normal rectangles. Find the area of each rectangle and add them up. Then, multiply this total area by the length of the solid to find its volume.

🎯 Exam Tip: For solids with uniform cross-sections, the volume is always the area of the cross-section multiplied by its length. Look for ways to divide complex cross-sections into simpler shapes.

Question 7. A swimming pool is 50 m long and 15 m wide. Its shallow and deep ends are \( 1\frac{1}{2} \) m and \( 4\frac{1}{2} \) m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.
Answer:
Given: Length of swimming pool (\(l\)) \( = 50 \text{ m} \)
Width of swimming pool (\(b\)) \( = 15 \text{ m} \)
Depth at one side \( = 1\frac{1}{2} \text{ m} = 1.5 \text{ m} \)
Depth at the other side \( = 4\frac{1}{2} \text{ m} = 4.5 \text{ m} \)
Since the bottom slopes uniformly, we can use the average depth for calculations.
Mean depth (\(h\)) \( = \frac{1}{2} \left( \text{Depth}_1 + \text{Depth}_2 \right) \)
\( = \frac{1}{2} \left( 1.5 \text{ m} + 4.5 \text{ m} \right) \)
\( = \frac{1}{2} \times 6 \text{ m} \)
\( = 3 \text{ m} \)
Now, calculate the volume of the pool:
Volume of the pool \( = l \times b \times h \)
\( = 50 \text{ m} \times 15 \text{ m} \times 3 \text{ m} \)
\( = 2250 \text{ m}^3 \)
To find the amount of water in litres, we use the conversion \( 1 \text{ m}^3 = 1000 \text{ litres} \):
Volume of water in litres \( = 2250 \times 1000 \text{ litres} \)
\( = 2,250,000 \text{ litres} \)
The pool requires \( 2,250,000 \text{ litres} \) of water to fill it. The concept of average depth is very useful for uniformly sloping bodies.
In simple words: Find the average depth of the pool. Then, multiply the pool's length, width, and average depth to get its volume in cubic meters. Finally, convert this volume into litres.

🎯 Exam Tip: For pools with a uniformly sloping bottom, you can use the average of the shallow and deep end depths to calculate the volume as if it were a rectangular prism. Remember the conversion factor for cubic meters to litres.

Question 8. The area of cross-section of a pipe is 5.4 cm² and water is pumped out of it the rate of 27 km / h. Find in litres the volume of water which flows out of the pipe in one minute.
Answer:
Given:
Area of cross-section of pipe \( = 5.4 \text{ cm}^2 \)
Rate of water flow \( = 27 \text{ km/h} \)
We need to find the volume of water in litres per minute. Let's convert all units to be consistent, preferably to meters and minutes.
Convert area from \( \text{cm}^2 \) to \( \text{m}^2 \):
Area \( = 5.4 \text{ cm}^2 = \frac{5.4}{100 \times 100} \text{ m}^2 = \frac{5.4}{10000} \text{ m}^2 = 0.00054 \text{ m}^2 \)
Convert flow rate from \( \text{km/h} \) to \( \text{m/min} \):
Rate \( = 27 \text{ km/h} = \frac{27 \times 1000 \text{ m}}{60 \text{ min}} = \frac{27000}{60} \text{ m/min} = 450 \text{ m/min} \)
Now, calculate the volume of water flowing out per minute:
Volume of water \( = \text{Area of cross-section} \times \text{Flow rate} \)
\( = 0.00054 \text{ m}^2 \times 450 \text{ m/min} \)
\( = 0.243 \text{ m}^3/\text{min} \)
Finally, convert the volume from cubic meters to litres, using \( 1 \text{ m}^3 = 1000 \text{ litres} \):
Volume in litres \( = 0.243 \times 1000 \text{ litres} \)
\( = 243 \text{ litres} \)
Thus, 243 litres of water flow out of the pipe in one minute. This step-by-step unit conversion ensures accuracy.
In simple words: First, change the pipe's area to square meters and the water's speed to meters per minute. Then, multiply these two numbers to get the volume of water that flows out in cubic meters per minute. Lastly, convert this volume into litres.

🎯 Exam Tip: Be meticulous with unit conversions. A common mistake is forgetting to convert length units for area (e.g., cm to m) or time units for rates (e.g., hours to minutes).

Question 9. The horizontal cross-section of a water tank is in the shape of a rectangle with a semicircle at one end, as shown in figure. The water is 2.4 m deep in the tank. Calculate the volume of water in the tank in gallons. (Take \( \pi \) to be \( \frac{22}{7} \) and 1 gallon to be 4.5 litres)
Answer:
The cross-section of the water tank consists of a rectangle and a semicircle.

From the figure, the dimensions are:
Length of the rectangular part \( = 16 \text{ m} \)
Width of the rectangular part (also diameter of semicircle) \( = 7 \text{ m} \)
Radius of the semicircle (\(r\)) \( = \frac{7}{2} \text{ m} = 3.5 \text{ m} \)
Area of the cross-section \( = \text{Area of rectangle} + \text{Area of semicircle} \)
\( = (16 \times 7) + \left( \frac{1}{2} \times \pi \times r^2 \right) \)
\( = (112) + \left( \frac{1}{2} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \right) \)
\( = 112 + \left( \frac{1}{2} \times \frac{22}{7} \times \frac{49}{4} \right) \)
\( = 112 + \left( \frac{11 \times 7}{4} \right) \)
\( = 112 + \frac{77}{4} \)
\( = 112 + 19.25 = 131.25 \text{ m}^2 \)
This can also be written as \( \frac{448+77}{4} = \frac{525}{4} \text{ m}^2 \).
The depth of water in the tank is \( 2.4 \text{ m} \).
Volume of water \( = \text{Area of cross-section} \times \text{Depth} \)
\( = \frac{525}{4} \text{ m}^2 \times 2.4 \text{ m} \)
\( = 525 \times 0.6 \text{ m}^3 \)
\( = 315 \text{ m}^3 \)
Now, convert the volume from cubic meters to litres, then to gallons:
We know \( 1 \text{ m}^3 = 1000 \text{ litres} \).
Volume in litres \( = 315 \times 1000 \text{ litres} = 315,000 \text{ litres} \)
We are given that \( 1 \text{ gallon} = 4.5 \text{ litres} \).
Volume in gallons \( = \frac{315,000 \text{ litres}}{4.5 \text{ litres/gallon}} \)
\( = 70,000 \text{ gallons} \)
The total volume of water in the tank is \( 70,000 \text{ gallons} \). This multi-step conversion is typical for real-world problems.
In simple words: First, find the area of the tank's top shape by adding the rectangle's area and the semicircle's area. Then, multiply this total area by the water's depth to get the volume in cubic meters. Convert this volume to litres, and then divide by 4.5 to get the volume in gallons.

🎯 Exam Tip: Break down complex area calculations into simpler geometric shapes. Be careful with calculations involving fractions and decimals, and always keep track of units throughout the problem to ensure the final answer is in the desired unit.

Question 10. In the figure, shows the end face of a triangular girder which is 5 m long. Find the volume of metal used in casting it. (Units in cm).
Answer:
The end face of the girder is a triangle. The problem states "Units in cm" for the volume, implying all dimensions should be in centimeters for consistent calculation.

From the diagram and problem description, assuming the units are consistent:
Base of the triangular cross-section (\(b\)) \( = 60 \text{ cm} \) (interpreting `60 m` in the prompt as `60 cm` for consistency with `90 cm` and the final volume unit).
Height of the triangular cross-section (\(h\)) \( = 90 \text{ cm} \)
Length of the girder (\(L\)) \( = 5 \text{ m} = 5 \times 100 \text{ cm} = 500 \text{ cm} \)
First, calculate the area of the triangular cross-section:
Area of the face of the girder \( = \frac{1}{2} \times \text{base} \times \text{height} \)
\( = \frac{1}{2} \times 60 \text{ cm} \times 90 \text{ cm} \)
\( = 30 \times 90 \text{ cm}^2 \)
\( = 2700 \text{ cm}^2 \)
Now, calculate the volume of metal used:
Volume \( = \text{Area of cross-section} \times \text{Length} \)
\( = 2700 \text{ cm}^2 \times 500 \text{ cm} \)
\( = 1,350,000 \text{ cm}^3 \)
To convert the volume to cubic meters, since \( 1 \text{ m} = 100 \text{ cm} \), then \( 1 \text{ m}^3 = (100 \text{ cm})^3 = 1,000,000 \text{ cm}^3 \).
Volume \( = \frac{1,350,000}{1,000,000} \text{ m}^3 = 1.35 \text{ m}^3 \)
The volume of metal used in casting is \( 1.35 \text{ m}^3 \). This calculation depends on careful unit interpretation.
In simple words: First, find the area of the triangle face by using the formula half times base times height, making sure all measurements are in centimeters. Then, multiply this area by the girder's length (also in centimeters) to get the total volume. Finally, convert this volume into cubic meters.

🎯 Exam Tip: Always pay attention to specified units in the question and ensure all dimensions are converted to the same unit before starting calculations. If units are mixed (e.g., m and cm), make an assumption for consistency and state it if needed.

Question 11. In the figure shows the end-view of a swimming pool which is 10 m wide. Calculate the volume of water when it is full of water \( \frac{1}{2} \) m from the top.
Answer:
The end-view of the swimming pool is a trapezoid. The pool is \( 10 \text{ m} \) wide. The water level is \( \frac{1}{2} \text{ m} \) (or \( 0.5 \text{ m} \)) below the top.

From the diagram:
Top width of pool (length of the trapezoid's top base) \( = 20 \text{ m} \)
Height of shallow end \( = 2 \text{ m} \)
Height of deep end \( = 3 \text{ m} \)
The water is \( 0.5 \text{ m} \) from the top. So, the actual depth of water will be:
Height of water from shallow end \( = 2 \text{ m} - 0.5 \text{ m} = 1.5 \text{ m} \)
Height of water from deep end \( = 3 \text{ m} - 0.5 \text{ m} = 2.5 \text{ m} \)
The average height of the water in the trapezoidal cross-section is:
Average height \( = \frac{1}{2} (\text{shallow water depth} + \text{deep water depth}) \)
\( = \frac{1}{2} (1.5 \text{ m} + 2.5 \text{ m}) \)
\( = \frac{1}{2} \times 4 \text{ m} = 2 \text{ m} \)
The length of the pool (width in cross-section view) is \( 20 \text{ m} \).
The width of the pool is given as \( 10 \text{ m} \).
Now, calculate the volume of water:
Volume of water \( = \text{Length} \times \text{Width} \times \text{Average height} \)
\( = 20 \text{ m} \times 10 \text{ m} \times 2 \text{ m} \)
\( = 400 \text{ m}^3 \)
The volume of water in the pool is \( 400 \text{ m}^3 \). Using the average height for a sloping bottom simplifies the volume calculation significantly.
In simple words: First, find the actual water depths at both ends by subtracting the empty space from the total depths. Then, find the average of these two water depths. Finally, multiply the pool's length, given width, and this average depth to find the total volume of water.

🎯 Exam Tip: When a problem specifies a water level relative to the top (e.g., "half a meter from the top"), always adjust the depths before calculating the average or area of the water-filled section. Ensure all dimensions are in consistent units.

Question 12. In the figure shows the end-face of a garage which is 20 m long.
(ii) Find the volume of air-space of the garage in m³.
(iii) If 40 m³ of air is required per worker, how many workers can be employed in this garage?
Answer:
The end-face of the garage consists of a rectangle and a triangle.

Length of the garage \( = 20 \text{ m} \).
The end-face is divided into a rectangle (BCEF) and a triangle (ABF).
Dimensions of the rectangle BCEF: Width \( CE = 4+4 = 8 \text{ m} \), Height \( BC = 5 \text{ m} \).
Area of rectangle BCEF \( = 8 \text{ m} \times 5 \text{ m} = 40 \text{ m}^2 \)
Dimensions of the triangle ABF: Base \( BF = 8 \text{ m} \), Height \( AG = 3 \text{ m} \).
Area of triangle ABF \( = \frac{1}{2} \times \text{base} \times \text{height} \)
\( = \frac{1}{2} \times 8 \text{ m} \times 3 \text{ m} = 12 \text{ m}^2 \)
(i) Total area of the end-face \( = \text{Area of rectangle} + \text{Area of triangle} \)
\( = 40 \text{ m}^2 + 12 \text{ m}^2 = 52 \text{ m}^2 \)
(ii) To find the volume of air-space of the garage:
Volume of air-space \( = \text{Total area of end-face} \times \text{Length of garage} \)
\( = 52 \text{ m}^2 \times 20 \text{ m} = 1040 \text{ m}^3 \)
The volume of air inside the garage is \( 1040 \text{ m}^3 \). This space is vital for ventilation.
(iii) If \( 40 \text{ m}^3 \) of air is required per worker, we can find how many workers can be employed:
Number of workers \( = \frac{\text{Total volume of air-space}}{\text{Air required per worker}} \)
\( = \frac{1040 \text{ m}^3}{40 \text{ m}^3/\text{worker}} \)
\( = 26 \text{ workers} \)
Therefore, 26 workers can be employed in this garage. Calculating capacity is essential for safety and planning.
In simple words: First, find the total area of the garage's front wall by adding the area of its rectangular base and its triangular roof. Then, multiply this total area by the garage's length to find its total air volume. Finally, divide this total air volume by the amount of air each worker needs to find how many workers can fit.

🎯 Exam Tip: For composite shapes, calculate the area of each simpler component separately before summing them. When calculating capacity (like the number of workers), ensure you divide the total available resource by the amount needed per unit.