RBSE Solutions Class 9 Maths Chapter 12 Surface Area and Volume of Cube and Cuboid More Ques

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 12 Surface Area and Volume of Cube and Cuboid Miscellaneous Exercise

Multiple Choice Questions

Question 1. The volume of a cube is 125 cubic metres, its side is:
(a) 7 m
(b) 6 m
(c) 5 m
(d) 2 m
Answer: (c) 5 m
In simple words: To find the side of a cube when you know its volume, you take the cube root of the volume. The cube root of 125 cubic metres is 5 metres, because \(5 \times 5 \times 5 = 125\). This shows how the dimensions connect to the total space inside.

🎯 Exam Tip: Remember that volume is measured in cubic units (like \(m^3\)) and side length in linear units (like m).

Question 2. The volume of a cube is 1331 cubic centimetre, then its surface area is:
(a) 762 sq. cm
(b) 726 sq. cm
(c) 426 sq. cm
(d) 468 sq. cm
Answer: (b) 726 sq. cm
In simple words: First, find the side length of the cube by taking the cube root of 1331, which is 11 cm. Then, calculate the total surface area using the formula \(6 \times \text{side}^2\). So, \(6 \times 11^2 = 6 \times 121 = 726\) square centimetres. The surface area is the total area of all the faces of the cube.

🎯 Exam Tip: Always perform calculations step-by-step to avoid errors, especially when converting between volume and surface area.

Question 3. The length, breadth and height of a cuboid are 4 m, 3 m and 2 m respectively, then surface area of cuboid:
(a) 25 sq. m
(b) 52 sq. m
(c) 64 sq. m
(d) 72 sq. m
Answer: (b) 52 sq. m
In simple words: To find the surface area of a cuboid, use the formula \(2(\text{length} \times \text{breadth} + \text{breadth} \times \text{height} + \text{height} \times \text{length})\). Putting in the numbers, you get \(2(4 \times 3 + 3 \times 2 + 2 \times 4) = 2(12 + 6 + 8) = 2(26) = 52\) square metres. This formula calculates the area of all six faces of the cuboid.

🎯 Exam Tip: Remember to calculate the area of each pair of identical faces and then sum them up, which the formula \(2(lb + bh + hl)\) helps achieve efficiently.

Question 4. The diagonal of a cuboid having dimension 8m x 7m x 6m is:
(a) 12.2 m
(b) 12.02 m
(c) 14.2 m
(d) 14.02 m
Answer: (a) 12.2 m
In simple words: The diagonal of a cuboid goes from one corner to the opposite far corner. You can find its length using the formula \( \sqrt{\text{length}^2 + \text{breadth}^2 + \text{height}^2} \). So, for dimensions 8m, 7m, and 6m, the diagonal is \( \sqrt{8^2 + 7^2 + 6^2} = \sqrt{64 + 49 + 36} = \sqrt{149} \). Calculating this, we get approximately 12.2 metres.

🎯 Exam Tip: The diagonal formula for a cuboid is a direct extension of the Pythagorean theorem in three dimensions. Ensure your calculations are accurate, especially with square roots.

Question 5. The edge of a cube is 5 cm, its diagonal will be:
(a) \( 4\sqrt{3} \) cm
(b) \( 2\sqrt{3} \) cm
(c) \( 5\sqrt{3} \) cm
(d) 5 cm
Answer: (c) \( 5\sqrt{3} \) cm
In simple words: For any cube, the length of its space diagonal (the one passing through the centre of the cube) is always the side length multiplied by \( \sqrt{3} \). Since the edge of this cube is 5 cm, its diagonal will be \( 5 \times \sqrt{3} = 5\sqrt{3} \) cm. This shortcut formula is useful for cubes.

🎯 Exam Tip: Remember the special relationship between a cube's edge 'a' and its space diagonal ' \( a\sqrt{3} \) '. This is a common formula that can save time in exams.

Question 6. The volume of cuboid is 400 cubic centimetre, and area of its base is 80 sq. cm, then its height is:
(a) 7 cm
(b) 6 cm
(c) 4 cm
(d) 5 cm
Answer: (d) 5 cm
In simple words: The volume of a cuboid is found by multiplying its base area by its height. So, if you divide the total volume by the base area, you will get the height. Here, the volume is 400 cubic centimetres and the base area is 80 square centimetres. Therefore, the height is \( 400 \div 80 = 5 \) cm. This is a basic way to find one dimension when others are known.

🎯 Exam Tip: Always remember the fundamental formula: Volume of a cuboid = Base Area × Height. This relationship is crucial for solving problems involving missing dimensions.

Question 7. A cuboid measuring 15 cm x 12 cm x 6 cm is melted. How many new cubes of sides 3 cm can be made?
Answer: First, calculate the volume of the cuboid: \( \text{Length} \times \text{Breadth} \times \text{Height} = 15 \times 12 \times 6 = 1080 \) cubic cm.
Next, find the volume of one small cube: \( \text{Side}^3 = 3^3 = 27 \) cubic cm.
To find how many small cubes can be made, divide the volume of the cuboid by the volume of one small cube: \( 1080 \div 27 = 40 \).
Thus, 40 new cubes can be prepared. When a solid is melted and recast, its total volume remains the same.
In simple words: We find the total space of the big cuboid first. Then we find the space of one small cube. To know how many small cubes fit, we divide the big space by the small space. So, 40 small cubes can be made from the melted cuboid.

🎯 Exam Tip: In melting and recasting problems, the key principle is that the total volume of the material remains constant. Always calculate volumes for both original and new shapes.

Question 8. Two cubical dice having edge 2 cm are joined end to end. Find the total surface area of the solid so formed.
Answer: When two cubical dice with an edge of 2 cm are joined end to end, they form a cuboid.
The length of the new solid will be \( 2 \, \text{cm} + 2 \, \text{cm} = 4 \, \text{cm} \).
The breadth will remain 2 cm.
The height will also remain 2 cm.
Now, calculate the total surface area of this new cuboid using the formula \( 2(\text{lb} + \text{bh} + \text{hl}) \):
\( \text{Surface area} = 2(4 \times 2 + 2 \times 2 + 2 \times 4) \)
\( = 2(8 + 4 + 8) \)
\( = 2(20) \)
\( = 40 \, \text{cm}^2 \).
Joining two cubes affects only their length, creating a cuboid with an increased length.
In simple words: When two cubes of 2 cm side are put together, they make a longer box, which is a cuboid. This box will be 4 cm long, 2 cm wide, and 2 cm high. We use the cuboid surface area formula to find the total area, which comes out to be 40 square centimetres.

🎯 Exam Tip: When combining solids, carefully determine the new dimensions of the resulting shape and identify which faces are no longer exposed, as these are not included in the new total surface area.

Question 9. An empty tank is 4 m long and 3 m wide. How many cubic metre of water must be filled in it so that depth of water becomes 2 m?
Answer: The tank is a cuboid shape with:
Length \( = 4 \, \text{m} \)
Breadth \( = 3 \, \text{m} \)
To make the depth (or height) of water 2 m, we need to find the volume of water required.
Volume of water \( = \text{Length} \times \text{Breadth} \times \text{Depth} \)
\( = 4 \times 3 \times 2 \)
\( = 24 \, \text{m}^3 \).
Therefore, 24 cubic metres of water must be filled in the empty tank to reach a depth of 2 m. The volume represents the amount of space the water occupies.
In simple words: The tank is 4 meters long and 3 meters wide. We want the water to be 2 meters deep. To find how much water is needed, we multiply the length, width, and depth. So, \(4 \times 3 \times 2 = 24\) cubic meters of water must be filled in the tank.

🎯 Exam Tip: Volume calculations are straightforward when all three dimensions (length, breadth, height/depth) are known. Ensure all units are consistent before multiplying.

Question 10. A cubical vessel contains 8 litres of water. Find the total surface area of the vessel.
Answer: First, convert the volume from litres to cubic centimetres:
We know that 1 litre \( = 1000 \, \text{cm}^3 \).
So, 8 litres \( = 8 \times 1000 = 8000 \, \text{cm}^3 \).
Let the side of the cubical vessel be \( l \). The volume of a cube is \( l^3 \).
\( l^3 = 8000 \, \text{cm}^3 \)
\( l = \sqrt[3]{8000} \)
\( l = 20 \, \text{cm} \).
Now, calculate the total surface area of the cube using the formula \( 6l^2 \):
Total surface area \( = 6 \times (20)^2 \)
\( = 6 \times 400 \)
\( = 2400 \, \text{cm}^2 \).
Thus, the total surface area of the cubical vessel is 2400 square centimetres. This involves converting units and then using the cube's properties.
In simple words: We start by changing 8 litres to 8000 cubic centimetres. Since it's a cube, we find its side length by taking the cube root of 8000, which is 20 cm. Then, we use the formula for the surface area of a cube, \(6 \times \text{side}^2\), to get \(6 \times 20^2 = 2400\) square centimetres.

🎯 Exam Tip: It is crucial to remember the conversion factor between litres and cubic centimetres (1 L = \(1000 \, \text{cm}^3\)) and to work with consistent units throughout the problem.

Question 11. A godown measures 60 m x 25 m x 10 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown.
Answer: First, calculate the volume of the godown:
Volume of godown \( = \text{Length} \times \text{Breadth} \times \text{Height} \)
\( = 60 \, \text{m} \times 25 \, \text{m} \times 10 \, \text{m} = 15000 \, \text{m}^3 \).
Next, calculate the volume of one wooden crate:
Volume of one wooden crate \( = 1.5 \, \text{m} \times 1.25 \, \text{m} \times 0.5 \, \text{m} = 0.9375 \, \text{m}^3 \).
To find the maximum number of crates that can be stored, divide the volume of the godown by the volume of one crate:
Required number of wooden crates \( = \frac{\text{Volume of the godown}}{\text{Volume of one wooden crate}} \)
\( = \frac{15000}{0.9375} \)
\( = 16000 \).
Thus, a maximum of 16000 crates can be stored in the godown. This calculation assumes the crates can be arranged without any wasted space.
In simple words: We find the total space inside the godown. Then, we find the space taken by one wooden box. By dividing the godown's space by the box's space, we find that 16,000 such boxes can fit.

🎯 Exam Tip: When dealing with packing problems, ensure all dimensions are in the same units. The maximum number of items is found by dividing the total volume by the volume of a single item.

Question 12. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Answer: The depth of the river is 3 m and its width is 40 m.
The rate of flow is 2 km per hour.
First, convert the flow rate to metres per minute:
\( 2 \, \text{km/hour} = 2 \times 1000 \, \text{m /} (60 \, \text{minutes}) = \frac{2000}{60} \, \text{m/minute} = \frac{100}{3} \, \text{m/minute} \).
This means that in one minute, a column of water \( \frac{100}{3} \) metres long flows into the sea.
The volume of water falling into the sea in one minute is:
Volume \( = \text{Length} \times \text{Width} \times \text{Depth} \)
\( = \frac{100}{3} \, \text{m} \times 40 \, \text{m} \times 3 \, \text{m} \)
\( = 4000 \, \text{m}^3 \).
Therefore, 4000 cubic metres of water will fall into the sea in one minute. Converting units properly is a crucial step here.
In simple words: The river is 3m deep and 40m wide, flowing at 2 km/hr. We change the speed to meters per minute, which is \( \frac{100}{3} \) m/min. Then, we multiply this length by the width and depth of the river to find the volume of water flowing out in one minute, which is 4000 cubic metres.

🎯 Exam Tip: Always ensure all units are consistent (e.g., all in metres and minutes) before performing calculations for volume flow rates. Carefully convert speed units.

Question 13. The dimension of rectangular parallelopiped are in the ratio 6 : 5 : 4 and its total surface area is 33300 square metres. Find its volume.
Answer: Let the length, breadth, and height of the cuboid (rectangular parallelopiped) be \( 6x \), \( 5x \), and \( 4x \) respectively.
The total surface area of a cuboid is given by \( 2(\text{lb} + \text{bh} + \text{hl}) \).
Given total surface area \( = 33300 \, \text{m}^2 \).
So, \( 2( (6x)(5x) + (5x)(4x) + (4x)(6x) ) = 33300 \)
\( 2( 30x^2 + 20x^2 + 24x^2 ) = 33300 \)
\( 2( 74x^2 ) = 33300 \)
\( 148x^2 = 33300 \)
\( x^2 = \frac{33300}{148} \)
\( x^2 = 225 \)
\( x = \sqrt{225} \)
\( x = 15 \, \text{m} \).
Now, find the actual dimensions:
Length \( = 6x = 6 \times 15 = 90 \, \text{m} \)
Breadth \( = 5x = 5 \times 15 = 75 \, \text{m} \)
Height \( = 4x = 4 \times 15 = 60 \, \text{m} \).
Finally, calculate the volume of the cuboid:
Volume \( = \text{Length} \times \text{Breadth} \times \text{Height} \)
\( = 90 \times 75 \times 60 \)
\( = 405000 \, \text{m}^3 \).
The volume of the cuboid is 405000 cubic metres. Using the ratio and surface area, we can find the individual dimensions first.
In simple words: We are given the ratio of the cuboid's sides and its total surface area. We use a variable 'x' to represent the actual side lengths based on the ratio. By using the surface area formula, we solve for 'x'. Once 'x' is found, we get the real length, breadth, and height. Then, we multiply these to find the cuboid's volume, which is 405000 cubic metres.

🎯 Exam Tip: When dimensions are given in a ratio, introduce a common multiplier (like 'x') to represent the actual lengths. This allows you to set up equations for area or volume in terms of 'x' and solve for it.

Question 14. A field is 20 metres long and 15 metres wide. A pit (outside the field) 10 metres long and 6 metres wide is dug to a depth of 5 m and the earth is spread uniformly in the field. By how much the level of field is raised?
Answer: The dimensions of the field are:
Length \( = 20 \, \text{m} \)
Breadth \( = 15 \, \text{m} \).
First, calculate the volume of the earth dug out from the pit:
Length of pit \( = 10 \, \text{m} \)
Width of pit \( = 6 \, \text{m} \)
Depth of pit \( = 5 \, \text{m} \)
Volume of earth \( = \text{Length} \times \text{Width} \times \text{Depth} \)
\( = 10 \times 6 \times 5 = 300 \, \text{m}^3 \).
This volume of earth is spread uniformly over the field. Let 'h' be the height by which the level of the field is raised.
The volume of earth spread on the field will be equal to the volume of the earth dug out.
Volume of earth spread \( = \text{Area of field} \times \text{Height raised} \)
\( = (\text{Length of field} \times \text{Breadth of field}) \times h \)
\( = (20 \times 15) \times h \)
\( = 300h \).
Since the volume of earth dug out is 300 \( \text{m}^3 \), we have:
\( 300h = 300 \)
\( h = \frac{300}{300} \)
\( h = 1 \, \text{m} \).
Hence, the level of the field will be raised by 1 metre. This problem connects the volume of displaced earth to the change in height of a different area.
In simple words: First, we find the volume of soil taken out from the pit, which is \(10 \times 6 \times 5 = 300\) cubic metres. This soil is spread over the field, which is \(20 \times 15 = 300\) square metres. To find how much the field's level goes up, we divide the volume of soil by the area of the field. So, \(300 \div 300 = 1\) metre.

🎯 Exam Tip: In problems involving spreading excavated earth, the volume of the pit is equal to the volume of the layer spread on the field. Ensure you use the correct area for spreading, not just the pit's area.