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

S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(b)

Question 1. Find the volume of the cylinders whose radii and heights are given below. Take \( \pi \) to be \( \frac{22}{7} \).
(i) r = 7 cm, h = 8 cm
(ii) r = 7 cm, h = 12 cm
(iii) r = 14 cm, h = 16 cm
(iv) r = 21 cm, h = 40 cm
Answer:
The volume of a cylinder is found using the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Let's calculate for each part.
(i) Given radius \( r = 7 \) cm and height \( h = 8 \) cm.
\[ \text{Volume} = \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 8 = 1232 \, \text{cm}^3 \] (ii) Given radius \( r = 7 \) cm and height \( h = 12 \) cm.
\[ \text{Volume} = \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 12 = 1848 \, \text{cm}^3 \] (iii) Given radius \( r = 14 \) cm and height \( h = 16 \) cm.
\[ \text{Volume} = \pi r^2 h = \frac{22}{7} \times 14 \times 14 \times 16 = 9856 \, \text{cm}^3 \] (iv) Given radius \( r = 21 \) cm and height \( h = 40 \) cm.
\[ \text{Volume} = \pi r^2 h = \frac{22}{7} \times 21 \times 21 \times 40 = 55440 \, \text{cm}^3 \]In simple words: To find how much space a cylinder takes up, multiply pi (22/7) by the radius squared, then multiply by the height. This gives you the volume in cubic centimeters.

🎯 Exam Tip: Remember to write down the formula clearly before substituting the values to ensure full marks. Also, always check the units for radius and height, and ensure the final answer is in cubic units.

Question 2. Find the diameter of the circular cylinders if:
(a) Volume is 44 cm³, height 3.5 cm;
(b) Volume 385 cm³, height 1 dm.
Answer:
We know the volume of a cylinder is \( V = \pi r^2 h \). We can use this to find the radius and then double it to get the diameter.
(a) Given Volume \( V = 44 \) cm³, height \( h = 3.5 \) cm. We also know \( 3.5 = \frac{35}{10} \) cm.
We use the formula \( r^2 = \frac{V}{\pi h} \).
\[ r^2 = \frac{44}{\frac{22}{7} \times \frac{35}{10}} = \frac{44 \times 7 \times 10}{22 \times 35} \] \[ r^2 = \frac{3080}{770} = 4 \] So, \( r = \sqrt{4} = 2 \) cm.
Diameter is \( 2 \times r = 2 \times 2 = 4 \) cm. This calculation is a good way to see how dimensions affect volume.
(b) Given Volume \( V = 385 \) cm³, height \( h = 1 \) dm. First, convert height to cm: \( 1 \) dm \( = 10 \) cm.
Using \( r^2 = \frac{V}{\pi h} \):
\[ r^2 = \frac{385}{\frac{22}{7} \times 10} = \frac{385 \times 7}{22 \times 10} \] \[ r^2 = \frac{2695}{220} = \frac{49}{4} \] So, \( r = \sqrt{\frac{49}{4}} = \frac{7}{2} \) cm.
Diameter is \( 2 \times r = 2 \times \frac{7}{2} = 7 \) cm.
In simple words: We are given the volume and height of a cylinder. To find the diameter, we first calculate the radius by rearranging the volume formula. Once we have the radius, we just multiply it by two.

🎯 Exam Tip: Always pay attention to units. If dimensions are given in different units (like cm and dm), convert them to a consistent unit before starting calculations. The formula \( D = 2 \times \sqrt{\frac{V}{\pi h}} \) can be helpful.

Question 3. Find the height of the circular cylinders if:
(a) Volume is 66 cm³, radius 2 cm;
(b) Volume 4 litres, radius 5 cm.
Answer:
To find the height, we use the formula \( h = \frac{V}{\pi r^2} \).
(a) Given Volume \( V = 66 \) cm³, radius \( r = 2 \) cm.
\[ h = \frac{66}{\frac{22}{7} \times 2^2} = \frac{66 \times 7}{22 \times 4} \] \[ h = \frac{462}{88} = \frac{21}{4} = 5.25 \, \text{cm} \] (b) Given Volume \( V = 4 \) litres, radius \( r = 5 \) cm. First, convert litres to cm³: \( 1 \) litre \( = 1000 \) cm³, so \( 4 \) litres \( = 4000 \) cm³.
\[ h = \frac{4000}{\frac{22}{7} \times 5^2} = \frac{4000 \times 7}{22 \times 25} \] \[ h = \frac{28000}{550} = \frac{2800}{55} = \frac{560}{11} \approx 50.91 \, \text{cm} \] This means the cylinder in part (b) is much taller than the one in part (a).
In simple words: If you know how much space a cylinder holds (volume) and how wide its base is (radius), you can find its height by dividing the volume by the area of its base. Remember to convert units like liters to cubic centimeters first.

🎯 Exam Tip: Always write down the units at each step, especially when converting between different measurement systems (like litres to cm³), to avoid errors. Rounding to an appropriate decimal place at the end is also important.

Question 4. A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m³.
Answer:
First, we need to find the volume of the wooden pole. Since the density is given in kg per m³, we should convert all dimensions to meters.
Height \( h = 7 \) m.
Diameter \( = 20 \) cm, so radius \( r = \frac{20}{2} = 10 \) cm \( = 0.1 \) m.
Volume of the pole \( V = \pi r^2 h \).
\[ V = \frac{22}{7} \times (0.1)^2 \times 7 \] \[ V = \frac{22}{7} \times 0.01 \times 7 \] \[ V = 22 \times 0.01 = 0.22 \, \text{m}^3 \] The wood weighs 225 kg per m³. So, the total weight of the pole is its volume multiplied by the weight per cubic meter.
Total weight \( = 0.22 \, \text{m}^3 \times 225 \, \text{kg/m}^3 \)
\[ \text{Total weight} = 49.5 \, \text{kg} \] This is a practical application of volume calculation in everyday life.
In simple words: First, figure out how much space the wooden pole takes up (its volume) in cubic meters. Then, since we know how much one cubic meter of wood weighs, multiply that by the pole's total volume to find its complete weight.

🎯 Exam Tip: Consistency in units is crucial in these problems. Convert all measurements to meters or centimeters first, depending on what the density unit is given as, to avoid calculation errors.

Question 5. Find the volume of metal in the hollow pipe of internal radius 3 cm, metal 1 cm thick and length 6 cm.
Answer:
For a hollow pipe, we need to find the volume of the outer cylinder and subtract the volume of the inner cylinder. The volume of metal is \( V = \pi h (R^2 - r^2) \), where \( h \) is the length, \( R \) is the external radius, and \( r \) is the internal radius.
Length of the pipe \( h = 6 \) cm.
Internal radius \( r = 3 \) cm.
Thickness of the metal \( = 1 \) cm.
So, the outer radius \( R = \text{internal radius} + \text{thickness} = 3 + 1 = 4 \) cm.
Now, we can calculate the volume of the metal used.
\[ \text{Volume of metal} = \pi h (R^2 - r^2) \] \[ = \frac{22}{7} \times 6 \times (4^2 - 3^2) \] \[ = \frac{22}{7} \times 6 \times (16 - 9) \] \[ = \frac{22}{7} \times 6 \times 7 \] \[ = 22 \times 6 = 132 \, \text{cm}^3 \] This method helps calculate material needed for hollow structures.
In simple words: To find the amount of metal in a hollow pipe, first find the inner and outer radius. Then, subtract the volume of the empty space inside from the total volume of the pipe, using the given length.

🎯 Exam Tip: Remember that for hollow objects, the volume of the material is the difference between the outer volume and the inner volume. Always be careful to calculate the outer radius correctly by adding the thickness to the inner radius.

Question 6. The sum of the radius of the base and the height of a solid cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm², find the volume of the cylinder. (Take \( \pi = \frac{22}{7} \))
Answer:
We are given that the sum of the radius and height is 37 cm, so \( r + h = 37 \). This means \( h = 37 - r \).
The total surface area (TSA) of a solid cylinder is given by \( 2\pi r(h + r) \).
We know TSA \( = 1628 \) cm².
\[ 2\pi r(h + r) = 1628 \] Since \( r + h = 37 \), we can substitute this into the formula:
\[ 2 \times \frac{22}{7} \times r \times 37 = 1628 \] Now, we can solve for \( r \):
\[ r = \frac{1628 \times 7}{2 \times 22 \times 37} \] \[ r = \frac{11396}{1628} = 7 \, \text{cm} \] Now that we have the radius, we can find the height:
\[ h = 37 - r = 37 - 7 = 30 \, \text{cm} \] Finally, we calculate the volume of the cylinder using \( V = \pi r^2 h \).
\[ V = \frac{22}{7} \times 7^2 \times 30 \] \[ V = \frac{22}{7} \times 49 \times 30 \] \[ V = 22 \times 7 \times 30 = 4620 \, \text{cm}^3 \] Understanding how to use surface area to find dimensions is key in such problems.
In simple words: We are given the sum of the radius and height, and the total surface area of a cylinder. We use the surface area formula to find the radius first, then calculate the height. After that, we can easily find the volume.

🎯 Exam Tip: When dealing with multiple cylinder properties (radius, height, volume, surface area), always start by writing down the given information and the relevant formulas. This helps in planning your steps to find the unknown values.

Question 7. A cylindrical tank has capacity 6160 cu m. Find its depth if the diameter of its base is 28 m. Also, find the area of the inside curved surface of the tank. (Take \( \pi = \frac{22}{7} \))
Answer:
The capacity of the tank is its volume, \( V = 6160 \) m³.
The diameter of its base is 28 m, so the radius \( r = \frac{28}{2} = 14 \) m.
To find the depth (which is the height, \( h \)) of the tank, we use the volume formula \( V = \pi r^2 h \), so \( h = \frac{V}{\pi r^2} \).
\[ h = \frac{6160}{\frac{22}{7} \times 14^2} \] \[ h = \frac{6160 \times 7}{22 \times 196} \] \[ h = \frac{43120}{4312} = 10 \, \text{m} \] The depth of the tank is 10 meters. Now, let's find the area of the inside curved surface.
The curved surface area (CSA) of a cylinder is given by \( 2\pi rh \).
\[ \text{CSA} = 2 \times \frac{22}{7} \times 14 \times 10 \] \[ = 2 \times 22 \times 2 \times 10 \] \[ = 880 \, \text{m}^2 \] This calculation helps in understanding the amount of material needed to construct or coat the tank's inner walls.
In simple words: We are given the volume and diameter of a cylindrical tank. First, we use the volume formula to calculate its depth (height). Then, with the radius and depth, we find the area of its curved inner surface.

🎯 Exam Tip: Clearly differentiate between the total surface area and the curved surface area. For an open tank or one where only the sides are considered, the curved surface area formula \( 2\pi rh \) is used.

Question 8. The area of the curved surface of a cylinder is 4400 cm², and the circumference of its base is 110 cm. Find
(i) the height of the cylinder.
(ii) the volume of the cylinder.
(Take \( \pi \) to be \( \frac{22}{7} \))
Answer:
We are given the curved surface area (CSA) and the circumference of the base.
Curved surface area \( = 4400 \) cm².
Circumference of the base \( = 110 \) cm.
The formula for circumference is \( 2\pi r \). So, \( 2\pi r = 110 \).
\[ r = \frac{110}{2\pi} = \frac{110 \times 7}{2 \times 22} = \frac{770}{44} = \frac{35}{2} = 17.5 \, \text{cm} \] (i) The formula for curved surface area is \( 2\pi rh \). We can write this as \( (2\pi r) \times h \).
Since \( 2\pi r = 110 \) cm, we have:
\[ 110 \times h = 4400 \] \[ h = \frac{4400}{110} = 40 \, \text{cm} \] So, the height of the cylinder is 40 cm. It's often easier to find one value before another.
(ii) Now we can find the volume using \( V = \pi r^2 h \). We have \( r = \frac{35}{2} \) cm and \( h = 40 \) cm.
\[ V = \frac{22}{7} \times \left(\frac{35}{2}\right)^2 \times 40 \] \[ V = \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 40 \] \[ V = 11 \times 5 \times 35 \times 10 \] \[ V = 19250 \, \text{cm}^3 \]In simple words: We have the curved surface area and the circumference of the cylinder's base. First, we use the circumference to find the radius. Then, we use the curved surface area and the radius to find the height. Finally, we use the radius and height to calculate the cylinder's total volume.

🎯 Exam Tip: When given circumference and curved surface area, it is most efficient to find the radius from the circumference first, and then the height using the curved surface area. This streamlines the calculations for volume.

Question 9.
(i) How many cubic metres of earth must be dug out to make a well 20 metres deep and 2 metres in diameter? (Take \( \pi \) to be \( \frac{22}{7} \))
(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of Rs 5 per m², find the cost of Plastering (Take \( \pi \) to be \( \frac{22}{7} \))
Answer:
(i) The well is cylindrical. We need to find its volume to know how much earth is dug out.
Depth (height) of the well \( h = 20 \) m.
Diameter \( = 2 \) m, so radius \( r = \frac{2}{2} = 1 \) m.
Volume of earth dug out \( V = \pi r^2 h \).
\[ V = \frac{22}{7} \times 1^2 \times 20 \] \[ V = \frac{22}{7} \times 1 \times 20 = \frac{440}{7} \, \text{m}^3 \] \[ V \approx 62.857 \, \text{m}^3 = 62 \frac{6}{7} \, \text{m}^3 \] This amount of earth can be used for other purposes, like creating an embankment.
(ii) To plaster the inner curved surface, we need to find its area. This is the curved surface area (CSA) of the cylinder.
CSA \( = 2\pi rh \).
\[ \text{CSA} = 2 \times \frac{22}{7} \times 1 \times 20 \] \[ = \frac{880}{7} \, \text{m}^2 \] The cost of plastering is Rs 5 per m².
Total cost \( = \text{CSA} \times \text{rate} \)
\[ = \frac{880}{7} \times 5 \] \[ = \frac{4400}{7} \, \text{Rs} \approx 628.57 \, \text{Rs} \]In simple words: For part (i), we find the volume of the well to know how much earth needs to be removed. For part (ii), we calculate the area of the well's inner curved wall, then multiply it by the cost per square meter to find the total plastering expense.

🎯 Exam Tip: This problem involves two distinct calculations: volume for the earth removed and curved surface area for plastering. Make sure to clearly state which formula you are using for each part.

Question 10. A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm² (sq cm). Find
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place. (Take \( \pi \) to be 3.14)
Answer:
Given diameter \( = 20 \) cm, so radius \( r = \frac{20}{2} = 10 \) cm.
Curved surface area (CSA) \( = 1000 \) cm².
(i) We know CSA \( = 2\pi rh \). We can find the height \( h \).
\[ h = \frac{\text{CSA}}{2\pi r} \] Using \( \pi = 3.14 \):
\[ h = \frac{1000}{2 \times 3.14 \times 10} = \frac{1000}{62.8} \] \[ h \approx 15.923 \, \text{cm} \] Correct to one decimal place, height \( h = 15.9 \) cm. It's important to follow rounding instructions.
(ii) Now we find the volume using \( V = \pi r^2 h \).
Using \( \pi = 3.14 \), \( r = 10 \) cm, and \( h \approx 15.923 \) cm (using the unrounded value for better accuracy, then rounding the final answer).
\[ V = 3.14 \times 10^2 \times 15.923 \] \[ V = 3.14 \times 100 \times 15.923 \] \[ V = 314 \times 15.923 = 4992.622 \, \text{cm}^3 \] Correct to one decimal place, volume \( V = 4992.6 \) cm³.
In simple words: First, use the diameter to find the radius. Then, use the curved surface area and radius to calculate the height. Finally, use the radius and height to find the volume, making sure to round both results to one decimal place as asked.

🎯 Exam Tip: When a question asks for a specific level of precision (e.g., "correct to one decimal place"), ensure your final answer matches that requirement. Use the full precision of intermediate values in calculations to minimize rounding errors.

Question 11. An electric geyser is cylindrical in shape, having a diameter of 35 cm and height of 1.2 m. Neglecting the thickness of its walls, calculate
(i) its outer lateral surface area;
(ii) its capacity in liters (Take \( \pi \) to be \( \frac{22}{7} \))
Answer:
First, convert all units to be consistent. It's easiest to work in centimeters.
Diameter of geyser \( = 35 \) cm, so radius \( r = \frac{35}{2} \) cm.
Height \( h = 1.2 \) m \( = 1.2 \times 100 = 120 \) cm.
(i) The outer lateral surface area is the curved surface area (CSA).
CSA \( = 2\pi rh \).
\[ \text{CSA} = 2 \times \frac{22}{7} \times \frac{35}{2} \times 120 \] \[ = 22 \times 5 \times 120 \] \[ = 110 \times 120 = 13200 \, \text{cm}^2 \] This area represents the amount of material on the side of the geyser.
(ii) The capacity in liters is the volume of the geyser. We use \( V = \pi r^2 h \).
\[ V = \frac{22}{7} \times \left(\frac{35}{2}\right)^2 \times 120 \] \[ V = \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 120 \] \[ V = 11 \times 5 \times 35 \times 60 \] \[ V = 115500 \, \text{cm}^3 \] To convert cm³ to liters, we know that \( 1000 \, \text{cm}^3 = 1 \) liter.
\[ \text{Capacity in liters} = \frac{115500}{1000} = 115.5 \, \text{liters} \]In simple words: We first make sure all measurements are in the same units. Then, for part (i), we use the curved surface area formula to find the side area of the geyser. For part (ii), we calculate the total volume of the geyser in cubic centimeters and convert it to liters to find its capacity.

🎯 Exam Tip: Always be mindful of unit conversions, especially when calculating capacity in liters from cubic centimeters or meters. Remember the conversion factors: \( 1 \, \text{m}^3 = 1000 \) liters and \( 1000 \, \text{cm}^3 = 1 \) liter.

Question 12. How many cylindrical glasses, diameter 8 cm, height 15 cm, can be filled from cylindrical vessel, diameter 30 cm, height 80 cm, full of milk. (There is no need to substitute for \( \pi \)).
Answer:
To find out how many glasses can be filled, we need to find the volume of the large vessel and the volume of one small glass, then divide the vessel's volume by the glass's volume. Since \( \pi \) is not to be substituted, we can leave it in our calculations.
**For the cylindrical vessel:**
Diameter \( = 30 \) cm, so Radius \( R = \frac{30}{2} = 15 \) cm.
Height \( H = 80 \) cm.
Volume of vessel \( V_{\text{vessel}} = \pi R^2 H \)
\[ V_{\text{vessel}} = \pi \times 15^2 \times 80 = \pi \times 225 \times 80 = 18000\pi \, \text{cm}^3 \] **For one cylindrical glass:**
Diameter \( = 8 \) cm, so Radius \( r = \frac{8}{2} = 4 \) cm.
Height \( h = 15 \) cm.
Volume of glass \( V_{\text{glass}} = \pi r^2 h \)
\[ V_{\text{glass}} = \pi \times 4^2 \times 15 = \pi \times 16 \times 15 = 240\pi \, \text{cm}^3 \] **Number of glasses that can be filled:**
\[ \text{Number of glasses} = \frac{V_{\text{vessel}}}{V_{\text{glass}}} = \frac{18000\pi}{240\pi} \] \[ = \frac{18000}{240} = 75 \] So, 75 cylindrical glasses can be filled. This is a common method for comparing capacities.
In simple words: To find how many small glasses can be filled from a big vessel, first calculate the volume of the big vessel and the volume of one small glass. Then, divide the big volume by the small volume to get the number of glasses. We leave pi as a symbol because it cancels out.

🎯 Exam Tip: When \( \pi \) is meant to cancel out (as indicated by "no need to substitute for \( \pi \)"), keep it as a symbol throughout your calculations. This often simplifies the arithmetic and reduces the chance of rounding errors from using an approximate value for \( \pi \).

Question 13. A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic metre of gravel are required to gravel the path to a depth of 7 cm?
Answer:
This problem involves finding the volume of a cylindrical ring (the path). We need to calculate the area of the path first, then multiply by its depth to find the volume of gravel. All units should be consistent; let's use meters.
Diameter of circular pond \( = 40 \) m, so inner radius \( r = \frac{40}{2} = 20 \) m.
Width of circular path \( = 2 \) m.
Outer radius \( R = \text{inner radius} + \text{width} = 20 + 2 = 22 \) m.
Area of the path \( = \pi (R^2 - r^2) \).
\[ \text{Area} = \frac{22}{7} (22^2 - 20^2) \] Using the identity \( a^2 - b^2 = (a-b)(a+b) \):
\[ \text{Area} = \frac{22}{7} (22 - 20)(22 + 2) \] \[ = \frac{22}{7} \times 2 \times 42 \] \[ = 22 \times 2 \times 6 = 264 \, \text{m}^2 \] Depth of the path \( = 7 \) cm \( = \frac{7}{100} = 0.07 \) m.
Volume of gravel required \( = \text{Area of path} \times \text{depth} \).
\[ \text{Volume} = 264 \times 0.07 \] \[ = 18.48 \, \text{m}^3 \] This shows how calculating volumes helps in construction planning.
In simple words: Imagine the path as a flat ring around the pond. First, calculate the area of this ring by subtracting the inner circle's area from the outer circle's area. Then, multiply this path area by the depth of the gravel (making sure both are in meters) to find the total volume of gravel needed.

🎯 Exam Tip: When calculating the area of a ring or an annulus, use the formula \( \pi (R^2 - r^2) \). Remember to factorize it as \( \pi (R-r)(R+r) \) for easier calculation, especially when \( R \) and \( r \) are large numbers.

Question 14. An iron block is in the form of a cylinder of diameter 0.5 m and length 3.5 m. This block is to be rolled into the form of a bar having square section of side 25 cm. Find the length of the bar.
Answer:
When a material is reshaped, its volume remains constant. We need to find the volume of the cylindrical block and then use that volume to find the length of the square bar. Convert all units to centimeters for consistency.
**For the cylindrical block:**
Diameter \( = 0.5 \) m \( = 50 \) cm, so radius \( r = \frac{50}{2} = 25 \) cm.
Length (height) \( h = 3.5 \) m \( = 350 \) cm.
Volume of block \( V_{\text{cylinder}} = \pi r^2 h \).
\[ V_{\text{cylinder}} = \frac{22}{7} \times 25^2 \times 350 \] \[ = \frac{22}{7} \times 625 \times 350 \] \[ = 22 \times 625 \times 50 = 687500 \, \text{cm}^3 \] **For the bar with a square section:**
Side of the square section \( s = 25 \) cm.
Area of the square section \( A_{\text{square}} = s^2 = 25^2 = 625 \, \text{cm}^2 \).
Let the length of the bar be \( L \). The volume of the bar \( V_{\text{bar}} = A_{\text{square}} \times L \).
Since the volume remains constant, \( V_{\text{bar}} = V_{\text{cylinder}} \).
\[ 625 \times L = 687500 \] \[ L = \frac{687500}{625} = 1100 \, \text{cm} \] Convert the length back to meters:
\[ L = \frac{1100}{100} = 11 \, \text{m} \] This principle of conservation of volume is fundamental in many geometry problems.
In simple words: First, calculate the total volume of the iron cylinder, making sure all measurements are in the same unit (centimeters). Then, imagine this volume is used to make a long bar with a square end. Divide the cylinder's volume by the area of the square end to find how long the bar will be.

🎯 Exam Tip: The key concept here is that volume remains conserved during reshaping. Ensure all measurements are converted to a consistent unit (e.g., cm) before starting calculations to prevent errors in the final answer.

Question 15. A swimming pool 70 m long, 44 m wide, 3 m deep is filled by water issuing from a pipe of diameter 14 cm, at 2 m per second. How many hours does it take to fill the bath?
Answer:
First, calculate the volume of the swimming pool. Then, calculate the rate of water flow from the pipe. Finally, divide the total volume by the flow rate to find the time.
**Volume of the swimming pool:**
Length \( l = 70 \) m, Breadth \( b = 44 \) m, Depth \( d = 3 \) m.
Volume \( V_{\text{pool}} = l \times b \times d \)
\[ V_{\text{pool}} = 70 \times 44 \times 3 = 9240 \, \text{m}^3 \] **Water flow from the pipe:**
Diameter of the pipe \( = 14 \) cm \( = 0.14 \) m, so radius \( r = \frac{0.14}{2} = 0.07 \) m.
Speed of water flow \( = 2 \) m per second.
The volume of water flowing per second is the area of the pipe's mouth multiplied by the speed.
Area of pipe's mouth \( A = \pi r^2 = \frac{22}{7} \times (0.07)^2 \)
\[ A = \frac{22}{7} \times 0.0049 = 22 \times 0.0007 = 0.0154 \, \text{m}^2 \] Volume of water flowing per second \( = A \times \text{speed} \)
\[ V_{\text{flow/sec}} = 0.0154 \times 2 = 0.0308 \, \text{m}^3/\text{sec} \] **Time taken to fill the pool:**
\[ \text{Time} = \frac{V_{\text{pool}}}{V_{\text{flow/sec}}} = \frac{9240}{0.0308} = 300000 \, \text{seconds} \] Now, convert seconds to hours.
\( 1 \) hour \( = 60 \) minutes \( = 60 \times 60 = 3600 \) seconds.
\[ \text{Time in hours} = \frac{300000}{3600} = \frac{3000}{36} = \frac{250}{3} = 83 \frac{1}{3} \, \text{hours} \] This problem shows how rates of flow relate to volumes and time.
In simple words: First, calculate the total amount of water the swimming pool can hold. Then, figure out how much water flows out of the pipe every second. Finally, divide the pool's total volume by the amount of water flowing per second to find the total time needed, then convert that time into hours.

🎯 Exam Tip: This is a classic rate problem. Always ensure all units are consistent (e.g., meters and seconds) before performing calculations. Convert the final time into the requested unit (hours) to complete the answer.

Question 16. What length of solid cylinder 2 cm in diameter must be taken to be cast into a hollow cylinder of external diameter 12 cm, 0.25 cm thick, 15 cm long?
Answer:
The volume of the material remains the same when it is reshaped. We need to find the volume of the hollow cylinder first, which is the volume of the metal used. This volume will then be equal to the volume of the solid cylinder.
**For the hollow cylinder:**
External diameter \( = 12 \) cm, so External Radius \( R = \frac{12}{2} = 6 \) cm.
Thickness of sheet \( = 0.25 \) cm.
Inner Radius \( r = R - \text{thickness} = 6 - 0.25 = 5.75 \) cm.
Length (height) \( h_{\text{hollow}} = 15 \) cm.
Volume of metal used \( V_{\text{metal}} = \pi h_{\text{hollow}} (R^2 - r^2) \).
\[ V_{\text{metal}} = \frac{22}{7} \times 15 \times (6^2 - 5.75^2) \] \[ = \frac{330}{7} \times (36 - 33.0625) \] \[ = \frac{330}{7} \times 2.9375 \, \text{cm}^3 \] \[ = \frac{330}{7} \times \frac{47}{16} = \frac{15510}{112} \approx 138.482 \, \text{cm}^3 \] **For the solid cylinder:**
Diameter \( = 2 \) cm, so Radius \( r_{\text{solid}} = \frac{2}{2} = 1 \) cm.
Let the length (height) of the solid cylinder be \( L \).
Volume of solid cylinder \( V_{\text{solid}} = \pi r_{\text{solid}}^2 L \).
\[ V_{\text{solid}} = \frac{22}{7} \times 1^2 \times L = \frac{22}{7} L \, \text{cm}^3 \] Since \( V_{\text{solid}} = V_{\text{metal}} \):
\[ \frac{22}{7} L = \frac{330}{7} \times 2.9375 \] \[ L = \frac{330}{22} \times 2.9375 \] \[ L = 15 \times 2.9375 \] \[ L = 44.0625 \, \text{cm} \] This demonstrates volume conservation during metalworking.
In simple words: First, calculate the amount of metal in the hollow cylinder by finding the volume of the outer cylinder and subtracting the volume of the inner empty space. Since this metal is reshaped into a solid cylinder, its volume stays the same. Finally, use the solid cylinder's radius and the metal's total volume to find how long the solid cylinder must be.

🎯 Exam Tip: Remember to calculate the volume of the actual material (metal) in the hollow cylinder, not just the overall external volume. This involves subtracting the internal empty space. Ensure consistent units throughout the problem.

Question 17. A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm, and its length is 21 cm. The metal everywhere is 0.4 cm thick. Calculate the volume of the metal to 1 place of decimal. (Take \( \pi=\frac{22}{7} \))
Answer:
To find the volume of metal in a hollow cylinder (tube), we use the formula \( V = \pi h (R^2 - r^2) \), where \( R \) is the external radius, \( r \) is the internal radius, and \( h \) is the length.
Length of the tube \( h = 21 \) cm.
Internal diameter \( = 11.2 \) cm, so Internal Radius \( r = \frac{11.2}{2} = 5.6 \) cm.
Thickness of metal \( = 0.4 \) cm.
External Radius \( R = \text{internal radius} + \text{thickness} = 5.6 + 0.4 = 6.0 \) cm.
Now, calculate the volume of the metal used:
\[ V_{\text{metal}} = \pi h (R^2 - r^2) \] \[ = \frac{22}{7} \times 21 \times (6.0^2 - 5.6^2) \] \[ = 22 \times 3 \times (36 - 31.36) \] \[ = 66 \times 4.64 \] \[ = 306.24 \, \text{cm}^3 \] Correct to one decimal place, the volume of metal is \( 306.2 \) cm³. This is how the actual material in a hollow structure is measured.
In simple words: First, find the inner and outer radius of the tube by using the given internal diameter and metal thickness. Then, use the formula for the volume of a hollow cylinder (outer volume minus inner volume) along with the tube's length to calculate the total amount of metal. Finally, round the answer to one decimal place.

🎯 Exam Tip: Be careful with decimal calculations, especially when squaring numbers like 5.6. Always double-check the subtraction \( R^2 - r^2 \) and the final rounding to the specified decimal place.

Question 18. Water flows along a pipe of radius 0.6 cm at 8 cm per second. This pipe is draining the water from a tank which holds 1000 litres of water when full. How long would it take to completely empty the tank?
Answer:
To find the time taken, we need the total volume of water in the tank and the rate at which water flows out of the pipe. Let's make sure all units are consistent, preferably in cm.
**Volume of water in the tank:**
Tank capacity \( = 1000 \) litres.
We know \( 1 \) litre \( = 1000 \) cm³, so \( 1000 \) litres \( = 1000 \times 1000 = 1000000 \) cm³.
**Rate of water flow from the pipe:**
Radius of pipe \( r = 0.6 \) cm.
Speed of water flow \( = 8 \) cm per second.
Volume of water flowing per second \( = (\text{Area of pipe's mouth}) \times (\text{speed}) \).
Area of pipe's mouth \( A = \pi r^2 = \frac{22}{7} \times (0.6)^2 \)
\[ A = \frac{22}{7} \times 0.36 = \frac{7.92}{7} \, \text{cm}^2 \] Volume of water flowing per second \( = \frac{7.92}{7} \times 8 = \frac{63.36}{7} \, \text{cm}^3/\text{sec} \] **Time taken to empty the tank:**
\[ \text{Time} = \frac{\text{Volume of water in tank}}{\text{Volume of water flowing per second}} \] \[ \text{Time} = \frac{1000000}{\frac{63.36}{7}} = \frac{1000000 \times 7}{63.36} \] \[ \text{Time} = \frac{7000000}{63.36} \approx 110479.80 \, \text{seconds} \] Now, convert seconds to hours.
\( 1 \) hour \( = 3600 \) seconds.
\[ \text{Time in hours} = \frac{110479.80}{3600} \approx 30.688 \, \text{hours} \] Rounding to two decimal places, it takes approximately \( 30.69 \) hours. This problem highlights the practical application of volume and flow rates.
In simple words: First, convert the tank's volume from liters to cubic centimeters. Then, calculate how much water flows out of the pipe each second by multiplying the pipe's cross-sectional area by the water's speed. Finally, divide the tank's total volume by the flow rate per second to find the total time, and convert that time into hours.

🎯 Exam Tip: Pay close attention to unit conversions (litres to cm³, seconds to hours) as these are common sources of error. Also, keep track of \( \pi \) and its numerical approximation, using a calculator for accuracy with decimals.

Question 19. A cylindrical bucket 28 cm in diameter; 72 cm high and full of water, is emptied into a rectangular tank 66 cm long, 28 cm wide. Find the height of the water-level in the tank. (Take \( \pi=\frac{22}{7} \))
Answer:
When water from one container is poured into another, the volume of water remains constant. We first find the volume of water in the cylindrical bucket, then use this volume to find the height of the water in the rectangular tank.
**Volume of water in the cylindrical bucket:**
Diameter of bucket \( = 28 \) cm, so Radius \( r = \frac{28}{2} = 14 \) cm.
Height of bucket \( h = 72 \) cm.
Volume of water \( V_{\text{bucket}} = \pi r^2 h \).
\[ V_{\text{bucket}} = \frac{22}{7} \times 14^2 \times 72 \] \[ = \frac{22}{7} \times 196 \times 72 \] \[ = 22 \times 28 \times 72 \] \[ = 44352 \, \text{cm}^3 \] **Height of water in the rectangular tank:**
Length of tank \( l = 66 \) cm, Breadth of tank \( b = 28 \) cm.
Let the height of the water-level in the tank be \( H \).
Volume of water in tank \( V_{\text{tank}} = l \times b \times H \).
Since the volume of water is the same:
\[ 66 \times 28 \times H = 44352 \] \[ 1848 \times H = 44352 \] \[ H = \frac{44352}{1848} = 24 \, \text{cm} \] The water-level in the rectangular tank will be 24 cm high. This is a common problem in fluid transfer.
In simple words: First, calculate the total amount of water in the cylindrical bucket. Then, imagine pouring all that water into the rectangular tank. Since the amount of water is the same, divide the volume of water by the tank's length and width to find how high the water level will be in the rectangular tank.

🎯 Exam Tip: The principle of conservation of volume is key here. Ensure you correctly calculate the volume of the water in the first container and then use it as the volume in the second container to find the unknown dimension.

Question 20. There is some water in a cylindrical vessel of base diameter 14 cm. When an iron cube is entirely immersed in it, the height of the water rises by \( 8\frac{9}{14} \) cm. Find the length of the edge of the cube (Take \( \pi=\frac{22}{7} \))
Answer:
When an object is immersed in water, the volume of water displaced is equal to the volume of the object. In this case, the rise in water level in the cylindrical vessel indicates the volume of the iron cube.
**Volume of water displaced (which is the volume of the cube):**
Diameter of cylindrical vessel \( = 14 \) cm, so Radius \( r = \frac{14}{2} = 7 \) cm.
Height the water rises \( h_{\text{rise}} = 8\frac{9}{14} \) cm \( = \frac{8 \times 14 + 9}{14} = \frac{112 + 9}{14} = \frac{121}{14} \) cm.
The volume of this displaced water is the volume of a cylinder with radius \( r \) and height \( h_{\text{rise}} \).
Volume of water displaced \( V_{\text{displaced}} = \pi r^2 h_{\text{rise}} \).
\[ V_{\text{displaced}} = \frac{22}{7} \times 7^2 \times \frac{121}{14} \] \[ = \frac{22}{7} \times 49 \times \frac{121}{14} \] \[ = 22 \times 7 \times \frac{121}{14} \] \[ = 11 \times 7 \times 121 \] \[ = 1331 \, \text{cm}^3 \] **Length of the edge of the cube:**
The volume of the cube \( V_{\text{cube}} = (\text{edge})^3 \).
Since \( V_{\text{cube}} = V_{\text{displaced}} \):
\[ (\text{edge})^3 = 1331 \] To find the edge, we take the cube root of 1331.
\[ \text{edge} = \sqrt[3]{1331} \] We know that \( 11 \times 11 \times 11 = 1331 \).
So, the length of the edge of the cube \( = 11 \) cm. This principle is known as Archimedes' principle.
In simple words: When the iron cube is put into the water, the water level goes up. The extra volume of water that rises is exactly the same as the volume of the cube. We use the vessel's radius and how much the water rose to find this volume. Then, we find the side length of the cube by taking the cube root of that volume.

🎯 Exam Tip: Remember the principle of displacement: the volume of a submerged object equals the volume of the fluid it displaces. Also, be careful when calculating with mixed fractions; convert them to improper fractions first.

Question 21. If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one?
Answer:
Let's consider the original cylinder and the new cylinder after the change.
**Original Cylinder:**
Let the original radius be \( r \).
Let the height be \( h \).
Volume of the original cylinder \( V_{\text{original}} = \pi r^2 h \).
**Reduced Cylinder:**
The radius is halved, so the new radius \( r' = \frac{r}{2} \).
The height remains the same, so the new height is \( h \).
Volume of the reduced cylinder \( V_{\text{reduced}} = \pi (r')^2 h \).
\[ V_{\text{reduced}} = \pi \left(\frac{r}{2}\right)^2 h \] \[ = \pi \frac{r^2}{4} h \] \[ = \frac{1}{4} \pi r^2 h \] Now, we need to find the ratio of the volume of the reduced cylinder to the volume of the original cylinder.
\[ \text{Ratio} = \frac{V_{\text{reduced}}}{V_{\text{original}}} = \frac{\frac{1}{4} \pi r^2 h}{\pi r^2 h} \] \[ = \frac{1}{4} \] So, the ratio is \( 1:4 \). This shows that changes in dimensions can have a squared or cubed effect on area and volume.
In simple words: We start with a cylinder's volume using its original radius and height. When the radius is cut in half but the height stays the same, the new volume becomes one-fourth of the original volume. So, the new volume compared to the old volume is like 1 to 4.

🎯 Exam Tip: When dealing with ratios of volumes or areas after a dimensional change, always express the new dimensions in terms of the original ones. Then, apply the formula to both the original and new figures and simplify the ratio. Remember that radius is squared in the volume formula, so halving it will reduce the volume by a factor of 4.

Question 22. A rectangular sheet of paper 22 cm long and 12 cm broad can be formed into the curved surface of a cylinder in two ways. Find the difference between the volumes of the two cylinders which can be formed.
Answer:
A rectangular sheet can form a cylinder's curved surface in two ways: by rolling it along its length or along its breadth. The dimensions of the sheet become the circumference and height of the cylinder.
**Case 1: Rolled along its length (length becomes circumference)**
Circumference \( C_1 = 22 \) cm.
Height \( h_1 = 12 \) cm.
From \( C_1 = 2\pi r_1 \): \( 22 = 2 \times \frac{22}{7} \times r_1 \)
\[ r_1 = \frac{22 \times 7}{2 \times 22} = \frac{7}{2} = 3.5 \, \text{cm} \] Volume of Cylinder 1 \( V_1 = \pi r_1^2 h_1 \)
\[ V_1 = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 12 \] \[ = \frac{22}{7} \times \frac{49}{4} \times 12 \] \[ = 22 \times 7 \times 3 = 462 \, \text{cm}^3 \] **Case 2: Rolled along its breadth (breadth becomes circumference)**
Circumference \( C_2 = 12 \) cm.
Height \( h_2 = 22 \) cm.
From \( C_2 = 2\pi r_2 \): \( 12 = 2 \times \frac{22}{7} \times r_2 \)
\[ r_2 = \frac{12 \times 7}{2 \times 22} = \frac{84}{44} = \frac{21}{11} \, \text{cm} \] Volume of Cylinder 2 \( V_2 = \pi r_2^2 h_2 \)
\[ V_2 = \frac{22}{7} \times \left(\frac{21}{11}\right)^2 \times 22 \] \[ = \frac{22}{7} \times \frac{441}{121} \times 22 \] \[ = \frac{2 \times 441}{11} = \frac{882}{11} \approx 80.18 \, \text{cm}^3 \] *Self-correction: The provided solution has \( V_2 = 252 \) cm³ from \( \frac{22}{7} \times \frac{21}{11} \times \frac{21}{11} \times 22 \). Let's re-evaluate. \( \frac{22}{7} \times \frac{21}{11} \times \frac{21}{11} \times 22 = (22 \times \frac{21 \times 21}{11 \times 11}) \times \frac{22}{7} = (2 \times 21 \times 21 / 11) \times \frac{2}{7} = (42 \times 21 / 11) \times \frac{2}{7} \). This is incorrect. The original OCR has a value of 252 cm^3 for V2 which I need to match. Let me check the calculation for \( V_2 \) carefully.* \( V_2 = \pi r_2^2 h_2 = \frac{22}{7} \times \left(\frac{21}{11}\right)^2 \times 22 \) \( V_2 = \frac{22}{7} \times \frac{21 \times 21}{11 \times 11} \times 22 \) \( V_2 = \frac{22 \times 21 \times 21 \times 22}{7 \times 11 \times 11} \) \( V_2 = (22/11) \times (22/11) \times (21/7) \times 21 \) \( V_2 = 2 \times 2 \times 3 \times 21 \) \( V_2 = 4 \times 63 = 252 \, \text{cm}^3 \). *The OCR solution is correct. My mental simplification was off. I will use 252 cm³ for \( V_2 \).*
Difference between the volumes \( = V_1 - V_2 \)
\[ = 462 - 252 = 210 \, \text{cm}^3 \] This shows that how you roll the paper affects the final volume, even if the surface area is the same.
In simple words: A rectangular paper can be rolled into a cylinder in two ways. We calculate the volume for both cases by finding the radius and height for each. Then, we subtract the smaller volume from the larger one to find the difference.

🎯 Exam Tip: Clearly distinguish between the two cases of cylinder formation. Remember that the length of the sheet becomes the circumference of the base in one case, and the breadth becomes the circumference in the other. Be careful with fractional calculations.

Question 23. A 20 m deep well with diameter 7 m is dug up and the earth from digging is spread evenly to form a platform 22 m \( \times \) 14 m. Determine the height of the platform.
Answer:
The volume of earth dug out from the well will be equal to the volume of the earth used to form the platform. We need to calculate the volume of the well first.
**Volume of earth dug out from the well:**
Depth of well \( h_{\text{well}} = 20 \) m.
Diameter of well \( = 7 \) m, so Radius \( r = \frac{7}{2} \) m.
Volume of well \( V_{\text{well}} = \pi r^2 h_{\text{well}} \).
\[ V_{\text{well}} = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 20 \] \[ = \frac{22}{7} \times \frac{49}{4} \times 20 \] \[ = 22 \times \frac{7}{4} \times 20 \] \[ = 22 \times 7 \times 5 = 770 \, \text{m}^3 \] **Height of the platform:**
The platform is a cuboid with length \( l = 22 \) m and breadth \( b = 14 \) m.
Let the height of the platform be \( H \).
Volume of platform \( V_{\text{platform}} = l \times b \times H \).
Since \( V_{\text{platform}} = V_{\text{well}} \):
\[ 22 \times 14 \times H = 770 \] \[ 308 \times H = 770 \] \[ H = \frac{770}{308} = \frac{770 \div 77}{308 \div 77} = \frac{10}{4} = 2.5 \, \text{m} \] The height of the platform is 2.5 meters. This is a practical example of volume conservation and reshaping.
In simple words: First, calculate the volume of earth removed from the cylindrical well. Then, this same volume of earth is used to create a rectangular platform. Divide this volume by the length and width of the platform to find its height.

🎯 Exam Tip: The core idea is that the volume of the dug-out earth is conserved. Make sure to calculate this volume correctly and then use it with the dimensions of the platform to find the unknown height.

Question 24. A certain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one fifth of a litre? (Answer correct to the nearest 100 words) (Take \( \pi=\frac{22}{7} \))
Answer:
This problem involves calculating volumes and using proportions to find the total number of words that can be written. We need to convert all units to centimeters first.
**Volume of ink in one pen barrel:**
Length of pen barrel \( h = 7 \) cm.
Diameter \( = 5 \) mm \( = 0.5 \) cm, so Radius \( r = \frac{0.5}{2} = 0.25 \) cm \( = \frac{1}{4} \) cm.
Volume of ink \( V_{\text{barrel}} = \pi r^2 h \).
\[ V_{\text{barrel}} = \frac{22}{7} \times \left(\frac{1}{4}\right)^2 \times 7 \] \[ = \frac{22}{7} \times \frac{1}{16} \times 7 \] \[ = \frac{22}{16} = \frac{11}{8} \, \text{cm}^3 \] **Volume of ink in the bottle:**
Volume of bottle \( = \frac{1}{5} \) litre.
We know \( 1 \) litre \( = 1000 \) cm³, so \( \frac{1}{5} \) litre \( = \frac{1}{5} \times 1000 = 200 \) cm³.
**Total number of pen barrels that can be filled from the bottle:**
\[ \text{Number of barrels} = \frac{V_{\text{bottle}}}{V_{\text{barrel}}} = \frac{200}{\frac{11}{8}} \] \[ = 200 \times \frac{8}{11} = \frac{1600}{11} \] **Total number of words:**
One barrel writes 310 words.
Total words \( = \text{Number of barrels} \times 310 \).
\[ = \frac{1600}{11} \times 310 = \frac{496000}{11} \approx 45090.909 \, \text{words} \] Rounding to the nearest 100 words:
The number is 45090.909. The digit in the tens place is 9, so we round up the hundreds place.
Total words \( = 45100 \) words (approx). This problem combines volume calculation with real-world proportion.
In simple words: First, calculate the tiny amount of ink a single pen barrel holds. Then, find out how much ink is in the bottle (one-fifth of a liter, converted to cubic centimeters). Divide the bottle's ink volume by the pen's ink volume to see how many pens' worth of ink the bottle contains. Finally, multiply this number by the words one pen can write, and round the final answer to the nearest 100 words.

🎯 Exam Tip: Pay close attention to unit conversions (mm to cm, litres to cm³) and rounding instructions. For "nearest 100 words", look at the tens digit: if it's 50 or above, round up the hundreds digit; otherwise, round down.

Question 25. A closed rectangular box 40 cm long, 30 cm wide and 25 cm deep, has the same volume as that of a cylindrical tin of radius 17.5 cm. Calculate the height of the cylindrical tin correct to 1 decimal place. (Take \( \pi \) to be 3.14)
Answer:
Since the volumes are equal, we first calculate the volume of the rectangular box. Then, we use this volume and the radius of the cylindrical tin to find its height.
**Volume of the rectangular box:**
Length \( l = 40 \) cm, Width \( b = 30 \) cm, Depth \( h_{\text{box}} = 25 \) cm.
Volume of box \( V_{\text{box}} = l \times b \times h_{\text{box}} \).
\[ V_{\text{box}} = 40 \times 30 \times 25 \] \[ = 1200 \times 25 = 30000 \, \text{cm}^3 \] **Height of the cylindrical tin:**
Volume of cylindrical tin \( V_{\text{tin}} = V_{\text{box}} = 30000 \) cm³.
Radius of cylindrical tin \( r = 17.5 \) cm.
Let the height of the cylindrical tin be \( H \). We know \( V_{\text{tin}} = \pi r^2 H \).
So, \( H = \frac{V_{\text{tin}}}{\pi r^2} \).
Using \( \pi = 3.14 \):
\[ H = \frac{30000}{3.14 \times (17.5)^2} \] \[ H = \frac{30000}{3.14 \times 306.25} \] \[ H = \frac{30000}{961.625} \approx 31.197 \, \text{cm} \] Correct to 1 decimal place, the height of the cylindrical tin is \( 31.2 \) cm. This problem demonstrates volume equivalence between different shapes.
In simple words: First, find the volume of the rectangular box by multiplying its length, width, and depth. Since the cylindrical tin has the same volume, use this volume along with the tin's radius to calculate its height. Finally, round the answer to one decimal place as requested.

🎯 Exam Tip: Pay attention to the specified value of \( \pi \) (here, 3.14). Also, ensure your calculations for squares of decimals are accurate and that you round the final answer correctly to the required decimal place.

Question 26. Earth taken out on digging a circular tank of diameter 17.5 m is spread all around the tank uniformly to a width of 4 m, to form an embankment of height 2 m. Calculate the depth of the tank correct to two decimal places.
Answer:
The volume of earth dug out from the circular tank is equal to the volume of the earth forming the embankment.
**Dimensions of the embankment:**
Diameter of circular tank \( = 17.5 \) m, so Inner Radius of embankment \( r = \frac{17.5}{2} = 8.75 \) m.
Width of embankment \( = 4 \) m.
Outer Radius of embankment \( R = r + \text{width} = 8.75 + 4 = 12.75 \) m.
Height of embankment \( h_{\text{embankment}} = 2 \) m.
Volume of embankment \( V_{\text{embankment}} = \pi h_{\text{embankment}} (R^2 - r^2) \).
\[ V_{\text{embankment}} = \frac{22}{7} \times 2 \times (12.75^2 - 8.75^2) \] Using \( a^2 - b^2 = (a-b)(a+b) \):
\[ = \frac{44}{7} \times (12.75 - 8.75)(12.75 + 8.75) \] \[ = \frac{44}{7} \times 4 \times 21.5 \] \[ = \frac{44}{7} \times 86 = \frac{3784}{7} \, \text{m}^3 \] **Depth of the tank:**
Let the depth of the tank be \( D \). The volume of earth dug out from the tank \( V_{\text{tank}} = \pi r^2 D \).
The radius of the tank is the inner radius of the embankment, \( r = 8.75 \) m.
Since \( V_{\text{tank}} = V_{\text{embankment}} \):
\[ \pi r^2 D = \frac{3784}{7} \] \[ \frac{22}{7} \times (8.75)^2 \times D = \frac{3784}{7} \] \[ 22 \times (8.75)^2 \times D = 3784 \] \[ 22 \times 76.5625 \times D = 3784 \] \[ 1684.375 \times D = 3784 \] \[ D = \frac{3784}{1684.375} \approx 2.2464 \, \text{m} \] Correct to two decimal places, the depth of the tank is \( 2.25 \) m. This problem applies volume conservation to a real-world construction scenario.
In simple words: The volume of earth removed from the circular tank is used to create a raised ring (embankment) around it. First, calculate the volume of this embankment using its inner and outer radii and its height. This volume is the same as the earth dug from the tank. Then, use this volume and the tank's radius to calculate the tank's depth, rounding the final answer to two decimal places.

🎯 Exam Tip: Visualizing the embankment as a hollow cylinder helps. Use \( \pi (R^2 - r^2)h \) for its volume. Remember that the tank's radius is the inner radius of the embankment, not the outer. Always round the final answer as specified.

Question 27. Water is flowing at the rate of 7 metres per second through a circular pipe whose internal diameter is 2 cm into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the water level in \( \frac{1}{2} \) hour.
Answer:
We need to find the volume of water flowing into the tank in \( \frac{1}{2} \) hour and then use this volume to determine the rise in the water level in the cylindrical tank. All units should be consistent; let's use centimeters.
**Volume of water flowing through the pipe:**
Speed of waterflow \( = 7 \) m/second \( = 7 \times 100 = 700 \) cm/second.
Internal diameter of pipe \( = 2 \) cm, so Radius of pipe \( r_{\text{pipe}} = \frac{2}{2} = 1 \) cm.
Time \( = \frac{1}{2} \) hour \( = \frac{1}{2} \times 60 \times 60 = 1800 \) seconds.
The volume of water flowing through the pipe in \( 1800 \) seconds is the length of the water column (speed \( \times \) time) multiplied by the pipe's cross-sectional area.
Length of water column \( L = 700 \, \text{cm/s} \times 1800 \, \text{s} = 1260000 \) cm.
Volume of water flowed \( V_{\text{flow}} = \pi r_{\text{pipe}}^2 L \).
\[ V_{\text{flow}} = \frac{22}{7} \times 1^2 \times 1260000 \] \[ = 22 \times 180000 = 3960000 \, \text{cm}^3 \] **Increase in water level in the cylindrical tank:**
Radius of the base of the cylindrical tank \( r_{\text{tank}} = 40 \) cm.
Let the increase in water level (height) be \( H_{\text{increase}} \).
The volume of water in the tank due to this increase is \( \pi r_{\text{tank}}^2 H_{\text{increase}} \).
Since this volume is equal to the volume of water flowed in:
\[ \pi r_{\text{tank}}^2 H_{\text{increase}} = V_{\text{flow}} \] \[ \frac{22}{7} \times 40^2 \times H_{\text{increase}} = 3960000 \] \[ \frac{22}{7} \times 1600 \times H_{\text{increase}} = 3960000 \] \[ H_{\text{increase}} = \frac{3960000 \times 7}{22 \times 1600} \] \[ H_{\text{increase}} = \frac{27720000}{35200} = \frac{277200}{352} = 787.5 \, \text{cm} \] This is a significant rise in water level. This problem connects flow rate to volume accumulation.
In simple words: First, calculate the total volume of water that flows out of the pipe in half an hour, making sure all units are in centimeters. Then, imagine this volume of water filling up the cylindrical tank. Divide this volume by the area of the tank's base to find how much the water level will rise.

🎯 Exam Tip: This problem requires careful unit conversions (meters/second to cm/second, hours to seconds). The volume of water flowed is a cylinder whose base is the pipe's cross-section and whose height is the distance the water travels in the given time.

Question 28. Water flow through a cylindrical pipe of internal diameter 7 cm at 36 km/hr. Calculate time in minutes it would take to fill a cylindrical tank, the radius of whose base is 35 cm and height is 1 m.
Answer: First, we find the volume of water flowing from the pipe per minute. The internal radius of the pipe (\(r_1\)) is \( \frac{7}{2} \) cm or 0.035 m. The water flows at a speed of 36 km/hr, which is \( 36 \times \frac{1000}{60} = 600 \) meters per minute. The cross-sectional area of the pipe is \( \pi r_1^2 = \pi \times (0.035)^2 \text{ m}^2 = 0.001225 \pi \text{ m}^2 \). So, the volume of water flowing per minute is \( 0.001225 \pi \times 600 = 0.735 \pi \text{ m}^3 \).
Now, we find the volume of the cylindrical tank. Its radius (\(r_2\)) is 35 cm or 0.35 m, and its height (\(h_2\)) is 1 m. The volume of the tank is \( \pi r_2^2 h_2 = \pi \times (0.35)^2 \times 1 \text{ m}^3 = 0.1225 \pi \text{ m}^3 \).
To find the time needed to fill the tank, we divide the tank's volume by the volume of water flowing per minute.
Time \( = \frac{\text{Volume of tank}}{\text{Volume of water per minute}} = \frac{0.1225 \pi}{0.735 \pi} \)
\( = \frac{0.1225}{0.735} = 0.1666... \) minutes
\( \implies \) Time \( = 0.166 \text{ minutes} \)
In simple words: We first figured out how much water the pipe sends out each minute. Then we found out how much water the tank can hold. By dividing the tank's total volume by the water flow rate, we found the time it takes to fill the tank, which is a little less than one-fifth of a minute.

🎯 Exam Tip: Remember to convert all units to be consistent (e.g., meters and minutes) before starting calculations to avoid errors in your final answer.

Question 29. Find the number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer: First, we calculate the volume of one coin. The diameter of a coin is 1.5 cm, so its radius (\(r_c\)) is \( \frac{1.5}{2} = 0.75 \) cm. The thickness (height) of the coin (\(h_c\)) is 0.2 cm. The volume of one coin is \( \pi r_c^2 h_c = \pi \times (0.75)^2 \times 0.2 \text{ cm}^3 = 0.1125 \pi \text{ cm}^3 \).
Next, we calculate the volume of the target cylinder. Its diameter is 4.5 cm, so its radius (\(R_t\)) is \( \frac{4.5}{2} = 2.25 \) cm. Its height (\(H_t\)) is 10 cm. The volume of the target cylinder is \( \pi R_t^2 H_t = \pi \times (2.25)^2 \times 10 \text{ cm}^3 = 50.625 \pi \text{ cm}^3 \).
To find the number of coins needed, we divide the volume of the target cylinder by the volume of one coin.
Number of coins \( = \frac{\text{Volume of target cylinder}}{\text{Volume of one coin}} = \frac{50.625 \pi}{0.1125 \pi} \)
\( = \frac{50.625}{0.1125} = 450 \)
In simple words: We found out how much space one small coin takes up and how much space the bigger cylinder needs. By dividing the big cylinder's space by the small coin's space, we learned that 450 coins are needed to make the big cylinder. This shows how metal can be reshaped.

🎯 Exam Tip: When melting and recasting, the total volume of the material remains constant. Always calculate the volume of both the original objects and the new object to find the number or dimensions.

Question 30. If 1 cubic cm of cast iron weighs 21 g, then find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.
Answer: First, we determine the dimensions of the cast iron pipe. The length (height) of the pipe (\(h\)) is 1 m, which is 100 cm. The internal diameter (bore) is 3 cm, so the internal radius (\(r\)) is \( \frac{3}{2} = 1.5 \) cm. The metal thickness is 1 cm, so the external radius (\(R\)) is \( 1.5 + 1 = 2.5 \) cm.
Next, we calculate the volume of the iron metal in the pipe. This is a hollow cylinder, so its volume is given by \( \pi h (R^2 - r^2) \).
Volume \( = \frac{22}{7} \times 100 \times ( (2.5)^2 - (1.5)^2 ) \text{ cm}^3 \)
\( = \frac{2200}{7} \times (6.25 - 2.25) \text{ cm}^3 \)
\( = \frac{2200}{7} \times 4 \text{ cm}^3 \)
\( = \frac{8800}{7} \text{ cm}^3 \)
Since 1 cubic cm of cast iron weighs 21 g, we can find the total weight by multiplying the volume by the weight per cubic cm.
Total weight \( = \frac{8800}{7} \times 21 \text{ g} \)
\( = 8800 \times 3 \text{ g} = 26400 \text{ g} \)
Converting this to kilograms, we get \( \frac{26400}{1000} = 26.4 \text{ kg} \). The weight of the pipe is 26.4 kg.
In simple words: We found the total space taken by the iron in the hollow pipe. Because we know how much 1 cubic cm of iron weighs, we could then calculate the total weight of the pipe. It's like finding the volume of a donut and then knowing its weight based on the material.

🎯 Exam Tip: For hollow cylinders, remember to calculate the volume of the metal using the difference between the squares of the external and internal radii, multiplied by \(\pi\) and height. Always convert units (like grams to kilograms) if required in the final answer.