Volume & Surface area of Solids

Q1: Find the volume of the cylinder in which:

i. height = 42 cm and base radius = 10 cm.

Step 1: Write given data:
Height \(h = 42\, cm\), Radius \(r = 10\, cm\)
Step 2: Volume \(V\) of a cylinder formula: \[ V = \pi r^2 h \]Step 3: Substitute values (use \(\pi = \frac{22}{7}\)): \[ V = \frac{22}{7} \times 10^2 \times 42 = \frac{22}{7} \times 100 \times 42 \]Step 4: Simplify: \[ V = 22 \times 100 \times 6 = 22 \times 600 = 13200\, cm^3 \]Answer: Volume of the cylinder = \(9240\, cm^3\)

ii. diameter = 70 cm and height = 20 cm

Step 1: Write given data:
Diameter \(d = 70\, cm\), so Radius \(r = \frac{d}{2} = \frac{70}{2} = 35\, cm\), Height \(h = 20\, cm\)
Step 2: Volume formula: \[ V = \pi r^2 h \]Step 3: Substitute values: \[ V = \frac{22}{7} \times 35^2 \times 20 = \frac{22}{7} \times 1225 \times 20 \]Step 4: Simplify: \[ V = 22 \times 175 \times 20 = 22 \times 3500 = 77000\, cm^3 \]Answer: Volume of the cylinder = \(77000\, cm^3\)

Q2: Find the curved surface area and the total surface area of the cylinder for which:

i. \(h = 30\, cm\) and \(r = 14\, cm\)

Step 1: Given height \(h = 30\, cm\), radius \(r = 14\, cm\)
Step 2: Curved surface area (CSA) formula: \[ CSA = 2 \pi r h \]Step 3: Total surface area (TSA) formula: \[ TSA = 2 \pi r (h + r) \]Step 4: Substitute values (use \(\pi = \frac{22}{7}\)): \[ CSA = 2 \times \frac{22}{7} \times 14 \times 30 = 2 \times \frac{22}{7} \times 14 \times 30 \] Simplify: \[ CSA = 2 \times 22 \times 2 \times 30 = 2640\, cm^2 \\ TSA = 2 \times \frac{22}{7} \times 14 \times (30 + 14) = 2 \times \frac{22}{7} \times 14 \times 44 \] Simplify: \[ TSA = 2 \times 22 \times 2 \times 44 = 3872\, cm^2 \]Answer: Curved Surface Area = \(2640\, cm^2\)
Total Surface Area = \(3872\, cm^2\)

ii. \(h = 32\, cm\) and \(r = 10.5\, cm\)

Step 1: Given height \(h = 32\, cm\), radius \(r = 10.5\, cm\)
Step 2: Calculate CSA: \[ CSA = 2 \pi r h = 2 \times \frac{22}{7} \times 10.5 \times 32 \] Simplify: \[ CSA = 2 \times 22 \times 1.5 \times 32 = 2112\, cm^2 \]Step 3: Calculate TSA: \[ TSA = 2 \pi r (h + r) = 2 \times \frac{22}{7} \times 10.5 \times (32 + 10.5) \\ = 2 \times \frac{22}{7} \times 10.5 \times 42.5 \\ TSA = 2805\, cm^2 \]Answer: Curved Surface Area = \(2112\, cm^2\)
Total Surface Area = \(2805\, cm^2\)

iii. \(h = 35\, cm\) and \(r = 21\, cm\)

Step 1: Given height \(h = 35\, cm\), radius \(r = 21\, cm\)
Step 2: Calculate CSA: \[ CSA = 2 \pi r h = 2 \times \frac{22}{7} \times 21 \times 35 \] Simplify: \[ CSA = 2 \times 22 \times 3 \times 35 = 4620\, cm^2 \]Step 3: Calculate TSA: \[ TSA = 2 \pi r (h + r) = 2 \times \frac{22}{7} \times 21 \times (35 + 21) \\ = 2 \times \frac{22}{7} \times 21 \times 56 \] Simplify: \[ TSA = 2 \times 22 \times 3 \times 56 = 7392\, cm^2 \]Answer: Curved Surface Area = \(4620\, cm^2\)
Total Surface Area = \(7392\, cm^2\)

iv. \(h = 5\, m\) and \(r = 35\, cm\)

Step 1: Convert units to same unit (convert height to cm):
\(5\, m = 500\, cm\)
Radius \(r = 35\, cm\)
Step 2: Calculate CSA: \[ CSA = 2 \pi r h = 2 \times \frac{22}{7} \times 35 \times 500 \] Simplify: \[ CSA = 2 \times 22 \times 5 \times 500 = 110000\, cm^2 \]Step 3: Calculate TSA: \[ TSA = 2 \pi r (h + r) = 2 \times \frac{22}{7} \times 35 \times (500 + 35) \\ = 2 \times \frac{22}{7} \times 35 \times 535 \] Simplify: \[ TSA = 2 \times 22 \times 5 \times 535 = 117700\, cm^2 \]Answer: Curved Surface Area = \(110000\, cm^2\)
Total Surface Area = \(117700\, cm^2\)

Q3: Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm, if the material of the cylinder weighs 5 grams per cm³.

Step 1: Write the given data:
Radius \(r = 10.5\, cm\), Height \(h = 60\, cm\), Density (weight per unit volume) \(= 5\, g/cm^3\)
Step 2: Calculate volume \(V\) of the cylinder: \[ V = \pi r^2 h \] Substitute \(\pi = \frac{22}{7}\): \[ V = \frac{22}{7} \times (10.5)^2 \times 60 \]Step 3: Simplify the volume: \[ (10.5)^2 = 110.25 \\ V = \frac{22}{7} \times 110.25 \times 60 \\ = \frac{22}{7} \times 6615 = 22 \times 945 = 20790\, cm^3 \]Step 4: Calculate weight of the cylinder: \[ \text{Weight} = \text{Volume} \times \text{Density} = 20790 \times 5 = 103950\, grams \] Convert to kilograms: \[ 103950\, grams = \frac{103950}{1000} = 103.95\, kg \]Answer: Weight of the cylinder = \(103.95\, kg\)

Q4: A cylindrical tank has a capacity of 6160 m³. Find its depth, if its radius is 14 m. Also find the cost of painting its curved surface at ₹30 per m².

Step 1: Write the given data:
Volume \(V = 6160\, m^3\), Radius \(r = 14\, m\), Cost of painting = ₹30 per \(m^2\)
Step 2: Volume of cylinder formula: \[ V = \pi r^2 h \] Here, \(h\) is the depth (height) of the tank.
Step 3: Solve for \(h\): \[ h = \frac{V}{\pi r^2} = \frac{6160}{\pi \times 14^2} \]Substitute \(\pi = \frac{22}{7}\): \[ h = \frac{6160}{\frac{22}{7} \times 196} = \frac{6160 \times 7}{22 \times 196} \]Simplify denominator: \[ 22 \times 196 = 4312 \]Calculate \(h\): \[ h = \frac{43120}{4312} = 10\, m \]Step 4: Calculate curved surface area (CSA) of the cylinder: \[ CSA = 2 \pi r h = 2 \times \frac{22}{7} \times 14 \times 10 = 2 \times 22 \times 2 \times 10 = 880\, m^2 \]Step 5: Calculate cost of painting: \[ \text{Cost} = \text{CSA} \times \text{rate} = 880 \times 30 = ₹26400 \]Answer: Depth of the tank = \(10\, m\)
Cost of painting the curved surface = ₹\(26400\)

Q5: The curved surface area of a cylinder is 4400 cm² and the circumference of its base is 110 cm. Find the height and the volume of the cylinder.

Step 1: Write the given data:
Curved surface area (CSA) = 4400 cm²
Circumference of base \(C = 110\, cm\)
Step 2: Recall formulas:
Circumference of base: \[ C = 2 \pi r \] Curved surface area: \[ CSA = 2 \pi r h \]Step 3: Find radius \(r\) from circumference: \[ 2 \pi r = 110 \\ \Rightarrow r = \frac{110}{2 \pi} \] Substitute \(\pi = \frac{22}{7}\): \[ r = \frac{110}{2 \times \frac{22}{7}} = \frac{110 \times 7}{44} = \frac{770}{44} = 17.5\, cm \]Step 4: Find height \(h\) using CSA formula: \[ CSA = 2 \pi r h \\ 4400 = 2 \times \frac{22}{7} \times 17.5 \times h \] Simplify: \[ 4400 = 2 \times \frac{22}{7} \times 17.5 \times h = 2 \times \frac{22}{7} \times \frac{35}{2} \times h = \frac{22}{7} \times 35 \times h = 110 \times h \] Therefore, \[ h = \frac{4400}{110} = 40\, cm \]Step 5: Calculate volume \(V\): \[ V = \pi r^2 h = \frac{22}{7} \times (17.5)^2 \times 40 \\ (17.5)^2 = 306.25 \\ V = \frac{22}{7} \times 306.25 \times 40 = \frac{22}{7} \times 12250 = 22 \times 1750 = 38500\, cm^3 \]Answer: Height of the cylinder = \(40\, cm\)
Volume of the cylinder = \(38500\, cm^3\)

Q6: The total surface area of a solid cylinder is 462 cm² and its curved surface is one-third of its total surface area. Find the volume of the cylinder.

Step 1: Write given data:
Total Surface Area (TSA) = 462 cm²
Curved Surface Area (CSA) = \(\frac{1}{3}\) of TSA = \(\frac{1}{3} \times 462 = 154\, cm^2\)
Step 2: Recall formulas: \[ TSA = 2 \pi r (h + r) \\ CSA = 2 \pi r h \]Step 3: Using CSA and TSA: \[ CSA = \frac{1}{3} TSA \\ \Rightarrow 2 \pi r h = \frac{1}{3} \times 2 \pi r (h + r) \]Simplify: \[ 2 \pi r h = \frac{2 \pi r}{3} (h + r) \\ 3 \times 2 \pi r h = 2 \pi r (h + r) \\ 6 \pi r h = 2 \pi r h + 2 \pi r^2 \\ 6 \pi r h – 2 \pi r h = 2 \pi r^2 \\ 4 \pi r h = 2 \pi r^2 \] Divide both sides by \(2 \pi r\): \[ 2 h = r \\ \Rightarrow h = \frac{r}{2} \]Step 4: Use the TSA formula to find \(r\): \[ TSA = 2 \pi r (h + r) = 462 \] Substitute \(h = \frac{r}{2}\): \[ 462 = 2 \pi r \left(\frac{r}{2} + r\right) = 2 \pi r \times \frac{3r}{2} = 3 \pi r^2 \]Substitute \(\pi = \frac{22}{7}\): \[ 462 = 3 \times \frac{22}{7} \times r^2 = \frac{66}{7} r^2 \] Multiply both sides by 7: \[ 462 \times 7 = 66 r^2 \\ 3234 = 66 r^2 \\ r^2 = \frac{3234}{66} = 49 \\ r = \sqrt{49} = 7\, cm \]Step 5: Find height \(h\): \[ h = \frac{r}{2} = \frac{7}{2} = 3.5\, cm \]Step 6: Find volume \(V\) of the cylinder: \[ V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 3.5 = \frac{22}{7} \times 49 \times 3.5 \] Simplify: \[ V = 22 \times 7 \times 3.5 = 22 \times 24.5 = 539\, cm^3 \]Answer: Volume of the cylinder = \(539\, cm^3\)

Q7: The sum of the radius of the base and the height of a solid cylinder is 37 m. If the total surface area of the cylinder is 1628 m², find its volume.

Step 1: Let the radius of the base be \(r\) m and the height be \(h\) m.
Given: \[ r + h = 37 \quad \Rightarrow \quad h = 37 – r \]Step 2: Total surface area (TSA) formula: \[ TSA = 2 \pi r (h + r) = 1628 \]Substitute \(h = 37 – r\): \[ 2 \pi r (37 – r + r) = 1628 \\ 2 \pi r \times 37 = 1628 \\ 74 \pi r = 1628 \]Substitute \(\pi = \frac{22}{7}\): \[ 74 \times \frac{22}{7} \times r = 1628 \\ \frac{1628 \times 7}{74 \times 22} = r \] Calculate denominator: \[ 74 \times 22 = 1628 \] Calculate numerator: \[ 1628 \times 7 = 11396 \] Therefore, \[ r = \frac{11396}{1628} = 7\, m \]Step 3: Find height \(h\): \[ h = 37 – r = 37 – 7 = 30\, m \]Step 4: Find volume \(V\): \[ V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 30 = \frac{22}{7} \times 49 \times 30 \\ = 22 \times 7 \times 30 = 4620\, m^3 \]Answer: Volume of the cylinder = \(4620\, m^3\)

Q8: The total surface area of a circular cylinder is 1320 cm² and its radius is 10 cm. Find the height of the cylinder.

Step 1: Write the given data:
Total Surface Area (TSA) = 1320 cm²
Radius \(r = 10\, cm\)
Step 2: Use the formula for total surface area of a cylinder: \[ TSA = 2 \pi r (h + r) \]Substitute values: \[ 1320 = 2 \times \frac{22}{7} \times 10 \times (h + 10) \\ 1320 = \frac{440}{7} (h + 10) \]Step 3: Solve for \(h + 10\):
Multiply both sides by 7: \[ 1320 \times 7 = 440 (h + 10) \\ 9240 = 440 (h + 10) \] Divide both sides by 440: \[ h + 10 = \frac{9240}{440} = 21 \\ h = 21 – 10 = 11\, cm \]Answer: Height of the cylinder = \(11\, cm\)

Q9: A swimming pool 70 m × 44 m × 3 m is filled by water issuing from a pipe of diameter 35 cm at 6 m per second. How many hours does it take to fill the pool?

Step 1: Write the given data:
Length of pool \(L = 70\, m\)
Breadth of pool \(B = 44\, m\)
Depth of pool \(H = 3\, m\)
Diameter of pipe \(d = 35\, cm = 0.35\, m\)
Speed of water \(v = 6\, m/s\)
Step 2: Calculate volume of swimming pool: \[ V_{pool} = L \times B \times H = 70 \times 44 \times 3 = 9240\, m^3 \]Step 3: Calculate cross-sectional area of pipe:
Radius \(r = \frac{d}{2} = \frac{0.35}{2} = 0.175\, m\) \[ A = \pi r^2 = \frac{22}{7} \times (0.175)^2 = \frac{22}{7} \times 0.030625 = 0.09625\, m^2 \]Step 4: Calculate volume flow rate of water: \[ Q = A \times v = 0.09625 \times 6 = 0.5775\, m^3/s \]Step 5: Calculate time taken to fill the pool: \[ t = \frac{V_{pool}}{Q} = \frac{9240}{0.5775} = 16000\, seconds \]Step 6: Convert time to hours: \[ t = \frac{16000}{3600} = \frac{40}{9} = 4 \frac{4}{9}\, hours \]Answer: Time taken to fill the pool = \(4 \frac{4}{9}\) hours

Q10: 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 × 14 m. Determine the height of the platform.

Step 1: Write the given data:
Depth of well \(h = 20\, m\)
Diameter of well \(d = 7\, m\), so radius \(r = \frac{7}{2} = 3.5\, m\)
Platform dimensions: length \(L = 22\, m\), breadth \(B = 14\, m\)
Step 2: Calculate volume of earth dug (volume of the well): \[ V_{well} = \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 20 \\ = \frac{22}{7} \times 12.25 \times 20 = \frac{22}{7} \times 245 = 22 \times 35 = 770\, m^3 \]Step 3: Let height of platform be \(H\). Volume of earth spread to form platform is: \[ V_{platform} = L \times B \times H = 22 \times 14 \times H = 308 H \]Step 4: Equate volume of earth dug to volume of platform: \[ 770 = 308 H \\ \Rightarrow H = \frac{770}{308} = 2.5\, m \]Answer: Height of the platform = \(2.5\, m\)

Q11: A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.

Step 1: Write the given data:
Diameter of well \(d = 10\, m\), radius \(r = \frac{10}{2} = 5\, m\)
Depth of well \(h = 8.4\, m\)
Width of embankment \(w = 7.5\, m\)
Step 2: Calculate volume of earth dug (volume of well): \[ V_{well} = \pi r^2 h = \frac{22}{7} \times 5^2 \times 8.4 = \frac{22}{7} \times 25 \times 8.4 \\ = \frac{22}{7} \times 210 = 22 \times 30 = 660\, m^3 \]Step 3: Calculate outer radius of embankment: \[ R = r + w = 5 + 7.5 = 12.5\, m \]Step 4: Let height of embankment be \(H\). Volume of embankment is volume of larger cylinder minus volume of well: \[ V_{embankment} = \pi H (R^2 – r^2) = \frac{22}{7} \times H \times (12.5^2 – 5^2) \\ = \frac{22}{7} \times H \times (156.25 – 25) = \frac{22}{7} \times H \times 131.25 \\ = \frac{22}{7} \times 131.25 \times H = 412.5 H \]Step 5: Equate volume of earth dug to volume of embankment: \[ 660 = 412.5 H \\ \Rightarrow H = \frac{660}{412.5} = 1.6\, m \]Answer: Height of the embankment = \(1.6\, m\)