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Question 1: Find the curved surface area of a cone with height $15 \text{ cm}$ and base diameter $16 \text{ cm}$.
Solution: $r = 8 \text{ cm}, h = 15 \text{ cm}$. Slant height $l = \sqrt{r^2 + h^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ cm}$. CSA $= \pi rl = \pi(8)(17) = 136\pi \text{ cm}^2$.
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Question 2: Find the total surface area of a hemisphere of radius $10 \text{ cm}$. (Use $\pi = 3.14$)
Solution: TSA of hemisphere $= 3\pi r^2 = 3 \times 3.14 \times 10^2 = 3 \times 3.14 \times 100 = 942 \text{ cm}^2$.
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Question 3: The radii of two cylinders are in the ratio $2:3$ and their heights are in the ratio $5:3$. Calculate the ratio of their volumes.
Solution: $V_1/V_2 = (\pi r_1^2 h_1) / (\pi r_2^2 h_2) = (r_1/r_2)^2 \times (h_1/h_2) = (2/3)^2 \times (5/3) = 4/9 \times 5/3 = 20/27$. Ratio is $20:27$.
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Question 4: If the volume of a sphere is numerically equal to its surface area, find its diameter.
Solution: $\frac{4}{3}\pi r^3 = 4\pi r^2 \Rightarrow \frac{r}{3} = 1 \Rightarrow r = 3$. Diameter $= 2r = 6$ units.
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Question 5: A cuboidal water tank is $6 \text{ m}$ long, $5 \text{ m}$ wide and $4.5 \text{ m}$ deep. How many litres of water can it hold? ($1 \text{ m}^3 = 1000 \text{ L}$)
Solution: Volume $= l \times b \times h = 6 \times 5 \times 4.5 = 135 \text{ m}^3$. Capacity $= 135 \times 1000 = 135000$ litres.
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Question 6: Find the height of a cylinder whose volume is $1.54 \text{ m}^3$ and diameter of base is $140 \text{ cm}$.
Solution: $d = 140 \text{ cm} = 1.4 \text{ m}, r = 0.7 \text{ m}$. $V = \pi r^2 h \Rightarrow 1.54 = \frac{22}{7} \times (0.7)^2 \times h \Rightarrow 1.54 = 22 \times 0.1 \times 0.7 \times h \Rightarrow 1.54 = 1.54h \Rightarrow h = 1 \text{ m}$.
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Question 7: A conical pit of top diameter $3.5 \text{ m}$ is $12 \text{ m}$ deep. What is its capacity in kilolitres?
Solution: $r = 1.75 \text{ m}, h = 12 \text{ m}$. $V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12 = 38.5 \text{ m}^3$. $1 \text{ m}^3 = 1 \text{ kL}$. Capacity $= 38.5 \text{ kL}$.
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Question 8: The radius of a spherical balloon increases from $7 \text{ cm}$ to $14 \text{ cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution: Ratio $= (4\pi r_1^2) / (4\pi r_2^2) = (r_1/r_2)^2 = (7/14)^2 = (1/2)^2 = 1/4$. Ratio is $1:4$.
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Question 9: A river $3 \text{ m}$ deep and $40 \text{ m}$ wide is flowing at the rate of $2 \text{ km/h}$. How much water will fall into the sea in a minute?
Solution: Rate $= 2000 \text{ m/60 min} = 100/3 \text{ m/min}$. Volume per min $= \text{Area} \times \text{Rate} = (3 \times 40) \times (100/3) = 120 \times 100/3 = 4000 \text{ m}^3$.
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Question 10: Find the cost of digging a cuboidal pit $8 \text{ m}$ long, $6 \text{ m}$ broad and $3 \text{ m}$ deep at the rate of ₹ $30$ per $\text{m}^3$.
Solution: Volume $= 8 \times 6 \times 3 = 144 \text{ m}^3$. Cost $= 144 \times 30 = ₹ 4320$.