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Question 1: Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour?
Solution:
Step 1: Radius of pipe $r = 1$ cm. Rate of flow $= 80$ cm/s. Time $t = 30$ min $= 1800$ s.
Step 2: Volume of water flowing in = Area of cross-section $\times$ Rate $\times$ Time $= \pi (1)^2 \times 80 \times 1800 = 144000\pi$ cm$^3$.
Step 3: Let rise in water level be $h$. Radius of tank $R = 40$ cm.
Step 4: Volume in tank $= \pi R^2 h = \pi (40)^2 h = 1600\pi h$.
Step 5: $1600\pi h = 144000\pi \Rightarrow h = \frac{144000}{1600} = 90$ cm.
Answer: 90 cm. -
Question 2: A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Step 1: Volume of earth dug $= \pi r^2 h = \pi (1.5)^2 (14) = 31.5\pi$ m$^3$.
Step 2: Embankment is a hollow cylinder. Inner radius $r_1 = 1.5$ m. Width $= 4$ m. Outer radius $r_2 = 1.5 + 4 = 5.5$ m.
Step 3: Area of base of embankment $= \pi (r_2^2 - r_1^2) = \pi (5.5^2 - 1.5^2) = \pi (30.25 - 2.25) = 28\pi$ m$^2$.
Step 4: Volume $= \text{Area} \times \text{Height} \Rightarrow 28\pi H = 31.5\pi \Rightarrow H = \frac{31.5}{28} = 1.125$ m.
Answer: 1.125 m. -
Question 3: A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Step 1: Pipe radius $r = 10$ cm $= 0.1$ m. Rate $= 3000$ m/h. Tank radius $R = 5$ m, Depth $H = 2$ m.
Step 2: Volume of tank $= \pi R^2 H = \pi (5)^2 (2) = 50\pi$ m$^3$.
Step 3: Volume flowing per hour $= \pi r^2 \times \text{Rate} = \pi (0.1)^2 (3000) = 30\pi$ m$^3$/h.
Step 4: Time $= \frac{\text{Volume of Tank}}{\text{Rate of Flow}} = \frac{50\pi}{30\pi} = \frac{5}{3}$ hours $= 1$ hour 40 minutes.
Answer: 100 minutes. -
Question 4: A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the solid.
Solution:
Step 1: Radius $r = 3.5$ cm. Height of cylinder $h = 19 - 3.5 - 3.5 = 12$ cm.
Step 2: Volume $= \text{Vol(Cyl)} + 2 \times \text{Vol(Hemi)} = \pi r^2 h + \frac{4}{3}\pi r^3 = \pi (3.5)^2 (12) + \frac{4}{3}\pi (3.5)^3 = 147\pi + 57.16\pi = 204.16\pi \approx 641.66$ cm$^3$.
Step 3: TSA $= \text{CSA(Cyl)} + 2 \times \text{CSA(Hemi)} = 2\pi rh + 4\pi r^2 = 2\pi r(h+2r) = 2 \times \frac{22}{7} \times 3.5 (12 + 7) = 22 \times 19 = 418$ cm$^2$.
Answer: Volume $\approx 641.66$ cm$^3$, TSA $= 418$ cm$^2$. -
Question 5: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:
Step 1: $h=10$ cm, $r=3.5$ cm.
Step 2: TSA $= \text{CSA(Cylinder)} + 2 \times \text{CSA(Hemisphere)}$.
Step 3: TSA $= 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r)$.
Step 4: TSA $= 2 \times \frac{22}{7} \times 3.5 (10 + 7) = 22 \times 17 = 374$ cm$^2$.
Answer: 374 cm$^2$. -
Question 6: A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?
Solution:
Step 1: $r = 4.5$ m, $h = 3.5$ m. Volume $= \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (4.5)^2 \times 3.5 = 74.25$ m$^3$.
Step 2: Slant height $l = \sqrt{r^2 + h^2} = \sqrt{4.5^2 + 3.5^2} = \sqrt{20.25 + 12.25} = \sqrt{32.5} \approx 5.7$ m.
Step 3: Canvas area $= \pi rl = \frac{22}{7} \times 4.5 \times 5.7 \approx 80.61$ m$^2$.
Answer: Volume $\approx 74.25$ m$^3$, Canvas $\approx 80.61$ m$^2$. -
Question 7: A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Step 1: Vol(Cylinder) = Vol(Cone). $\pi (18)^2 (32) = \frac{1}{3} \pi r^2 (24)$.
Step 2: $324 \times 32 = 8 r^2 \Rightarrow r^2 = \frac{324 \times 32}{8} = 324 \times 4 = 1296$.
Step 3: $r = \sqrt{1296} = 36$ cm.
Step 4: Slant height $l = \sqrt{r^2 + h^2} = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872} = 12\sqrt{13}$ cm.
Answer: Radius 36 cm, Slant height $12\sqrt{13}$ cm. -
Question 8: Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Step 1: Volume of water in 30 mins $= \text{Width} \times \text{Depth} \times \text{Length}$.
Step 2: Length $= 10$ km/h $\times 0.5$ h $= 5$ km $= 5000$ m.
Step 3: Volume $= 6 \times 1.5 \times 5000 = 45000$ m$^3$.
Step 4: Area $\times$ Height $= \text{Volume}$. Height $= 8$ cm $= 0.08$ m.
Step 5: Area $= \frac{45000}{0.08} = 562500$ m$^2$.
Answer: 56.25 hectares. -
Question 9: A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of ₹ 2 per square metre, if the radius of the base is 14 metres.
Solution:
Step 1: Cylinder: $r=14, h_1=3$. Cone: $r=14$, height above ground 13.5 $\Rightarrow$ cone height $h_2 = 13.5 - 3 = 10.5$ m.
Step 2: Slant height $l = \sqrt{14^2 + 10.5^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5$ m.
Step 3: Total Area $= 2\pi r h_1 + \pi r l = \pi r (2h_1 + l) = \frac{22}{7} \times 14 (6 + 17.5) = 44 \times 23.5 = 1034$ m$^2$.
Step 4: Cost $= 1034 \times 2 = 2068$.
Answer: ₹ 2068. -
Question 10: The rain water from a roof of 22 m $\times$ 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the vessel is just full, find the rainfall in cm.
Solution:
Step 1: Volume of water in vessel $= \pi r^2 h = \frac{22}{7} \times (1)^2 \times 3.5 = 11$ m$^3$.
Step 2: Volume of rain on roof $= \text{Area} \times \text{Height} = 22 \times 20 \times H$.
Step 3: $440 H = 11 \Rightarrow H = \frac{11}{440} = \frac{1}{40}$ m.
Step 4: $H = \frac{1}{40} \times 100 = 2.5$ cm.
Answer: 2.5 cm.