Ganit Prakash - Class-X - Right Circular Cylinder
Let us work out 8 Solid Geometry
📘 Let us work out 8 Solutions
Right Circular Cylinder(i) Solid has [ ] surfaces.
(ii) Number of curved surface is [ ] and number of plane surface is [ ].
Answer:
The given picture is a Right Circular Cylinder.
(i) Solid has 3 surfaces.
(ii) Number of curved surface is 1 and number of plane surface is 2.
Answer:
Five solid objects of my house, the shapes of which are right circular cylinder:
- 1. Tube light
- 2. Water pipe
- 3. Gas cylinder
- 4. Cylindrical water glass / jar
- 5. Unsharpened pencil / Rolling pin (middle part)
Given:
Let the height of the drum = \(h\) cm.
Diameter of the drum = 28 cm.
Therefore, radius of the drum \((r)\) = \( \frac{28}{2} \) = 14 cm.
Total surface area of the drum (with lid) = 2816 sq cm.
As per the question, total surface area \(2\pi r(r + h) = 2816\)
\[\begin{array}{l} \Rightarrow 2\pi r(r + h) = 2816 \\ \Rightarrow 2 \times \frac{22}{7} \times 14(14 + h) = 2816 \\ \Rightarrow 88(14 + h) = 2816 \\ \Rightarrow (14 + h) = \frac{2816}{88} \\ \Rightarrow (14 + h) = 32 \\ \Rightarrow h = 32 - 14 = 18 \end{array}\]Therefore, the height of the drum is 18 cm.
Let us write by calculating the cost of plastering the two pillars at ₹ 125 per square metre.
Given:
Radius of each pillar \((r)\) = \( \frac{5.6}{2} \) decimetre = 2.8 decimetre = 0.28 metre
Height \((h)\) = 2.5 metre = \(2.5 \times 10\) decimetre = 25 decimetre
Part 1: Volume of concrete
Total volume of the two pillars \(\rightarrow 2 \times \pi r^2h\)
\[\begin{array}{l} = 2 \times \frac{22}{7} \times (2.8)^2 \times 25 \text{ cubic decimetre} \\ = 2 \times \frac{22}{7} \times 2.8 \times 2.8 \times 25 \text{ cubic decimetre} \\ = 1232 \text{ cubic decimetre} \end{array}\]Therefore, the required concrete material is 1232 cubic decimetre.
Part 2: Cost of plastering
Curved surface area of the two pillars \(\rightarrow 2 \times 2\pi rh\)
We use radius in metre (0.28 m) and height in metre (2.5 m) for area in sq. metre.
\[\begin{array}{l} = 2 \times 2 \times \frac{22}{7} \times 0.28 \times 2.5 \text{ square metre} \\ = 8.8 \text{ square metre} \end{array}\]Cost of plastering per square metre = ₹ 125
Therefore, required cost = \(8.8 \times 125\) = ₹ 1100.
Given:
Inner radius of the cylinder \((r)\) = \( \frac{2.8}{2} \) dcm = 1.4 dcm
Height (length) \((h)\) = 7.5 dcm
Inner volume of the cylinder, \(\pi r^2h\)
\[\begin{array}{l} = \frac{22}{7} \times 1.4 \times 1.4 \times 7.5 \text{ cubic decimetre} \\ = 46.2 \text{ cubic decimetre} \end{array}\]Now, total weight of gas = 15.015 kg = 15015 gram.
Therefore, weight of gas in 46.2 cubic decimetre = 15015 gram
Weight of gas per 1 cubic decimetre = \( \frac{15015}{46.2} \) gram = 325 gram.
Therefore, the weight of the gas per cubic decimetre is 325 gram.
Given:
Let the height of each of the three equal jars = \( h \) dcm.
Diameter of each of the three jars = 1.4 dcm
Radius of each of the three jars \((r_1)\) = \( \frac{1.4}{2} \) = 0.7 dcm
Volume of the 1st jar = \( \pi r_1^2h \)
Total volume of acid in the three jars:
\[\begin{array}{l} = \left( \frac{2}{3} + \frac{5}{6} + \frac{7}{9} \right) \times \pi (0.7)^2 h \\ = \left( \frac{12 + 15 + 14}{18} \right) \times \pi \times (0.7)^2 \times h \\ = \frac{41}{18} \times \pi \times (0.7)^2 \times h \text{ cubic dcm} \end{array}\]Now, for the large jar:
Diameter = 2.1 dcm
Radius \((R)\) = \( \frac{2.1}{2} \) = 1.05 dcm
Height of acid \((H)\) = 4.1 dcm
Volume of acid in the large jar = \( \pi R^2H = \pi \times (1.05)^2 \times 4.1 \) cubic dcm.
According to the problem, the volume of acid remains the same:
\[\begin{array}{l} \frac{41}{18} \times \pi \times (0.7)^2 \times h = \pi \times (1.05)^2 \times 4.1 \\ \Rightarrow \frac{41}{18} \times 0.49 \times h = 1.1025 \times 4.1 \\ \Rightarrow h = \frac{1.1025 \times 4.1 \times 18}{0.49 \times 41} \\ \Rightarrow h = \frac{1.1025}{0.49} \times \frac{4.1}{41} \times 18 \\ \Rightarrow h = 2.25 \times 0.1 \times 18 \\ \Rightarrow h = 0.225 \times 18 \\ \Rightarrow h = 4.05 \end{array}\]Therefore, the height of each of the three jars is 4.05 dcm.
Given:
Diameter of the base = 14 cm
Radius of the base \((r)\) = \( \frac{14}{2} \) = 7 cm
Total surface area of the pot (open at one end) = 2002 sq cm
Let the height of the pot be \( h \) cm.
Since the pot is open at one end, its total surface area includes the curved surface area and the area of only one base.
\[\begin{array}{l} \text{Total Surface Area} = 2\pi rh + \pi r^2 \\ \Rightarrow 2\pi rh + \pi r^2 = 2002 \\ \Rightarrow \pi r(2h + r) = 2002 \\ \Rightarrow \frac{22}{7} \times 7 \times (2h + 7) = 2002 \\ \Rightarrow 22 \times (2h + 7) = 2002 \\ \Rightarrow 2h + 7 = \frac{2002}{22} \\ \Rightarrow 2h + 7 = 91 \\ \Rightarrow 2h = 91 - 7 \\ \Rightarrow 2h = 84 \\ \Rightarrow h = 42 \text{ cm} \end{array}\]Now, we calculate the volume (capacity) of the drum:
\[\begin{array}{l} \text{Volume} = \pi r^2h \\ = \frac{22}{7} \times 7^2 \times 42 \\ = 22 \times 7 \times 42 \\ = 154 \times 42 \\ = 6468 \text{ cubic cm} \end{array}\]We know that 1000 cubic cm = 1 liter.
Therefore, water contained in the drum = \( \frac{6468}{1000} \) liters = 6.468 liters.
Given:
Diameter of the pipe = 14 cm
Radius of the pipe \((r)\) = \( \frac{14}{2} \) cm = 7 cm = 0.7 dcm
Length of water drained in 1 minute \((h)\) = 2500 metre = 25000 dcm
Volume of water drained per minute can be considered as the volume of a cylinder of radius 0.7 dcm and length 25000 dcm.
\[\begin{array}{l} \text{Volume per minute} = \pi r^2h \\ = \frac{22}{7} \times (0.7)^2 \times 25000 \text{ cubic dcm} \\ = \frac{22}{7} \times 0.49 \times 25000 \\ = 22 \times 0.07 \times 25000 \\ = 22 \times 7 \times 250 \\ = 154 \times 250 \\ = 38500 \text{ cubic dcm} \end{array}\]Since 1 cubic dcm = 1 liter, the pump drains 38500 liters of water per minute.
Now, we need to find the water drained per hour (60 minutes):
\[\begin{array}{l} \text{Water drained per hour} = 38500 \times 60 \text{ liters} \\ = 2310000 \text{ liters} \end{array}\]We know that 1000 liters = 1 kiloliter.
Therefore, water drained in kiloliters = \( \frac{2310000}{1000} \) = 2310 kiloliters.
Given:
Diameter of the gas jar \((D_1)\) = 7 cm
Radius of the gas jar \((R_1)\) = \( \frac{7}{2} \) = 3.5 cm
Diameter of the iron pipe \((D_2)\) = 5.6 cm
Radius of the iron pipe \((R_2)\) = \( \frac{5.6}{2} \) = 2.8 cm
Length of the iron pipe \((H_2)\) = 5 cm
Let the rise in the water level be \( h \) cm.
According to Archimedes' principle, the volume of water displaced will be equal to the volume of the solid immersed.
\[\begin{array}{l} \text{Volume of displaced water} = \text{Volume of the iron pipe} \\ \pi \times (R_1)^2 \times h = \pi \times (R_2)^2 \times H_2 \\ \Rightarrow (3.5)^2 \times h = (2.8)^2 \times 5 \\ \Rightarrow 12.25 \times h = 7.84 \times 5 \\ \Rightarrow 12.25 \times h = 39.2 \\ \Rightarrow h = \frac{39.2}{12.25} \\ \Rightarrow h = 3.2 \end{array}\]Therefore, the level of water will rise by 3.2 cm.
Given:
Let the radius of the pillar be \( r \) metre and its height be \( h \) metre.
Curved surface area = \( 2\pi rh \) = 264 sq. metre ... (Equation 1)
Volume = \( \pi r^2h \) = 924 cubic metre ... (Equation 2)
Dividing Equation 2 by Equation 1, we get:
\[\begin{array}{l} \frac{\pi r^2h}{2\pi rh} = \frac{924}{264} \\ \Rightarrow \frac{r}{2} = 3.5 \\ \Rightarrow r = 3.5 \times 2 = 7 \text{ metre} \end{array}\]Since the radius is 7 metre, the diameter of the pillar = \( 2 \times 7 \) = 14 metre.
Now, substituting the value of \( r \) in Equation 1:
\[\begin{array}{l} 2\pi rh = 264 \\ \Rightarrow 2 \times \frac{22}{7} \times 7 \times h = 264 \\ \Rightarrow 44h = 264 \\ \Rightarrow h = \frac{264}{44} \\ \Rightarrow h = 6 \text{ metre} \end{array}\]Therefore, the diameter of the pillar is 14 metre and its height is 6 metre.
Given:
Height of the cylindrical tank \((H)\) = 9 metre
Let the radius of the tank be \( R \) metre.
Diameter of the pipe = 6 cm = 0.06 metre
Radius of the pipe \((r)\) = \( \frac{0.06}{2} \) = 0.03 metre
Speed of water flow = 225 metre/minute
Time taken to empty the tank = 36 minutes
Total length of the water column flowing out in 36 minutes \((h)\):
\[\begin{array}{l} h = 225 \times 36 \text{ metre} = 8100 \text{ metre} \end{array}\]Total volume of water drained by the pipe in 36 minutes:
\[\begin{array}{l} = \pi r^2h \\ = \pi \times (0.03)^2 \times 8100 \text{ cubic metre} \end{array}\]This volume must be equal to the total volume of water initially in the tank:
\[\begin{array}{l} \pi R^2H = \pi r^2h \\ \Rightarrow \pi \times R^2 \times 9 = \pi \times (0.03)^2 \times 8100 \\ \Rightarrow 9R^2 = 0.0009 \times 8100 \\ \Rightarrow 9R^2 = 7.29 \\ \Rightarrow R^2 = \frac{7.29}{9} \\ \Rightarrow R^2 = 0.81 \\ \Rightarrow R = \sqrt{0.81} = 0.9 \text{ metre} \end{array}\]So, the radius of the tank is 0.9 metre.
Diameter of the tank = \( 2R = 2 \times 0.9 \) = 1.8 metre (or 180 cm).
Given:
Let the radius of the log be \( r \) dcm and its height be \( h \) dcm.
Curved surface area = \( 2\pi rh \) = 440 sq. dcm ... (Equation 1)
Weight of the log = 9.24 quintals = 9.24 \( \times \) 100 kg = 924 kg
Weight of 1 cubic dcm of wood = 1.5 kg
Therefore, total volume of the log:
\[\begin{array}{l} \text{Volume} = \frac{\text{Total weight}}{\text{Weight per cubic dcm}} \\ \pi r^2h = \frac{924}{1.5} \\ \pi r^2h = 616 \text{ cubic dcm} \dots \text{(Equation 2)} \end{array}\]Dividing Equation 2 by Equation 1, we get:
\[\begin{array}{l} \frac{\pi r^2h}{2\pi rh} = \frac{616}{440} \\ \Rightarrow \frac{r}{2} = 1.4 \\ \Rightarrow r = 1.4 \times 2 = 2.8 \text{ dcm} \end{array}\]Since the radius is 2.8 dcm, the diameter of the log = \( 2 \times 2.8 \) = 5.6 dcm.
Now, substituting the value of \( r \) in Equation 1:
\[\begin{array}{l} 2\pi rh = 440 \\ \Rightarrow 2 \times \frac{22}{7} \times 2.8 \times h = 440 \\ \Rightarrow 2 \times 22 \times 0.4 \times h = 440 \\ \Rightarrow 17.6 \times h = 440 \\ \Rightarrow h = \frac{440}{17.6} \\ \Rightarrow h = 25 \text{ dcm} \end{array}\]Therefore, the length of the diameter of the log is 5.6 dcm and its height is 25 dcm.
Given:
Outer diameter = 30 cm \(\Rightarrow\) Outer radius \((R)\) = 15 cm = 1.5 dcm
Inner diameter = 26 cm \(\Rightarrow\) Inner radius \((r)\) = 13 cm = 1.3 dcm
(Note: The question states inner and outer diameters as 30 cm and 26 cm, which is practically impossible. We logically assume outer is 30 cm and inner is 26 cm.)
Length/Height \((h)\) = 14.7 metre = 147 dcm
Total surface area to be painted = Outer curved surface area + Inner curved surface area + Area of the two circular ring ends
\[\begin{array}{l} = 2\pi Rh + 2\pi rh + 2\pi(R^2 - r^2) \\ = 2\pi h(R + r) + 2\pi(R + r)(R - r) \\ = 2\pi(R + r)(h + R - r) \\ = 2 \times \frac{22}{7} \times (1.5 + 1.3) \times (147 + 1.5 - 1.3) \\ = 2 \times \frac{22}{7} \times 2.8 \times 147.2 \\ = 2 \times 22 \times 0.4 \times 147.2 \\ = 17.6 \times 147.2 \\ = 2590.72 \text{ sq dcm} \end{array}\]Cost of painting per sq dcm = ₹ 2.25
Total cost = \( 2590.72 \times 2.25 \) = ₹ 5829.12.
Given:
Height \((h)\) = 2.8 metre = 28 dcm
Inner diameter = 4.6 dcm \(\Rightarrow\) Inner radius \((r)\) = \( \frac{4.6}{2} \) = 2.3 dcm
Volume of iron = 84.48 cubic dcm
Let the outer radius be \( R \) dcm.
Volume of the hollow cylinder (volume of iron) = \( \pi(R^2 - r^2)h \)
\[\begin{array}{l} \pi(R^2 - 2.3^2) \times 28 = 84.48 \\ \Rightarrow \frac{22}{7} \times (R^2 - 5.29) \times 28 = 84.48 \\ \Rightarrow 22 \times 4 \times (R^2 - 5.29) = 84.48 \\ \Rightarrow 88(R^2 - 5.29) = 84.48 \\ \Rightarrow R^2 - 5.29 = \frac{84.48}{88} \\ \Rightarrow R^2 - 5.29 = 0.96 \\ \Rightarrow R^2 = 5.29 + 0.96 \\ \Rightarrow R^2 = 6.25 \\ \Rightarrow R = \sqrt{6.25} = 2.5 \text{ dcm} \end{array}\]Outer diameter = \( 2R = 2 \times 2.5 \) = 5 dcm.
Given:
Let the radius of the cylinder be \( r \) dcm.
Initial height of the cylinder \((h_1)\) = \( 2r \)
Initial volume \((V_1)\) = \( \pi r^2h_1 = \pi r^2(2r) = 2\pi r^3 \)
If height \((h_2)\) = \( 6r \)
New volume \((V_2)\) = \( \pi r^2h_2 = \pi r^2(6r) = 6\pi r^3 \)
According to the problem, the volume would be greater by 539 cubic dcm:
\[\begin{array}{l} V_2 - V_1 = 539 \\ \Rightarrow 6\pi r^3 - 2\pi r^3 = 539 \\ \Rightarrow 4\pi r^3 = 539 \\ \Rightarrow 4 \times \frac{22}{7} \times r^3 = 539 \\ \Rightarrow r^3 = \frac{539 \times 7}{4 \times 22} \\ \Rightarrow r^3 = \frac{49 \times 7}{4 \times 2} \quad [\text{Dividing 539 and 22 by 11}] \\ \Rightarrow r^3 = \frac{343}{8} \\ \Rightarrow r = \sqrt[3]{\frac{343}{8}} = \frac{7}{2} = 3.5 \text{ dcm} \end{array}\]Height of the cylinder \( = 2r = 2 \times 3.5 \) = 7 dcm.
Given (For the Pipes):
Number of pipes = 3
Diameter of each pipe = 2 cm \(\Rightarrow\) Radius \((r)\) = 1 cm = 0.1 dcm
Speed of water flow = 420 metre/minute = 4200 dcm/minute
Time = 40 minutes
(i) Volume of water spent:
Length of the water column flowing out of one pipe in 40 minutes \((h)\):
\[\begin{array}{l} h = 4200 \times 40 \text{ dcm} = 168000 \text{ dcm} \end{array}\]Total volume of water pumped out by 3 pipes:
\[\begin{array}{l} = 3 \times \pi r^2h \\ = 3 \times \frac{22}{7} \times (0.1)^2 \times 168000 \text{ cubic dcm} \\ = 3 \times \frac{22}{7} \times 0.01 \times 168000 \\ = 3 \times \frac{22}{7} \times 1680 \\ = 3 \times 22 \times 240 \\ = 15840 \text{ cubic dcm} \quad [\text{since } 1 \text{ cubic dcm} = 1 \text{ liter}] \end{array}\]Therefore, 15840 cubic dcm (or liters) of water has been spent.
(ii) Volume of water remaining in the tank:
Given (For the Tank):
Diameter of the tank = 2.8 metre = 28 dcm \(\Rightarrow\) Radius \((R)\) = 14 dcm
Length/Height of the tank \((H)\) = 6 metre = 60 dcm
Total initial volume of water in the tank:
\[\begin{array}{l} = \pi R^2H \\ = \frac{22}{7} \times (14)^2 \times 60 \text{ cubic dcm} \\ = \frac{22}{7} \times 196 \times 60 \\ = 22 \times 28 \times 60 \\ = 36960 \text{ cubic dcm (or liters)} \end{array}\]Volume of water remaining:
\[\begin{array}{l} = \text{Total initial volume} - \text{Volume spent} \\ = 36960 - 15840 \\ = 21120 \text{ cubic dcm} \end{array}\]Therefore, 21120 cubic dcm (or liters) of water still remains in the tank.
(i) If each pillar is of 3 metre height, let us write by calculating how many cubic dcm of plaster materials will be needed.
(ii) If the ratio of sand and cement in the plaster material be 4:1, let us write how many cubic dcm of cement will be needed.
Given:
Number of pillars = 4
Height of each pillar \((h)\) = 3 metre = 30 dcm
Diameter of each pillar = 17.5 cm = 1.75 dcm
Inner radius \((r)\) = \( \frac{1.75}{2} \) = 0.875 dcm
Thickness of plaster = 3.5 cm = 0.35 dcm
Outer radius with plaster \((R)\) = \( 0.875 + 0.35 \) = 1.225 dcm
(i) Volume of plaster materials needed:
Volume of plaster for one pillar is the volume of the hollow cylindrical shell.
\[\begin{array}{l} \text{Volume of plaster for 1 pillar} = \pi(R^2 - r^2)h \\ = \frac{22}{7} \times (1.225^2 - 0.875^2) \times 30 \\ = \frac{22}{7} \times (1.225 + 0.875)(1.225 - 0.875) \times 30 \\ = \frac{22}{7} \times 2.1 \times 0.35 \times 30 \\ = 22 \times 0.3 \times 0.35 \times 30 \\ = 22 \times 9 \times 0.35 \\ = 198 \times 0.35 \\ = 69.3 \text{ cubic dcm} \end{array}\]Total volume of plaster for 4 pillars = \( 4 \times 69.3 \) = 277.2 cubic dcm.
(ii) Volume of cement needed:
Ratio of sand to cement = 4 : 1
Total parts in the mixture = 4 + 1 = 5 parts
Part of cement in the mixture = \( \frac{1}{5} \)
\[\begin{array}{l} \text{Volume of cement needed} = \frac{1}{5} \times \text{Total volume of plaster} \\ = \frac{1}{5} \times 277.2 \\ = 55.44 \text{ cubic dcm} \end{array}\]Therefore, 55.44 cubic dcm of cement will be needed.
Given for the hollow cylinder:
Outer diameter = 16 cm \(\Rightarrow\) Outer radius \((R)\) = 8 cm
Inner diameter = 12 cm \(\Rightarrow\) Inner radius \((r)\) = 6 cm
Height \((H)\) = 36 cm
Volume of material in the hollow cylinder:
\[\begin{array}{l} V_1 = \pi(R^2 - r^2)H \\ = \pi \times (8^2 - 6^2) \times 36 \\ = \pi \times (64 - 36) \times 36 \\ = \pi \times 28 \times 36 \text{ cubic cm} \end{array}\]Given for the small solid cylinder:
Diameter = 2 cm \(\Rightarrow\) Radius \((r_1)\) = 1 cm
Height \((h_1)\) = 6 cm
Volume of one small solid cylinder:
\[\begin{array}{l} V_2 = \pi r_1^2h_1 \\ = \pi \times (1)^2 \times 6 \\ = 6\pi \text{ cubic cm} \end{array}\]Let the number of solid cylinders obtained be \( n \).
\[\begin{array}{l} n = \frac{\text{Volume of hollow cylinder } (V_1)}{\text{Volume of one solid cylinder } (V_2)} \\ = \frac{\pi \times 28 \times 36}{6\pi} \\ = \frac{28 \times 36}{6} \\ = 28 \times 6 \\ = 168 \end{array}\]Therefore, 168 solid cylinders can be obtained.
📘 19. Very short answer type question (V.S.A.)
(A) Multiple Choice QuestionsExplanation:
Let the radii of the two cylinders be \( r_1 \) and \( r_2 \), and their heights be \( h_1 \) and \( h_2 \).
Given, \( \frac{r_1}{r_2} = \frac{2}{3} \) and \( \frac{h_1}{h_2} = \frac{5}{3} \)
Area of lateral surface of a cylinder = \( 2\pi rh \)
Ratio of their lateral surface areas:
\[\begin{array}{l} = \frac{2\pi r_1 h_1}{2\pi r_2 h_2} \\ = \left(\frac{r_1}{r_2}\right) \times \left(\frac{h_1}{h_2}\right) \\ = \frac{2}{3} \times \frac{5}{3} \\ = \frac{10}{9} = 10 : 9 \end{array}\]Correct Option: (c) 10:9
Explanation:
Let the radii of the two cylinders be \( r_1 \) and \( r_2 \), and their heights be \( h_1 \) and \( h_2 \).
Given, \( \frac{r_1}{r_2} = \frac{2}{3} \) and \( \frac{h_1}{h_2} = \frac{5}{3} \)
Volume of a cylinder = \( \pi r^2 h \)
Ratio of their volumes:
\[\begin{array}{l} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} \\ = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) \\ = \left(\frac{2}{3}\right)^2 \times \frac{5}{3} \\ = \frac{4}{9} \times \frac{5}{3} \\ = \frac{20}{27} = 20 : 27 \end{array}\]Correct Option: (b) 20:27
Explanation:
Let the radii be \( r_1, r_2 \) and heights be \( h_1, h_2 \).
Given, Volumes are equal: \( V_1 = V_2 \)
\[\begin{array}{l} \Rightarrow \pi r_1^2 h_1 = \pi r_2^2 h_2 \\ \Rightarrow \frac{r_1^2}{r_2^2} = \frac{h_2}{h_1} \end{array}\]Given ratio of heights \( h_1 : h_2 = 1 : 2 \), which means \( \frac{h_2}{h_1} = \frac{2}{1} \).
\[\begin{array}{l} \Rightarrow \frac{r_1^2}{r_2^2} = \frac{2}{1} \\ \Rightarrow \frac{r_1}{r_2} = \sqrt{\frac{2}{1}} = \frac{\sqrt{2}}{1} \end{array}\]Ratio of radii = \( \sqrt{2} : 1 \)
Correct Option: (b) \(\sqrt{2} : 1\)
Explanation:
Let original radius = \( r \) and original height = \( h \).
Original Volume \( V = \pi r^2 h \)
New radius \( r' = \frac{r}{2} \), New height \( h' = 2h \)
New Volume \( V' \):
\[\begin{array}{l} V' = \pi (r')^2 h' \\ = \pi \left(\frac{r}{2}\right)^2 (2h) \\ = \pi \left(\frac{r^2}{4}\right) (2h) \\ = \frac{1}{2} \pi r^2 h \\ = \frac{1}{2} V \end{array}\]The volume becomes half of the original volume.
Correct Option: (c) half
Explanation:
Let original radius = \( r \) and original height = \( h \).
Original lateral surface area \( A = 2\pi rh \)
New radius \( r' = 2r \), New height \( h' = \frac{h}{2} \)
New lateral surface area \( A' \):
\[\begin{array}{l} A' = 2\pi r' h' \\ = 2\pi (2r) \left(\frac{h}{2}\right) \\ = 2\pi rh \\ = A \end{array}\]The lateral surface area remains equal to the original area.
Correct Option: (a) equal
📘 19. Very short answer type question (V.S.A.)
(B) True or FalseExplanation:
The total volume of the right circular drum is \( \pi r^2h \) cubic cm.
If half part of the drum is filled with water, the volume of the water will be half of the total volume.
Therefore, Volume of water = \( \frac{1}{2} \pi r^2h \) cubic cm.
The statement claims it is \( \pi r^2h \), which is incorrect.
Answer: False
Explanation:
Let the radius \( r = 2 \) units and height be \( h \) units.
Numerical value of Volume \( (V) = \pi r^2h = \pi (2)^2 h = 4\pi h \).
Numerical value of Lateral/Curved Surface Area \( (S) = 2\pi rh = 2\pi (2) h = 4\pi h \).
(Note: In such context questions in the textbook, "surface area" usually refers to the "curved surface area". If it meant total surface area, it would be \( 2\pi r(r+h) = 4\pi(2+h) = 8\pi + 4\pi h \), which is not equal to \( 4\pi h \). Given standard textbook conventions for this specific problem, it implies curved surface area.)
Since Volume = \( 4\pi h \) and Curved Surface Area = \( 4\pi h \), they are numerically equal for any height \( h \).
Answer: True
📘 19. Very short answer type question (V.S.A.)
(C) Fill in the blanksExplanation:
When the rectangular paper is rolled such that its perimeter (circumference of the base) is equal to the length (\( l \)) of the paper, the height of the cylinder formed becomes the breadth (\( b \)) of the paper.
Lateral surface area of a cylinder = Perimeter of base \( \times \) Height
Lateral surface area = \( l \times b = lb \)
Answer: \( lb \) (or \( l \times b \))
Explanation:
The longest rod that can be placed inside a cylinder corresponds to the diagonal of its vertical cross-section (which is a rectangle with sides equal to diameter and height).
Length of the longest rod = \( \sqrt{\text{diameter}^2 + \text{height}^2} \)
\[\begin{array}{l} = \sqrt{3^2 + 4^2} \\ = \sqrt{9 + 16} \\ = \sqrt{25} \\ = 5 \text{ cm} \end{array}\]Answer: 5
Explanation:
Let the radius of the cylinder be \( r \) and height be \( h \).
Volume of the cylinder = \( \pi r^2h \)
Lateral surface area = \( 2\pi rh \)
Given that their numerical values are equal:
\[\begin{array}{l} \pi r^2h = 2\pi rh \end{array}\]Assuming \( r > 0 \) and \( h > 0 \), dividing both sides by \( \pi rh \):
\[\begin{array}{l} r = 2 \text{ units} \end{array}\]Therefore, the diameter of the cylinder = \( 2r = 2 \times 2 = 4 \text{ units} \).
Answer: 4
📘 20. Short answer type question (S.A.)
Right Circular CylinderGiven:
Let the radius of the base be \( r \) metre and height be \( h \) metre.
Lateral surface area \(= 2\pi rh =\) 264 ... (i)
Volume \(= \pi r^2h =\) 924 ... (ii)
Dividing equation (ii) by equation (i):
\[\begin{array}{l} \frac{\pi r^2h}{2\pi rh} = \frac{924}{264} \\ \Rightarrow \frac{r}{2} = \frac{7}{2} \quad [\text{Since } \frac{924}{264} = 3.5 = \frac{7}{2}] \\ \Rightarrow r = 7 \end{array}\]Therefore, the length of the radius of the base is 7 metre.
Given:
Let the height of the cylinder be \( h \) unit.
Lateral surface area, \( c = 2\pi rh \)
Volume, \( v = \pi r^2h \)
We need to find the value of \(\frac{cr}{v}\):
\[\begin{array}{l} \frac{cr}{v} = \frac{(2\pi rh) \times r}{\pi r^2h} \\ = \frac{2\pi r^2h}{\pi r^2h} \\ = 2 \end{array}\]Therefore, the value of \(\frac{cr}{v}\) is 2.
Given:
Height \( (h) \) = 14 cm
Lateral surface area = 264 sq cm
Let the radius of the base be \( r \) cm.
We know, Lateral surface area = \( 2\pi rh \)
\[\begin{array}{l} 2\pi rh = 264 \\ \Rightarrow 2 \times \frac{22}{7} \times r \times 14 = 264 \\ \Rightarrow 2 \times 22 \times 2 \times r = 264 \\ \Rightarrow 88r = 264 \\ \Rightarrow r = \frac{264}{88} \\ \Rightarrow r = 3 \text{ cm} \end{array}\]Now, Volume of the cylinder = \( \pi r^2h \)
\[\begin{array}{l} V = \frac{22}{7} \times (3)^2 \times 14 \\ = 22 \times 9 \times 2 \\ = 396 \text{ cubic cm} \end{array}\]Therefore, the volume of the cylinder is 396 cubic cm.
Given:
Let the radii of the two cylinders be \( r_1 \) and \( r_2 \), and their heights be \( h_1 \) and \( h_2 \).
Ratio of heights, \(\frac{h_1}{h_2} = \frac{1}{2}\)
Ratio of perimeters (circumferences of base) = \(\frac{2\pi r_1}{2\pi r_2} = \frac{3}{4}\) \(\Rightarrow \frac{r_1}{r_2} = \frac{3}{4}\)
Let their volumes be \( V_1 \) and \( V_2 \).
Ratio of their volumes:
\[\begin{array}{l} \frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} \\ = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) \\ = \left(\frac{3}{4}\right)^2 \times \left(\frac{1}{2}\right) \\ = \frac{9}{16} \times \frac{1}{2} \\ = \frac{9}{32} \end{array}\]Therefore, the ratio of their volumes is 9:32.
Explanation:
Let original radius = \( r \) and original height = \( h \).
Original Volume \( (V_1) = \pi r^2h \)
New radius \( (r') = r - 50\% \text{ of } r = r - 0.5r = 0.5r = \frac{r}{2} \)
New height \( (h') = h + 50\% \text{ of } h = h + 0.5h = 1.5h = \frac{3h}{2} \)
New Volume \( (V_2) \):
\[\begin{array}{l} V_2 = \pi (r')^2h' \\ = \pi \left(\frac{r}{2}\right)^2 \left(\frac{3h}{2}\right) \\ = \pi \left(\frac{r^2}{4}\right) \left(\frac{3h}{2}\right) \\ = \frac{3}{8}\pi r^2h = \frac{3}{8}V_1 \end{array}\]Decrease in volume = \( V_1 - V_2 = V_1 - \frac{3}{8}V_1 = \frac{5}{8}V_1 \)
Percentage decrease in volume:
\[\begin{array}{l} = \frac{\text{Decrease in volume}}{\text{Original Volume}} \times 100\% \\ = \frac{\frac{5}{8}V_1}{V_1} \times 100\% \\ = \frac{5}{8} \times 100\% \\ = \frac{500}{8}\% \\ = 62.5\% \end{array}\]Therefore, the volume will be decreased by 62.5%.
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