Given: volume of reservoir = 108 m³ Rate of pouring water into cuboidal reservoir = 60 liters/minute = 60/100m³/minute [∵1l=1/100 m³] = 60x60/1000 m³/hour ∴ 1 m³ water filled in reservoir will take = 1000x60x60 hours ∴ 108 m³ water filled in reservoir will take = 108/1000/60x60 hours = 30 hours It wRead more
Given: volume of reservoir = 108 m³
Rate of pouring water into cuboidal reservoir = 60 liters/minute
= 60/100m³/minute [∵1l=1/100 m³]
= 60×60/1000 m³/hour
∴ 1 m³ water filled in reservoir will take = 1000x60x60 hours
∴ 108 m³ water filled in reservoir will take = 108/1000/60×60 hours = 30 hours
It will take 30 hours to fill the reservoir.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
(i) Let the edge of cube be l. Since, Surface area of the cube (A)=6l² When edge of cube is doubled, then Surface area of the cube (A’)= 6(2l)² = 6x4l² = 4x6l² A’ = 4 x A, Hence, the surface area will increase four times. (ii) Volume of cube (V) = l³ When edge of cube is doubled, then volume of cubeRead more
(i) Let the edge of cube be l.
Since, Surface area of the cube (A)=6l²
When edge of cube is doubled, then
Surface area of the cube (A’)= 6(2l)² = 6x4l² = 4x6l²
A’ = 4 x A, Hence, the surface area will increase four times.
(ii) Volume of cube (V) = l³
When edge of cube is doubled, then volume of cube (V’)=(2l)³=8l³
V’ = 8 x V, Hence, the volume will increase 8 times.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Given: Radius of cylindrical tank (r) = 1.5 m And Height of cylindrical tank (h) = 7 m Volume of cylindrical tank = πr²h = 22/7x1.5x1.5x7=49.5 cm³ = 49.5 x 1000 liters [∵1 m³ = 1000 liters] = 49500 liters Hence, the required quantity of milk is 49500 liters. Class 8 Maths Chapter 11 Exercise 11.4 SoRead more
Given: Radius of cylindrical tank (r) = 1.5 m
And Height of cylindrical tank (h) = 7 m
Volume of cylindrical tank = πr²h
= 22/7×1.5×1.5×7=49.5 cm³
= 49.5 x 1000 liters [∵1 m³ = 1000 liters]
= 49500 liters
Hence, the required quantity of milk is 49500 liters.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Given: Volume of cylinder = 1.54 m³ and Diameter of cylinder = 140cm ∴ Radius (r) = d/2=140/2=70cm Volume of cylinder = πr²h ⇒ 1.54=22/7x0.7x0.7xh ⇒ h= 1.54x7/22x0.7x0.7 ⇒ h=154x7x10x10/22x7x7x100=1m Hence, the height of the cylinder is 1 m. Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video fRead more
Given: Volume of cylinder = 1.54 m³ and Diameter of cylinder = 140cm
∴ Radius (r) = d/2=140/2=70cm
Volume of cylinder = πr²h
⇒ 1.54=22/7×0.7×0.7xh ⇒ h= 1.54×7/22×0.7×0.7
⇒ h=154x7x10x10/22x7x7x100=1m
Hence, the height of the cylinder is 1 m.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Given: Length of cuboid (l) = 60 cm, Breadth of cuboid (b) = 54 cm and Height of cuboid (h) = 30 cm We know that, Volume of cuboid = lxbxh = 60 x 54 x 30 cm³ And Volume of cube = (Side)³ = 6x6x6cm³ ∴ Number of small cubes =Volume of cuboid/Volume of cube = 60x54x30/6x6x6=450 Hence, the required cubeRead more
Given: Length of cuboid (l) = 60 cm, Breadth of cuboid (b) = 54 cm and
Height of cuboid (h) = 30 cm
We know that, Volume of cuboid = lxbxh = 60 x 54 x 30 cm³
And Volume of cube = (Side)³ = 6x6x6cm³
∴ Number of small cubes =Volume of cuboid/Volume of cube = 60x54x30/6x6x6=450
Hence, the required cubes are 450.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Given: Base area of cuboid = 180 cm² and Volume of cuboid = 900 cm³ We know that, Volume of cuboid = lxbxh ⇒ 900 = 180xh [∵Base area = lxb=180(given)] ⇒ h=900/180=5m Hence, the height of cuboid is 5 m. Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video for more answers vist to: https://www.tiwRead more
Given: Base area of cuboid = 180 cm² and Volume of cuboid = 900 cm³
We know that, Volume of cuboid = lxbxh
⇒ 900 = 180xh [∵Base area = lxb=180(given)]
⇒ h=900/180=5m
Hence, the height of cuboid is 5 m.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is greater than that of cylinder A (and square of radius gives more value than previous). Diameter of cylinder A = 7 cm ⇒ Radius of cylinder A = 7/2cm And Height of cylinder A = 14 cm ∴ Volume of cylinder A = πr²h = 22/Read more
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is
greater than that of cylinder A (and square of radius gives more value than
previous).
Diameter of cylinder A = 7 cm
⇒ Radius of cylinder A = 7/2cm
And Height of cylinder A = 14 cm
∴ Volume of cylinder A = πr²h = 22/7×7/2×7/2×14
= 539 cm³
Now Diameter of cylinder B = 14 cm
⇒ Radius of cylinder B = 14/2 = 7cm
And Height of cylinder B = 7 cm
∴ Volume of cylinder A = πr²h = 22/7x7x7x7
= 1078 cm³
Total surface area of cylinder A = πr(2h+r)
[∵ It is open from top]
22/7×7/2(2×14+7/2)=11x(28+7/2)
= 11×63/2=346.5 cm²
Total surface area of cylinder B = 𝜋r(2h+r) [∵It is open from top]
= 22/7×7(2×7+7)
22x (14+7) = 22×21 = 462cm²
Yes, cylinder with greater volume also has greater surface area.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Here, Length of cabinet (l) = 2 m, Breadth of cabinet (b) = 1 m And Height of cabinet (h) = 1.5 m ∴ Surface area of cabinet = lb+2(bh+hl) = 2 x 1 + 2 (1 x 1.5 + 1.5 x 2) = 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 m2 Hence, the required surface area of cabinet is 11 m². Class 8 Maths Chapter 11 Exercise 11.3Read more
Here,
Length of cabinet (l) = 2 m,
Breadth of cabinet (b) = 1 m
And Height of cabinet (h) = 1.5 m
∴ Surface area of cabinet = lb+2(bh+hl)
= 2 x 1 + 2 (1 x 1.5 + 1.5 x 2)
= 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 m2
Hence, the required surface area of cabinet is 11 m².
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
Here Surface area of cube = 600 cm² ⇒ 6l² = 600 ⇒ l²=100 ⇒ l = 10 cm Hence the side of cube is 10 cm Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Here Surface area of cube = 600 cm²
⇒ 6l² = 600 ⇒ l²=100 ⇒ l = 10 cm
Hence the side of cube is 10 cm
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor. When the amount of space occupied by an object such as water, milk, coffee, tea, etc. Then we have to find out volume of the object. (a) Volume (b) SurRead more
We find area when a region covered by a boundary, such as outer and inner surface
area of a cylinder, a cone, a sphere and surface of wall or floor.
When the amount of space occupied by an object such as water, milk, coffee, tea, etc. Then we have to find out volume of the object.
(a) Volume (b) Surface are (c) Volume
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.
Given: volume of reservoir = 108 m³ Rate of pouring water into cuboidal reservoir = 60 liters/minute = 60/100m³/minute [∵1l=1/100 m³] = 60x60/1000 m³/hour ∴ 1 m³ water filled in reservoir will take = 1000x60x60 hours ∴ 108 m³ water filled in reservoir will take = 108/1000/60x60 hours = 30 hours It wRead more
Given: volume of reservoir = 108 m³
Rate of pouring water into cuboidal reservoir = 60 liters/minute
= 60/100m³/minute [∵1l=1/100 m³]
= 60×60/1000 m³/hour
∴ 1 m³ water filled in reservoir will take = 1000x60x60 hours
∴ 108 m³ water filled in reservoir will take = 108/1000/60×60 hours = 30 hours
It will take 30 hours to fill the reservoir.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
If each edge of a cube is doubled, (i) how many times will its surface area increase? (ii) how many times will its volume increase?
(i) Let the edge of cube be l. Since, Surface area of the cube (A)=6l² When edge of cube is doubled, then Surface area of the cube (A’)= 6(2l)² = 6x4l² = 4x6l² A’ = 4 x A, Hence, the surface area will increase four times. (ii) Volume of cube (V) = l³ When edge of cube is doubled, then volume of cubeRead more
(i) Let the edge of cube be l.
Since, Surface area of the cube (A)=6l²
When edge of cube is doubled, then
Surface area of the cube (A’)= 6(2l)² = 6x4l² = 4x6l²
A’ = 4 x A, Hence, the surface area will increase four times.
(ii) Volume of cube (V) = l³
When edge of cube is doubled, then volume of cube (V’)=(2l)³=8l³
V’ = 8 x V, Hence, the volume will increase 8 times.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in liters that can be stored in the tank.
Given: Radius of cylindrical tank (r) = 1.5 m And Height of cylindrical tank (h) = 7 m Volume of cylindrical tank = πr²h = 22/7x1.5x1.5x7=49.5 cm³ = 49.5 x 1000 liters [∵1 m³ = 1000 liters] = 49500 liters Hence, the required quantity of milk is 49500 liters. Class 8 Maths Chapter 11 Exercise 11.4 SoRead more
Given: Radius of cylindrical tank (r) = 1.5 m
And Height of cylindrical tank (h) = 7 m
Volume of cylindrical tank = πr²h
= 22/7×1.5×1.5×7=49.5 cm³
= 49.5 x 1000 liters [∵1 m³ = 1000 liters]
= 49500 liters
Hence, the required quantity of milk is 49500 liters.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Find the height of the cylinder whose volume if 1.54 m3 and diameter of the base is 140
Given: Volume of cylinder = 1.54 m³ and Diameter of cylinder = 140cm ∴ Radius (r) = d/2=140/2=70cm Volume of cylinder = πr²h ⇒ 1.54=22/7x0.7x0.7xh ⇒ h= 1.54x7/22x0.7x0.7 ⇒ h=154x7x10x10/22x7x7x100=1m Hence, the height of the cylinder is 1 m. Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video fRead more
Given: Volume of cylinder = 1.54 m³ and Diameter of cylinder = 140cm
∴ Radius (r) = d/2=140/2=70cm
Volume of cylinder = πr²h
⇒ 1.54=22/7×0.7×0.7xh ⇒ h= 1.54×7/22×0.7×0.7
⇒ h=154x7x10x10/22x7x7x100=1m
Hence, the height of the cylinder is 1 m.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
A cuboid is of dimensions 60 cm x 54 cm x 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Given: Length of cuboid (l) = 60 cm, Breadth of cuboid (b) = 54 cm and Height of cuboid (h) = 30 cm We know that, Volume of cuboid = lxbxh = 60 x 54 x 30 cm³ And Volume of cube = (Side)³ = 6x6x6cm³ ∴ Number of small cubes =Volume of cuboid/Volume of cube = 60x54x30/6x6x6=450 Hence, the required cubeRead more
Given: Length of cuboid (l) = 60 cm, Breadth of cuboid (b) = 54 cm and
Height of cuboid (h) = 30 cm
We know that, Volume of cuboid = lxbxh = 60 x 54 x 30 cm³
And Volume of cube = (Side)³ = 6x6x6cm³
∴ Number of small cubes =Volume of cuboid/Volume of cube = 60x54x30/6x6x6=450
Hence, the required cubes are 450.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³?
Given: Base area of cuboid = 180 cm² and Volume of cuboid = 900 cm³ We know that, Volume of cuboid = lxbxh ⇒ 900 = 180xh [∵Base area = lxb=180(given)] ⇒ h=900/180=5m Hence, the height of cuboid is 5 m. Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video for more answers vist to: https://www.tiwRead more
Given: Base area of cuboid = 180 cm² and Volume of cuboid = 900 cm³
We know that, Volume of cuboid = lxbxh
⇒ 900 = 180xh [∵Base area = lxb=180(given)]
⇒ h=900/180=5m
Hence, the height of cuboid is 5 m.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Diameter of cylinder A is 7 cm and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area.
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is greater than that of cylinder A (and square of radius gives more value than previous). Diameter of cylinder A = 7 cm ⇒ Radius of cylinder A = 7/2cm And Height of cylinder A = 14 cm ∴ Volume of cylinder A = πr²h = 22/Read more
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is
greater than that of cylinder A (and square of radius gives more value than
previous).
Diameter of cylinder A = 7 cm
⇒ Radius of cylinder A = 7/2cm
And Height of cylinder A = 14 cm
∴ Volume of cylinder A = πr²h = 22/7×7/2×7/2×14
= 539 cm³
Now Diameter of cylinder B = 14 cm
⇒ Radius of cylinder B = 14/2 = 7cm
And Height of cylinder B = 7 cm
∴ Volume of cylinder A = πr²h = 22/7x7x7x7
= 1078 cm³
Total surface area of cylinder A = πr(2h+r)
[∵ It is open from top]
22/7×7/2(2×14+7/2)=11x(28+7/2)
= 11×63/2=346.5 cm²
Total surface area of cylinder B = 𝜋r(2h+r) [∵It is open from top]
= 22/7×7(2×7+7)
22x (14+7) = 22×21 = 462cm²
Yes, cylinder with greater volume also has greater surface area.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Rukshar painted the outside of the cabinet of measure 1 m x 2 m x 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?
Here, Length of cabinet (l) = 2 m, Breadth of cabinet (b) = 1 m And Height of cabinet (h) = 1.5 m ∴ Surface area of cabinet = lb+2(bh+hl) = 2 x 1 + 2 (1 x 1.5 + 1.5 x 2) = 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 m2 Hence, the required surface area of cabinet is 11 m². Class 8 Maths Chapter 11 Exercise 11.3Read more
Here,
Length of cabinet (l) = 2 m,
Breadth of cabinet (b) = 1 m
And Height of cabinet (h) = 1.5 m
∴ Surface area of cabinet = lb+2(bh+hl)
= 2 x 1 + 2 (1 x 1.5 + 1.5 x 2)
= 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 m2
Hence, the required surface area of cabinet is 11 m².
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Find the side of a cube whose surface area id 600 cm².
Here Surface area of cube = 600 cm² ⇒ 6l² = 600 ⇒ l²=100 ⇒ l = 10 cm Hence the side of cube is 10 cm Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Here Surface area of cube = 600 cm²
⇒ 6l² = 600 ⇒ l²=100 ⇒ l = 10 cm
Hence the side of cube is 10 cm
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Given a cylindrical tank, in which situation will you find surface are and in which situation volume. (a) To find how much it can hold. (b) Number of cement bags required to plaster it. (c) To find the number of smaller tanks that can be filled with water from it.
We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor. When the amount of space occupied by an object such as water, milk, coffee, tea, etc. Then we have to find out volume of the object. (a) Volume (b) SurRead more
We find area when a region covered by a boundary, such as outer and inner surface
area of a cylinder, a cone, a sphere and surface of wall or floor.
When the amount of space occupied by an object such as water, milk, coffee, tea, etc. Then we have to find out volume of the object.
(a) Volume (b) Surface are (c) Volume
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/