RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS

Other Exercises

Question 1.
Find the surface area of a sphere of radius 14 cm.
Solution:
Radius of a sphere (r) = 14 cm
∴ Surface area = 4πr2 = 4 x \(\frac { 22 }{ 7 }\) x 14 x 14 cm2
= 2464 cm3

Question 2.
Find the total surface afea of a hemisphere of radius 10 cm.
Solution:
Radius of hemisphere (r) = 10 cm
∴ Total surface area = 3πr2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 2.1

Question 3.
Find the radius of a sphere whose surface area is 154 cm2.
Solution:
Surface area of a sphere = 154 cm2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 3.1

Question 4.
The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m. Find the area available to the motor cyclist for riding.
Solution:
Diameter of hollow sphere = 7 m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 4.1

Question 5.
Find the volume of a sphere whose surface area is 154 cm2.
Solution:
Surface area of a sphere = 154 cm2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 5.1

Question 6.
How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?
Solution:
Edge of a solid cube = 44 cm
∴ Volume = a2 = (44)2 cm2
= 44 × 44 × 44 cm3
Diameter of a spherical bullet = 4 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 6.1

Question 7.
If a sphere of radius 2r has the same volume as that of a cone with circular base of radius r, then find the height of the cone.
Solution:
Radius of a sphere (R) = 2r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 7.1

Question 8.
If a hollow sphere of intefnal and external diameters 4 cm and 8 cm respectively melted into a cone of base diameter 8 cm, then find the height of the cone.
Solution:
Internal diameter of a hollow sphere = 4cm
∴ Internal radius = \(\frac { 4 }{ 2 }\) = 2 cm
Similarly the outer radius (R) = \(\frac { 8 }{ 2 }\) = 4 cm
∴ Volume of melted used in hollow sphere
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 8.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 8.2

Question 9.
The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height of the cone.
Solution:
Radius of a sphere (r) = 5 cm
∴ Surface area = 4πr2
= 4π x 5 x 5 = 100π cm2
Radius of cone (r1) = 4 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 9.1

Question 10.
If a sphere is inscribed in a cube, find the ratio of the volume of cube to the volume of the sphere.
Solution:
Let edge of a cube = a
Then its volume = a3
∵ A sphere is inscribed in the cube
∴ Diameter of sphere = a
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 10.1

Hope given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS are helpful to complete your math homework.

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NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation

NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation

MULTIPLE CHOICE QUESTIONS

Question 1.
An object is put one by one in three liquids having different densities. The object floats with 1/9, 2/11 and 3/7 parts of their volumes outside the liquid surface in liquids of densities d1, d2 and d3 respectively. Which of the following statement is correct ?
(a) d1 > d> d3
(b) d1 > d< d3
(c) d< d> d3
(d) d1 < d< d3
Answer:
(d). Upthrust due to liquid on an object is directly proportional to the density of the liquid.

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Question 2.
An obj ect weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12 N.
Answer:
(a). Explanation : Weight of liquid displaced by an object = Weight of object in air – weight in liquid.

Question 3.
A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(a) maximum when length and breadth form the base
(b) maximum when breadth and width form the base
(c) maximum when width and length form the base
(d) the same in all the above three cases.
Answer:
(b). Explanation :
NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation image - 1

SHORT ANSWER QUESTIONS

Question 4.
(a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force.
If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cubeas compared to the first case for water. Give reason for each case. (CBSE 2012)
(b) A ball weighing 4 kg of density 4000 kg m-3 is completely immersed in water of density 103 kg m-3. Find the force of buoyancy on it. (Given g = 10 ms-2.)
Answer:
(a) Buoyant force = Vρg. Since, density (ρ) of saturated salt solution is more than that of water. So, the cube will experience greater buoyant force in saturated salt solution. When size of cube is reduced, its volume (V) also reduces. Hence, it will experience less buoyant force then in first case.
NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation image - 2

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2

Other Exercises

Question 1.
Find the volume of a sphere whose radius is
(i) 2 cm
(ii) 3.5 cm
(iii) 10.5 cm
Solution:
(i) Radius of sphere (r) = 2 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 1.1

Question 2.
Find the volume of a sphere whose diameter is,
(i) 14 cm
(ii) 3.5 dm
(iii) 2.1 m
Solution:
(i) Diameter of a sphere = 14 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 2.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 2.2

Question 3.
A hemspherical tank has inner radius of 2.8 m. Find its capacity in litres.
Solution:
Radius of hemispherical tank (r) = 2.8 m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 3.1

Question 4.
A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.
Solution:
Thickness of steel = 0.25 cm = \(\frac { 1 }{ 4 }\)cm
Inside radius of the hemispherical bowl (r) = 5 cm
∴ Outer radius (R) = 5 + 0.25 = 5.25 cm
∴ Volume of the steel used = \(\frac { 1 }{ 4 }\)π(R3 – r3)
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 4.1

Question 5.
How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?
Solution:
Edge of cube (r) = 22 cm
∴ Volume = a3 = (22)3 cm3
= 22 x 22 x 22 = 10648 cm3
Diameter of a bullet = 2 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 5.1

Question 6.
A shopkeeper has one laddoo of radius 5 cm. With the same material how many laddoos of radius 2.5 cm can be made?
Solution:
Radius of bigger laddoo (R) = 5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 6.1

Question 7.
A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. It the diameters of two balls be \(\frac { 3 }{ 2 }\) cm and 2 cm, find the diameter of the third ball.
Solution:
Diameter of a spherical ball of lead = 3 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 7.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 7.2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 7.3

Question 8.
A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises \(\frac { 5 }{ 3 }\) cm. Find the radius of the cylinder.
Solution:
Radius of sphere (r1) = 5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 8.1
Level of water rises in the cylinder after immersing the sphere in it
∴ Height of water level = \(\frac { 5 }{ 3 }\) cm
Let r be radius of the cylinder, then Volume of water = Volume of the sphere
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 8.2

Question 9.
If the radius of a sphere is doubled, what is the ratio of the volumes of the first sphere to that of the second sphere?
Solution:
Let r2 be the radius of the given sphere
then volume = \(\frac { 4 }{ 3 }\) πr3
By doubling the radius the radius of the new sphere = 2r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 9.1

Question 10.
A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.
Solution:
Radius of hemispherical bowl (r) = 3.5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 10.1

Question 11.
A cylinder whose height is two thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.
Solution:
Radius of a sphere (r) = 4 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 11.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 11.2

Question 12.
A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.
Solution:
Radius of hemispherical bowl (r) = 6 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 12.1

Question 13.
The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.
Solution:
Diameter of a copper sphere = 18 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 13.1

Question 14.
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.
Solution:
Diameter of a sphere = 6 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 14.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 14.2

Question 15.
The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height 2 \(\frac { 2 }{ 3 }\) cm. Find the diameter of the cylinder.
Solution:
Internal radius of the hollow spherical shell (r) = 3 cm
and external radius (R) = 5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 15.1

Question 16.
A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere (r) = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 16.1

Question 17.
A hollow sphere of internal and external radius 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.
Solution:
Internal radius of a hollow sphere (r) = 2 cm
and external radius (R) = 4 cm
∴ Volume of the metal used
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 17.1

Question 18.
A metallic sphere of radius 10.5 cm is melted and thus recast into small cones each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.
Solution:
Radius of a metallic sphere (R) = 10.5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 18.1

Question 19.
A cone and a hemisphere have equal bases and equal volumes. Find the ratio Of their heights.
Solution:
Let r be the radius and h be the height of the cone, hemisphere
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 19.1

Question 20.
The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.
Solution:
By carving a largest sphere out of the cube, the diameter of the sphere = 10.5
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 20.1

Question 21.
A cube, of side 4 cm, contains a sphere touching its sides. Find the volume of the gap in between.
Solution:
Side of cube = 4 cm
∴ Volume = (side)3 = 4x4x4 = 64 cm3
Diameter of the largest sphere touching its sides = 4 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 21.1

Question 22.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank. (NCERT)
Solution:
Thickness of hemispherical tank = 1 cm
Inner radius (r) = 1 m = 100 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 22.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 22.2

Question 23.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule? (NCERT)
Solution:
Diameter of a medicine spherical capsule = 3.5 mm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 23.1

Question 24.
The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? (NCERT)
Solution:
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 24.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 24.2

Question 25.
A cone and a hemisphere have equal bases and equal volumes. Find the ratio in their heights.
Solution:
Let r be the radius of cone and hemisphere and let h be the height of the cone then
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 25.1

Question 26.
A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?
Solution:
Radius of cylinderical tub (r) = 16 cm
Height of water in it (h) = 30 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 26.1

Question 27.
A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use π = 22/7).
Solution:
Radius of cylinder (r) = 12 cm
Depth of water in it (h) = 20 cm
By dropping a ball, the water level rose by 6.75 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 27.1

Question 28.
A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres?
Solution:
Radius of cylinderical jar (r) = 6 cm
Level of oil in it (h) = 2 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 28.1

Question 29.
A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm eacfy are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?
Solution:
Diameter of measuring jar = 10 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 29.1
Now after swing the ball in the water of jar Let volume of water raised, by h cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 29.2

Question 30.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2:3.
Solution:
∵ Bases and heights of a cones hemisphere and a cylinder are equal
Let r be the radius and h be their heights
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 30.1

Question 31.
A cylinderical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?
Solution:
Radius of the cylinderical tub (r) = 12 cm
Depth of water in it (h) = 20 cm
By dropping a spherical ball in it, the water raised by 6.75 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 31.1

Question 32.
A sphere, a cylinder and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.
Solution:
Diameter of a sphere, cylinder and a cone are equal
Let each as diameter = 2r
Then radius of each = r
Height of cylinder = diameter = 2r
and height of cone = 2r
Now volume of sphere = \(\frac { 4 }{ 3 }\)πr3
Volume of cylinder = πr2h
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 32.1

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RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

Other Exercises

Question 1.
Find the surface area of a sphere of radius.
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
In a sphere,
(i) Radius (r) = 10.5 cm
Surface area = 4πr2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 1.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 1.2

Question 2.
Find the surface area of a sphere of diameter
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
Solution:
(i) Diameter of a sphere = 14 cm
Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 2.1

Question 3.
Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. [Use π = 3.14]
Solution:
(i) Radius of hemisphere = 10 cm
∴ Total surface area of hemisphere = 2πr2
= 2 x 3.14 x 10 x 10 cm2
= 628 cm2
(ii) Total surface area of solid hemisphere
= 3πr2 = 3 x 3.14 x 10 x 10 cm2
= 942 cm2

Question 4.
The surface area of a sphere in 5544 cm2, find the diameter.
Solution:
Let r be the radius of a sphere, then Surface area = 4πr2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 4.1

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹4 per 100 cm2. [NCERT]
Solution:
Inner diameter of a hemispherical bowl = 10.5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 5.1

Question 6.
The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of ₹2 per sq. m.
Solution:
Radius of dome (hemispherical) = 63 dm
Area of curved surface
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 6.1

Question 7.
Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earth’s surface is covered by water?
Solution:
Radius of earth (sphere) = 6370 km
Water on the earth = \(\frac { 3 }{ 4 }\) % total area
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 7.1

Question 8.
A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape be 7 cm.
Solution:
Total height of the so formed shape = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 8.1

Question 9.
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Diameter of moon = \(\frac { 1 }{ 4 }\) of diameter of earth
Let radius of earth = r km
Then radius of moon = \(\frac { 1 }{ 4 }\) r km
Now surface area of earth = 4πr2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 9.1

Question 10.
A hemi-spherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹5 per 100 cm2. [NCERT]
Solution:
Circumference of the base of dome (r) = 17.6 m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 10.1

Question 11.
A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at ₹7 per 100 cm2.
Solution:
Diameter of toy = 16 cm
Radius (r) = \(\frac { 16 }{ 2 }\) = 8 cm
Height of conical part (h) = 15 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 11.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 11.2

Question 12.
A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of ₹10 per m2.
Solution:
Diameter of the tank = 1.4 m
∴ Radius (r) = \(\frac { 1.4 }{ 2 }\) m = 0.7 m
and height of cylindrical portion = 8m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 12.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 12.2

Question 13.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2. [NCERT]
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.1
Solution:
Diameter of each spheres = 21 cm
∴ Radius (R) = \(\frac { 21 }{ 2 }\) cm
Radius of each cylinder (r) = 1.5 cm
and height (h) = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.2
Now surface area of one sphere = 4πR2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.3
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.4

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Value Based Questions in Science for Class 9 Chapter 16 Floatation

Value Based Questions in Science for Class 9 Chapter 16 Floatation

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 16 Floatation

VALUE BASED QUESTIONS

Question 1.
Anita and Suneeta were good friends. They had gone to a coastal side for walking. Anita was wearing flat surfaces chapal and Suneeta was wearing a high heel chapal. On the sandy surface, Suneeta was feeling uncomfortable while walking. Anita helped her to walk for some time on the sandy surface. Thereafter, both of them returned back. Anita asked Suneeta not to wear high heel chapal while walking on a sandy surface.
Answer the following questions based on the above paragraph.

  1. Why was Suneeta feeling uncomfortable while walking on the sandy surface ?
  2. Comment on the attitude of Anita.

Answer:

  1. The pressure exerted by Suneeta on the sandy surface was large as pressure = weight of Suneeta/area of heel of the chapal. Due to large pressure, her feet were sinking in sand. Therefore, she was feeling uncomfortable while walking.
  2. Anita is concerned about Suneeta. She was a good friend. Anita had high degree of general awareness.

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Question 2.
Amit is a student of class IX. His neighbour used to complain that he was not getting pure milk. But the man, who was supplying the milk said that the milk was pure. Amit knew how to test the purity of milk. He brought the lactometer and tested the milk. He told the man supplying milk not to cheat his neighbour.
Answer the following questions based on the above paragraph.

  1. On what principle, lactometer works ?
  2. What values are shown by Amit ?

Answer:

  1. Archimedes principle.
  2. Amit has high degree of general awareness. He is concerned about his neighbour. On the basis of his knowledge, he proved that the man supplying milk was wrong.

Question 3.
Some students were swimming in a swimming pool. Aayan, a student of class V was sitting sad near the swimming pool. Suneel came to Aayan and asked the reason of his sadness. Aayan told Suneel that he did not know how to swim. However, he also wanted to swim. Suneel asked Aayan to wear the life saving jacket. Then Suneel helped Aayan to swim.
Answer the following questions based on the above paragraph.

  1. What is the basic principle, on which life saving jacket works ?
  2. Comment on the attitude of Suneel.

Answer:

  1. When a person wears a life saving jacket and enters into water, the weight of water displaced by jacket is more than the weight of the person. Hence, the person can float in water easily.
  2. Suneel feels concerned for others. He could not see the sadness of Aayan. He is helpful. He used his knowledge to fulfill the desire of Aayan.

Question 4.
A milkman used to sell milk in the city and always carried lactometer with him. The customers trusted him and his business flourished.

  1. What is lactometer ?
  2. What is the principle of lactometer ?
  3. Carrying a lactometer by a milkman shows high values. List any two. (CBSE 2015)

Answer:

  1. Lactometer is a device used to test the purity of milk.
  2. Lactometer is based on Archimedes principle.
    1. Milkman is honest,
    2. He is a good business man.
    3. He is concerned with the health of his customers.

Hope given Value Based Questions in Science for Class 9 Chapter 16 Floatation are helpful to complete your science homework.

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