Maths-
General
Easy

Question

A hemisphere has 3 cm radius , find its volume?

Hint:

Volume of hemisphere is 2 over 3 pi r cubed.

The correct answer is: 56.54


    Explanation:
    • We have given radius of hemisphere 3cm.
    • We have to find the volume of the hemisphere.
    Step 1 of 1:
    We have given the radius of the hemisphere 3cm
    So, The Volume will be

    V equals 2 over 3 pi 3 cubed

    equals 2 pi cross times 9
    equals 56.54

    Related Questions to study

    General
    Maths-

    Bloom’s category: Understanding
    Level: Medium
    Identify and name three collinear points.

    ANS :- Point F, point O ,point G are three collinear points.
    The  points that lie on the same straight line are Collinear points.
    From the diagram we get points F,O,G lie on the straight line AC .so,they are collinear.
    We get Point F, point O ,point G are three collinear points .
    ∴ Point F, point O ,point G are three collinear points.

    Bloom’s category: Understanding
    Level: Medium
    Identify and name three collinear points.

    Maths-General
    ANS :- Point F, point O ,point G are three collinear points.
    The  points that lie on the same straight line are Collinear points.
    From the diagram we get points F,O,G lie on the straight line AC .so,they are collinear.
    We get Point F, point O ,point G are three collinear points .
    ∴ Point F, point O ,point G are three collinear points.
    General
    Maths-

    A dome of a building is in the form of a hemisphere. The total cost of white washing it from inside, was '1330.56. The rate at which it was white washed is '3 per square metre. Find the volume of the dome approximately

    Explanation:
    • We have given total cost of white washing the dome is 1330.56
    • We have to find the volume of dome.
    We have given total cost of white washing the dome is 1330.56
    And the rate of white washing is 3 per square meter
    So, The surface area of the dome is
    equals fraction numerator 1330.56 over denominator 3 end fraction
    = 443.52m2
    We know that white washing is only done on curved surface area
    So,
    2 pi r squared equals 443.52 m squared
    r squared equals 696.679
    r equals 26.39 m
    Now the volume of dome will be
    V equals 2 over 3 pi r cubed
    equals 2 over 3 pi left parenthesis 26.39 right parenthesis cubed
    equals 38492.55 m cubed

    A dome of a building is in the form of a hemisphere. The total cost of white washing it from inside, was '1330.56. The rate at which it was white washed is '3 per square metre. Find the volume of the dome approximately

    Maths-General
    Explanation:
    • We have given total cost of white washing the dome is 1330.56
    • We have to find the volume of dome.
    We have given total cost of white washing the dome is 1330.56
    And the rate of white washing is 3 per square meter
    So, The surface area of the dome is
    equals fraction numerator 1330.56 over denominator 3 end fraction
    = 443.52m2
    We know that white washing is only done on curved surface area
    So,
    2 pi r squared equals 443.52 m squared
    r squared equals 696.679
    r equals 26.39 m
    Now the volume of dome will be
    V equals 2 over 3 pi r cubed
    equals 2 over 3 pi left parenthesis 26.39 right parenthesis cubed
    equals 38492.55 m cubed
    General
    Maths-

    Bloom’s category: Understanding
    Level: Medium
    Through any three points not on the same line,

    ANS :- Option B
    If the three points do not lie on the same line then they are coplanar. We can say that any two points are co linear so we get 2 lines by connecting them in pairs of two points ,but we know that only 1 plane passes through any two lines.
    i.e three points lie on same plane
    ∴ Option B

    Bloom’s category: Understanding
    Level: Medium
    Through any three points not on the same line,

    Maths-General
    ANS :- Option B
    If the three points do not lie on the same line then they are coplanar. We can say that any two points are co linear so we get 2 lines by connecting them in pairs of two points ,but we know that only 1 plane passes through any two lines.
    i.e three points lie on same plane
    ∴ Option B
    parallel
    General
    Maths-

    A hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the volume of steel used.

    Explanation:
    • We have given a hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm.
    • We have to find the volume of the steel
    We have given thickness of steel is 0.25cm
    Inner radius is equal to 5cm
    So, Outer radius is equal to 5 + 0.25 = 5.25cm
    Outer Volume of the hemisphere will be

    V subscript o equals 2 over 3 pi r cubed

    equals 2 over 3 pi left parenthesis 5.25 right parenthesis cubed
    equals 303.065 cm cubed
    Inner volume will be

    V subscript i equals 2 over 3 pi r cubed

    equals 2 over 3 pi left parenthesis 5 right parenthesis cubed
    equals 261.799 cm cubed
    Therefore, the volume of steel used will be

    equals V subscript o minus V subscript i

    equals 303.065 minus 261.799
    equals 41.265 cm cubed

    A hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the volume of steel used.

    Maths-General
    Explanation:
    • We have given a hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm.
    • We have to find the volume of the steel
    We have given thickness of steel is 0.25cm
    Inner radius is equal to 5cm
    So, Outer radius is equal to 5 + 0.25 = 5.25cm
    Outer Volume of the hemisphere will be

    V subscript o equals 2 over 3 pi r cubed

    equals 2 over 3 pi left parenthesis 5.25 right parenthesis cubed
    equals 303.065 cm cubed
    Inner volume will be

    V subscript i equals 2 over 3 pi r cubed

    equals 2 over 3 pi left parenthesis 5 right parenthesis cubed
    equals 261.799 cm cubed
    Therefore, the volume of steel used will be

    equals V subscript o minus V subscript i

    equals 303.065 minus 261.799
    equals 41.265 cm cubed

    General
    Maths-

    Bloom’s category: Understanding
    Level: Medium
    A plane has ____ dimensions.

    ANS :- Option B
    A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.It has 2 dimension (i.e length and breadth are infinite and no height dimension)
    ∴ Option B


    Bloom’s category: Understanding
    Level: Medium
    A plane has ____ dimensions.

    Maths-General
    ANS :- Option B
    A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.It has 2 dimension (i.e length and breadth are infinite and no height dimension)
    ∴ Option B


    General
    Maths-

    If the Diameter of the metallic sphere is 6 cm , it melted and a wire of diameter 0.2 cm is drawn. Then , what would be the length of the wire ?

    Explanation:
    • We have given diameter of the metallic sphere is 6 cm , it melted and a wire of diameter 0.2 cm is drawn.
    • We have to find the length of the wire.
    Step 1 of 1:
    We have  diameter of metallic sphere 6cm
    Radius of metallic sphere 3cm
    Also we have diameter of cross-section of cylinder wire 0.2cm
    So, Radius of cross-section of cylinder wire 0.1cm
    Let the length of the wire be hcm
    Since the metallic sphere is converted into cylindrical shaped wire of length hcm
    Volume of metal used in wire is equal to volume of sphere
    So,
    pi r squared h equals 4 over 3 pi r cubed
    pi left parenthesis 0.1 right parenthesis squared h equals 4 over 3 pi left parenthesis 3 right parenthesis cubed
    0.01h = 36
    h = 3600 cm
    Therefore, The length of the wire is 3600cm.

    If the Diameter of the metallic sphere is 6 cm , it melted and a wire of diameter 0.2 cm is drawn. Then , what would be the length of the wire ?

    Maths-General
    Explanation:
    • We have given diameter of the metallic sphere is 6 cm , it melted and a wire of diameter 0.2 cm is drawn.
    • We have to find the length of the wire.
    Step 1 of 1:
    We have  diameter of metallic sphere 6cm
    Radius of metallic sphere 3cm
    Also we have diameter of cross-section of cylinder wire 0.2cm
    So, Radius of cross-section of cylinder wire 0.1cm
    Let the length of the wire be hcm
    Since the metallic sphere is converted into cylindrical shaped wire of length hcm
    Volume of metal used in wire is equal to volume of sphere
    So,
    pi r squared h equals 4 over 3 pi r cubed
    pi left parenthesis 0.1 right parenthesis squared h equals 4 over 3 pi left parenthesis 3 right parenthesis cubed
    0.01h = 36
    h = 3600 cm
    Therefore, The length of the wire is 3600cm.
    parallel
    General
    Maths-

    Bloom’s category: Understanding
    Level: Medium
    A point has ____ dimension.

    ANS :- Option C
    A point is the fundamental object in geometry. It is represented by a dot and point represents position only.It has no dimension (i.e 0 dimension)
    ∴ Option C

    Bloom’s category: Understanding
    Level: Medium
    A point has ____ dimension.

    Maths-General
    ANS :- Option C
    A point is the fundamental object in geometry. It is represented by a dot and point represents position only.It has no dimension (i.e 0 dimension)
    ∴ Option C
    General
    Maths-

    A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine is needed to fill this capsule ?

    Explanation:
    • We have given a capsule with diameter 3.5mm
    • We have to find the how much medicine require to fill the capsule.
    Step 1 of 1:
    Since the capsule is spherical in shape, the amount of medicine needed to fill the capsule is the volume of a sphere.
    The volume of sphere = 4 over 3 pi r cubed
    Diameter of the spherical capsule, d = 3.5mm
    Radius of the spherical capsule, r = 3.5 / 2mm = 1.75 mm
    Now, the medicine is needed to fill this capsule = The volume of the spherical capsule
    The medicine needed to fill this capsule = 4 over 3 pi r cubed
    = 4/3 cross times 22/7 cross times 1.75mm cross times 1.75mm cross times 1.75mm
    22 .46mm3 (approx)

    A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine is needed to fill this capsule ?

    Maths-General
    Explanation:
    • We have given a capsule with diameter 3.5mm
    • We have to find the how much medicine require to fill the capsule.
    Step 1 of 1:
    Since the capsule is spherical in shape, the amount of medicine needed to fill the capsule is the volume of a sphere.
    The volume of sphere = 4 over 3 pi r cubed
    Diameter of the spherical capsule, d = 3.5mm
    Radius of the spherical capsule, r = 3.5 / 2mm = 1.75 mm
    Now, the medicine is needed to fill this capsule = The volume of the spherical capsule
    The medicine needed to fill this capsule = 4 over 3 pi r cubed
    = 4/3 cross times 22/7 cross times 1.75mm cross times 1.75mm cross times 1.75mm
    22 .46mm3 (approx)
    General
    Maths-

    Prove that the sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of its sides.

    Solution :-
    Aim  :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. d subscript 1 squared plus d subscript 2 squared equals 2 open parentheses l squared plus b squared close parentheses where , d subscript 1 and d subscript 2 are lengths of diagonals
    Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.
    Explanation(proof ) :-
    text  Let length of  end text A D equals L text  and side  end text C D equals b
    As we know opposite sides of parallelogram are equal i.e AD = BC = L
    text  Consider  end text straight triangle ADX text  and  end text straight triangle BCY
    AX and BY are perpendicular to line DC  and intersect at X and Y respectively.
    Consider ΔADX and ΔBCY ,
    AX = BY (distance between two parallel lines is always equal)
    AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)
    Both have right angles at X and Y .
    By RHS congruence rule,
    ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

    Let DX = CY = a and AC = d subscript 1  ; BD =  d subscript 2
    Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

    Applying pythagoras theorem in ΔACX ,Wetext  get  end text A C squared equals A X squared plus X C squared not stretchy rightwards double arrow A C squared equals h squared plus

    left parenthesis b minus a right parenthesis squared not stretchy rightwards double arrow d subscript l superscript 2 equals h squared plus left parenthesis b minus a right parenthesis squared minus Eq 1
    Applying pythagoras theorem in ΔBYD, we get
    B D squared equals B Y squared plus D Y squared not stretchy rightwards double arrow B D squared equals h squared plus left parenthesis b plus a right parenthesis squared not stretchy rightwards double arrow d subscript 2 superscript 2 equals h squared plus left parenthesis b plus a right parenthesis squared minus Eq 2 — Eq2
    Adding equation 2 and 1 we get ,
     d subscript 1 superscript 2 plus d subscript 2 superscript 2 equals 2 h squared plus 2 open parentheses b squared plus a squared close parentheses Eq3

    Applying pythagoras theorem in ΔADX,
    A D squared equals A X squared plus D X squared not stretchy rightwards double arrow L squared equals a squared plus h squared—- Eq4
    Substitute Eq4 in Eq3
    We get d subscript 1 squared plus d subscript 2 squared equals 2 h squared plus 2 b squared plus 2 a squared not stretchy rightwards double arrow d subscript 1 squared plus d subscript 2 squared equals 2 open parentheses h squared plus a squared close parentheses plus 2 b squared
    not stretchy rightwards double arrow d subscript 1 superscript 2 plus d subscript 2 superscript 2 equals 2 open parentheses l squared plus b squared close parentheses

    Prove that the sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of its sides.

    Maths-General
    Solution :-
    Aim  :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. d subscript 1 squared plus d subscript 2 squared equals 2 open parentheses l squared plus b squared close parentheses where , d subscript 1 and d subscript 2 are lengths of diagonals
    Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.
    Explanation(proof ) :-
    text  Let length of  end text A D equals L text  and side  end text C D equals b
    As we know opposite sides of parallelogram are equal i.e AD = BC = L
    text  Consider  end text straight triangle ADX text  and  end text straight triangle BCY
    AX and BY are perpendicular to line DC  and intersect at X and Y respectively.
    Consider ΔADX and ΔBCY ,
    AX = BY (distance between two parallel lines is always equal)
    AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)
    Both have right angles at X and Y .
    By RHS congruence rule,
    ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

    Let DX = CY = a and AC = d subscript 1  ; BD =  d subscript 2
    Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

    Applying pythagoras theorem in ΔACX ,Wetext  get  end text A C squared equals A X squared plus X C squared not stretchy rightwards double arrow A C squared equals h squared plus

    left parenthesis b minus a right parenthesis squared not stretchy rightwards double arrow d subscript l superscript 2 equals h squared plus left parenthesis b minus a right parenthesis squared minus Eq 1
    Applying pythagoras theorem in ΔBYD, we get
    B D squared equals B Y squared plus D Y squared not stretchy rightwards double arrow B D squared equals h squared plus left parenthesis b plus a right parenthesis squared not stretchy rightwards double arrow d subscript 2 superscript 2 equals h squared plus left parenthesis b plus a right parenthesis squared minus Eq 2 — Eq2
    Adding equation 2 and 1 we get ,
     d subscript 1 superscript 2 plus d subscript 2 superscript 2 equals 2 h squared plus 2 open parentheses b squared plus a squared close parentheses Eq3

    Applying pythagoras theorem in ΔADX,
    A D squared equals A X squared plus D X squared not stretchy rightwards double arrow L squared equals a squared plus h squared—- Eq4
    Substitute Eq4 in Eq3
    We get d subscript 1 squared plus d subscript 2 squared equals 2 h squared plus 2 b squared plus 2 a squared not stretchy rightwards double arrow d subscript 1 squared plus d subscript 2 squared equals 2 open parentheses h squared plus a squared close parentheses plus 2 b squared
    not stretchy rightwards double arrow d subscript 1 superscript 2 plus d subscript 2 superscript 2 equals 2 open parentheses l squared plus b squared close parentheses
    parallel
    General
    Maths-

    Twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’. Find the i) Radius r’ of the new sphere ii) ratio of S and S’

    Explanation:
    • We have given twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’.
    • We have to find the Radius of new sphere and the ratio of S and S’
    Step 1 of 2:
    We have given 27 solid spheres each of radius r and surface area s
    So,
    (i)
    Volume of each sphere will be
    V equals 27 cross times 4 over 3 pi r cubed
    Volume of big sphere
    V equals 4 over 3 pi r to the power of 13
    So,
    4 over 3 pi r to the power of straight prime 3 end exponent equals 27 cross times 4 over 3 pi r cubed
    r to the power of straight prime equals 3 r
    Step 2 of 2:
    Surface area is 4 pi r squared
    s equals 4 pi r squared and
    equals 4 pi left parenthesis 3 r right parenthesis squared
    equals 36 pi r squared
    Therefore,
    s over s to the power of straight prime equals fraction numerator 4 pi r squared over denominator 36 pi r squared end fraction
    1 over 9
    Or
    1 : 9

    Twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’. Find the i) Radius r’ of the new sphere ii) ratio of S and S’

    Maths-General
    Explanation:
    • We have given twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’.
    • We have to find the Radius of new sphere and the ratio of S and S’
    Step 1 of 2:
    We have given 27 solid spheres each of radius r and surface area s
    So,
    (i)
    Volume of each sphere will be
    V equals 27 cross times 4 over 3 pi r cubed
    Volume of big sphere
    V equals 4 over 3 pi r to the power of 13
    So,
    4 over 3 pi r to the power of straight prime 3 end exponent equals 27 cross times 4 over 3 pi r cubed
    r to the power of straight prime equals 3 r
    Step 2 of 2:
    Surface area is 4 pi r squared
    s equals 4 pi r squared and
    equals 4 pi left parenthesis 3 r right parenthesis squared
    equals 36 pi r squared
    Therefore,
    s over s to the power of straight prime equals fraction numerator 4 pi r squared over denominator 36 pi r squared end fraction
    1 over 9
    Or
    1 : 9
    General
    Maths-

    Bloom’s category: Application
    Level: Medium
    Identify points, lines and planes in the given image of a floor.

    HINT :- Identify points, lines and planes in the given image using the definitions of them .
    ANS :- 15 points(intersections points) ,8 lines and 1 plane
    Explanation :-
    A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.
    A point is the fundamental object in geometry. It is represented by a dot and point represents position only.
    A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

    We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and
    8 lines (= 5+3)
    Here all these lie on the plane (surface of floor)
    ∴ 15 points(intersections points) ,8 lines and 1 plane

    Bloom’s category: Application
    Level: Medium
    Identify points, lines and planes in the given image of a floor.

    Maths-General
    HINT :- Identify points, lines and planes in the given image using the definitions of them .
    ANS :- 15 points(intersections points) ,8 lines and 1 plane
    Explanation :-
    A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.
    A point is the fundamental object in geometry. It is represented by a dot and point represents position only.
    A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

    We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and
    8 lines (= 5+3)
    Here all these lie on the plane (surface of floor)
    ∴ 15 points(intersections points) ,8 lines and 1 plane
    General
    Maths-

    A dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of RS 4989.60. If the cost of white washing is Rs 20 per square metre , find the i) inside surface area of the dome ii) Volume of the air inside the dome

    Explanation:
    • We have given a dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of . If the cost of white washing is per square metre.
    • We have to find the (i) Inside surface area of dome. (ii) Volume of the air inside the dome
    Step 1 of 1:
    It is given that the total cost of white washing is Rs 4989.60.
    The cost of white washing per unit square meter is Rs 20.
    So,
    Total inside surface area will be
    fraction numerator T o t a l space cos t over denominator C o s t space p e r space u n i t space a r e a end fraction
    fraction numerator 4929.60 over denominator 20 end fraction
    = 246.48m2
    Step 2 of 2:
    Now inside surface area is 246.48m2
    2 pi r squared equals 246.48 m squared
    r2 = 39.22
    r = 6.26
    So, The volume will be
    V equals 2 over 3 pi r cubed
    equals 2 over 3 pi left parenthesis 6.26 right parenthesis cubed
    = 82.07
    Hence, The inside surface area is 246.48m2 and the inside volume is 82.07m3

    A dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of RS 4989.60. If the cost of white washing is Rs 20 per square metre , find the i) inside surface area of the dome ii) Volume of the air inside the dome

    Maths-General
    Explanation:
    • We have given a dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of . If the cost of white washing is per square metre.
    • We have to find the (i) Inside surface area of dome. (ii) Volume of the air inside the dome
    Step 1 of 1:
    It is given that the total cost of white washing is Rs 4989.60.
    The cost of white washing per unit square meter is Rs 20.
    So,
    Total inside surface area will be
    fraction numerator T o t a l space cos t over denominator C o s t space p e r space u n i t space a r e a end fraction
    fraction numerator 4929.60 over denominator 20 end fraction
    = 246.48m2
    Step 2 of 2:
    Now inside surface area is 246.48m2
    2 pi r squared equals 246.48 m squared
    r2 = 39.22
    r = 6.26
    So, The volume will be
    V equals 2 over 3 pi r cubed
    equals 2 over 3 pi left parenthesis 6.26 right parenthesis cubed
    = 82.07
    Hence, The inside surface area is 246.48m2 and the inside volume is 82.07m3
    parallel
    General
    Maths-

    Find the volume of a sphere whose surface area is 154 sq.cm

    Explanation:
    • We have given surface area 154 cm squared.
    • We have to find the volume of the sphere.
    Step 1 of 1:
    We have given surface area 154cm2.
    So,
    4 pi r squared equals 154 cm squared
    r squared equals fraction numerator 154 over denominator 4 pi end fraction
    r squared equals 12.25
    r = 3.5
    Volume of sphere
    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 3.5 right parenthesis cubed
    equals 179.67 cm cubed
    Thus, the volume of the sphere is 179.67cm3.

    Find the volume of a sphere whose surface area is 154 sq.cm

    Maths-General
    Explanation:
    • We have given surface area 154 cm squared.
    • We have to find the volume of the sphere.
    Step 1 of 1:
    We have given surface area 154cm2.
    So,
    4 pi r squared equals 154 cm squared
    r squared equals fraction numerator 154 over denominator 4 pi end fraction
    r squared equals 12.25
    r = 3.5
    Volume of sphere
    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 3.5 right parenthesis cubed
    equals 179.67 cm cubed
    Thus, the volume of the sphere is 179.67cm3.
    General
    Maths-

    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 ?

    Explanation:
    • We have given a hemisphere with thickness 1cm.
    • We have to find the volume of iron used to make the tank.
    Step 1 of 1:
    Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.
    The Volume of hemisphere of base radius r is equal to 2 over 3 pi r cubed
    The inner radius of the tank r = 1m
    Thickness of iron = 1cm = 1/100 m =0.01m
    Outer radius of the tank, R = 1m + 0.01m = 1.01m
    The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.
    Volume of the iron used to make the tank 2 over 3 pi R cubed minus 2 over 3 pi r cubed
    equals 2 divided by 3 pi open parentheses R cubed minus r squared close parentheses
    equals 2 divided by 3 cross times 22 divided by 7 cross times open square brackets left parenthesis 1.01 m right parenthesis cubed minus left parenthesis 1 m right parenthesis cubed close square brackets
    equals 2 divided by 3 cross times 22 divided by 7 cross times open square brackets 1.030301 m cubed minus 1 m cubed close square brackets
    equals 2 divided by 3 cross times 22 divided by 7 cross times 0.030301 straight m cubed
    equals 0.06348 straight m cubed text  (approx.)  end text
    0.06348 mof iron used to make the tank

    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 ?

    Maths-General
    Explanation:
    • We have given a hemisphere with thickness 1cm.
    • We have to find the volume of iron used to make the tank.
    Step 1 of 1:
    Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.
    The Volume of hemisphere of base radius r is equal to 2 over 3 pi r cubed
    The inner radius of the tank r = 1m
    Thickness of iron = 1cm = 1/100 m =0.01m
    Outer radius of the tank, R = 1m + 0.01m = 1.01m
    The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.
    Volume of the iron used to make the tank 2 over 3 pi R cubed minus 2 over 3 pi r cubed
    equals 2 divided by 3 pi open parentheses R cubed minus r squared close parentheses
    equals 2 divided by 3 cross times 22 divided by 7 cross times open square brackets left parenthesis 1.01 m right parenthesis cubed minus left parenthesis 1 m right parenthesis cubed close square brackets
    equals 2 divided by 3 cross times 22 divided by 7 cross times open square brackets 1.030301 m cubed minus 1 m cubed close square brackets
    equals 2 divided by 3 cross times 22 divided by 7 cross times 0.030301 straight m cubed
    equals 0.06348 straight m cubed text  (approx.)  end text
    0.06348 mof iron used to make the tank
    General
    Maths-

    How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

    Explanation:
    • We have given a hemispherical bowl of diameter 10.5cm
    • We have to find how many litre of milk this bowl can hold.
    Step 1 of 1:
    The diameter of the hemispherical bowl is
    Radius will be fraction numerator 10.5 cm over denominator 2 end fraction equals 5.25 cm
    So, The volume of the hemispherical bowl will be
    V equals 2 over 3 pi r cubed
    equals 2 over 3 pi left parenthesis 5.25 right parenthesis cubed
    equals 303.065 cm cubed
    Also we know that 1 cm cubed equals 0.001 text  liter  end text
    So,
    303.065 cm cubed equals 0.3030 text  liter  end text

    How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

    Maths-General
    Explanation:
    • We have given a hemispherical bowl of diameter 10.5cm
    • We have to find how many litre of milk this bowl can hold.
    Step 1 of 1:
    The diameter of the hemispherical bowl is
    Radius will be fraction numerator 10.5 cm over denominator 2 end fraction equals 5.25 cm
    So, The volume of the hemispherical bowl will be
    V equals 2 over 3 pi r cubed
    equals 2 over 3 pi left parenthesis 5.25 right parenthesis cubed
    equals 303.065 cm cubed
    Also we know that 1 cm cubed equals 0.001 text  liter  end text
    So,
    303.065 cm cubed equals 0.3030 text  liter  end text
    parallel

    card img

    With Turito Academy.

    card img

    With Turito Foundation.

    card img

    Get an Expert Advice From Turito.

    Turito Academy

    card img

    With Turito Academy.

    Test Prep

    card img

    With Turito Foundation.