Maths-
General
Easy

Question

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 ?

Hint:

The volume of sphere = 4 over 3 pi r cubed

The correct answer is: 22.46mm3 ( approx )


    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)

    Related Questions to study

    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
    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
    parallel
    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
    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
    parallel
    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
    General
    Maths-

    straight angle A of an isosceles triangle ABC is acute in which A B equals A C text  and  end text B D perpendicular A C and . Prove that
    B C squared equals 2 cross times A C cross times C D

    Solution :-
    Aim  :- proveB C squared equals 2 ∗ A C ∗ C D
    Hint :- Using pythagoras theorem, find the length of  side AB and BC .
    Substitute the RHS of B C squared equals C D squared plus B D squared in terms of AC and CD Explanation(proof 
    Using pythagoras theorem,in triangle BDC
    B C squared equals C D squared plus B D squared— Eq1
    Using pythagoras theorem,in triangle BDA                                                                                                      A B squared equals B D squared plus A D squared not stretchy rightwards double arrow B D squared equals A B squared minus A D squared— Eq2
    Substitute Eq2 in Eq1
    B C squared equals A B squared minus A D squared plus C D squared
    Substitute AD = AC - CD (from diagram)
    B C squared equals A B squared minus left parenthesis A C minus C D right parenthesis squared plus C D squared not stretchy rightwards double arrow B C squared
    equals A B squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared
    text  As  end text A B equals A C left parenthesis text  given  end text right parenthesis text  we get  end text comma B C squared equals A C squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared
    not stretchy rightwards double arrow B C squared equals 2 ∗ A C ∗ C D
    Hence proved

     

    straight angle A of an isosceles triangle ABC is acute in which A B equals A C text  and  end text B D perpendicular A C and . Prove that
    B C squared equals 2 cross times A C cross times C D

    Maths-General
    Solution :-
    Aim  :- proveB C squared equals 2 ∗ A C ∗ C D
    Hint :- Using pythagoras theorem, find the length of  side AB and BC .
    Substitute the RHS of B C squared equals C D squared plus B D squared in terms of AC and CD Explanation(proof 
    Using pythagoras theorem,in triangle BDC
    B C squared equals C D squared plus B D squared— Eq1
    Using pythagoras theorem,in triangle BDA                                                                                                      A B squared equals B D squared plus A D squared not stretchy rightwards double arrow B D squared equals A B squared minus A D squared— Eq2
    Substitute Eq2 in Eq1
    B C squared equals A B squared minus A D squared plus C D squared
    Substitute AD = AC - CD (from diagram)
    B C squared equals A B squared minus left parenthesis A C minus C D right parenthesis squared plus C D squared not stretchy rightwards double arrow B C squared
    equals A B squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared
    text  As  end text A B equals A C left parenthesis text  given  end text right parenthesis text  we get  end text comma B C squared equals A C squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared
    not stretchy rightwards double arrow B C squared equals 2 ∗ A C ∗ C D
    Hence proved

     
    General
    Maths-

    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?

    Explanation:
    • We have given the diameter of moon is approximately one fourth of the diameter of the earth.
    • We have to find the fraction of the volume of the earth is the volume of the moon
    Step 1 of 1:
    Let the diameter of moon be ,Then the diameter of earth will be 4d
    So, Radius of moon will be d over 2. And the radius of earth will be fraction numerator 4 d over denominator 2 end fraction equals 2 d
    Since, both earth and moon are sphere
    We can use formula of volume of sphere 4 over 3 pi r cubed.
    V subscript m equals 4 over 3 pi r cubed
    equals 4 over 3 pi open parentheses d over 2 close parentheses cubed
    equals fraction numerator pi d cubed over denominator 6 end fraction
    And volume of earth will be
    V subscript E equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 2 d right parenthesis cubed
    equals fraction numerator 32 pi d cubed over denominator 3 end fraction
    Therefore, the fraction will be
    V subscript m over V subscript E equals fraction numerator pi d cubed divided by 6 over denominator 32 pi d cubed divided by 3 end fraction
    equals 1 over 64

    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?

    Maths-General
    Explanation:
    • We have given the diameter of moon is approximately one fourth of the diameter of the earth.
    • We have to find the fraction of the volume of the earth is the volume of the moon
    Step 1 of 1:
    Let the diameter of moon be ,Then the diameter of earth will be 4d
    So, Radius of moon will be d over 2. And the radius of earth will be fraction numerator 4 d over denominator 2 end fraction equals 2 d
    Since, both earth and moon are sphere
    We can use formula of volume of sphere 4 over 3 pi r cubed.
    V subscript m equals 4 over 3 pi r cubed
    equals 4 over 3 pi open parentheses d over 2 close parentheses cubed
    equals fraction numerator pi d cubed over denominator 6 end fraction
    And volume of earth will be
    V subscript E equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 2 d right parenthesis cubed
    equals fraction numerator 32 pi d cubed over denominator 3 end fraction
    Therefore, the fraction will be
    V subscript m over V subscript E equals fraction numerator pi d cubed divided by 6 over denominator 32 pi d cubed divided by 3 end fraction
    equals 1 over 64
    parallel
    General
    Maths-

    The diameter of a metallic ball is 4.2 cm. What is the mass of the ball , if the density of the metal is 8.9 g per cubic cm ?

    Explanation:
    • We have given diameter of metallic ball .and the density of the metal is
    • We have to find the mass of the ball.
    Step 1 of 1:
    We will first find the volume of the metallic ball
    V equals 4 over 3 pi r cubed
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell equals 4 over 3 pi left parenthesis 4.2 right parenthesis cubed end cell row cell equals 310.339 cm cubed end cell end table
    Now we know that
    Density = fraction numerator M a s s over denominator V o l u m e end fraction
    Put the value of density and volume in formula, we will get mass
    Density = fraction numerator M a s s over denominator V o l u m e end fraction
    8.9 = fraction numerator M a s s over denominator 310.399 end fraction
    Mass = 8.9 cross times 310.399gm
    = 2762.5511gm

    The diameter of a metallic ball is 4.2 cm. What is the mass of the ball , if the density of the metal is 8.9 g per cubic cm ?

    Maths-General
    Explanation:
    • We have given diameter of metallic ball .and the density of the metal is
    • We have to find the mass of the ball.
    Step 1 of 1:
    We will first find the volume of the metallic ball
    V equals 4 over 3 pi r cubed
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell equals 4 over 3 pi left parenthesis 4.2 right parenthesis cubed end cell row cell equals 310.339 cm cubed end cell end table
    Now we know that
    Density = fraction numerator M a s s over denominator V o l u m e end fraction
    Put the value of density and volume in formula, we will get mass
    Density = fraction numerator M a s s over denominator V o l u m e end fraction
    8.9 = fraction numerator M a s s over denominator 310.399 end fraction
    Mass = 8.9 cross times 310.399gm
    = 2762.5511gm
    General
    Maths-

    Find the amount of water displaced by a solid spherical ball of diameter 0.21m

    Explanation:
    • We have given a solid spherical ball with diameter
    • We have to find the volume of water displaced by the given sphere.
    Step 1 of 1:
    We know that the volume of water displaced by the given sphere will be equal to volume of sphere.
    So,
    Volume of sphere will be

    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi open parentheses fraction numerator 0.21 over denominator 2 end fraction close parentheses cubed
    equals 0.00484 m cubed

    Find the amount of water displaced by a solid spherical ball of diameter 0.21m

    Maths-General
    Explanation:
    • We have given a solid spherical ball with diameter
    • We have to find the volume of water displaced by the given sphere.
    Step 1 of 1:
    We know that the volume of water displaced by the given sphere will be equal to volume of sphere.
    So,
    Volume of sphere will be

    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi open parentheses fraction numerator 0.21 over denominator 2 end fraction close parentheses cubed
    equals 0.00484 m cubed

    General
    Maths-

    Write truth table for the converse of p → q.

    Hint:
    If a conditional statement of the form "If p then q" is given. The converse is "If q then p." Symbolically, the converse of p not stretchy rightwards arrow q is
    q not stretchy rightwards arrow p
    Solution
    The converse of the conditional statement p not stretchy rightwards arrow q is q not stretchy rightwards arrow  p

    Final Answer:
    Hence, we have drawn the truth table above.

     

    Write truth table for the converse of p → q.

    Maths-General
    Hint:
    If a conditional statement of the form "If p then q" is given. The converse is "If q then p." Symbolically, the converse of p not stretchy rightwards arrow q is
    q not stretchy rightwards arrow p
    Solution
    The converse of the conditional statement p not stretchy rightwards arrow q is q not stretchy rightwards arrow  p

    Final Answer:
    Hence, we have drawn the truth table above.

     
    parallel
    General
    Maths-

    Find the volume of a sphere whose radius is 0.63 m

    Explanation:
    • We have given a sphere with radius .
    • We have to find the volume of sphere.
    Step 1 of 1:
    We know that the volume of sphere with radius  is .
    Now put the value of
    So,
    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 0.63 right parenthesis cubed
    equals 1.0473 straight m cubed

    Find the volume of a sphere whose radius is 0.63 m

    Maths-General
    Explanation:
    • We have given a sphere with radius .
    • We have to find the volume of sphere.
    Step 1 of 1:
    We know that the volume of sphere with radius  is .
    Now put the value of
    So,
    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 0.63 right parenthesis cubed
    equals 1.0473 straight m cubed
    General
    Maths-

    Make a valid conclusion in the situation.
    If two points lie in a plane, then the line containing them lies in the plane.
    Points A and B lie in plane PQR.

    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: Two points lie in a plane
    q: Line containing the two points is in the plane
    So we can write the given statement “If two points lie in a plane, then the line containing them lies in the plane” as
    not stretchy rightwards arrow q
    We are given that two points A and B lie in plane PQR so we can conclude that the line containing the points A and B  lie in the plane PQR.
    Final Answer:
    Hence we can conclude that the line containing points A and B  lie in the plane PQR.

    Make a valid conclusion in the situation.
    If two points lie in a plane, then the line containing them lies in the plane.
    Points A and B lie in plane PQR.

    Maths-General
    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: Two points lie in a plane
    q: Line containing the two points is in the plane
    So we can write the given statement “If two points lie in a plane, then the line containing them lies in the plane” as
    not stretchy rightwards arrow q
    We are given that two points A and B lie in plane PQR so we can conclude that the line containing the points A and B  lie in the plane PQR.
    Final Answer:
    Hence we can conclude that the line containing points A and B  lie in the plane PQR.
    General
    Maths-

    The conditional p → q is only false when
    a. p = T, q = T
    b. p = T, q = F
    c. p = F, q = F
    d. p = F, q = T

    Hint:
    p → q is a conditional statement which mean if p then q. The conditional statement is saying that if p is true, then q will immediately follow and thus be true
    Solution
    The truth table of p → q is given as

    Hence, we can see that p → q is only false when p is True and q is False.
    Final Answer:
    So, p → q is only false when p = T and q = F. Hence option b is correct.

    The conditional p → q is only false when
    a. p = T, q = T
    b. p = T, q = F
    c. p = F, q = F
    d. p = F, q = T

    Maths-General
    Hint:
    p → q is a conditional statement which mean if p then q. The conditional statement is saying that if p is true, then q will immediately follow and thus be true
    Solution
    The truth table of p → q is given as

    Hence, we can see that p → q is only false when p is True and q is False.
    Final Answer:
    So, p → q is only false when p = T and q = F. Hence option b is correct.
    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.