A juice seller was serving his customers using glass. The inner diameter of the cylindrical glass was 5 cm , but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was 10 cm, find what the apparent capacity of the glass was and what the actual capacity of the glass was

Answer:

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Step-by-step explanation:

★ Area of 4 walls = 2400/20

→ Area of 4 walls = 120 m²

★ Area of 4 walls = 120

→ 2(l + b)h = 120

→ (l + b)h = 120/2

→ (l + b)h = 60

★ Let l = b

→ 2lh = 60

→ lh = 60/2

→ lh = 30

★ Volume of a square room = 180

→ lbh = 180

→ l(30) = 180

→ l = 180/30

→ l = 6

★ Now put the values,

lh = 30

→ 6h = 30

→ h = 30/6

→ h = 5

Hence,theheightoftheroomis5m.

[tex]\large\underline{\sf{Solution-}}[/tex]

Dimensions of solid wooden cylinder

Radius, r = 6 cm

Height, h = 16 cm

Dimensions of cone

Radius, R = 2 cm

Height, H = 6 cm

According to statement, a solid wooden cylinder is of radius 6 cm and height 16 cm. Two cones each of radius 2 cm and height 6 cm are drilled out of the cylinder.

[tex]\bf \: \: Volume_{(Remaining\:solid)} \\ \\ [/tex]

[tex]\sf \: = \: Volume_{(Cylinder)} \: - \: 2 \: Volume_{(Cone)} \\ \\ [/tex]

[tex]\sf \: = \: \pi {(r)}^{2} h \: - \: 2 \times \dfrac{1}{3} \: \pi {(R)}^{2} H \\ \\ [/tex]

[tex]\sf \: = \: \pi {(r)}^{2} h \: - \: \dfrac{2}{3} \: \pi {(R)}^{2} H \\ \\ [/tex]

[tex]\sf \: = \: \pi \bigg({(r)}^{2} h \: - \: \dfrac{2}{3} \: {(R)}^{2} H\bigg) \\ \\ [/tex]

[tex]\sf \: = \: \pi \bigg({(6)}^{2} (16) \: - \: \dfrac{2}{3} \: {(2)}^{2} (6)\bigg) \\ \\ [/tex]

[tex]\sf \: = \: \pi \bigg(36 \times 16 \: - \: 16\bigg) \\ \\ [/tex]

[tex]\sf \: = \: 16 \: \pi \bigg(36 - 1\bigg) \\ \\ [/tex]

[tex]\sf \: = \: 16 \times \frac{22}{7} \times 35 \\ \\ [/tex]

[tex]\sf \: = \: 16 \times 22 \times 5\\ \\ [/tex]

[tex]\sf \: = \: 16 \times 110\\ \\ [/tex]

[tex]\sf \: = \: 1760 \: {cm}^{3} \\ \\ [/tex]

Hence,

[tex]\sf \:\bf\implies \:Volume_{(Remaining\:solid)} = \: 1760 \: {cm}^{3} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

Answer:

The correct answer of this question is [tex]1760cm^{3}[/tex]

Step-by-step explanation:

Given - A solid wooden cylinder is of radius 6 cm and height 16 cm. Two cones each of radius 2 cm and height 6 cm are drilled out of the cylinder.

To Find - Find the volume of the remaining solid.

Radius a solid wooden cylinder is  r = 6 cm

Height a solid wooden cylinder is h = 16 cm

A solid wooden cylinder with a radius of 6 cm and a height of 16 cm has been created. The cylinder is drilled with two cones, each with a radius of 2 cm and a height of 6 cm.

[tex]Volume_{cylinder} - 2 Volume_{cone} \\\pi r^{2}h - 2 . \frac{1}{3} \pi R^{2} H\\\pi r^{2} h - \frac{2}{3} \pi R^{2} H\\\pi (r^{2} h - \frac{2}{3} 2^{2} . 6)\\\pi (36 . 16 - 16 )\\16\pi ( 36 - 1)\\16 . \frac{22}{7} . 35\\16 . 22 . 5\\16 . 110 = 1760cm^{3}[/tex]

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Explanation:

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