Question

A test tube has diameter 20mm and height is 15 cm . The lower portion is a hemisphere . Find the capacity of the test tube.​

Answer 1

❏ Question:-

A test tube has diameter 20mm and height is 15 cm . The lower portion is a hemisphere . Find the capacity of the test tube.

❏ Solution:-

✏ Given:-

• diameter (d)= 20 mm .
• radius (r)= 10 mm= 1 cm.
• height (h)=15 cm.

✏ To Find:-

• Capacity of this test tube = ?

✏ Explanation:-

To find the capacity of this test tube ...

first:- we have to calculate the volume of the cylindrical portion

second:- we have to calculate the volume of the

hemispherical portion,

Let's calculate:-

$\setlength{\unitlength}{1 cm}\begin{picture}(12,4)\thicklines\put(2.7,2.7){ . }\put(6,3){\circle{2}}\put(6,6){\circle{2}}\put(6,3){\line(0,1){3}}\put(5.3,3){\line(0,1){3}}\put(6.7,3){\line(0,1){3}}\put(6,3){\line(1,0){0.7}}\put(6,6){\line(1,0){0.7}}\put(6.2,6.13){ 1\:cm }\put(6.8,4.5){ 15\:cm }\end{picture}$

• Volume of the cylindrical portion

$\sf\longrightarrow Volume_{cylinder}=\pi r^2 h$

• Volume of the hemispherical portion

$\sf\longrightarrow Volume_{hemisphere}=\frac{2}{3}\pi r^3$

Hence,

Total volume of the test tube is .

$\sf\longrightarrow Volume_{test\:tube}=\pi r^2 h+\frac{2}{3}\pi r^3$

$\sf\longrightarrow Volume_{test\:tube}=\pi r^2( h+\frac{2r}{3})$

(putting the values);

$\sf\longrightarrow Volume_{test\:tube}=\frac{22}{7} \times1^2( 15+\frac{2\times1}{3})$

$\sf\longrightarrow Volume_{test\:tube}=\frac{22}{7} \times1( 15+\frac{2}{3})$

$\sf\longrightarrow Volume_{test\:tube}=\frac{22}{7} \times1( 15+\frac{2}{3})$

$\sf\longrightarrow Volume_{test\:tube}=49.24\:cm^3$ (approx)

∴ Capacity of that test tube is = 49.24 cm^3

Answer 2

Step-by-step explanation:

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