CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
The volume of any three-dimensional object is the space occupied within the object's boundaries. There are some different formulas for calculating the volume of different objects. Similarly, volumes of the combination of two or more objects can be determined. Because sometimes, two or more different or the same types of objects are combined to form a new shape. The volume of the combination of an object is the sum of the volume of the combined object. For example, a toy is made with a right circular cone at the top and a hemisphere at the base, then the sum of the volumes of the right circular cone and the hemisphere is the total volume of the toy.
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This exercise has eight questions as per the NCERT Books that require students to visualise the shapes as described in the problem statements, then break up the figure, especially those types of figures whose volume or the formula to find its volume is known to us. So that we can very easily determine their volumes and either add them or subtract them to get the final volume of the resultant figure. 10th class Maths exercise 12.2 NCERT solutions are designed as per the students' demand, covering comprehensive, step-by-step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in-depth understanding of concepts. Students can find all exercises together using the link provided below.
Answer:
The volume of the solid is given by :
The volume of the solid = Volume of the cone + Volume of a hemisphere
The volume of the cone :
$=\ \frac{1}{3} \pi r^2h$
or $=\ \frac{1}{3} \pi \times 1^2\times 1$
or $=\ \frac{\pi}{3}\ cm^3$
And the volume of the hemisphere :
$=\ \frac{2}{3}\pi r^3$
or $=\ \frac{2}{3}\pi \times 1^3$
or $=\ \frac{2\pi}{3}\ cm^3$
Hence, the volume of the solid is :
$=\ \frac{\pi}{3}\ +\ \frac{2\pi}{3}\ =\ \pi\ cm^3$
Answer:
The volume of air present = Volume of cylinder + 2 (Volume of a cone)
Now, the volume of a cylinder: $=\ \pi r^2h$
or $=\ \pi \left ( \frac{3}{2} \right )^2\times 8$
or $=\ 18\pi \ cm^3$
And the volume of a cone is :
$=\ \frac{1}{3} \pi r^2h$
or $=\ \frac{1}{3} \pi \times \left ( \frac{3}{2} \right )^2\times 2$
or $=\ \frac{3}{2} \pi \ cm^3$
Thus, the volume of air is :
$=\ 18 \pi\ +\ 2\times \frac{3}{2} \pi \ =\ 21\pi$
or $=\ 66\ cm^3$
Answer:
It is clear from the figure that gulab jamun has one cylindrical part and two hemispherical parts.
Thus, the volume of gulab jamun is = the Volume of the cylindrical part + 2 (Volume of the hemisphere )
Now, the volume of the cylinder is $=\ \pi r^2h$
or $=\ \pi\times 1.4^2\times 2.2$
or $=\ 13.55\ cm^3$
And the volume of a hemisphere is :
$\\=\ \frac{2}{3}\pi r^3\\\\=\ \frac{2}{3}\times \pi \times (1.4)^3\\\\=\ 5.75\ cm^3$
Thus, the volume of 1 gulab jamun is $= 13.55 + 2 (5.75) = 25.05 cm^3.$
Hence the volume of 45 gulab jamun $=\ 45(25.05)\ cm^3\ =\ 1127.25\ cm^3$
Further, it is given that one gulab jamun contains sugar syrup up to $30$ $\%$.
So, the total volume of sugar present:
$=\ \frac{30}{100}\times 1127.25\ =\ 338\ cm^3$
Answer:
The volume of wood is given by = volume of the cuboid - the volume of four cones.
Firstly, the volume of a cuboid: $=\ lbh$
or $=\ 15\times 10\times 3.5$
or $=\ 525\ cm^3$
And, the volume of the cone :
$\\=\ \frac{1}{3} \pi r^2h\\=\ \frac{1}{3} \pi \times (0.5)^2\times 1.4\\\\=\ 0.3665\ cm^3$
Thus the volume of wood is $= 525 + 4 (0.3665) = 523.53 \:cm^3$
Answer:
According to the question :
Water spilt from the container = Volume of lead balls.
Let us assume the number of lead balls to be n.
Thus, the equation becomes :
$\frac{1}{4}\times Volume_{cone}\ =\ n\times \frac{4}{3} \pi r^3$
or $\frac{1}{4}\times \frac{1}{3}\pi\times5^2\times 8\ =\ n\times \frac{4}{3} \pi\times 0.5^3$
or $n\ =\ \frac{25\times 8}{16\times \left ( \frac{1}{2} \right )^3}$
or $n\ =\ 100$
Hence, the number of lead shots dropped is 100.
Answer:
The pole can be divided into one large cylinder and one small cylinder.
Thus, the volume of the pole = volume of the large cylinder + volume of the small cylinder
$=\ \pi r_l^2h_l\ +\ \pi r_s^2h_s$
or $=\ \pi \times 12^2\times 220\ +\ \pi \times 8^2\times 60$
or $=\ \pi \times \left ( 144\times 220\ +\ 64\times 60 \right )$
or $=\ 3.14\times 35520$
or $=\ 111532.5\ cm^3$
Now, according to the question mass of the pole is :
$=\ 8\times 111532.5$
or $=\ 892262.4\ g\ =\ 892.262\ Kg$
Answer:
It is clear from the question that the required volume is :
The volume of water (left) =Volume of a cylinder - Volume of solid
Now the volume of the cylinder is $=\ \pi r^2h$
or $=\ \pi\times (60)^2\times 180\ cm^3$
And the volume of the solid is :
$\\=\ \frac{1}{3} \pi r^2h\ +\ \frac{2}{3}\pi r^3\\\\=\ \frac{1}{3} \times \pi \times (60)^2\times 120\ +\ \frac{2}{3}\times \pi \times (60)^3\\\\=\ \pi (60)^2\times 80\ cm^3$
Thus, the volume of water left :
$\\=\ \pi (60)^2\times 180\ -\ \pi (60)^2\times 80\\\\=\ \pi (60)^2\times 100\\\\=\ 1131428.57\ cm^3\\\\=\ 1.131\ m^3$
Answer:
The volume of the vessel is given by :
The volume of the vessel = Volume of sphere + Volume of the cylindrical part
Now, the volume of the sphere is :
$\\=\ \frac{4}{3}\pi r^3\\\\=\ \frac{4}{3}\pi \left ( \frac{8.5}{2} \right )^3\\\\=\ 321.55\ cm^3$
And the volume of the cylinder is:-
$\\=\ \pi r^2h\\\\=\ \pi \times (1)^2\times 8\\\\=\ 25.13\ cm^3$
Thus the volume of the vessel is $= 321.55 + 25.13 = 346.68\: cm^3$
Also Read:
Also See:
Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.
Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.
Frequently Asked Questions (FAQs)
The core concept behind this exercise is to find the volume combination of solids. We have to remember the formula for finding the volume of cuboid, sphere, cylinder etc
On Question asked by student community
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