NCERT Exemplar Problems and Solutions for Class 10 Mathematics chapter 12, Surface Areas and Volumes is available here. This part brings solutions to questions numbered from 11 to 14 from exercise 12.4 that includes only the Long Answer Type Questions. Every question has been provided with a detailed solution.

 

ind below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Surface Areas and Volumes:

Exercise 12.4

Long Answer Type Questions (Q. No. 11-14):

Question. 11 16 glass spheres each of radius 2cm are packed into a cuboidal box of internal dimensions 16cm × 8cm × 8cm and then the box is filled with water. Find the volume of water filled in the box.

Solution.

Question. 12 A milk container of height 16cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8cm and 20cm, respectively. Find the cost of milk at the rate of ` 22 per L

which the container can hold.

Solution.              

Given, radius of lower end of milk container, r = 8 cm

Radius of upper end of milk container, = 20cm

And, height of milk container, h = 16cm

Now, as the container is in the shape of a a frustum of a cone,

Now, given that cost of 1 Lmilk = Rs.22

⟹ Cost of 10.45942 L milk = 22 × 1045942 = Rs.230.12

Hence, the required cost of milk contained in the container is Rs.230.12.

Question. 13 A cylindrical bucket of height 32cm and base radius 18cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24cm, find the radius and slant height of the heap.

Solution.

Given, radius of the base of the bucket = 18cm

And, height of the bucket = 32cm

Question. 14 A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a

cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6cm and 12cm,

respectively. If the slant height of the conical portion is 5cm, then find the total surface area and volume of the

rocket, (use p = 3.14)

Solution.              

Since, rocket is the combination of a right circular cylinder and a cone.

And total surface area of the rocket = CSA of cone + CSA of cylinder + Area of base of cylinder

                                                    = 47.1 + 226.08+28.26 = 301.44cm2