In this chapter, we provide KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 pdf, free KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 10 Maths Solutions** **Chapter 15 Surface Areas and Volumes Ex 15.2**

(Unless stated otherwise, take n =227

Question 1.

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of ‘n’.

Solution:

Radius of hemisphere, r = 1 cm.

Volume of hemisphere, V =23πr3

=2π3×(1)3

Radius of Cone, r = 1 cm,

Height of Cone, h = 1 cm.

∴ Volume of a cone, V =13πr2h

∴ Volume of Total cube =2π3+π3cm3

= πcm^{3}

Question 2.

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

solution:

Volume of cylinder, V = πr^{2}h

= π × (1.5)^{2} × 8

= 18π cm^{3}.

Volume of Cylinder V =13πr2h

=13π×(1.5)2×2

=32πcm3

∴ The volume of air contained in the model that Rachel made.

= 66 cm^{3}.

Question 3.

A Gulab Jamun contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (see figure given aside).

Solution:

Length of cylindrical Gulab jamun,

l = 5 – (1.4 + 1.4) = 2.2 cm.

Radius, r = 1.4 cm.

∴ Volume of Cylinder (1 Gulab Jamun), = Left hemisphere + Volume of Cylinder + Right hemisphere

∴ Total volume of such 45 Gulab jamun,

=45×223×0.28×12.2

= 22 × 0.28 × 183

= 1127.28 cm^{3}.

∴ Gulab Jamun contains sugar syrup upto about 30% of its volume.

∴ 1127.28×30100

= 338 cm^{3}.

Question 4.

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see

Figure)

Solution:

Radius of conical depression, r = 0.5 cm.

(Depth) height, h = 1.4 cm.

Volume of 1 depression which is conical, =13πr2h

Volume of such 4 depressions

=4×1130cm3

=4430cm3

∴ Total volume of wooden pen-stand, =(15×10×3.5)−4430

= 525 – 1.47

= 523.53 cm^{3}

Question 5

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution:

Volume of water in a Cone,

Volume of lead shot =43πr3

=43π×(0.5)3

=π6cm3

Let the number of lead balls kept in vessel 14 of the water flows out,

⇒n×π6=14×200π3

⇒n×π6=1006π

∴ n = 100

∴ Number of lead balls =100.

Question 6.

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. (Find the mass of the pole, given that 1 cm^{3} of iron has approximately 8g mass. (Use π = 3.14).

Solution:

Height of solid iron pole, h = 220 cm.

Base diameter of pole, d = 24 cm.

∴ Radius, r = 12 cm.

Height of cylinder, h = 60 cm.

Radius of cylinder, r= 8 cm.

Mass of the pole = ?

Volume of 1^{st} Cylinder, V = πr^{2}h

= π × (12)^{2} × 220

Volume of 2^{nd} cylinder, V = πr^{2}h

= 77 × (8)^{2} × 60

∴ Total Volume = π × (12)^{2} × 220 + π × (8)^{2} × 60

= {144 × 220 + 64 × 60}π

= 35520π

= 111532.8 cm^{3}.

Mass of iron about 1 ccm is 8 gm.

∴ Mass of iron 111532.8 ……. ??

= 111.5328 × 8 kg.

= 892.26 kg.

Question 7.

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm. and its height is 180 cm.

Solution:

(i) Radius of cylinder GHEF, r = 60 cm.

Height, h = 120 + 60

= 180 cm.

Volume of cylinderical vessel =πr2h

= π × (60)^{2} × 180

(ii) Volume of hemisphere + Volume of cone

(iii) Quality of water remained in cylinder.

Question 8.

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm^{3}. Check whether she is correct, taking the above as the inside measurements and n = 3.14.

Solution:

Diameter of spherical glass vessel = 8.5 cm.

∴ r = 4.25 cm.

Length of cylindrical neck = 8 cm.

Diameter = 2 cm.

∴ Radius, r = 1 cm.

Volume of water she found = 345 cm^{3}.

(i) Volume of Cylinder neck, V

(ii) Volume of cylindrical vessel =43πr3

=43×π×(4.25)3

∴ Total volume of water, V

∴ Her answer is not correct.

**All Chapter KSEEB Solutions For Class 10 Maths**

—————————————————————————–**All Subject KSEEB Solutions For Class 10**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share kseebsolutionsfor.com to your friends.

**Best of Luck!!**