Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.3 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.3 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.3: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.3 – Mensuration on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 7 Exercise 7.3 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 7.3

TN Samacheer Kalvi 10th Maths Solutions Chapter 7 – Mensuration

Board TNSCERT Class 10th Maths
Class 10
Subject Maths
Chapter 7 (Exercise 7.3)
Chapter Name Mensuration

TNSCERT Class 10th Maths Pdf | all Exercise Solution

 

Exercise – 7.3

 

(1) A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 14 cm and the height of the vessel is 13 cm. Find the capacity of the vessel.

Solution:

Given that hemisphere diameter (2r) = 14cm

Hemisphere radius (r) = 7cm

Given the height of the vessel = 13cm

∴The height of the cylinder (h) = (13 – 7) cm

= 6 cm

The cylinder radius (r) = 7cm

Capacity of the vessel = volume of the cylinder + volume of the hemisphere

= πr2h + 2/3 πr3 cm3

= πr2 (h + 2/3 r) cm3

= 22/7 × 7 × 7 (6+ 2/3 × 7) cm3

= 22 × 7 (18+14/3) cm3

= 22 × 7 × 32/3 cm3

= 4928/3 cm3

= 1642.67 cm3

Therefore, the capacity of the vessel is 1642.67 cm3

 

(2) Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. 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 the model that Nathan made.

Solution:

Given model total length = 12 cm

∴ One cone height (h1) = 2cm

∴ Cylinder height (h2) = [12 – (2×2)} cm

= 8 cm

Cylinder and cone diameter [2r] = 3 cm

Cylinder and cone radius (r) = 3/2 cm.

Volume of the model = volume of the cylinder + volume of the two cone

= πr2 h2 + 2 × 1/3 π r2 h1 cm3

= πr2 (h2 + 2/3 h1) cm3

= 22/7 × 3/2 × 3/2 (8 + 2/3 × 2) cm3

= 11×3×3/7×2 × 24+4/3 cm3

= 11×9×28/7×2×3 cm3

= 66 cm3

Therefore, the model volume is 66 cm3.

 

(3) From a solid cylinder whose height is 2.4 cm and the diameter 1.4 cm, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest cm3.

Solution:

Given cylinder and cone diameter (2r) = 1.4 cm

Cylinder and cone radius (r) = 7cm

Cylinder and cone height (h) = 2.4 cm

Volume of the cylinder (v1) = πr2h cm3

= π × .7 × .7 × 2.4 cm3

= π × .49 × 2.4 cm3

Volume of the cone (v2) = 1/3 π r2h cm3

= 1/3 × π × .7 × .7 × 2.4 cm3

= 1/3 π × .49 × 2.4 cm3

Thus, volume of the remaining solid

= (v1 – v2) cm3

= (π .49 × 2.4 – 1/3 π × 4.9 × 2.4) cm3

= 2/3 × π × .49 × 2.4 cm

= 2/3 × 22/7 × .49 × 24 cm

= 2.46 cm3

Therefore, the volume of the remaining solid is 2.464 cms.

 

(4) A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in aright circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if theradius of the cylinder is 6 cm and height is 18 cm.

Solution:

Given that, line cylinder and hemisphere radius (r) = 6cm

Cylinder height (h) = 18 cm

Cone height (h1) 12cm

Hemisphere height (h2) = 6cm

Now, volume of the water displayed = volume of the cone + volume the hemisphere

= 1/3 πr2 h1 + 2/3 πr3 cm3

= 1/3 πr2(h1 + 2r) cm3

= 1/3 × 22/7 × 6 × 6 (12 + 2 × 6) cm3

= 1/3 × 22/7 × 6 × 6 × 24 cm3

= 6336/7 cm3

= 905.14 cm3

Therefore, the volume of the water displace 905.14 cm3.

 

(5) A capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends. If the length of the entire capsule is 12 mm and the diameter of the capsule is 3 mm, how much medicine it can hold?

Solution:-

Cylinder and hemisphere radius (r) = 3/2 mm.

Height of cylinder (h) (12 – 2 × 3/2) mm

= 9 mm

Now, volume of the capsule = volume of the cylinder + volume of the two hemisphere

= πr2h + 2/3 × 2πr3 mm3

= πr2 (h + 2×2/3 r) mm3

=

= 22×3×3×11/7×2×2 mm3

= 1089/14 mm3

= 77.78 mm3

Therefore, the volume of the capsule 77.78 mm3.

 

(6) As shown in figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.

Solution:

Cube side length (a) = 7cm.

Hemisphere radius (r) = 7/2 cm.

Now surface Area of the cube

= 6 × a2 cm2

= 6 × 72 cm2

= 294 cm2

Surface Area of the hemisphere

= 2πr2

= 2 × 22/7 × 7 × 7 cm2

= 77 cm2

Area of the base of the hemisphere

= πr2 cm2

= 22/7 ×7/2 × 7/2 cm2

= 77/2 cm2

Required surface area

= Surface area of the cube + surface area of the hemisphere – base area of the hemisphere

= [294 + 77 – 77/2] cm2

= [294 + 77/2] cm2

= (294 + 38.5) cm2

= 332.5 cm2

Therefore, the required surface area is 332.5 cm2

 

(7) A right circular cylinder just enclose a sphere of radius r units. Calculate

(i) The surface area of the sphere

(ii) The curved surface area of the cylinder

(iii) The ratio of the areas obtained in (i) and (ii).

Solution:

(i) Sphere radius = r units.

∴The surface area of the sphere = 4πr2 sq. units.

(ii) Cylinder radius = r units

Cylinder height (h) = 2r [∵ right circular cylinder]

The curved surface area of the cylinder = 2πr × h sq. units

= 2πr × 2r sq. units

= 4πr2sq. units

(iii) Surface area of the sphere/curved surface area of the cylinder = 4πr2/4πr2 = 1/1

∴ Surface area of the sphere : curved surface are of the cylinder  = 1:1

 

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 7 Mensuration Exercise 7.3

Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.

Updated: March 3, 2022 — 2:26 pm

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