**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

= πr^{2}h + 2/3 πr^{3} cm^{3}

= πr^{2} (h + 2/3 r) cm^{3}

= 22/7 × 7 × 7 (6+ 2/3 × 7) cm^{3}

= 22 × 7 (18+14/3) cm^{3}

= 22 × 7 × 32/3 cm^{3}

= 4928/3 cm^{3}

= 1642.67 cm^{3}

Therefore, the capacity of the vessel is 1642.67 cm^{3}

** **

**(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 (h_{1}) = 2cm

∴ Cylinder height (h_{2}) = [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

= πr^{2} h_{2} + 2 × 1/3 π r^{2 }h_{1} cm^{3}

= πr^{2} (h_{2} + 2/3 h_{1}) cm^{3}

= 22/7 × 3/2 × 3/2 (8 + 2/3 × 2) cm^{3}

= 11×3×3/7×2 × 24+4/3 cm^{3}

= 11×9×28/7×2×3 cm^{3}

= 66 cm^{3}

Therefore, the model volume is 66 cm^{3}.

** **

**(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 cm ^{3}. **

**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 (v_{1}) = πr^{2}h cm^{3}

= π × .7 × .7 × 2.4 cm^{3}

= π × .49 × 2.4 cm^{3}

Volume of the cone (v_{2}) = 1/3 π r^{2}h cm^{3}

= 1/3 × π × .7 × .7 × 2.4 cm^{3}

= 1/3 π × .49 × 2.4 cm^{3}

Thus, volume of the remaining solid

= (v_{1} – v_{2}) cm^{3}

= (π .49 × 2.4 – 1/3 π × 4.9 × 2.4) cm^{3}

= 2/3 × π × .49 × 2.4 cm

= 2/3 × 22/7 × .49 × 24 cm

= 2.46 cm^{3}

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 (h_{1}) 12cm

Hemisphere height (h_{2}) = 6cm

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

= 1/3 πr^{2} h_{1} + 2/3 πr^{3 }cm^{3}

= 1/3 πr^{2}(h_{1} + 2r) cm^{3}

= 1/3 × 22/7 × 6 × 6 (12 + 2 × 6) cm^{3}

= 1/3 × 22/7 × 6 × 6 × 24 cm^{3}

= 6336/7 cm^{3}

= 905.14 cm^{3}

Therefore, the volume of the water displace 905.14 cm^{3}.

** **

**(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

= πr^{2}h + 2/3 × 2πr^{3} mm^{3}

= πr^{2} (h + 2×2/3 r) mm^{3}

=

= 22×3×3×11/7×2×2 mm^{3}

= 1089/14 mm^{3}

= 77.78 mm^{3}

Therefore, the volume of the capsule 77.78 mm^{3}.

** **

**(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 × a^{2} cm^{2}

= 6 × 7^{2} cm^{2}

= 294 cm^{2}

Surface Area of the hemisphere

= 2πr^{2}

= 2 × 22/7 × 7 × 7 cm^{2}

= 77 cm^{2}

Area of the base of the hemisphere

= πr^{2} cm^{2}

= 22/7 ×7/2 × 7/2 cm^{2}

= 77/2 cm^{2}

Required surface area

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

= [294 + 77 – 77/2] cm^{2}

= [294 + 77/2] cm^{2}

= (294 + 38.5) cm^{2}

= 332.5 cm^{2}

Therefore, the required surface area is 332.5 cm^{2}

** **

**(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 = 4πr^{2} 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

= 4πr^{2}sq. units

(iii) Surface area of the sphere/curved surface area of the cylinder = 4πr^{2}/4πr^{2} = 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

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