**Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.5 Pdf**

TN Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.5: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.5 – Mensuration on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 7 Exercise 7.5 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 7.5

**TN Samacheer Kalvi 10th Maths Solutions Chapter 7 – Mensuration**

Board |
TNSCERT Class 10th Maths |

Class |
10 |

Subject |
Maths |

Chapter |
7 (Exercise 7.5) |

Chapter Name |
Mensuration |

**TNSCERT Class 10th Maths Pdf | all Exercise Solution**

__Exercise – 7.5__

__(*) Multiple choice questions__

**(1) The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is.**

**(A) 60 π cm2**

**(B) 68 π cm2**

**(C) 120 π cm2**

**(D) 136 π cm2**

**Solution: **

Cone diameter (2r) = 16cm

Cone radius (r) = 8cm

Cone height = 15cm

l = √r^{2} + h^{2}= √8^{2} + 152 = √64+2225

l = √289 + 7cm

Cone curved surface area = πrl cm^{2}

= π × 80 × 17 cm

= 136π cm^{2} (D)

** **

**(2) If two solid hemispheres of same base radius r units are joined together along their bases, then curved surface area of this new solid is**

**(A) 4πr ^{2} sq. units **

**(B) 6πr ^{2} sq. units **

**(C) 3πr ^{2} sq. units **

**(D) 8πr ^{2} sq. units **

**Solution: **

We know that, hemisphere curves surface area = 2πr^{2}, where r = radius of hemisphere.

Two hemisphere curved surface area = 2×2πr^{2 }sq. units

= 4πr^{2} sq. units (A)

** **

**(3) The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be**

**(A) 12cm **

**(B) 10cm **

**(C) 13cm **

**(D) 5 cm **

**Solution: **

We know that,

r = radius = 5cm

h = height

l = slant height = 1.3cm

l = √r^{2} + h^{2}

h = √l^{2} – r^{2}

= √(13)^{2} – (15)^{2} = √169 – 25

h = √144 = 12cm (A)

** **

**(4) If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is**

**(A) 1:2 **

**(B) 1:4 **

**(C) 1:6 **

**(D) 1:8**

**Solution: **

Let, cylinder radius = r radius

Let, cylinder height = h units

New cylinder radius = r/2 units

Original cylinder volume = πr^{2}h sq. units

New cylinder volume = π × (r/2)^{2} × h sq. units

= πr^{2}h/h sq. units

New cylinder volume/orginal cylinder = (πr^{2}h/h)/πr^{2}h = 1/4 = 1:4 (B)

** **

**(5) The total surface area of a cylinder whose radius is 1/3 of its height is **

**(A) 9****πh ^{2}/8 sq. units **

**(B) 24πh ^{2} sq. units **

**(C) 8πh ^{2}/9 sq. units **

**(D) 56πh ^{2}/9 sq. units **

**Solution: **

Let, cylinder height = h units

radius (r) = h/3 units.

Cylinder total surface area = 2πr (h + r) sq. units

= 2π h/3 (h + h/3) sq. units

= 2π h/3 (4h/3)

= 8πh^{2}/9 sq. units (C)

** **

**(6) In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is**

**(A) 5600π cm ^{3}**

**(B) 11200π cm ^{3}**

**(C) 56π cm ^{3}**

**(D) 3600π cm ^{3 }**

**Solution: **

Let external radius = R cm

Let internal radius = r cm

Given R + r = 14 —- (i) R – r = 4 —– (ii)

(i) + (ii), we get

R + r + R – r = 14+4

2R = 18

R = 9 cm

From (i), R = 9 putting, we get

R + r = 14

r = 14 – R

r = 14 – 9

r = 5 cm

∴ Volume of the hollow cylinder

= π (R^{2} – r^{2}) h cm^{3}

= π (9)^{2} – (5)^{2} 20 cm^{3} [h = 20cm]

= π {81 – 25} × 20 cm^{3}

= π × 56 × 20 cm^{3}

= 11200 π cm^{3} (B)

** **

**(7) If the radius of the base of a cone is tripled and the height is doubled then the volume is**

**(A) Made 6 times **

**(B) made 18 times **

**(C) made 12 times **

**(D) Unchanged**

**Solution: **

Let, cone radius = r units

Cone height = h units

Volume of cone = 1/3 πr^{2}h cube. Units

New cone radius = 3r units

New cone height = 2h units.

New cone volume = 1/3 π× (3r)^{2} × 2h cube unit

= 1/3 π × 3r × 3r × 2h cube unit

= 18 × 1/3 πr^{2}h cube unit

= 18 times original cone volume (B)

** **

**(8) The total surface area of a hemi-sphere is how much times the square of its radius.**

**(A) ****π**

**(B) 4π**

**(C) 3π**

**(D) 2π**

**Solution: **

Total surface area of hemisphere

= 3πr^{2} sq. units

= 3π × (radius)^{2} sq. units

= 3π times the square of radius

** **

**(9) A solid sphere of radius x cm is melted and cast into a shape of a solid cone of same radius. The height of the cone is**

**(A) 3x cm **

**(B) x cm **

**(C)4x cm **

**(D)2x cm**

**Solution:**

Given, sphere radius (r) = x cm.

∴ Volume of sphere = 4/3 πr^{3} cm^{3}

Cone radius (r) = x cm

Cone height = h cm

Now, volume of cone = volume of sphere

1/3 πr^{2}h = 4/3 πr^{3}

r^{2}h = 4r^{3}

h = 4r

h = 4x cm (C)

** **

**(10) A frustum of a right circular cone is of height 16cm with radii of its ends as 8cm and 20cm. Then, the volume of the frustum is**

**(A) 3328π cm ^{3}**

**(B) 3228π cm ^{3}**

**(C) 3240π cm ^{3}**

**(D) 3340π cm ^{3}**

**Solution: **

Frustum of a right circular cone height (h) = 16cm

r = 8cm and R = 20cm

Volume of the frustum = 1/3 πh (R^{2} + Rr + r^{2}) cm^{3}

= 1/3 π × 16 × (400 + 160 + 64) cm^{3 }

= 1/3π × 16 × 624 cm^{3}

= 3328 π cm^{3} (A)

** **

**(11) A shuttle cock used for playing badminton has the shape of the combination of**

**(A) A cylinder and a sphere **

**(B) A hemisphere and a cone**

**(C) A sphere and a cone **

**(D) Frustum of a cone and a hemisphere**

**Solution: **

(d) Frustum of a cone and a hemisphere

** **

**(12) A spherical ball of r _{1} units is melted to make 8 new identical balls each of radius r_{2} units. Then r_{1} : r_{2} is **

**(A) 2:1 **

**(B) 1:2**

**(C) 4:1 **

**(D) 1:4 **

**Solution: **

Volume of the sphere/Volume of each sphere = 8/1

(4/3 π r_{1}^{3})/(y/3π r_{2}^{3} = 8/1

r_{1}^{3}/r_{2}^{3} = 8/1

(r_{1}/r_{2})^{3} = (2/1)^{3}

r_{1}:r_{2} = 2:1 (A)

** **

**(13) The volume (in cm3) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is**

**(A) 4/3 π**

**(B) 5π**

**(C) 5π**

**(D) 20/3π**

**Solution: **

Sphere radius (r) = 1cm

Volume of the sphere = 4/3π r^{3} cube units

= 4/3 π× 1 × 1 × 1 cm^{3}

= 4/3π cm^{3} (A)

** **

**(14) The height and radius of the cone of which the frustum is a part are h _{1} units and r_{1} units respectively. Height of the frustum is h_{2} units radius of the smaller base is r_{2} units. If h_{2}:h_{1} = 1:2 then r_{2}:r_{1} is **

**(A) 1:3 **

**(B) 1:****2 **

**(C) 2:1 **

**(D) 3:1 **

**Solution: **

Given h_{2}:h_{1} = 1:2

Ratio of their volumes

= 1/3 π h_{1} (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}): 1/3 π h_{2} (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

= 1/3 πh_{1} : 1/3 π h_{2}

h_{1} : h_{2 }= 2:1

Then r_{1} : r_{2} = 2:1 [∵ volume ratio and radius ratio same]

r_{2} : r_{1} = 1:2 (B)

** **

**(15) The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is **

**(A) 1:2:3 **

**(B) 2:1:3 **

**(C) 1:3:2 **

**(D) 3:1:2 **

**Solution: **

Cylinder : cone : sphere = πr^{2}h : 1/3 πr^{2}h : 4/3 πr^{3}

= r^{2}h : 1/3 r^{2}h : 4/3 r^{3}

= h: h/3 : 4r/3

=h : h/3 : 4/3 × h/2 [r = h/2]

= 3 : 1 : 2 (D)

*Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 7 Mensuration *Exercise 7.5

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