Samacheer Kalvi 10th Maths Solutions Chapter 7 Unit Exercise 7 Pdf
TN Samacheer Kalvi 10th Maths Solutions Chapter 7 Unit Exercise 7: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 7 Unit Exercise 7 – Mensuration on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 7 Unit Exercise 7 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Unit Exercise 7
TN Samacheer Kalvi 10th Maths Solutions Chapter 7 – Mensuration
Board | TNSCERT Class 10th Maths |
Class | 10 |
Subject | Maths |
Chapter | 7 (Unit Exercise 7) |
Chapter Name | Mensuration |
TNSCERT Class 10th Maths Pdf | all Unit Exercise Solution
Unit Exercise – 7
(1) The barrel of a fountain-pen cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one fifth of a litre ?
Solution:
Given that the pen length (h) = 7cm
Pen radius (r) = 5/2 mm
5/2×10 cm
= 1/4 cm
Now, volume the fountain pen cylinder –
= πr2h cube units
= 22/7 × 1/11 × 1/4 × 7 cm3
= 11/8 cm3
Now, 11/8 cm3 = 11/8×1000 litre
Now, 4/8000 litre writing words = 330
1 litre writing words = 350/(11/8000)
1/5 litre writing words = 330×8000/11 × 1/8
= 48000
Therefore, the required words is 48000 words.
(2) A hemi-spherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litre per second. How much time will it take to empty the tank completely ?
Solution:
Given the hemi-sphere radius (r) = 1.75 m
∴ Volume of the hemisphere = 2/3 πr3
= 2/3 × 22/7 × 1.15 × 1.75 × 1.75 m3
= 2/3 × 22 × 25/100 × 125/100 × 135/100 m3
= 11×49/3×16 m3
Now, 11×49/3×16 m3 = 11×49/3×16 = 1000 litres
= 11×49×125/6 litres
= 11229.16 litres
Now, time taken = 11229.16/7 seconds
= 1604.17 seconds
= 26.74 minutes
= 27 minutes (approx.)
Therefore the pipe fully empty taken time 27 minutes.
(3) Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Solution:
Now, hemisphere radius = r units.
So, cone radius = r units.
Hemisphere radius = cone height
∴ Cone height (h) = r units.
Now, maximum volume of cone
= 1/3 πr2h cu units
= 1/3 πr2× r cu units
= 1/3 πr3 cu units
Thus, the cone maximum volume is 1/3πr3cube units.
(4) An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion be 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Solution:
Oil funnel total height = 22cm
Cylinder height (h) = 10cm
Frustum height (h1) = (22 – 10) cm
= 12cm
Now, radius of the cylinder (r) = 4cm
Radius of the frustum (R) = 9cm
Tin sheet area
= Curved surface area of frustum + curved surface area cylinder.
= π (R + r) l + 2πrh1 sq units.
= π {(9+4) √(12)2 + (9-4)2 + 2×4×10} cm [∵R = √h2 + (R-r)2]
= π {13 √144 + 25 + 80} cm2
= π {13×13 + 80} cm2
= π× 249 cm2
= 22/7 × 249 cm2
= 782.57 cm2
Therefore, the area of the Tin sheet is 782.57 cm2.
(5) Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:-
Given that the cylinder radius (r1) = 4.5/2 cm
The cylinder height (h1) = 10cm
Volume of the cylinder = πr12h1 cube units
= π× (4.5/2)2× 10 cm3
Coin radius (r2) = 1.5/2 cm
Coin thickness (h2) = 2mm = 2/10 cm
Volume of the coin
= πr22 h2 cu units
= π× (1.5/2)2× 2/10 cm3
Number of coins
= volume of cylinder/volume of one coin
= π×4.5×4.5×10×2×2×10/2×2×π1.5×1.5×2
= 450 coins
Therefore, the number of coin 450.
(6) A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of solid cylinder.
Solution:
Given that hollow cylinder external radius (a) = 4.8 cm
Hollow cylinder internal radius (r) = 1.1 cm
Height of the cylinder (h) = 4cm
Solid cylinder height (h2) = 12cm
Let solid cylinder radius = r cm
Now, volume of solid cylinder = volume of hollow cylinder
πr2 h2 = πh (R2 – r2)
r2 12 = 4 {(4-3)2 – (1.1)2}
12r2 = 4 (18.49 – 1.21)
r2 = 4×7.28/12
r2 = 5.76
r = 2.4 cm
Thus, the diameter of solid cylinder is (2×2.4) = 4.8 cm
(7) The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at₹100 per sq. m.
Solution:
Given slant height of frustum (l) = 4m
Let the top part radius = R m
Let the bottom part radius = r m
Now, 2πR = 18
R = 18×7/2×22
R = 63/12 m
and 2πr = 16
r = 16×7/2×22
r = 56/22 m
Now, curved surface area of the frustum
= π (R + r) l sq. units
= 22/7 × (63/22 + 56/22) × 4 m2
= 22/7 × 119/22 × 4 m2
= 68 m2
Cost of painting = 100 × 68
= 6800
Therefore the frustum painting cost 6800.
(8) A hemi-spherical hollow bowl has material of volume 436π/3 cubic cm. Its external diameter is 14cm. Find its thickness.
Solution:-
Given hemisphere external radius (R) = 7cm
Let, hemisphere internal radius = r cm
Now, volume of hemisphere hollow bowl = 436π/3
2/3 π (R3 – r3) = 436π/3
(7)3 – r3 = 218
r3 = 343 – 218 = 125
r = 5
∴ Internal radius (r) = 5cm
Thus, the thickness = (7 – 5) cm = 2cm
Therefore, the hemisphere hollow bowl thickness is 2cm
(9) The volume of a cone is1005 5/7 cu.cm. The area of its base is 201 1/7 sq. cm. Find the slant height of the cone.
Solution:
Let, cone radius = r cm
Let cone height = h cm
Now, πr2 = 201 1/7
πr2 = 1408/7 —– (i)
Now, r2 = 1408/7 × 7/22
r2 = 64
r = 8
Now, 1/2 πr2h = 1005 5/7
1/ πr2h = 7040/7
h = 7040×7×3/7×1408 [By equation (i)]
h = 15cm
Now, slant height of cone (l) = √h2 + r2
l = √(15)2 + (8)2
l = √225 + 64
l = √289
l = 17 cm
Therefore, the cone slant height is 17 cm.
(10) A metallic sheet in the form of a sector of a circle of radius 21 cm has central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Solution:
Given cone radius (r) = 21cm
Central angle (θ) = 216°
Let, base radius of the cone = R cm
Perimeter of the base of cone = are length of the sector
2πR = θ/360 × 2πr
R = θ/360 × r
R = 216/360 × 21 cm
R = 12.6 cm
Slant height of a cone (l) = 21cm
h = √l2 – r2 [∵ l2 = h2 + r2]
h = √(21)2 – (12.6)2
h = √441 – 158.76
h = √282.24
h = 16.8
Now, volume of the cone = 1/3 πR2h cu. units
= 1/3 × 22/7 × 12.6 × 12.6 × 16.8 cm3
= 22 × 4.2 × 12.6 × 2.4 cm3
= 2794.18 cm3
Therefore, the volume of the cone is 2794.18 cm3.
Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 7 Mensuration Unit Exercise 7
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.