Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.4 Pdf
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TN Samacheer Kalvi 10th Maths Solutions Chapter 7 – Mensuration
Board | TNSCERT Class 10th Maths |
Class | 10 |
Subject | Maths |
Chapter | 7 (Exercise 7.4) |
Chapter Name | Mensuration |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise – 7.4
(1) An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder.
Solution:
Sphere radius (r1) = 12cm
Cylinder radius (r2) = 8cm
Cylinder height = h cm
Now, volume of cylinder = volume of sphere
πr22 h = 4/3 πr13
r22 h = 4/3 r13
h = 4r13/3r22
h = 4×12×12×12/3×8×8
h = 36cm
Therefore the height of the cylinder is 36cm.
(2) Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks will rise by 21 cm.
Solution:
Given that rectangular tank length (l) = 50m
= 5000 cm
Rectangular tank wide (b) = 44m = 4400 cm
Level of water in the tanks (h1) = 21 cm
Now, volume of the tank = l × b × h1 cm3
= 5000 × 4400 × 21 cm3
Now, radius of the pipe (r) = 7cm
Speed of water (h2) 15cm /h = 1500000 cm/h
Now, one hour volume of water following
= πr2 h2 cm3
= π × 7 × 7 × 1500000 cm3
Time taken = volume of tank/volume of pipe (one hour)
= 5000×4400×21/π×7×7×15 h
= 7×5×44×21×7/22×7×7×15 h
= 2h
Therefore the levels of water in the tanks will rise by 2h.
(3) A conical flask is full of water. The flask has base radius r units and height h units, the water is poured into a cylindrical flask of base radius xr units. Find the height of water in the cylindrical flask.
Solution:
Given conical flask radius = r units
Given, conical flask height = h units
∴ Volume of conical flask = 1/3 πr2h cube units
Given, cylindrical flask radius = xr units
Let, cylindrical flask height = h2 units
Now, volume of cylindrical flaks = volume of conical flask
π× (xr)2 × h2 = 1/3 πr2h
x2 h2 = 1/3 h
h2 = h/3x2 units
Therefore the cylindrical flask height is h/3×2 units.
(4) A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter.
Solution:
Given cone diameter (2r) = 14cm
Cone radius (r) = 7cm
Cone height (h) = 8cm
Given hollow sphere external radius (r2) = 5cm
Let hollow sphere internal radius = R cm
Now, volume of hollow sphere = Volume of cone
4/3 π (r23 – R3) = 1/3 πr2 h
4 (y (5)3 – R3) = 7×7×8
125 – R3 = 98
R3 = 125 – 98
R3 = 27
R = 3cm
Therefore, the international of the hollow sphere is = 3×2 cm = 6cm
(5) Seenu’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 2m×1.5m×1m. The overhead tank has its radius of 60 cm and height 105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially.
Solution:
Given, the sump dimensions
= 2m × 1.5m × 1m
= 200cm × 150cm × 100 cm
∴ Volume of the sump = 200 × 150 × 100 cm3
= 3000000 cm3
Now, given tank radius (r) = 60cm
Given tank height (h) = 105 cm.
Volume of the tank = πr2h cube units
= 22/7 × 60 × 60 × 105 cm3
= 1188000 cm3
Volume of water left in the sump
= volume of the sump – volume of the tank
= 3000000 – 1188000 cm3
= 1812000 cm3
Therefore the volume of water left in the sump is 1812000 cm3
(6) The internal and external diameter of a hollow hemispherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, then find the height of the cylinder.
Solution:
Given hollow hemisphere internal radius (r) = 3cm
Hollow hemisphere internal radius (R) radius (R) = 5cm
Cylinder diameter (2r) = 14cm
Cylinder radius (r1) = 7cm
Let the cylinder height = h. cm.
Now, volume of the cylinder = Volume of the hollow hemisphere,
πr12 h = 2/3 π (R3 – r3)
7 × 7 × h = 2/3 (53 – 33)
7×7×h = 2/3 (125 – 27)
h = 2×98/3×7×7
h = 4/3
h = 1.33 cm
Therefore the cylinder height is 1.33 cm.
(7) A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder.
Solution:
Given that the solid sphere radius (r1) = 6cm
Cylinder height (h) = 32 cm
Let cylinder internal radius = r cm
Now, Volume of the hollow cylinder = value of the solid sphere
π (R2 – r2) h = 4/3 π r13
(52 – r2) × 32 = 4/3 × 6 × 6 × 6
25 – r2 = 9
r2 = 25 – 9 = 16
r = 4cm
∴ Thus, the cylinder internal radius = 4cm
Therefore, the thickness of the cylinder is = (5 – 4) cm = 1 cm
(8) A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel.
Solution:
Let the cylinder vessel height = h units
∴ Cylinder radius (r) 50/100 h + h unit
= h/2 + h units
= 34/2 units
∴ Volume of the cylindrical vessel
= πr2h cube units
= π× (3h/2)2 × h cube units
= π × 5h2/4 × 4 cube units
= 9 πh3/4 cube units
Now, hemisphere diameter = cylinder diameter
So, hemisphere radius (r) = 34/2 units.
Volume of the hemisphere
= 2/3 × π cube units
= 2/3 × π × (34/2)3 cube units
= 2×π×3π×3h×3h/3×2×2×2 cube units
= 9πh3/4 cube units
∴ Now, volume of cylinder and volume of hemisphere same.
So, 100% of the juice can be transferred.
Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 7 Mensuration Exercise 7.4
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