Unit Test (Solutions): Surface Areas and Volumes | Mathematics (Maths) Class 10 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: Two identical solid cubes of side a are joined end to end. Then the total surface area of the resulting cuboid is  (1 Mark)

(a) 12a²
(b) 10a²
(c) 8a²
(d) 11a²

Ans: (b)

The total surface area of a cube having side a = 6a²
If two identical faces of side a are joined together, then the total surface area of the cuboid so formed is 10a².

Q2: A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter  (1 Mark)

(a) r cm 
(b) 2r cm 
(c) h cm 
(d) 2h cm

Ans: (b)

Radius of cylinder = r
Height = h > 2r
Since the sphere is enclosed by a cylinder, the radius of the cylinder will be the radius of the sphere.
Therefore, diameter of sphere = 2r cm

Q3: If we join two hemispheres of same radius along their bases, then we get a; (1 Mark)

(a) Cone
(b) Cylinder
(c) Sphere
(d) Cuboid

Ans: (c)

If we join two hemispheres of same radius along their bases, then we get a Sphere.

Q4: The radius of the top and bottom of a bucket of slant height 35 cm are 25 cm and 8 cm. The curved surface of the bucket is _________________________________. (1 Mark) 

Ans: 3630 sq.cm

Curved surface of bucket = π(R1 + R2) x slant height (l)
Curved Surface = (22/7) x (25 + 8) x 35
CSA = 22 x 33 x 5 = 3630 sq.cm.

Q5: If a cone is cut parallel to the base of it by a plane in two parts, then the shape of the top of the cone will be a: (1 Mark)

(a) Sphere
(b) Cube
(c) Cone itself
(d) Cylinder

Ans: (c)

If we cut a cone into two parts parallel to the base, then the shape of the upper part remains the same.

Q6: 2 cubes each for the volume of 64 cm3 are joined end to end. Find out the surface area of the resulting cuboid. (2 Marks)

Ans: Given that,
The Volume (V) for each cube is = 64 cm³
This implies that a3 = 64 cm³
The side for the cube, i.e., a = 4 cm
And the breadth and length of the resulting cuboid will be 4 cm each, while its height will be 8 cm.
Thus, the surface area of the cuboid (TSA) = 2(lb + bh + lh)
On putting the values, we observe,
= 2(8×4 + 4×4 + 4×8) cm²
= (2 × 80) cm²
Therefore, TSA of the cuboid = 160 cm²     

Q7: A cubical block of the side 7 cm is surmounted with a hemisphere. What is the greatest diameter the hemisphere could have? Find out the surface area of the solid. (2 Marks)

Ans: It is given that each side for the cube is 7 cm. Hence, the radius will be 7/2 cm.
We get,
The total surface area of the solid (TSA) = surface area of the cubical block + CSA for the hemisphere – area of the base of the hemisphere
So, TSA of the solid = 6×(side)² + 2πr² - πr²
= 6 × (side)² + πr²
= 6 × (7)² + (22/7) × (7/2) × (7/2)
= (6×49) + (77/2)
= 294+38.5 = 332.5 cm2
Hence, the surface area of the solid is 332.5 cm2

Q8: Two cones have their heights in the ratio of 1 : 3 as well as radii in the ratio of 3: 1. What is the ratio for their volumes? (2 Marks)

Ans: Given that,
Ratio for the heights of two cones = 1:3
Ratio for the radii = 3:1
Let h and 3h be the height of the two cones.
And, 3r and r be the corresponding radii for the cones.
Hence, r1 = 3r, h1 = h, r2 = r, h2 = 3h.
Ratio for the volumes = [(1/3)πr1²h1]/ [(1/3)πr2²h2]

= [(3r)² h]/[r² (3h)]

= (9r²h)/(3r²h)

= 3/1
Thus, the ratio of volumes = 3:1

Q9: Water is flowing at the rate of 15 km/h passing through a pipe of the diameter of 14 cm into a cuboidal pond which is 50 m long and 44 m wide. At what time will the level for the water in the pond rise by 21 cm? (3 Marks)

Ans: Assume time taken by pipe to fill pond = t hours
Water flows 15 km in 1 hour, so it will flow 15t metres in t hours.
We get, volume of the cuboidal pond up to the height of 21 cm = volume of  the water that passes through the pipe in “t” hours
Considering cuboidal pond,
For the Length, l = 50 m
For the Breadth, b = 44 m
For the Height, h = 21 cm = 0.21 m
We get, 
Volume of the tank = lbh
Volume of the water = 50(44)(0.21) = 462 m3
Considering cylindrical pipe
For the Base diameter = 14 cm
For the Base radius, r = 7 cm = 0.07 m
For Height, h = 15t km = 15000t m
We get,
Volume of the cylinder = πr2h
Volume of the water passed in pipe = π(0.07)²(15000t)
=22/7 × 0.07 × 0.07 × 15000t
= 231t cm³
Thus, we have
231t = 462
t = 2 hours
The time required to fill the tank up to a height of 25 cm is 2 hours.

Q10: A heap of rice is in the form for the cone of diameter of 9 m and height of 3.5 m. Find out the volume of the rice. How much canvas cloth is required to cover the heap only? (3 Marks)

Ans: As per the question,
Consider conical heap,
For the Base Diameter = 9 cm
Therefore, for the base radius, r = 4.5 cm
For Height, h = 3.5 cm
We get,
Slant height,
l = √(r2+h2)

l = √(4.52+3.52)

l = √(20.25+12.25)

l = √32.5

l = 5.7cm

The equation for the volume of cone = 1/3πr²h
We get,
volume of the rice = volume of the conical heap
volume of the rice = 1/3π(4.5)²(3.5) =74.25cm³ 
We also get,
Canvas required to just cover heap = Curved surface area of the conical heap
Curved surface area of the cone = πrl
Thus, the canvas required = π(4.5)(5.7) = 80.61 cm² [appx]

Q11: Rasheed gets a playing top (lattu) as his birthday present, in which, surprisingly, he found no colour on it. He wanted to colour it with his crayons. The top is shaped as like a cone surmounted with a hemisphere. The entire top of 5 cm in height, as well as the diameter of the top, is 3.5 cm. Find out the area he has to colour. (Take π = 22/7) (3 Marks)

Ans:

TSA for the toy = CSA for the hemisphere + CSA for the cone
Curved surface area of the hemisphere = 1/ 2 (4πr²) = 2πr² = 2(22/7) × (3.5/2) × (3.5/2) = 19.25 cm²
Height of the cone = height of the top – radius of the hemispherical part
= (5 – 3.5/2) cm = 3.25 cm
Slant height of the cone (l): The slant height l can be found using the Pythagorean theorem:

Thus, CSA of cone = πrl = (22/7) × (3.5/2) × 3.7 = 20.35 cm²
So, the surface area of the top = [2(22/7) × (3.5/2) × (3.5/2) + (22/7) × (3.5/2) × 3.7] cm²

= (22/7) × (3.5/2) (3.5+3.7) cm²

= (11/2) × (3.5 + 3.7) cm²

= 39.6 cm² (approximately)

Q12: Water flows at the rate of 10m/minute and passes through a cylindrical pipe 5 mm in diameter. How long will it take to fill a conical vessel which has a diameter at the base is 40 cm and a depth of 24 cm? (5 Marks)

Ans: Assume the time taken by pipe to fill vessel = t minutes
As water flows 10 m in 1 minute, it will flow 10t meters in t minutes.
As per the question,
Volume of the conical vessel = volume of the water that passes through the pipe in t minutes
Consider the conical pipe,
For the Base Diameter = 40 cm
For the Base radius, r = 20 cm
For Height, h = 24 cm
We know that the volume of the cone = 1/3πr2h
Volume of the conical vessel = 1/3π(20)²(24) = 3200 π cm³
Consider cylindrical pipe,
For the Base diameter = 5 mm = 0.5 cm
For the Base radius, r = 0.25 cm
Water covers 10t m distance in pipe,
So, we get,
For Height, h = 10t m = 1000t cm
We get,
Volume of the cylinder = πr2h
Volume of the water passed in pipe = π(0.25)²(1000t) = 62.5tπ cm³
Hence, we find,
62.5tπ = 3200
62.5t = 3200
t = 51.2 minutes
We get,
0.2 minutes = 0.2(60) seconds = 12 seconds
Hence, t = 51 minutes 12 seconds

Q13: From a solid cylinder those height is 2.4 cm as well as diameter of 1.4 cm. A conical cavity for the same height as well as same diameter is hollowed out. Find out the total surface area of the remaining solid to the nearest cm². (5 Marks)

Ans:

From the question we have:
The Diameter of the cylinder = Diameter of the conical cavity = 1.4 cm
Hence, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7
And the height of the cylinder = height of the conical cavity = 2.4 cm
Therefore, Slant height of the conical cavity (l) = √(h2 + r2)

= √(2.42 + 0.72) 

= √(5.76 +0.49) 

= √(6.25) 

= l = 2.5 cm

 TSA for the remaining solid = surface area of the conical cavity + TSA for the as the cylinder= πrl+(2πrh + πr²)

= πr(l+2h+r)

= (22/7)× 0.7(2.5+4.8+0.7)

= 2.2×8 = 17.6 cm²

Hence, the total surface area of the remaining solid is 17.6 cm²