Exercise 6.1 Cylinder and Sphere

The above-mentioned topic is mentioned in the vedanta Excel in Mathematics book in Unit 6 - the second part of Mensuration where we deal with the surface area and volumes of Cylinder and Sphere exercises.

Here is a complete solution of exercise 6.1 from vedanta's book for Class 10 students. All solutions are accurate but there might be bad handwriting or extra crossing in many solutions. In case you do not understand the handwriting, numerical or language, do let us know in the comment section so that we can make it clear to you.

We have almost all solutions from vedanta Excel in Mathematics Book 10 for Grade 10 students. We also have notes and other unit tests in it. If you want to see it, visit here:
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General Section


1 a) If x be the radius and y be the height of a cylinder, write the formulae to find:
  • curved surface area (CSA) = $2 \pi rh = 2 \pi xy$
  • total surface are (TSA) = $2 \pi r(r+h) = 2\pi x(x+y)$
  • volume = $\pi rh = \pi xy$

1 b) If p be the radius of a sphere, write the formula to find its
  • surface area = $4\pi r^2 = 4\pi  p^2$
  • volume = $\dfrac{4}{3} \pi r^3 = \dfrac{4}{3} \pi p^3$



1 c) A and a are the external and internal radii and b be the height of a hollow cylinder respectively. Write the formula to find its
  • curved surface area (CSA) = $2\pih(R+r) = 2\pib(A+a)$
  • total surface area (TSA) = $2\pi(R+r)(h+R-r) = 2\pi(A+a)(b+A-a)$
  • volume = $\pih(R+r)(R-r) = \pib(A+a)(A-a)$


2 a) The radius of the circular base of a cylinder is 7 cm and its height is 15 cm. Find its (i) curved surface area (CSA) and (ii) total surface area and (iii) volume.

Solution:
Here,
$r = 7cm$
$h = 15cm$
Now,
$CSA = 2\pi rh$
$= 2 * \dfrac{22}{7} * 7 * 15$

$= 2* 22 * 15$
$= 660 cm^2$
And,
$TSA = 2\pi r (r+h)$

$= 2 * \dfrac{22}{7} * 7 (7+15)$

$= 2 * 22 * 22$
$ = 968 cm^2$
Also,
$Volume = \pi r^2 h$
$= \dfrac{22}{7} * 7^2 * 15$
$= 22 * 7 * 15$
$= 2310 cm^3$

3 a) Find the (i) curved surface area (CSA) (ii) total surface area (TSA) and (iii) volume of a cylinder with height 30 cm and the perimeter of the circular base is 88 cm.

Solution:
For a cylinder,
height (h) = 30 cm
perimeter of circular base (P) = 88cm
$or, 2 \pi r = 88cm$
$\therefore r = 14 cm$
Now,
(i) Curved Surface Area (CSA) = $2 \pi r h$

$= 2 * \dfrac{22}{7} * 14 * 30$

$= 2640 cm^2$
And,
(ii) Total Surface Area (TSA) = $2 \pi r ( r + h)$

$= 2 * \dfrac{22}{7} * 14 * (14 + 20)$

$= 3872 cm^2$
Also,
(iii) Volume (V) = $\pi r^2 h$

$= \dfrac{22}{7} * 14^2 * 20$
$= 18 480 cm^3$

Creative Section - A

9 a) The perimeter of the circular base of a cylinder is 132 cm and it is 40 cm high. Find the following.
i) CSA
ii) TSA
iii) Volume
Solution:
Here, In a cylinder,
$\text{Perimeter} (P) = 132 cm$
$\implies 2 \pi r = 132 $
$\implies r = 132 * \dfrac{7}{22*2}$

$\therefore r = 21cm$
$\text{Height} (h) = 40 cm$
Now,
(i) CSA = $2 \pi r h$

$= P * h$
$= 132 * 40$
$= 5280 cm^2$
And,
(ii) TSA = $\text{CSA} +\text{Base area}$

$= 5280 + \pi r^2$
$= 5280 + \dfrac{22}{7} * 21^2$

$= 5280 + 1386$
$= 6666 cm^2$
Also,
(iii) Volume = $\text{ pi r^2 h }$

$= \dfrac{22}{7} * 21^2 * 40$
$= 55 440cm^3$

10 a) The area of curved surface of a solid cylinder is equal to 2/3 of the total surface area. If the total surface area is 924 cm^2, find the volume of the cylinder.

Solution:
Here,
$\text{Total Surface Area} (TSA) = 924 cm^2$

Given,
$\text{Curved Surface Area} = \dfrac{2}{3} \text{TSA}$

$or, 2 \pi r h = \dfrac{2}{3} * 924$

$or, 2 \pi r h = 616$ --- (i)
$or, rh = 616 * \dfrac{7}{44}$

$or, rh = 98$
$or, h = \dfrac{98}{r}$  ---- (ii)

We know,
$\text{TSA} = 2 \pi r h + 2\pi r^2$

$or, 924 = 616 + 2\pi r^2$
$or, 308 = 2\pi r^2$
$or, r^2 = 308 * \dfrac{7}{44}$

$or, r^2 = 49$
$\therefore r = 7 cm$
Put value of r in equation (ii), we get, $h = 14 cm$

Now,
$\text{Volume} = \pi r^2 h$
$= \dfrac{22}{7} * 7^2 * 14$
$= 2156 cm^3$
Hence, the required volume of the cylinder is 2156 cm^3.


15 a) If the surface area of a sphere is 5,544 cm^2, find its volume.

Solution:
Here,
$\text{Surface area of a sphere} = 5544 cm^2$

$or, 4 \pi r^2 = 5544$
$or, r^2 = 5544 * \dfrac{7}{22} * \dfrac{1}{4}$

$or, r^2 = 441$
$\therefore r = 21 cm$
Now,
$\text{Volume} = \dfrac{4}{3} \pi r^3$

$= \dfrac{4}{3} * \dfrac{22}{7} * 21^3$

$= 38,808 cm^3$
Hence, the required volume of the given sphere is 38,808 cm^3.


18 a) The surface area of a sphere is $\pi$ sq.cm. If its radius is doubled, by how much does its surface area increase?

Solution:
Here,
$\text{Surface area of a sphere} = \pi cm^2$

$or, 4 \pi r^2 = \pi $

$or, 4 r^2 = 1$
$or, r^2 = \dfrac{1}{4}$
$\therefore r= \dfrac{1}{2} cm$

Now, doubling the radius of the sphere as mentioned in the question, we get, $r = \dfrac{1}{2} * 2 = 1 cm$

Again,

$\text{Surface area of sphere with radius 1 cm} = 4 \pi * (1)^2$

$= 4 \pi cm^2$
Comparing the area of sphere now and then, we get, 

$\text{difference in surface area} = 4 \pi cm^2 - \pi cm^2 = 3 \pi cm^2$

Hence, the surface area of the sphere increases by 3 $\pi cm^2$ according to the question.



Creative Section - B

22 a)
Solution:
Here,
Total height of the solid (H) = 14 cm
Total surface area of the solid (TSA$_{total}$) = 770 $cm^2$

Let h be the height of the cylinder and r be the radius of the cylindrical base as well as that of the hemisphere.

We have,
$h + r = 14$

$or, h = 14 - r$
Also,
CSA of hemisphere + CSA of cylinder + Base Area = Total Surface Area of solid

$or, 2 \pi r^2 + 2 \pi r h + \pi r^2 = 770$

$or, \dfrac{22}{7} ( 2 r^2 + 2rh + r^2) = 770$

$or, 3r^2 + 2rh = 770 * \dfrac{7}{22}$

$or, r[3r + 2(14-r)] = 245$
$or, r[3r + 28 - 2r] = 245$
$or, r^2 + 28 r - 245 = 0$
$or, r^2 + 35r - 7r - 245 = 0$

$or, (r + 35)(r - 7) = 0$
Either,
$r + 35 = 0 \implies r = -35$
Or,
$r - 7 = 0 \implies r = 7$





About vedanta EXCEL in MATHEMATICS Book 10

Author: Hukum Pd. Dahal
Editor: Tara Bahadur Magar

Vanasthali, Kathmandu, Nepal
+977-10-4382404, 01-4362082
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Ex 6.1 - Cylinder and Sphere Class 10 Solutions | vedanta Excel in Mathematics is a collection of the solutions related to calculation of surface area and volume of cylinder and sphere from Mensuration (II): Cylinder and Sphere chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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