CBSE 10th Maths Chapter 13 - Surface Areas and Volumes-MCQs
58807.If r is the radius of the sphere, then the surface area of the sphere is given by
4 π r$^{2}$
2 π r$^{2}$
π r$^{2}$
$\dfrac{4}{3}$ π r$^{2}$
Explanation:
If r is the radius of the sphere, then the surface area of the sphere is given by 4 π r$^{2}$
58808.If we change the shape of an object from a sphere to a cylinder, then the volume of cylinder will
Increase
Decrease
Remains unchanged
Doubles
Explanation:
If we change the shape of a three-dimensional object, the volume of the new shape will be same.
58809.Fifteen solid spheres are made by melting a solid metallic cone of base diameter 2cm and height 15cm. The radius of each sphere is
$\dfrac{1}{2}$
$\dfrac{1}{4}$
$\dfrac{1}{3}\sqrt{2}$
$\dfrac{1}{3}\sqrt{4}$
Explanation:
Volume of 15 spheres = Volume of a cone
15 x ($\dfrac{4}{3}$) π r$^{3}$= $\dfrac{1}{3} πr$^{2}$h
5×4 π r$^{3}$ = $\dfrac{1}{3} π 1$^{2}$(15)
20r$^{3}$ = 5
r$^{3}$ = $\dfrac{5}{20}$ = $\dfrac{1}{4}$
r = $\dfrac{1}{3}\sqrt{4}$
58810.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
4000 sq.cm
3500 sq.cm
3630 sq.cm
3750 sq.cm
Explanation:
Curved surface of bucket = π($R_1 + R_2$) x slant height (l)
Curved Surface = ($\dfrac{22}{7}$) x (25 + 8) x 35
CSA = 22 x 33 x 5 = 3630 sq.cm.
58811.If a cylinder is covered by two hemispheres shaped lid of equal shape, then the total curved surface area of the new object will be
4πrh + 2πr$^{2}$
4πrh – 2πr$^{2}$
2πrh + 4πr$^{2}$
2πrh + 4πr
Explanation:
Curved surface area of cylinder = 2πrh
The curved surface area of hemisphere = 2πr$^{2}$
Here, we have two hemispheres.
So, total curved surface area = 2πrh + 2(2πr$^{2}$) = 2πrh + 4πr$^{2}$
58812.A tank is made of the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and radius is 30 cm. The total surface area of the tank is
30 m
3.3 m
30.3 m
3300 m
Explanation:
Total surface area of tank = CSA of cylinder + CSA of hemisphere
= 2πrh + 2πr$^{2}$= 2π r(h + r)
= 2 x $\dfrac{22}{7}$ x 30(145 + 30) cm$^{2}$
=33000 cm$^{2}$
= 3.3 m$^{2}$
58813.A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter
r cm
2r cm
h cm
2h cm
Explanation:
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
58814.Two identical solid cubes of side a are joined end to end. Then the total surface area of the resulting cuboid is
12a$^{2}$
10a$^{2}$
8a$^{2}$
11a$^{2}$
Explanation:
The total surface area of a cube having side a = 6a$^{2}$
If two identical faces of side a are joined together, then the total surface area of the cuboid so formed is 10a$^{2}$.
58815.A solid cylinder of radius r and height h is placed over another cylinder of same height and radius. The total surface area of the shape so formed is
4πrh + 4πr$^{2}$
2πrh + 4πr$^{2}$
2πrh + 2πr$^{2}$
4πrh + 2πr$^{2}$
Explanation:
We know that,
The total surface area of cylinder = 2πrh + 2πr$^{2}$
When one cylinder is placed over the other cylinder of same height and radius, then height of the new cylinder will be 2h and radius will be r.
Thus, the total surface area of the shape so formed = 2πr(2h) + 2πr$^{2}$ = 4πrh + 2πr$^{2}$
58816.The number of shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm is approximately equal to
84
90
92
80
Explanation:
Volume of cuboidal lead solid = 9 cm × 11 cm × 12 cm
