9th Grade > Mathematics
SURFACE AREAS AND VOLUMES MCQs
:
C
We know that,
Volume of a cylindrical container
=πr2h
1000 cm3=1 L
Hence, the volume of each cylindrical tumbler
=227×72×10
=1540 cm3
So, the total volume of juice required
=1540×25
=38500 cm3
= 38.5 L
:
D
Given that,
r:h = 3:4
Let r = 3x and h = 4x
From the relation
l2=h2+r2,
The Slant height:
l=√(3x)2+(4x)2
⇒l=5x
We know that,
Total surface area of cone
=πr(r+l)
Curved surface area of cone
=πrl
Hence, the required ratio
=Total surface area of coneCurved surface area
=πr(r+l)πrl
=π×3x(3x+5x)π×3x×5x
=8:5
:
Total Surface Area = 2( l x b + b x h + h x l)
= 2(26 x 14 + 14 x 6.5 + 6.5 x 26)
= 2(364 + 91 + 169
=1248 m2
:
C
Given that, the ratio of the height of the pillar to that of dome is 4:3 .
⇒hcylinderhcone=43
We know that,
volume of cylinder=πr2h.
volume of cone=13πr2h.
Now, ratio of volume of cylinder to the volume of cone
=πr2hcylinder13πr2hcone
=πr2×413πr2×3
=41
=4:1
:
D
The curved surface area of a cone = πrl
where, r is radius and l is slant height.
⇒308=πrl
⇒ 227×r×14=308
⇒r=308×722×14
⇒r=7 m
Base area
=πr2
=227×72
=154 m2
:
Volume of cube = (side)3
Volume of cuboid = (length) × (breadth) × (height)
So height of cuboidal tank can be found by equating the two
∴height=side3length×breadth
=1035×10
∴ Height = 20m
:
The volume will remain constant.
Let's assume x such hemispherical balls will be formed.
So 43πR3=x23πr3
(Where R is the radius of spherical ball and r is the radius of the hemispheres)
⇒43π(21)3=x23π(7)3
⇒x=2×21×21×217×7×7
⇒x=54
:
A and C
Volume of cylinder =π×r2×h
⇒231=227×r2×6
⇒r2=231×722×16
⇒r2=12.25
⇒r=3.5
So, a ball will fit inside the cylinder if its radius is less than or equal to 3.5 cm.