# Free Surface Areas and Volumes 01 Practice Test - 10th Grade

A solid iron rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe (in m).

A.

112

B.

221

C.

121

D.

211

#### SOLUTION

Solution : A

Volume of iron
=(440×260×100) cm3.

Internal radius of the pipe
=30 cm.

External radius of the pipe
=(30+5) cm=35 cm.

Let the length of the pipe be h cm.

Volume of iron in the pipe = (External volume) - (Internal volume)

=[π(35)2×hπ(30)2×h]cm3

=(πh)((35)2(30)2) cm3

=(65×5)πh cm3

Volume of iron in the pipe = Volume of iron block

(325πh) cm3=440×260×100

h=440×260×100325×722 cm

h=11200 cm

h=11200100=112 m

Hence, the length of the pipe is 112 m.

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter as that of cylinder is hollowed out. Find the total surface area of the remaining solid. (Use  π=227)

A.

19.60 cm2

B.

18.60 cm2

C.

17.60 cm2

D.

15.60 cm2

#### SOLUTION

Solution : C Height of Cylinder =h=AO=2.4 cm

Diameter of Cylinder =1.4 cm

Radius of Cone = Radius of Cylinder
OB=r=1.42=0.7 cm

Lets find Slant Height l of cone, using pythagoras theorem on AOB , we get

AB=l=h2+r2=(2.4)2+(0.7)2

l=5.76+0.49=6.25=2.5 cm

Surface Area of remaining Solid
= Surface Area of the Cylinder + Inner surface Area of the hollow Cone
=(2πrh+πr2)+πrl=πr(2h+r+l)=227×0.7(2×2.4+0.7+2.5)=2.2(4.8+0.7+2.5)=2.2(8)=17.60 cm2

Total surface area of the remaining solid is 17.60 cm2

Find the surface area of the given below figure having dimension in cm as shown. A.

900 cm2

B.

880 cm2

C.

650 cm2

D.

400 cm2

#### SOLUTION

Solution : B

Surface area = base circle + curved surface area of the cylinder + curved surface area of the cone.

=πr2+2πrh+πrl

=π(5)2+2π(5)(20)+π(5)(11)

=π×5(5+40+11)

=227×5×56

=880 cm2.

Anita buys a new salt cellar in the shape of a cylinder topped by a hemisphere as shown below. The cylinder has a diameter of 6 cm and a height of 10 cm. She pours the salt into the salt cellar, so that it takes up half the total volume of the cellar. Find the depth of the salt, marked with x in the diagram A.

3 cm

B.

9 cm

C.

6 cm

D.

12 cm

#### SOLUTION

Solution : C

Let the depth of the salt in the cellar be x

Volume of   =   volume of cylinder +
salt cellar          volume of hemisphere

=πr2h+23πr3=π×32×10+23π×33=π[90+18]=108π cm3

So height x will come on the cylinder.
Half of total volume =54π cm3
πr2x=54ππ×32×x=54π9x=54x=6

The salt will be 6 cm deep.

A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed. What fraction of water overflows?

A.

0.375

B.

0.625

C.

0.125

D.

0.875

#### SOLUTION

Solution : A Radius of the conical vessel, R=AC=6cm

Height of the conical vessel, h=OC=8cm

Radius of the sphere, PD=PC=r

PC=PD=rAC=AD=6 cm

[Since, lengths of two tangents from an external point to a circle are equal]

OCA & OPD are right triangle.
[ Tangent and radius  are perpendicular to each other]

OA=OC2+AC2=82+62      =100=10 cmOP2=OD2+PD2

OD=OAAD=106=4 cmOP=OCPC=8r (8r)2=42+r26416r+r2=16+r216r=48r=3 cm

Volume of water overflown =  Volume
of sphere
=43πr3=43π×(3)3=36π cm3

Original volume of water = volume of
cone
=13πr2h=13π×62×8=96π cm3

Fraction of water overflown =Volume of water overflownOriginal volume of water=36π96π=38=0.375

Fraction of water overflown is 0.375

The height of a cone is 40 cm. A small cone is cut off at the top by a plane parallel to the base. If the volume of the small cone be 164 of the volume of the given cone, at what height ( in cm) above the base is the section made ?

A.

20

B.

30

C.

40

D.

50

#### SOLUTION

Solution : B Let R be the radius of the given cone, r the radius of the small cone, h be the height of the frustum and h1 be the height of the small cone.

In the figure,ONCOMA

ONOM=NCMA    [Sides of similar triangles are proportional]

h140=rR

h1=(rR)×40 .......(i)

We are given that Volume of small conevolume of given cone=164

13πr2×h113πR2×40=164

r2R2×140×[(rR)40]=164       [ By (i) ]

(rR)3=164=(14)3

rR=14    ...... (ii)

From (i) and (ii) h1=14×40=10 cm

,h=40h1=(4010) cm

h=30 cm

Hence, the section is made at a height of 30 cm above the base of the cone  .

The surface area of a cuboid is 1372 cm2. If the ratio length : breadth : height is 4:2:1, then its length (in cm) is

A.

7

B.

14

C.

21

D.

28

#### SOLUTION

Solution : D

S.A. = 1372 cm2 (Given)

2(l×b+b×h+l×h)=1372

[ Surface area of a cuboid of dimensions l×b×h is given by 2(lb+bh+lh).]

Also, it is given that l:b:h=4:2:1.

Let l=4k, b=2k and h=k.

Then, 2(4k×2k+2k×k+4k×k)=1372

28k2=1372

k2=49

k=7

l=4k=4×7=28 cm

If the curved surface area of a cylinder of height 14 cm is 88 sq. cm, then find the diameter of the cylinder.

A. 2 cm
B. 1 cm
C. 4 cm
D. 3 cm

#### SOLUTION

Solution : A

CSA of a cylinder =2πrh

Given,
CSA =88 cm2 and h=14 cm
2πrh=882r=88πh    =88×722×114    =4×7×1142r=2 cm

Therefore, diameter = 2r = 2 cm.

What is the diameter (in cm) of a sphere whose surface area is 616 cm2?

A.

28

B.

21

C.

7

D.

14

#### SOLUTION

Solution : D

Surface area of sphere =
4πr2=616r2=616×722×14r2=49r=49r=7 cm

The diameter, D=2r=2×7=14 cm

Metallic spheres of radii 6cm, 8cm and 10cm, respectively are melted to form a single solid sphere. The radius of the resulting sphere(in cm) is

___

#### SOLUTION

Solution :

Radius of first Sphere = r1=6cm

Radius of second Sphere = r2=8cm

Radius of third Sphere = r3=10cm
Let radius of resulting sphere = r cm
According to given condition, we have
Volume of 1st sphere + Volume of 2nd sphere + Volume of 3rd sphere = Volume of resulting sphere
43.π.(r1)3+43.π.(r2)3+43.π.(r3)3=43.π.r3

43((r1)3+(r2)3+(r3)3)=43.r3

(63+83+(10)3)=r3

( 216 + 512 + 1000 ) = r3

r3 = 1728

r = 12 cm