Free Surface Areas and Volumes 01 Practice Test - 10th Grade
Question 1
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).
112
221
121
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×h−π(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 cmh=11200100=112 m
Hence, the length of the pipe is 112 m.
Question 2
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)
19.60 cm2
18.60 cm2
17.60 cm2
15.60 cm2
SOLUTION
Solution : C
Height of Cylinder =h=AO=2.4 cmDiameter 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 getAB=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
Question 3
Find the surface area of the given below figure having dimension in cm as shown.
900 cm2
880 cm2
650 cm2
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.
Question 4
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
3 cm
9 cm
6 cm
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=54⇒x=6
∴ The salt will be 6 cm deep.
Question 5
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?
0.375
0.625
0.125
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=OA−AD=10−6=4 cmOP=OC−PC=8−r (8−r)2=42+r264−16r+r2=16+r216r=48⇒r=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
Question 6
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 ?
20
30
40
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,△ONC∼△OMA
∴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=40−h1=(40−10) cm
⇒h=30 cm
Hence, the section is made at a height of 30 cm above the base of the cone .
Question 7
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
7
14
21
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
Question 8
If the curved surface area of a cylinder of height 14 cm is 88 sq. cm, then find the diameter of the cylinder.
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 cmTherefore, diameter = 2r = 2 cm.
Question 9
What is the diameter (in cm) of a sphere whose surface area is 616 cm2?
28
21
7
14
SOLUTION
Solution : D
Surface area of sphere =
4πr2=616⇒r2=616×722×14⇒r2=49⇒r=√49⇒r=7 cm∴ The diameter, D=2r=2×7=14 cm
Question 10
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