Mensuration 3D

7. A cuboidal block of 6 cm × 9 cm cuboid is :

Cube
1. Volume of a cube is 512 cubic
× 12 cm is cut up into exact fdlh ?kukHk ds rhu Øekxr i`"Bkssa dk {ks=kiQy
number of equal cube. The 12 l seh-
2
gSA rc ?kukHk dk vk;ru Kkr djsa\
cm, find its total surface area? least possible number of cubes
(a) 3600 (b) 100
, d ?ku dk vk;ru 512 ?ku ls-eh gS] blds dqy will be
(c) 80 (d) 24 3
lrg dk {ks=kiQy D;k gS\ 6cm×9cm×12cm foek okys ?kukHk ls
12. Two cubes of sides 6 cm each
(a) 768 cm² (b) 192 cm²
de ls de dqy fdrus ?ku cuk;s tk ldrs gSa\
are kept side to form a
(a) 6 (b) 9 (c) 24 (d) 30
(c) 384 cm² (d) 576 cm² rectangular parallelopiped. The
8. A cuboidal water tank has 216 area (in sq. cm) of the whole
2. A cube of edge 6 cm is painted
1 surface of the rectangular
on all sides and then cut into litres of water. Its depth is 3
of
unit cubes. The number of unit parallelopiped is
cubes with no sides painted is its length and breadth is
1
of nks6cm Hkqtk okys ?kuksa dks feykdj ,d
2
, d 6cm Hkqtk okyk jaxhu1cm
?ku]okys ?kukHk cuk;k x;kA ?kukHk dk lEiw.kZ i`"B
?kuks esa dkVk x;kA mu ?kuksa dh la[;k Kkr
1
3
djsa]
of the difference of length and {ks=kiQy Kkr djsa\
ftudh , d Hkh Hkqtk jaxh gqbZ uk gks\ breadth. The length of the tank (a) 432 (b) 360
(a) 0 (b) 64 (c) 186 (d) 108 is (c) 396 (d) 340
, d ?kukHkdkj ikuh dh Vadh
216esa
yhVj 13. A parallelopiped whose sides are
3. If diagonal of a cube is 12 cm, in ratio 2 : 4 :8 have the same
1
then its volume in cm3 is : ikuh gS mldh xgjkbZ mldh yEckbZ
3
gS dk volume as a cube. The ratio of
fdlh ?ku dk fod.kZ12 lseh- gSA ?ku dk their surface area is:
vkSj mldh pkSM+kbZ mldh yEckbZ vkSj xgjkbZ ds
vk;ru Kkr djsa\ ,d ?kukHk dh Hkqtk,sa
2 : 4 : 8 easa gSa] ftldk
A Vadh dh yEckbZ gSa\vk;ru ,d ?ku ds cjkcj gSA i`"B {ks=kiQyksa
1 1
vUrj ds 3 dk 2
gS
(a) 8 (b) 12 (c) 24 (d) 2 3
dk vuqikr Kkr djsa\
4. How many cubes, each of edge 3 (a) 72 dm (b) 18 dm
(a) 7 : 5 (b) 4 : 3
cm, can be cut from a cube of edge (d) 6 dm (d) 2 dm (c) 8 : 5 (d) 7 :6
15 cm? 9. The area of three adjacent
14. The area of the four walls of a
15 lseh- Hkqtk okys ?ku3esa
lsehls
. Hk
qtk faces of a cuboid are x, y, z
room is 660 m2 and its length
okys fdrus ?ku cuk;s tk ldrs gSa\ square units respectively. If
the volume of the cuboid by v is twice its breadth. If the
(a) 25 (b) 027 (c) 125 (d) 144 height of the room is 11 m, then
cube units. then the correct
Cuboid area of its floor
relation between v,x, y, z is
5. The length, breadth and height
of a room is 5m , 4 m and 3m fdlh ?kukHk ds lEeq[k i`"B {ks=kiQy
x, y, z (in m2) is
respectively. Find the length of gS
A ;fn bldk vk;ru v gks
rksv,x, y, z ds fdlh dejs dh pkjksa nhokjksa dk660
{ks=kiQy
the largest bamboo that can be chp lEcU/ LFkkfir djsa\ m² rFkk mldh yEckbZ] pkSM+kbZ dh nks xquh gS
kept inside the room. (a) v2 = xyz (b) v3 =xyz ;fn dejs dh m¡pkbZ
11 m gks] rc vk/kj dk
,d dejs dh yEckbZ] pkSM+kbZ vkSj mQ¡pkbZ 2 3 3
(c) v = x y z 3
(d) v3 =x2y2z2 {ks=kiQy Kkr djsaA
Øe'k%5 ehVj]4 ehVjvksj3 ehVjgSA ml 10. A rectangular sheet of metal is (a) 120 (b) 150
lcls cM+h ck¡l dh NM+h dh yEckbZ Kkr 40 cm by 15 cm . equal squares
(c) 200 (d) 330
dhft;s tks bl dejs esa iw.kZr;k j[kh tk of side 4cm are cut off at the 15. The volume of air in a room is
corners and the remaining is
ldrh gSA folded up to form an open 204 m3. The height of the room
(a) 5 m (b) 60 m rectangular box The volume of is 6 m. What is the floor area of
(c) 7 m (d) 5 2 m the box is the room?
6. A wooden box measures 20 cm fdlh 40 lseh- YkEch rFkk15 lseh- pkSM+h fdlh dejs esa gok dk vk;ru 204m³ gSA
by 12 cm by 10 cm . Thickness of vk;rkdkj pknj ds dksuksa ls dh pknj
4 lseh- dejs dh Å¡pkbZ6m gSA iQ'kZ dk {ks=kiQy
wood is 1 cm. Volume of wood to
dkV yh tkrh gS rFkk cps Hkkx dks eksM+dj Kkr djsa\
make the box ( in cubic cm) is
,d [kqyk cDlk cuk fy;k tkrk gSA cDls dk (a) 32 m
2
(b) 46 m2
ydM+h ds ,d fMCcs dh20 ekil seh-× 12 (c) 44 m 2
(d) 34 m2
vk;ru Kkr djsa\
l seh-× 10 l seh- gSA ydM+h dh1 eksVkbZ
l seh- 16. 2 cm of rain has fallen on a
(a) 896 cm3 (b) 986 cm3
gS bl cDls dks cukus esa iz;ksx gq;h ydM+h(c)dk600 cm3 (d) 916 cm3
square km of land. Assuming
vk;ru gSA 11. The areas of three consecutive
that 50% of the raindrops could
(a) 960 (b) 519 have been collected and con-
faces of a cuboid are 12 cm2,
(c) 2400 (d) 1120 3 tained in a pool having a 100
then the volume (in cm ) of the
m×10 m base, by what level

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 would the water level in the pool 22 portion is
have increased ? fdeh2- gS  7 ysrs gq, ioZr dh mQ¡pkbZ gSaA fdlh 'kadq ds vk/kj dh f=kT;k7 lseh- rFkk
fdlh 1km² Hk wfe ij2cm o"kkZ gqbZA ekuk (a) 2.2 km (b) 2.4 km Å¡pkbZ 24 lseh- gSA Å¡pkbZ ds vk/s Hkkx ls
fd dsoy 50% i kuh gh laxzfgr gks ik;k rFkk (c) 3 km (d) 3.11 km
'kadq dks dkVk x;kA Åijh Hkkx dk vk;ru
100m×10m vk/kj okys ,d laxzgky; esa 21. The base of a conical tent is
19.2 metres in diameter and Kkr djsa\
,df=kr fd;k x;kA ikuh dk mBk ry Kkr djsa\ the height is 2.8 metres . The (a) 169 cm3 (b) 154 cm3
(a) 1 km (b) 10 m area of the canvas required to (c) 1078 cm 3
(d) 800 cm3
(c) 10 cm (d) 1 m put up such a tent ( in square 26. Some solid metallic right cir-
17. Some bricks are arranged in 22 cular cones. each with radius
an area measuring 20 m³ If the meters ) (taking   ) is
7 of the base 3 cm and height 4
length, breadth and height of nearly. cm, are melted to form a solid
each brick is 25 cm, 12.5 cm fdlh 'kaDokdkj VSaV ds vk/kj dk O;klsphere of radius 6 cm. The num-
and 8 cm respectively, then the 19.2 eh-gS vkSj bldh mQ¡pkbZ2.8 ehVjgS ber of right circular cones is
number of bricks are (suppose
there is no gap in between two
bl izdkj dk VSaV yxkus ds fy, dSuol dk dqN 3 lseh- f=kT;k rFkk
4 lseh- Å¡pkbZ okys
bricks) {ks=kiQYk (oxZ ehVj esa) yxHkx gksxk\
'kadqvksa dks fi?kykdj 6 lseh-
,d f=kT;k
dqN b±Vksa20}kjkeh- dk ,d pcwrjk cuk;k
3 (a) 3017.1 (b) 3170 okyk xksyk cuk;k x;kA 'kadqvksa dh la[;k
(c) 301.7 (d) 30.17
x;kA ;fn ,d b±V dh yEckbZ 25 lseh- pkSM+kbZ Kkr djsa\
22. The volume of a right circular
12.5 lseh- rFkk Å¡pkbZ
8 lseh- gSA rc pcwrjs cone is 1232 cm3 and its verti- (a) 12 (b) 24 (c) 48 (d) 6
27. A right circular cylinder of
esa iz;qDr b±Vksa dh la[;k Kkr djsa\ cal height is 24 cm . Its curved
surface area is height 16 cm is covered by a
(a) 6,000 (b) 8,000
rectangular tin foil of size 16 cm
(c) 4,000 (d) 10,000 ,d YkEco`Rrh; 'kadq dk vk;ru 1232 ?k u
× 22 cm, The volume of the cyl-
Cylinder and Cone 24 lseh- gSA bldk
lseh- rFkk bldh mQapkbZ inder is
18. A cylindrical tank of diameter oØi`"B dk {ks=kiQy D;k gksxk\ 16 lseh- Å¡pkbZ okys csyu lseh-×
16 dks
35 cm is full of water. If 11 litres (a) 154 cm2 (b) 550 cm2
of water is drawn off, the water (c) 604 cm2 (d) 704 cm2 22 lseh- vk;rkdkj pknj ls iw.kZr% <dk
level in the tank will drop by : 23. The volume of a right circular x;kA csyu dk vk;ru Kkr djsa\
35 lseh- O;kl okys csyukdkj VSad ikuh lscylinder, 14 cm in height, is (a) 352 cm3 (b) 308 cm3
iwjk Hkjk gSA 11;fn
yhVjikuhdksfudky equal to that of a cube whose (c) 616 cm 3
(d) 176 cm3
fy;k tk;s] rc ikuh dk fxjk Lrj Kkr djsa\ edge is 11 cm. Find the radius 28. If the area of the base of a cone
of the base of the cylinder is is 770 cm2 and the area of its
1 6
(a) 10 cm (b) 12 cm curved surface is 814 cm². then
22
2 7 (Take   7 ) find its volume.
3
14cm mQ¡pkbZ okys flysaMj dk vk;ru]
fdlh 'kadq ds vk/kj dk {ks=kiQy
770 lseh-
(c) 14 cm (d) 11 cm
7
11cm Hk qtk okys ?ku ds vk;ru ds cjkcj
2
mlds fr;Zd i`"B dk {ks=kiQy
814 lseh-2 gSA
19. The height of a conical tank is vk;ru Kkr djsa\
60 cm and the diameter of its gSA vk/kj dh f=kT;k Kkr djsa\
base is 64cm. The cost of (a) 5.2 cm (b) 5.5 cm (a) 2 1 3 5 cm³ (b) 392 5 cm³
painting it from outside at the (c) 11.0 cm (d) 22.0 cm (c) 5 50 5 cm³ (d) 6 1 6 5 cm³
rate of ` 35 per sq. m. is : 24. A rectangular paper sheet of 29. The size of a rectangular piece
'ka 60 lseh-rFkk
Dokdkj ik=k dh mQ¡pkbZ mlds dimensions 22 cm × 12 cm is of paper is 100 cm × 44 cm. A
vk/kj dk O;kl64 lseh- gSA mlds fr;Zd folded in the form of a cylinder cylinder is formed by rolling the
Rs. 35 izfreh- dh nj ls jaxus dk
i`"B dks 2 along its length. What will be paper along its breadth. The
volume of the cylinder is
[kpZ Kkr djsa\ the volume of this cylinder?
22
(a) ` 52.00 approx, (Take  
22
) (Use   )
7 7
(b) ` 39.20 approx,
22cm×12cm dh vk;rkdkj pknj dks fdlh vk;rkdkj pknj dk {ks=kiQYk
(c) ` 35.20 approx,
(d) ` 23.94 approx, mldh yEckbZ ds ifjr% eksM+dj ,d csyu 100 lseh-× 44 lseh- gSA pknj dks mldh
20. The slant height of a conical cuk;k x;kA csyu dk vk;ru Kkr djsa\ pkSM+kbZZ ds ifjr% eksM+dj ,d csyu cuk;k
mountain is 2.5 km and the (a) 460 cm3 (b) 462 cm3 x;kA csyu dk vk;ru Kkr djsa\
area of its base is 1.54 km 2 . (c) 624 cm3 (d) 400 cm3 (a) 4400 cm3 (b) 15400 cm3
22 25. In a right circular cone, the ra- (c) 35000 cm3 (d) 144 cm3
Taking   , the height of the
7
dius of its base is 7 cm and its 30. The radius of the base and
mountain is height of a metallic solid
height 24 cm. A cross- section
,d 'kaDokdkj ioZr dh fr;Zd mQapkbZ
2.5 cylinder are r cm and 6 cm
is made through the midpoint
fdeh- gS vkSj mlds vk/kj dk {ks=kiQy
1.54 of the height parallel to the
respectively. It is melted and
recast into a solid cone of the
base. The volume of the upper

2 Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039

 same radius of base. The (c) 1 : 6 : 9 (d) 1 : 7 : 19
3
height of the cone is : cm gS
A ml 'kaDokdkj VSaV
427dks]
m²
fdlh csyu dh vk/kj dh f=kT;k rFkk Å¡pkbZ 7
Sphere
Øe'k%r lseh- rFkk
6 lseh- gSaA mls fi?kyk;k dSuokl ls cuk;k x;kA 'kadq dh fr;Zd Å¡pkbZ
39. The total surface area of a
x;k rFkk leku f=kT;k ds 'kadq ds :i esa <kykKkr djsa\ sphere is 8 square unit. The
x;kA 'kadq dh Å¡pkbZ Kkr djsa\ (a) 17 metre (b) 15 metre
(c) 19 metre (d) 8.5 metre volume of the sphere is
(a) 54 cm (b) 27 cm
(c) 18 cm (d) 9 cm 35. If the height of a given cone be fdlh xksys dk lEiw.kZ i`"B 8{ks=kiQy
π unit²
31. The height of the cone is 30 cm. doubled and radius of the base A xksys dk vk;ru Kkr djsa\
gS
A small cone is cut off at the remains the same the ratio of
top by a plane parallel to its the volume of the given cone to 8 2
(a)  cubic unit
1 that of the second cone will be 3
base. If its volume is of the ; fn fn;s x;s 'kadq dh mQ¡pkbZ dks nksxquk dj
27
8
volume of the cone. At what fn;k tk;s rFkk vk/kj dh f=kT;k dks leku (b)  cubic unit
3
height above the base, is the j[kk tk;s rks fn;s x;s 'kadq dk vk;ru] nwljs
section made ?
'kadq ds vk;ru ls fdl vuqikr esa gksxk\ (c) 8 3 cubic unit
fdlh 'k¡dq dh Å¡pkbZ A ,d NksVs
30 lseh-gS (a) 2 : 1 (b) 1 : 8
'k¡dq dks vk/kj ds lekukUrj dqN Å¡pkbZ (c)
ls 1 : 2 8 3
(d) 8 : 1 (d)  cubic unit
dkVk x;kA NksVs 'k¡dq dk vk;ru] cM+s
36.'k¡dq ds
A circus tent is cylinderical up 5
to a height of 3 m and conical 40. A solid metallic sphere of radius
vk;ru dk 1 27 gSA vk/kj ls fdruh Å¡pkbZ above it. If its diameter is 105m
8 cm is melted to form 64 equal
ij 'k¡dq dks dkVk x;k \ and the slant height of the
small solid spheres. The ratio of
conical part is 63 m, then the
(a) 6 cm (b) 8 cm total area of the canvas the surface area of this sphere
(c) 10 cm (d) 20 cm required to make the tent is to that of a small sphere is
32. The lateral surface area of a cyl- 22 8 l seh- f=kT;k okys xksys dks64
fi?kykdj
(take   )
inder is 1056 cm2 and its height 7 cjkcj xksyksa esa c¡kVk x;kA cM+s o`Rr rFkk Nk
is 16cm. Find its volume? fdlh ldZl dk VSUV
3m Å¡pkbZ rd csyukdkj o`Rr ds i`"B {ks=kiQy dk vuqikr Kkr djsa\
fdlh csyu dk vkarfjd i`"B {ks=kiQy
1056 rFkk muds Åij 'kadq gSA ;fn mldk O;kl (a) 4 : 1 (b) 1 :16
l seh
² rFkk Å¡pkbZ
16 l seh-
gSA vk;ru Kkr djsa\ 105 m rFkkfr;Zd m¡pkbZ63 m gks] rc (c) 16 :1 (d) 1 : 4
(a) 4545 cm3 (b) 4455 cm3 VSUV dks iwjk <+dus ds fy, iz;qDr 41.
diM+s
Thedk ratio of weights of two
(c) 5445 cm3 (d) 5544 cm3 {ks=kiQy Kkr djsa\ spheres of different materials
33. The radius of the base and (a) 11385 m2 (b) 10395 m2 is 8 : 17 and the ratio of weights
2
height of a right circular cone (c) 9900 m (d) 990 m2 per 1 cc of materials of each is
are in the ratio 5 : 12. If the 37. A right triangle with sides 9 cm, 289 : 64. The ratio of radii of
2
12 cm and 15 cm is rotated the two spheres is
volume of the cone is 3 1 4 about the side of 9 cm to form a
7
cone. The volume of the cone so nksfHkUu inkFkks± dh xsanksa ds Hkkj dk vuqIkk
cm3, the slant height (in cm) of
formed is: 8 : 17 rFkk ³ Hk
izfr lseh kj dk vuqikr289
the cone will be
9cm, 12 cm vkSj 15 cm dh Hkqtkvksa okys : 64 gS A nksuksa xsanks dh f=kT;kvksa dk vuqikr
fdlh 'k¡dq ds vk/kj dh f=kT;k rFkk
yac f=kHkqt dks ;fn Hkqtk ij ?kqek;k tk,
9cm Kkr djsa\
Å¡pkbZ
5:12 ds vuqikr esa gSA ;fn 'k¡dq
ftlls 'akdq cu ldsA bl izdkj cus 'kadq dk (a) 8 : 17 (b) 4 : 17
dk vk;ru 314
2
cm³ gks] rc 'k¡dq dh vk;ru fdruk gksxk\
7 (c) 17 : 4 (d) 17 : 8
(a) 432  cm³ (b) 327  cm³
fr;Zd Å¡pkbZ Kkr djsa\ (c) 334  cm³ (d) 324  cm³ 42. A sphere of radius 21cm is cut
(a) 12 (b) 13 (c) 15 (d) 17 38.If a right circular cone is into 8 identical parts by 3 cuts
34. The radius of the base of a (1 cut along each axis). What
separated into solids of
conical tent is 16 metres. If will be the total surface area (in
volumes V1, V2 , V3 by two planes
cm²) of each part?
3 parallel to the base which also
427
7
sq. metre canvas is trisect the altitude, then V1 : ,d 21 lseh- f=kT;k okys xksys dks 3 dVko
required to construct the tent, V2 : V3 is (izR;sd v{k ij 1 dVko) yxkdj 8 le:i
then the slant height of the tent fdlh 'kadq dks vk/kj ds lekukUrj] 'kh"kZ yEcHkkxksa esa dkVk tkrk gSA izR;sd Hkkx dk dqy
esa) D;k gksxk\
i`"Bh; {ks=kiQy 2(ls-eh-
22 dks rhu cjkcj Hkkx
V1, V2 rFkkV3 esa ck¡Vk
is: ( take   ) (a) 844.5 (b) 1732.5
7 x;kAV1 : V2 : V3 gS\
fdlh 'kaDokdkj VSaV dh vk/kj dh
16f=kT;k (a) 1 : 2 : 3 (b) 1 : 4 : 6 (c) 1039.5 (d) 1115.6

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43. A cone of radius 90 cm and , d 14cm O;kl okyh Bksl xsan dks fi?kykdkj {ks=kiQy Kkr djsa\
height 120 cm stands on its 14cm O ;kl okys rkj ds #i esa ifjofrZr (a) 75.43 sq. cm,
base. It cut into 3 parts by 2 cuts fd;k x;kA rkj dh yEckbZ Kkr djsa\ (b) 103.71 sq. cm,
parallel to its base such that the (c) 85.35 sq. cm,
height of the three parts (from
16
(a) 27 cm (b) cm (d) 120.71 sq. cm,
top to bottom) are in ratio of 1 : 3
52. If a metallic cone of radius 30
2 : 3. What is the total surface 28 cm and height 45 cm is melted
area (in cm2) of the middle part? (c) 15 cm (d) cm
3 and recast into metallic
, d 90 l s-eh- f=kT;k rFkk120
Å¡pkbZ
l seh- 48. A sphere of diameter 6 cm is spheres of radius 5 cm, find the
Å¡pkbZ okyk 'kadq vius vk/kj ij [kM+kdropped gSA in a right circular number of spheres.
bls vk/kj ds lekarj2 d Vko ls3 Hkkxksa esa cylindrical vessel partly filled fdlh 30cm f=kT;k rFkk 45cm m¡ps 'kadq
bl izdkj dkVk tkrk gS fd rhuksa Hkkxksa dh water. The diameter of the
with
Å¡pkbZ (Åij ls uhps dh vksj) dk vuqikr cylindrical vessel is 12 cm. If the 5cm f=kT;k okyh xsasnks esa ifjofrZr
dks fi?kykdkj
1 : 2 : 3 gSA eè; Hkkx dk dqy i`"Bh; sphere is just completely fd;k x;kA xsanks dh la[;k Kkr djsa\
{ks=kiQy (lseh-
2
esa) D;k gS\ submerged in water, then the (a) 81 (b) 41 (c) 80 (d) 40
rise of water level in the
(a) 14600 (b) 16500 53. A metallic sphere of radius
cylindrical vessel is:
10.5 cm is melted and then re-
(c) 17800 (d) 18500 fdlh 6cm O;kl okyh xsan dks dqN ikuh ls cast into small cones each of
44. Three small hemispheres of Hkjs gq, csyukdkj ik=k esa Mkyk x;kA csyukdkj
radius 3.5 cm and height 3 cm.
radii 1 cm, 2 cm and 3 cm are ik=k dk O;kl 12cm gSA ;fn ikuh dk ry The number of cones thus
melted to form a sphere. What bruk Å¡pk mBk fd og xsan dks dsoy Mqcks ik;s]
formed is
will be the approximate radius
of the new sphere?
rc mBs ikuh dk Lrj Kkr djsa\ 10.5cm f=kT;k okys /krq ds xksys dks
(a) 2 cm (b) 1 cm fi?kykdkj3.5cm f=kT; k okys rFkk
1 lseh-]2 lseh- vkSj 3 lseh- f=kT;kvksa ds 3cm
(c) 3 cm (d) 4 cm
rhu NksVs xksyk¼ksa dks fi?kykdkj ,d u;k Å¡pkbZ okys 'kadqvksa esa ifjofrZr fd;k x;kA
49. A solid metallic spherical ball of
xksyk cuk;k x;kA xksys dh f=kT;k yxHkx diameter 6 cm is melted and re- 'kadqvksa dh la[;k Kkr djsa\
fdruh gksxh\ cast into a cone with diameter (a) 140 (b) 132 (c) 112 (d) 126
(a) 2.6 cm (b) 3.2 cm of the base as 12 cm. The Prism
(c) 3.6 cm (d) 2.8 cm height of the cone is
54. A right prism has a triangular
45. A hemispherical cup of radius 6 lseh- O;kl okyh xasan dks fi?kykdj
12 base whose sides are 13 cm, 20
4 cm is filled to the brim with lseh- O;kl okys 'kadq esa ifjofrZr fd;k x;kAcm and 21 cm, If the altitude of
coffee. The coffee is then poured
'kadq dh Å¡pkbZ Kkr djks\ the prism is 9 cm, then its vol-
into a vertical cone of radius 8 ume is
cm and height 16 cm. The per- (a) 2 cm (b) 3 cm
centage of the volume of the (c) 4 cm (d) 6 cm ,d yac fizTe dk vk/kj f=kHkqtkdkj gS ftldh
50. A hemisphere and a cone have
cone that remains empty is : Hkqtk,¡13 cm, 20 cm vkSj21 cm gS aA
,d v/Zxksykdkj di ftldh f=kT;k 4 cm equal base . If their heights are
;fn fizTe dk 'kh"kZ9yEccm gS
] rks mldk
gSdkWiQh ls iw.kZ Hkjk gSA also equal to their base, the ra-
dkWiQh dks
8cm vk;ru fdruk gksxk\
k rFkk16cm Å¡ps'kaDokdkj di esa
f=kT; tio of their curved surface will be:
(a) 1143 cm³ (b) 1314 cm³
myV fn;k x;kA 'kadq ds [kkyh Hkkx % dk ,d v¼Zxksys vkSj ,d 'kadq ds vk/kj cjkcj
Kkrdjsa\ gS ;fn mudh mQ¡pkbZ Hkh cjkcj gks] rks(c) muds
1413 cm³ (d) 1134 cm³
(a) 87.5% (b) 80.5% oØi`"Bksa dk vuqikr gksxk\ 55. A prism has base a right angled
(c) 81.6% (d) 88.2% triangle whose sides adjacent
(a) 1 : 2 (b) 2 :1
46. If the radius of a sphere be
to the right angles are 10 cm
doubled. the area of its surface (c) 1 : 2 (d) 2 : 1
and 12 cm long. The height of
will become 51. A toy is in the form of a cone
the prism is 20 cm . the den-
fdlh xksys dh f=kT;k nks xquh dh tk;s] rcmounted on a hemisphere.
sity of the material of the prism
lEiw.kZ i`"B {ks=kiQy fdruk xquk gks tk,xk\ The radius of the hemisphere
is 6gm/cubic cm. the weight of
(a) Double and that of the cone is 3 cm and
height of the cone is 4 cm. The the prism is
(b) Three times
(c) Four times total surface area of the toy fdlh fizTe dk vk/kj ledks.k f=kHkqt
(d) Nine times 22 gS] ftldh yEcor~ Hkqtk,sa 10cm r Fkk
Miscellaneous (taking  
7
) is 12cm gSaA fizTe dh Å¡pkbZ
20cm
fdlh v/Zxksys ds Iysu Hkkx ij ,d 'kadq r Fkk fizTe dk ?kUkRo 6gm/cm³ gSA
47. A solid spherical copper ball
whose diameter is 14 cm is
j[kdj ,d f[kykSuk cuk;k x;kA 'kadq rFkk v/ fizTe dk Hkkj Kkr djsa\
melted and converted into a wire
having diameter equal to 14 cm. Zxksys ds vk/kj dh f=kT;k
3cm rFkk'kadq dh (a) 6.4 kg (b) 7.2 kg
The length of the wire is Å¡pkbZ 4 cm gS A f[kykSus dk lEiw.kZ i`"B (c) 3.4 kg (d) 4.8 kg

4 Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039

56. Height of a prism-shaped part of If the volume of the prism be 2070 equilateral triangle of side
a machine is 8 cm and its base cm3, then the area of the lateral
10 3 cm. If the total surface
is an isosceles triangle, whose surface is
each of the equal sides is 5 cm area of the pyramid is 270 3
fdlh fizTe dk vk/kj prqHkqZt
ABCD gS A
and remaining side is 6 cm. The sq. cm. its height is
volume of the part is fn;k x;k gS]AB = 9 lseh-]BC = 14
fdlh e'khu ds fizTe vkdkj ds Hkkx dh lseh-]CD = 13 lseh-]DA = 12 lseh- fdlh fi jkfeM dk vk/kj10 3 cm Hk
qtk
8 lseh gS vkSj mldk ry Hkkx ,d
Å¡pkbZ rFkk DAB = 90° gS A ;fn fizTe dk okyk leckgq f=kHkqt gSA ;fn fijkfeM dk
lef}ckgq f=kHkqt gS ftldh ,d cjkcj okyh vk;ru 2070 lseh-gks
3
] rc fizTe dk fr;Zd 270 3 cm² gS] rc
lEiw.kZ i`"B {ks=kiQy
5 lseh dh gSa vkSj 'ks"k
Hkqtk,¡ Hkqtk
6 lseh dh i`"B {ks=kiQy Kkr djsa\
Å¡pkbZZ Kkr djsa\
gS] ml Hkkx dk vk;ru crkb,A (a) 720 cm2 (b) 810 cm2
(a) 90 cm³ (b) 96 cm³ (c) 1260 cm2 (d) 2070 cm2 (a) 12 3 cm (b) 10 cm
(c) 120 cm³ (d) 86 cm³ 61. If the area of the base, height
57. Let ABCDEF be a prism whose and volume of a right prism be (c) 10 3 cm (d) 12 cm
base is a right angled triangle, 3 3  65. A right prism stands on a base
where sides adjacent to 90º are   of 6 cm side equilateral tri-
9 cm and 12 cm, If the cost of  2  p² cm², 100 3 cm and
  angle and its volume is
painting the prism is Rs. 151.20 7200 cm³ respectively, then 3
at the rate of 20 paisa per sq cm 81 3 cm . the height (in cm )
the value of P (in cm) will be ?
then the height of the prism is: of the prism is
; fn ,d ledks.kh; fizTe ds vk/kj dk {ks=kiQy]
ekuk fd ABCDEF ,d fizTe gS ftldk 6 lseh Hkqtk okys leckgq f=kHkqt dks vk/kj
vk/kj ledks.kh; f=kHkqt gS 90º
ftldhdh 3 3 
ekudj ,d fizTe cuk;k x;k rFkk vk;ru
Å¡pkbZ rFkk vk;ru Øe'k%
  P²
ly aXu Hkqtk,a
9 cm vkSj 12 cm gSA ;fn  2  81 3 cm³ gS
A fizTe dh Å¡pkbZ Kkr djsa\
20 iSls izfr oxZ
fizTe dks jaxus dh ykxr
cm², 100 3 cm vkSj7200 cm³ gS]
lseh dh nj ls 151.20 gSrks fizTe dh (a) 9 (b) 10 (c) 12 (d) 15
ÅapkbZ fdruh gS\ rksP ( ls-eh-)d k eku Kkr djsaA 66. The base of a right pyramid is
2 a square of side 40 cm long. If
(a) 16 cm (b) 17 cm 3
(a) 4 (b) (c) 3 (d) the volume of the pyramid is
(c) 18 cm (d) 15 cm 3 2 8000 cm3, then its height is:
58. The base of a right prism is an Pyramid fdlh fijkfeM dk vk/kj40 l seh-Hkqtk okyk
equilateral triangle of area 173 62. If the slant height of a right
2
cm and the volume of the pyramid with square base is 4
oxZ gSA ;fn fdlh fijkfeM dk vk;ru
8000
3
prism is 10380 cm . The area metre and the total slant sur- lseh- gS
3
A fijkfeM dh mQ¡pkbZ Kkr djsa\
of the lateral surface of the face of the pyramid is 12 (a) 5 cm (b) 10 cm
prism is square metre, then the ratio of (c) 15 cm (d) 20 cm
(use 3 =1.73) total slant surface area and 67. A right pyramid stands on a
area of the base is:
fdlh leckgq f=kHkqt vk/kj okys fizTe dk vk/ square base of diagonal 10 2
fdlh fijkfeM dh fr;Zd Å¡pkbZ
4 ehVj rFkk
173 cm² gS rFkk fizTe dk
kj dk {ks=kiQy cm. If the height of the pyra-
vk;ru 10380 cm³ gSA fizTe dk fr;Zd i`"B dqy fr;Zd i`"B {ks=kiQy
12m² rFkk vk/kj mid is 12 cm, the area (in cm2)
{ks=kiQy Kkr djsa\ ,d oxZ gSA rc fr;Zd i`"B {ks=kiQy rFkk of its slant surface is
vk/kj dk {ks=kiQy dk vuqikr Kkr djsa\
(a) 1200 cm2 (b) 2400 cm2 10 2 cm fod.kZ okys oxZ dks vk/kj
(c) 3600 cm 2
(d) 4380 cm 2 (a) 16 : 3 (b) 24 : 5
59. Base of a prism of height 10 cm
ekudj ,d fijkfeM cuk;k x;kA ;fn fijkfeM
(c) 32 : 9 (d) 12 : 3
is square. Total surface area of 63. If the base of a right pyramid is dh Å¡pkbZZ
12 l seh- gS] fr;Zd i`"B dk {ks=kiQy
the prism is 192 sq. cm. The triangle of sides 5 cm, 12 cm Kkr djsaA
volume of the prism is
and 13 cm and its volume is (a) 520 (b) 420 (c) 360 (d) 260
10 ls-eh- m¡pkbZ okys fizTe dk vk/kj oxkZdkj
330 cm, then its height (in cm) 68. A pyramid has a square base,
gSA fizTe dk dqy i`"Bh; {ks=kiQy
192 oxZ ls- will be whose side is 8 cm. I f the
eh- gSA fizTe dk vk;ru gS fdlh fijkfeM dk vk/kj5cm, 12cm height of pyramid is 16 cm,
(a) 120 cm3 (b) 640 cm3
3
rFkk13cm Hkqtk okyk f=kHkqt gSA fijkfeM
then what is the total surface
(c) 90 cm (d) 160 cm3
60. The base of a right prism is a
dk vk;ru 330cm³ gS] rc fijkfeM dh area (in cm2) of the pyramid?

quadrilateral ABCD, given that Å¡pkbZZ Kkr djksa\ , d fijkfeM dk vk/kj oxZ gS ftldh Hkqtk
AB = 9 cm, BC = 14 cm, CD = 13 (a) 33 (b) 32 (c) 11 (d) 22 8 l seh- gSA ;fn fijkfeM dh16
Å¡pkbZ
l seh-
cm, DA = 12 cm and DAB = 90°. 64. The base of a right pyramid is

Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039 5

 f=kT;k (ls-eh- esa) D;k gS\
gS] rks fijkfeM dk dqy i`"Bh; {ks=kiQy (lseh-
2
esa) D;k gS\ (a) 3 49 (b) 7

(a) 64  17  1 (b) 32  13  1 (c) 14 (d) 3

98
72. A right triangular pyramid XYZB
(c) 64  13  1 (d) 32  5  1 is cut from cube as shown in fig-
ure. The side of cube is 16 cm. X,
69. A pyramid has square base. The
Y and Z are mid points of the edges
side of square is 12 cm height
of the cube. What is the total sur-
of pyramid is 21 cm. The
face area (in cm²) of the pyramid?
pyramid is cut into 3 parts by 2
cuts parallel to its base. the t Slk fd vkÑfr esa n'kkZ;k x;k gS fd ,d ?ku
cuts are at height of 7 cm and ls ,d ledks.kh; f=kHkqtkdkj fijkfeM
XYZB
14 cm respectively from the d kVk x;k gSA ?ku dh16Hkqtkl seh gSA
X, Y
base. What is the difference (in rFkk Z ?ku ds 'kh"kks± ij eè; fcUnq gSaA fijkfeM
cm³) in the volume of top most e sa D;k gS\
dk dqy i`"Bh; {ks=kiQy²(ls-eh-
and bottom most part?
, d fijkfeM dk vk/kj ,d oxZ gSA oxZ dh Hkqtk Y
12 l seh- rFkk fijkfeM dh mQapkbZ
21 l seh- gSA X B
fijkfeM dks mlds vk/kj ij lekarj
2 dVkoksa ls
Z
3 Hkkxksa esa dkVk tkrk gSA dVko vk/kj ls Øe'k%
7 l seh- rFkk
14 dh mQapkbZ ij gSA lcls mQij
rFkk lcls uhps ds Hkkx ds vk;ru dk varj (a) 48[( 3)  1] (b) 24[4  ( 3)]
(lseh-³ esa) D;k gS\
(a) 672 (b) 944 (c) 28[6  ( 3)] (d) 32[3  ( 3)]
(c) 786 (d) 918 73. A hollow cylinder is made up of
70. There is a pyramid on a base metal. The difference between
which is a regular hexagon of outer and inner curved surface
side 2a cm. If every slant edge area of this cylinder is 352 cm2.
Height of the cylinder is 28cm.
5a
of this pyramid is of length If the total surface area of this
2
hollow cylinder is 2640 cm 2 ,
cm, then the volume of this then what are the inner and
pyramid is outer radius (in cm)?
fdlh 2a lseh- Hkqtk okys "kV~Hkqt dks vkèkkj /krq dk ,d [kks[kyk csyu cuk;k x;k gSA
ekurs gq, ,d fijkfeM gSA ;fn fijkfeM dk
csyu ds ckÞ; rFkk vkarfjd oØ i`"Bh;
5a
fr;Zd fdukjk2 l seh- gS] fijkfeM dk vk;ru {ks=kiQy ds 352 eè; lseh-2 dk varj gSA
Kkr djsa\ csyu dh Å¡pkbZ 28 lseh- gSA ;fn bl
(a) 3a3 cm3 (b) 3 2 a2 cm3 [kks[kys csyu dk dqy i`"Bh; {ks=kiQy
2640

(c) 3 3 a3 cm3 (d) 6a3 cm3 lseh-2 gS]rks csyu dh vkarfjd rFkk ckÞ;
f=kT;k (lseh- esa) D;k gS\
71. A regular pyramid has a square
(a) 4, 6 (b) 10, 12
base. The height of the pyra-
mid is 22 cm and side of its base (c) 8, 10 (d) 6, 8
is 14 cm. Volume of pyramid is
equal to the volume of a sphere.
What is the radius (in cm) of
the sphere?
,d le fijkfeM dk vk/kj ,d oXkZ gSA
fijkfeM dh mQ¡pkbZ 22 ls-eh- gSA rFkk mlds
vk/kj dh Hkqtk 14 ls-eh- gSA fijkfeM dk
vk;ru ,d xksys ds cjkcj gSA xksys dh

6 Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039

 Volume of cube a=
Solution 64 = 4 units
3
= 3 × 3× 3 cm³
1. (c) Given, Surface area of parallelopiped
6  9  12
volume of a cube = (side)³ =512  Number of cubes= Surface area of cube
333
Side = 3
512 = 3 8  8  8 = 8cm = 24 2 8  32  16
2(lb  bh  hl )  
 Total surface area of cube 8. (b) Let l = 9x, h = 3x, b = x = = 2
6a ² 6 4

= 6×(side)2 = 6×8×8 = 384cm2 l × b × h = 216 × 1000
(1 litre = 1000 cm³) 7
2. (b) No. of cubes with no side = =7:6
9x × 3x × x = 216000 6
painted = (n–2)3
27x³ = 216000 14. (c) Let breadth = b m
Where n is the side of the big-
x³ = 8000 length of room = 2b m
ger cube 
x = 20 (l = 2b)
Required number = (6–2)³ = 64
l = 180 cm = 18 dm Height = 11 m
3. (a) Let the side of cube = a cm
9. (a) Let the length, breadth and Area of four walls of room
a 3 height be l, b, h respectively
 660 m² (given)
lb = x 2(l + b)× h = 660
bh = y 2(2b +b)× 11 = 660
lh=z 3b × 22 = 660
Diagonal of cube = a 3 cm  l²b²h² = xyz b = 10
(l bh)² = xyz  Breadth = 10 m
a 3 = 12
Length = 20 m
on squaring  v² = xyz
Area of floor = l × b
a²(3) = 12 10. (a) Volume of the box
Length × breadth
a² = 4 =l×b×h 20 × 10 = 200 m²
a = 2cm = (40 – 8) × (15 – 8)× 4 15. (d) volume of air in room
Volume of cube = a³ = 2³ = 8 cm³ = 32 × 7 × 4
= 204 m³
15  ³ = 896 cm³ (area of floor) × height = 204
4. (c) Number of cubes= 11. (d) Let the three sides of the cuboid
3  ³  volume = area of base × height
be l, b and h (Area of floor) × 6 = 204
= 125
5. (d) Length of largest bomboo  lb = bh = hl = 12 204
l² b²h² = 12 × 12 × 12 = 1728 Area of floor = 6
= 34 m²
= 5²  4²  3² = 25  16  9

16. (b) Let the increase in level
=  lbh = 1 7 2 8 = 12 12
50 = 5 2m = xm
6. (a) The external dimensions of = 2 4 3 cm³
the box are l = 20 cm, b = 12  2  1
cm, h = 10 cm 12. (b) Whole surface area of cuboid  1000 1000   = 100×10× x
 100  2
External volume of the box = 2(whole surface area of cube) –
2 × area of one face  x = 10m
= 20×12×10 = 2400 cm³
( two faces of the two cubes 17. (b) Volume = 20 m³
Thickness of the wood = 1 cm
are not visible now) = 20 × (100)³ cm³
Internal length =20 – 2=18 cm
Internal breadth=12 –2=10 cm  Required area = 12a² – 2a² = 10a² Volume of one brick
Internal height ( vkarfjd mQ¡pkbZ) = 10×6² = 360 cm² = (25 × 12.5 × 8) cm³
= 10 – 2 = 8 cm 13. (d) Sides of parallelopiped are in
 Required number of bricks
ratio = 2 : 4 : 8
Internal volume of the box
Let length = 2 units 20  100  100  100
= 18×10×8 = 1440 cm³ = = 8000
breadth = 4 units 25  12.5  8
Volume of the wood
Height = 8 units
= (2400 – 1440) cm³ = 960cm³ 35
Let the side of cube = a unit 18. (d) Radius of tank = cm
7. (c) The number of cubes will be 2
volume of cube = volume of
least if each cube will be of
parallelopiped Let initial height = H
maximum edge
a³ = 2 × 4 × 8 Final height = h
Maximum possible length a³ = 64
= HCF of 6, 9, 12 = 3

Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039 7

 Height = 2.8 Volume of the cylinder = R² H
2 2
 35   35  l² = r² + h² = 9.6² + 2.8²
π   H – π  h = 11000 22 7 7
 2   2  = 92.16 + 7.84 = 100 =   × 12
7 2 2
cm³ l = 100 = 10 m = 22 × 7 × 3 = 462 cm³
Area of the canvas= π r l 25. (b)
2
 35 
π    H – h   11000 12 = h
 2 22
=  9.6  10 = 301.7
7 H 24 cm
12
11000  2  2  7 80 1
H–h= = 22. (b) Volume of a cone = r ²h
35  35  22 7 3
R = 7 cm
1
3 r ² 24 = 1232 cm ² Volume of bigger cone
= 11 cm 3
7 1
1232  3  7 = ×(7)²×24
19. (d) r² = 3
24  22
r² = 7 ×7 1 22
H= 60 cm =  × 7× 7 × 24
r = 3 7
7  7 = 7 cm
= 22 ×7 × 8 = 1232 cm³
l = r²  h²  7²  24²  625
3
R = 32 cm =25 volume of smaller cone h
= 3
We have to find the slant Curved surface area = rl volume of bigger cone H 
height
3
Take ratio of H and R 22 Volume of smaller cone 12
=  7  25 = 3
7 1232 24
= 60 : 32
15 : 8 = 550 cm² volume of smaller cone= 154 cm³
23. (b) 26. (b)
L = 15²  8² = 17
= 17 × 4 = 68 cm
Cost of painting
= Surface area of cone × 35 h
h=14 cm = 6 cm
= π R L × 35 11cm

22 32  68 3cm
=   35 Volume of the cylinder = vol-
7 10000
ume of cube Volume of sphere
= Rs. 23.94 (approx) n=
r ²h = (side)³
20. (b) l = 2.5 km volume of cone
22
Area of base = π r² = 1.54 × r²× 14 = 11 × 11× 11
7 4
6 ³
1.54  7 11  11  11 121 3
r² = r² = = n= 1 = 24
22 22  2 4 3 ²  4
11 3
1.54  7 r= cm = 5.5 c
2 27. (c) Height of cylinder
r = = 0.7 km
22 24. (b) = Breadth of tin foil
l² = r² + h² 2pR=22 cm  Circumference of the base of
22 cm cylinder= Length of the foil =
h² = l² – r ²
22 cm
12 cm 12 cm
= 2.5² – 0.7²  2r = 22
= 5.76 = 2.4 km 22  7 7
r= = cm
 Cylinder is folded along the 22  2 2
diameter
21. (c) Radius = length of rectangle 22 7 7
2
 2R = 22 Volume=r²h= × × × 16
7 2 2
19.2 =616cm³
= = 9.6 m 22 22  7 7
2 R= = = cm
2 2  22 2

8 Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039

28. (d) r² = 770  Curved surface area of cylinder
22 21 21
770  7 Volume = r²h = × × ×16 22 105
7 2 2
 r² = =2rh = 2 × × ×3
22 = 5544 cm³ 7 2

 r = 7 5 cm 33. (b) Let the radius and height be = 990 m²

5x and 12x  Total curved area of structure
rl = 814
1 2200  curved area of cone + curved
814  7 37  × × 25x² × 12x = area of cylinder = 10395 + 990
 l= = 3 7
22  7 5 5 = 11385 m²
l² = h² + r² 2200  7  3  Total area of canvas = 11385 m²
 x³ = 37. (a)
7  22  25  12
37  37
= = h² + 245  x =1
5

1369 144
 slant height= 5²  12² =13 cm
 h² = – 245 = 34.(d)radius of cone
5 5 r = 16 metre (given) 1
Let slant height = l metre Volume(vk;ru )  r²h
12 3
 h= Curved surface area = rl
5 1
3
= 427 m 2 (given)  12 ×12 × 9
1 7 3
Volume = r²h
3 22
 16  l 
2992  144 × 3  432
7 7
1 22 12 2992 38. (d) Ratio of volume of bigger cone
= × × 7 5 × 7 5 × l=  8.5 metre
3 7 5 22  16 and smaller cone = (Ratio of
35. (c) Let height of cone = h altitude) 3
= 6 1 6 5 cm³
Radius of cone = r = (1 : 2 : 3)3 = (1 : 8 : 27)
29. (b) In this case the breadth be- 1
π r² h
 Ratio of parts
comes the circumference of Volume of cone =
3 = 1 : 8–1 : 27–8
the base of the cylinder Now height is doubled
 2r = 44 = 1:7:19
Volume of new cone
44  7 1 1 39. (a)
 r= = 7 cm = π r² (2h) = π r² h
22  2 3 3 r
New volume of cylinder = r²h Required ratio= 1 : 2
22 36. (a)
= × 7 × 7 × 100
7 Total surface area of sphere
= 15400 cm³
= 8 square unit
1
30. (c) r²H = r²h 4r² = 8
3
r² = 2
1
H= h r= 2 units
3
 h = 3H = 3 × 6 = 18 cm 4
31. (d) Ratio of height Volume of sphere = r ³
3
= 3
Ratio of volume 105
 radius of cone = m 4 8 2
2 = 3   2³ = units
h 1 3
 = slant height of cone = 63 m
H 3 4 4
 Curved surface area of cone 40. (c) R 3  64  r 3
3 units  30 3 3
= (rl)
2 units  20 4
R ³
The cut is made 20 cm 22 105 3
 = × ×63 = 10395 m² = 64
above the base 7 2 4
 r³
32. (d) 2rh = 1056 3
105
= Radius of cylinder = m  8 3
1056  7 21 2   = (4)³
r= = r 
2  22  16 2 Height = 3 m (given)
 = 2 cm
r

Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039 9

Ratio of area = (Ratio of radius)² 1 : 3 : 6 R=7cm

= (8 : 2)² = 16 : 1 Height, radius and slant r h

height are all in same ratio. 7cm
41. (a)
r1 r2 FC = 90cm, EG = 45cm, DH
= 15cm 4
Volume of the solid sphere = r ³
3
AF = 120cm, AE = 60cm, AD
4
Ratio of volume of sphere × = 20cm =  × 7 × 7 × 7 cm³
ratio of weight per 1 cc. of ma- 3
AH = 25cm AG = 75cm
terial of each Let the length of wire = h cm
HG = 50cm
= Ratio of weight of two sphere 4
Area of middle part = π R²h =  ×7×7×7
3
4 (l (R + r) + R² + r²)
r1 ³ 4
3 289 8 = π (50 (60) + 15² + 45²) 7×7×h= × 7 ×7 × 7
  3
4 64 17
r2 ³ = 16500 28
3 44. (b) h= cm
3
4 22 4 22
r1 ³ 8 6 4 8 88 
3 7

r13  r23  r33  
3 7
 R3  48. (b)
= =
r2 ³ 1 7 2 8 9 1 7 1 7 1 7 r =3 H
4 22 4 22
r1 8   (1 + 8 + 27) =   R3
 3 7 3 7
r2 17 R
 R³ = 36cm³
 8 : 17 4
R = 3.3 cm Volume of sphere = r ³
3
1 = 3.2 cm (Approximate)
42.(b) Surface area of Sphere
8 45. (a) Volume of coffee 4
=  × 3× 3×3 = 36 cm³
3
1 r ² 2 3 2 22 3
= × 4 r² = = r =   4  If the water level rises by H cm
8 2 3 3 7
R²H = 36
Three Quadrant of Sphere 128 3
= cm 6 × 6 × h = 36
3
90 3 h = 1 cm
=3×  r² =  r²
360 4 1 2 49. (b) Volume of metallic sphere =
Volume of cone = r  h
3 volume of cone
Total Surface area of each part
Sphere 4 1
1 2 π×3×3×3 = π R² h
=  8   16 3 3
r ² 3 3
4 1
= +  r² π×3×3×3 = × π × 6 ×6 × h
2 4 1024 3 3
= 
3 108
5 22 Required percentage h= = 3 cm
= × × 21 × 21  66
4 7 1024 128 50. (b) Radius of both Hemisphere
–
55  63 3 3 and cone = R
=  100
= = 1732.5 cm² 1024 Also height of hemisphere is
2
3 equal to its Radius = R
A  height of both hemisphere
43. (b) 896
= ×100 = 87.5% and cone = R
1024
H
46. (c) Let the original radius be ‘r’ Now, In cone
D slant height,
 Area = 4r²
G l= R²  R² = 2R
E New area = 4(2r²) = 16r²
C .S .A o f h e m isp h e re
 New area is 4 times the old area
B C C .S .A o f co n e
F 47.(d)
Ratio of AD : DE : EF 2 π R² 2
1 : 2 : 3 = = = 2 :1
π R × 2 R 1
Ratio of AD : AE : AF

10 Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039

51. (b) 173
 9327 723 4
 a² =
1 .7 3
 997722
173
 972 =  4  100
173
Volume of Prism a² = 400
= (9  7  2)  9  1134cm 3 a = 20 cm.
Perimeter of base = 20 × 3
Surface area of hemishpere = 55. (b) Volume of prism = 60 cm
2r²
1  Volume of prism = 10380 cm³
=  10  12  20  1200 cm ³
22 2 (given)
=2× × 9 = 56.57 cm²
7 Area of base × height
 Weight of prism = 1200 × 6
height of cone = 4 cm = 7200 gm = 7.2 kg 10380
radius = 3 cm 56. (b) Volume of the part (prism) = height = = 60
173
 Slant height = 16  9 = 5 cm
Area of base × height LSA = Perimeter of base × height
 Surface area of cone = rl Area of base (Isosceles  ) LSA = 60 × 60 = 3600 cm²
b
59. (d) Let the side of the square
22
= × 3 × 5  47.14 cm² = 4a ² – b ² = a cm
7 4
6 T.S.A = C.S.A + 2 base area
 Total surface area of the toy = = 4  5  ² –  6  ² = 12 cm²
4 C.S.A= base perimeter × h
Area of cone + area of hemi- Volume of prism= 12 × 8
sphere = 96 cm³ Volume = base area × h
 47.14 + 56.57  103.71 cm² 57. (c) Painted Area of Prism  T.SA = base perimeter × h + 2
= 151.20 × 5 = 756.00 cm² base area
52. (a) Required number of spheres
AC = 15 192 = 4a × 10 +2a2
volume of metallic cone [By using pythagoras theorem]
= volume of a sphere Total surface Area = Perimeter 2a2 + 40a – 192 = 0
of base × Height + 2 × Area of a2 + 20a – 96 = 0
1 base a2 + 24a– 4a – 96 = 0
 30  30  45
= 3 = 81 1 a (a+ 24)–4 (a+24) = 0
4 = (15 + 9 + 12) × h + 2 × × 9 × 12
 5  5  5 2
3 (a+24) (a–4) = 0
756 = 36 × h + 108 a = 4, (–24)
53. (d) Number of cones 
36h = 756 – 108
 a=4
volume of sphere 648
= h= = 18 cm (Side can never be – ve)
volume of cone 36
Volume = base area × h
58. (c)
4 Volume = 16×10
10.5  ³
= 3
1 Volume = 160 cm3
3.5  ²  3
3 60. (a) 13 cm C
D
4  10.5  10.5  10.5
= = 126 15 cm
3.5  3.5  3 12 cm 14 cm
54.(d) Volume of prism = (area of base
× height A B
Let side of equilateral triangle 9 cm
Area of base (i.e area of be = a
triangle) In Δ ABD,
3
 Area of base  Area = a
2
BD = AB²  AD² = 9²  12²
4
= s s – a s – b s – c  = 81  144 = 225 = 15 cm
3 2
= ( By Heron's formula) = a  173 cm 2
4 1
13  20  21 54 Area of Δ ABD = × AB × AD
So, S=   27 173 2
2 2  a² = ×4
3
1
 2 7 2 7 – 1 3 2 7 – 2 0 2 7 – 2 1 ( 3  1 .7 3 ) = 2
 9  12 = 54 cm²

 27  14  7  6 In Δ BCD

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 Semiperimeter
1
13  14  15 42 × Base area × height = 330
= = = 21 3 66. (c)
2 2
Area of Δ BCD 1
× 30 × height = 330
= s s – a  s – b  s – c  3

= 2 1 2 1 – 1 3  2 1 – 1 4  2 1 – 1 5  330  3
height = = 33 cm
30
= 2 1  8  7  6 = 21 × 4 = 84 cm²  Area of base = 40 × 40
64. (d)
Area ABCD = 84 + 54 = 138 cm² = 1600 cm²
A
Height of prism = Let height of pyramid = h
10 3 1
volu m e 2070 10 3 o M M  Volume = × h × area of base
= = 15 cm 3
A rea of base 138
(h)
slant height
1
Perimeter of base B E
10 3
C o E × h × 1600 = 8000
3
= 9 + 14 + 13 + 12 = 48 cm h = 15 cm
Base is equilateral triangle
Area of lateral surface = 67. (d)
In radius of equilateral
perimeter × height = 48 × 15 triangle
2
= 720 cm² 10 12
slant height
side of equilateral 
= OE =
61. (a) Volume of Right Prism = 2 3 5
Area of the base × Height 10
= 2 3 = 5 cm
3 1
 1 0 2 = 10 cm
Side of square =
2
3 3 2 Slant length, l = h ²  O E ²
 7200  P  100 3 Slant height = 5²  12² = 13 cm
2 = h ²  25
 72×2 = 9P2 Total surface area = 270 3 Lateral surface area
 P2 = 16 1 1
(Perimeter of base × slant = × perimeter of base ×
 P = 4 2 2
62. (a) Total slant surface area height) + Base area = 270 3
Slant height
1 1 3
= 4
2
 4  a  12
2

30 3  h ²  25 
4 =
1
× 40 × 13 = 260 cm ²
2
(where a is the side of the 10 3  ² = 270 3
square base) 68. (a) slant height = 16²  4²
15 3 h ²  25  7 5 3
12 3
 a = = cm = 272  4 17
8 2 = 270 3
= 13
h ²  25
Total surface Area of pyramid
9
 area of base = cm ²
4 h² +25 = 169 1 
h² +169 – 25 = 144 =  2  (Perimeter of base)  (Slant height)
12
9 h = 1 4 4 = 12 cm
 Required ratio = = 16 : 3 (Area of base)
4 65. (a)
63. (a) 1
6 cm 6 cm =  32  4 17  64
2

6 cm = 64 17  64  64  17  1
Volume of prism =area of
base × height 69. (a) Volume of Pyramid
Clearly the base triangle is 3 2
= 6  × height 1
the right triangle 4 = × Area of base × Height
3
 Area of triangle ABC 3
× 6 × 6× height = 81 3
1 4 1
=  5  12 = 30 cm² = × 12 × 12 × 21 = 1008 cm²
2 81 3  4 3
Volume of the pyramid Height = = 9 cm
3 66 Side of square of upper
1 base
=  base area  × height
3 a1 7
= =
a3 21

12 Telegram Channel-Maths by Sultan Sir, Contact No.-8418077039

 70. (c) Side of regular hexagon = 2a
a1 1 1 7 3
= cm r³ = ×14×14×22× ×
12 3 3 22 4
3 2
a1 = 4 Area of hexagon = 6  × 2a  r = 7 cm
4
Volume of upper Pyramid 72. (d) BX = BY = 8cm
 6 3acm² 2

1 112  XY = YZ = XZ = 8 2
= ×4×4×7= cm³ Slant edge of pyramid
3 3
5a
Side of upper & middle  cm
2
Pyramid 
a2 14 5a
= = 2
a3 21
a2 = 8
Volume of upper & middle
Pyramid
1 896 5a
= × 8 × 8 × 14 = Slant edge 
3 3 2

Volume of only Bottom l2 = 8² – 4 2 ² = 32

 
(Given)
896 2128 5a l= 4 2
= 1008 – = cm³
3 3 2
1
2128 TSA = ×Perimeter of base ×
Req. Difference = – 2
3 O 2 h + Area of base
112 2016
= = 672 5a 1
3 3 HF = (slant height) = ×3 × 8 2 ×
2 2
Alternate: OH = Height (h)
(2a) (given) 3
4 2+ 4 8 2 ²  
 2 
 5a  – 2a 2
12
height = 2 
12 12
  = 96 + 32 3 = 32 3  3 cm²
 
12
2
25a 3a
= – 4a 2 =
4 2 73. (d) 2h (R–r) = 352
 Volume of pyramid
1 22
Volume of Pyramid = × 1 2× ×28(R–r) = 352
3 = × area of base × height 7
3
Area of Base × Height R–r = 2 ......(i)
1 3 TSA = 2Rh + 2rh + 2(R²–r²)
1 = × 6 3a 2 × a
= × 12 × 12 × 21 = 1008 3 2
3 2640 = 2[28(R+r)+(R²–r²)]
= 3 3a ³ cm³
Side 1 2 3 22
Volume 1 8 27 1 2640 = 2×
7
71. (b) Volume of pyramid = ×
× 112 3
3 [28 (R+r)+2 (R+r)]
1008
Area of Base × height
1 7 19 1
= ×14² × 22
3
diff. (19-1) = 18
difference between top and Volume of sphere = Volume of
bottom = 18 pyramid

18 112 4 1
 = 672  r³ = ×14×14×22
3 3 3

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 44
2640 = [30 (R+r)]
7

2640×7
= R+r
44  30
R+r = 14 ......(ii)
By solving eq. (i) & (ii), we get
2R = 14+2

16
R= =8
2
2r = 14 – 2

12
r= =6
2

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