File No.38/04/20/12/2014
VIII CLASS - IIT, NEET FOUNDATION - OLYMPIAD
HYDERABAD
2013-2014 PROGRAM M E
TOTAL BITS: 35
DATE : 22.12.14
TO : 27.12.14
WEEKLY WORKSHEET_22_KEY & SOLUTIONS
MATHEMATICS - MENSURATION - KEY
1) 2
2) 3
3) 2
4) 2
5) 3
6) 4
7) 1
8) 1
9) 1
10) 2 11) 3 12) 4 13) 3 14) 1 15) 3
PHYSICS - MAGNETISM - LEVEL - I - KEY
1) 1
2) 4
3) 3
4) 3
5) 4
6) 4
7) 1
8) 4
9) 3
10) 4
CHEMISTRY - CHEMICAL BONDING - LEVEL - I - KEY
1) 1
2) 4
3) 2
4) 2
5) 2
6) 2
7) 3
8) 2
9) 3
10) 1
MATHEMATICS - MENSURATION - SOLUTIONS
1Sol. Internal dimensions of the box are : Length = (12 - 2) cm = 10 cm
Breadth = (10 - 2) cm = 8 cm
Height = (8 - 2) cm = 6 cm
2
 Inner surface area = 2  10  8  8  6  6  10  cm
= 2   80  48  60  cm2
= 2188 cm2
= 376 cm2
2Sol. Volume of the new cube = 63  83  10 3
= 216 + 512 + 1000
= 1728 cm3
 Each edge of the new cube =
3
2013 - 2014
1728  12cm


 Diagonal of the new cube = 12  3 cm
= 12  1.732 cm
= 20.78 cm
3Sol. Let the width of the wall be x m. Then,
Height = 6x m and Length =  7  6x  m  42x m
Given, volume of wall = 16128
 42x  x  6x 16128
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 x3 
16128
 64  x  4
42  6
 Width = 4 m
4Sol. Let the sides of the cuboid be x, 2x and 4x units.
Let the length of each edge of the cube = a units
Then, a3   x  2x  4x 
CHALLENGER
1
WEEKLY WORKSHEET - 22
File No.38/04/20/12/2014
VIII CLASS - IIT, NEET FOUNDATION - OLYMPIAD
 a3  8x 3  a  2x
2
2
Length of diagonal of the cuboid =
x 2   2x    4x 
=
x 2  4x 2  16x 2 
21x 2
Length of diagonal of the cube = a 3
= 2x 3  12x 2
 Required ratio =
5Sol. Number of cubes
21x 2
12x
2

21
 1.75
12
=
Volume of cuboid
Volume of cube
=
8 4 2
 64
1 1 1
Surface area of cuboid = 2   8  4  4  2  2  8 cm2
= 2   32  8  16  cm2  112 cm2
Surface ara of 64 cube = 64 x 6 cm2  384cm2
 Required ratio =
112 7

 7 : 24
384 24
6Sol. Volume of water flown through the pipe in 1 hour = 1.5  1.25  20  1000  m3
= 37500 m3
2013 - 2014
Volume of water flown in the tank =  200  150  2 m3
= 60000 m3
 Time taken =
60000
8
hrs  hrs
37500
5
8

   60  min  96min
5

7Sol. A = ph
= 320 x 12 = 3840 sq. m
8Sol. TSA = 2(LB + BH + LH) = 2(16 x 14 + 14 x 10 + 16 x 10)
= 2(224 + 140 + 160) = 2 x 524 = 1048 m2
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9Sol. (a) LSA of a cuboid is 2h(l + b) sq. units
(b) Breadth of a cubois is
A
 l units
2h
(C) TSA of a cubois is 2(lb + bh + hl) sq. units
A
(D) The height of a cuboid is 2  l  b  units
CHALLENGER
2
WEEKLY WORKSHEET - 22
File No.38/04/20/12/2014
VIII CLASS - IIT, NEET FOUNDATION - OLYMPIAD
10Sol. Ratio = 36 : 49
4S12 : 4S22  36:49  S1 : S2  6:7
2
11Sol. T.S.A of second cube = 6  7   294 sq. units
2
12Sol. L.S.A of first cube = 4  6   144 sq. units
13Sol. Rs 50 is the cost of white washing for 100m2
Rs 1024 is the cost of ______?

1024  100
 2048m2
50
Let, length = 5k; breadth = 3k, height = 2k
2h l  b   2048 m2
2.2k  5k  3k   2048  32k 2  2048  k 2 
2048
 64  k  8
32
 l  5k  5  8  40m ; b  3k  3  8  24m and h  2k  2  8  16m
14Sol. Perimeter base = 2(l + b) = 2(40 + 24) = 128 m
15Sol. Area of four wass is = 2h  l  b   2 16   40  24  3264  2048 m2
PHYSICS - MAGNETISM - LEVEL - I - SOLUTIONS
1Sol. Since widths of both the sections are same, their poles strengths are equal.
2Sol. Amp-m2 is the unit for magnetic moment.
3Sol. Intensity along axis =
0 M
.
2 d3
2013 - 2014
4Sol. Magnetic lines of force run in an external field from north pole of magnet towards south pole and inside
magnet from south pole to north pole. Hence they have a direction. Moreover, the tangent at any point on
a line of force indicates direction of magnetic field at that point. A line of force is a vector quantity.
5Sol. Mgnetic flux density is expressed in tesla. Tesla =
6Sol. F 
Weber Newton

m2
amp x m
 0  6M 1M 2 
.
 along axis in a line.
4  d 4 
7Sol. Tesla represents units of magnetic induction in S.I. system
8Sol. Force between two magnets lying along same axis 
1
where r denotes distance between the centres of
r4
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two magnets.
4
F1 r2 4
4.8  2r 


    24  16
F2 r14
F2  r 
16 F2 = 4.8 = F2 = 0.3 N.
9Sol. At a neutral point,
CHALLENGER
3
WEEKLY WORKSHEET - 22
File No.38/04/20/12/2014
VIII CLASS - IIT, NEET FOUNDATION - OLYMPIAD
Magnetic fields due to magnet and due to Earth are equal in magnetic and opposite in direction. The
resultant magnetic field is zero.
Field intensity due to magnet =
0 M 107 x 2 x 6.75
. =
2 d3
d3
Field intensity due to Earth = 5 x 10-5

10 7 x 2 x 6.75
 5x105
d3
 d3 
107 x 2 x 6.75 2 x 6.75 x10 2

 27 x10 3
5 x105
5
 d3  (27 x10 3 )  d  3 x10 1  0.3m  30cm
10Sol. 1 weber = 108 maxwell
CHEMISTRY - CHEMICAL BONDING - LEVEL - I - SOLUTIONS
1Sol. Solid NaCl is bad conductor of electricity because there is no ions
2Sol. Yet the formula of sulphate of a metal (M) is M2(SO4)3, it is M3+ ion so formula of its phosphate would be
MPO4.
3Sol. The phosphate of a metal has the formula MHPO4 it means metal is divalent so its chloride would be MCl2.
4Sol. Perchlorate ion, ClO4- has a tetrahedral structure with sp3 hybridization. Cl and O are bonded through
covalent bonds ana bond between Na+ and ClO4- is ionic.
5Sol. Na is highly electropositive while Cl is highly electronegative so they will from ionic bond.
6Sol. In (C2H5)2OBH3 bond between Band ‘O’ is coordinates covalent.
7Sol. Covalent charecter is ionic compounds is governed by Faazan’s Rule. AlCl3 will show maximum covalent
covalent charecter on account of higher polarising poweer of Al3+ because of its having higher positive
charge and smaller size.
2013 - 2014
8Sol. Valency of phosphorous in H3PO4 is suspended ‘x’ then 3 + x - 8 =0, x - 5 = , x = 5.
9Sol. Octect rule is not valid for CO molecule.
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10Sol. Order of polarising power Be++ > Li+ > Na+ . Hence order of covalent charecter BeCl2 > LiCl > NaCl.
CHALLENGER
4
WEEKLY WORKSHEET - 22