NSTSE Sample Paper Solutions Class 12 PCM

UNIFIED COUNCIL
A n I S O 9 0 0 1 : 2008 C e r t i f i e d O r g a n i s a t i o n

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NATIONAL LEVEL SCIENCE TALENT SEARCH EXAMINATION
Solutions for Class : 12_PCM

MATHEMATICS 1 1 1 0 1 2
1. (C) Let T be the thickness of the sides, then  x 4 0 0 4 0 0
+
5 3 3 3 3 3 3
that of the top will be T.
4
0 xa xb
5 V 5 xa 0 xc 0
 S = (2  rh)T +  r2. T = 2  T.r. 2 + .T.r2 3. (C) Given,
4 r 4 xb xc 0
1 5 Expanding the given determinant, we have
= 2TV. + .T.r2 = f(r)
r 4 2x3 + 2x (ac – ab – bc) = 0
2TV 5 4V  x0
f '(r) = 2
 T.r  0  r3 =
r 2 5
 1 0  1 0  1 0
 5  r3 = 4V = 4  r2h 1  1   
4. (C) A =  1  1 = 2  1 1
2

r 4 2  2   2 
 
h 5
 1 0  1 0  1 0
     
= A2.A =  2  1 1   1 1  = 3  1 1 
3
x x x 1 x  2
3
2x  3x  1 3x 3x  3  2  2   2 
2. (A) xA + B = 3
x  2x  3 2x  1 2x  1 Continuing in this way, we get
 1 0
x3  x x 1 x  2    1 0
R 2  R1  R 3  4 0 0 A100 = 100  1 1  = 50 1 
3  2   
x  2x  3 2x  1 2x  1
5. (B) The required determinant is obtained by
x x 1 x 2 the successive operations
1 3 1
R1  x R2 R3  x 3R2  4 0 0
4 4 C1  2 C1 and C1  C1 + 3 C2 + 4C3
2x  2x  3 2x  1 2x  1
 The value of the determinant is multi-
x x 1 x 2 plied by 2 (since of the first operation),
R 3  2R1  4 0 0 second operation does not affect the
3 3 3 value of the determinant.

x x x 0 1 2 PHYSICS
 4 0 0
+
 4 0 0 6. (B) Net force on a current carrying loop in
3 3 3 3 3 3 uniform magnetic field is zero. Hence,
the loop cannot translate. From fleming’s
left hand rule we can see that if

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magnetic field is perpendicular to paper f v 20 v
inwards and current in the loop is Again, m = 4=
f 20
clockwise (as shown) the magnetic force
 or v = 20 – 80 = – 60 cm
Fm on each element of the loop is radially
outwards, or the loops will have a 10. (A) Four charges q1, q2, q3 and q4 are placed
tendency to expand. at the four corners of the square PQRS
y
as shown below.
i Fm × Here,
q1 2m q2
x P Q
r r

2m 2m
r O r
7. (D) According to the given information
substance ‘X’ is ferromagnetic in nature
S R
which is gadolinium. q4 2m q3
A paramagnetic substance is feebly q1 = 2 μ C = 2  10–6 C;
attracted by a magnet. When a rod of
q2 = – 2 μ C = – 2  10–6 C;
paramagnetic substance is suspended
in a magnetic field, it slowly sets itself q3 = – 3 μ C = – 3  10–6 C;
parallel to the direction of the magnetic q4 = 6 μ C = 6  10–6 C;
field. It also moves from a weaker part
of the magnetic field to its stronger part, and PQ = QR = RS = PS = 2 m
but it is feebly attracted by the magnet. Let r be the distance of each charge from
When a rod of diamagnetic substance the centre O of the square.
is suspended in a magnetic field, it Then,
slowly sets itself at right angles to the r2 + r2 2 or r = 1 m
direction of field. It moves from stronger Potential at point O due to charges at the
part of the magnetic field to its weaker four corners,
part i.e., it is feebly repelled by the
magnet. 1 q1 q2 q3 q4
V= + + +
4πε0 r r r r
Aluminium and oxygen are paramagnetic.
Gold is diamagnetic. Gadolinium is 1 1
ferromagnetic. So, substance X is q +q +q +q
.
4πε0 r 1 2 3 4
gadolinium.
Given I = 20 A, n = 9×1030 m–3 ; A = 10–4 m2 9 × 109
8. (A) 2 × 10 6 + 2 × 10 6 + 3× 10 6 + 6 × 10 6
1
and e = 1.6×10–19 C
= 2.7  104 V
I 20
Vd = neA = 9  1030  1.6  1019  104 CHEMISTRY
= 0.138 × 10-6 m s-1 11. (A) Addition of HCl to acetylene in the
presence of mercuric salt leads to
9. (B) As the image formed is erect and hence formation of vinyl chloride which in the
virtual, the magnification produced by monomer for poly vinyl chloride.
the lens is positive i.e. m = + 4.
12. (A) The ideal conditions for the manufacture
Also, f = + 20 cm of H2SO4 by contact process are low
f temperature, high pressure and high
Now, m = concentration of reactants.
u+ f
20 13. (B) The process of heating the quenched
 4= or u + 20 = 5 steel to a temperature much below
u + 20
redness and cooling it slowly.
or u = –15 cm

2

14. (C) Amorphous solids do not melt.
They simply soften on heating, and
gradually begin to flow on further
heating. These solids are, therefore
considered as super cooled liquids.
15. (D) [Fe(CN)6]3- is an octahedral complex ion
and is paramagnetic in nature. Secondly
it is an inner orbital complex ion with
the presence of only one unpaired
electron in it.
CRITICAL THINKING
16. (D)
17. (D)
18. (C)
19. (D)
20. (B)

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