JEE (MAIN) MOCK TEST # 03
JEE (Main) : MOCK TEST # 03
ANSWER KEY
PART-1 : PHYSICS
Q. 1 2 3 4 5 6 7 8 9 10
A. A D B B C A A D B B
SECTION-I
Q. 11 12 13 14 15 16 17 18 19 20
A. B C C C A C A C B D
Q. 1 2 3 4 5
SECTION-II
A. 100 1.59 0.57 8 0.48
PART-2 : CHEMISTRY
Q. 1 2 3 4 5 6 7 8 9 10
A. C B D B B C A A A D
SECT ION-I
Q. 11 12 13 14 15 16 17 18 19 20
A. D C A C B D A D C B
Q. 1 2 3 4 5
SECT ION-II
A. 13.7 630 6 2 2
PART-3 : MATHEMATICS
Q. 1 2 3 4 5 6 7 8 9 10
A. C C B A D B B B B C
SECTION-I
Q. 11 12 13 14 15 16 17 18 19 20
A. B B A D B C D D B C
Q. 1 2 3 4 5
SECTION-II
A. 800 32 –1 1 2
HINTS & SOLUTIONS
PART-1 : PHYSICS V 18
9. i= = = 2A
SECTION-I R 9
2. F sin q ³ f ...(1) 10. r<1
f = µ(mg – F cos q) ...(2) r>1
mgsinq ³ µmg(1-cosq) h
16. l= Since v is increasing in case (i), but it is
mv
æ qö
cot ç ÷ ³ µ not changing in case (ii). Hence, in the first case
è 2ø
de-Broglie wavelength will change, but it second
5. Strain = 0 case, it remain the same
6. R = R1 + R2 + R3
17. Here, P = 100 W, l1 = 1 nm, l2 = 500 nm
3L L L L Let n1 and n2 be the number of photons of X-rays
= + +
K eq A K A 5KA KA
and visible light emitted from the two sources
2
hc hc n n n l 1
\ n1 = n2 or 1 = 2 or 1 = 1 =
15 l1 l2 l1 l 2 n 2 l 2 500
Keq = K
16
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JEE (MAIN) MOCK TEST # 03
–
N Pb 1 5. Number of photo e emitted per second
18. = 1.5 ´ 10 -3 ´ 10 -3
N Ur 2 n=
æ 1240 ö V
N Ur çè 400 ÷ø ´ e ´ 5
=2
N Pb \ photo current = ne = 0.48µA
N Ur 2 PART-2 : CHEMISTRY
=
N0 3 SECTION-I
2 1 æ a ö
= 2. ç P + 2 ÷ (V - b) = RT
3 2n è V ø
At low P = V is high so V – b = V
3
= 2n æ a ö
2 ç P + 2 ÷ V = RT
è V ø
æ 3ö
ln ç ÷ = n ln 2 3. K=
K 2 ´ K 33
è2ø K1
Equation – (1) + (2)
æ 3ö t
ln ç ÷ = ln 2 Å 3 × (3)
è2ø T 4. In closed packing T.V. = 2Z = 2 × 6 = 12
l(3 / 2) 5. DG° = DH° – TDS°
t=T
l(2)
0 = DH° – TD S°
19. Photons per area per second at a distance r are
5.00 × 1018/4pr2. Photons per second entering the DH° 25 ´ 103
Þ T= = = 500K
eye, radius R is then this times pR2. Set this product DS ° 50
equal to 500 per second and solve for r. The result 5
is [B]. 6. (C) Mol. wt. of XY2 = = 100
0.05
20. The P-N junction will conduct only when it is forward
85
biased i.e. when – 5V is fed to it, so it will conduct Mol. wt. of X 3Y3 = ´ N A = 170
3.011´1023
only for 3rd quarter part of signal shown and when
Let molar mass of X and Y are a and b respectively
it conducts potential drop 5 volt will be across both
\ a + 2 b = 100
the resistors, so output voltage across R2 is 2.5V.
2a + 3b = 170
\ V0 = – 2.5V
a = 40 ;
SECTION-II b = 30
7. Ksp = 108 S 5
n Cp + n 2 Cp2 1
2. rmix = 1 1 = 1.59 æ 10-13 ö 5
n1Cv1 + n 2 Cv2 S= ç ÷
è 108 ø
4. Bird
y H+
14.
OH OH OH
Ring
Expansion
h = x + my
O
OH
dh mdy 4
= 0+ = ´6
dt dt 3
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JEE (MAIN) MOCK TEST # 03
Cl
PART-3 : MATHEMATICS
Zn SN2
16. + SECTION-I
2
2. Let the coordinates of B be (h, k)
Frankland reaction
Draw BL and CM perpendicular to the x-and the
CH3
H
Å + y-axis. Therefore, a cos q = CM = OD = AL = 12
NH3–C–COOH
CH3 and a sin q = DM = OA = BL = 5
Å 1
H
17. NH3–C–COO Y
H –
CH3
OH 1 12 C
–H2O NH2–C–COO M
H 5 a
D
12 a B
a
5
18. X
O 5 A 12 L
\ k = BL = DM = OM – OD
stable by resonance \ h = OL = OA + AL
= 5 + 12 = 17
Hence, point B is (17, 5)
19.
3. From geometry we know PA. PB = (PT)2 where
PT is the length of the tangent from P to the circle.
Hence PA.PB = (3)2 + (11)2 – 9 = 112 = 121
Alternately
Equation of any line through (3, 11) is
x -3 y - 11
20. Cannizaro reaction is an example of hydride ion = = r (say)
cos q sin q
(H1 ) transfer reaction. Then the coordinates of a point on this line at a
SECTION-II distance r from (3, 11) are (3 + rcosq, 11 + rsinq)
and if this lies on the given circle x2 + y2 = 9
1. pH = 13.7
then (3 + rcosq)2 + (11 + rsinq)2 = 9
31 Þ 9 + 121 + 2r(3cosq + 11sinq) + r2 = 9
2. Ew = P4 =
5 Þ r2 + 2r(3cosq + 11sinq) + 121 = 0
eq. of P4 = eq. of HNO3 which is quadratic in r, gives two values of r say
r1 and r2 and hence the distances of the points
62 weight
= Þ wt. = 630 A and B from P.
(31 / 5) 63
Thus, PA.PB = r1r2 = 121.
wt. of HNO3 required = 630 7. Let roots are a, –a, b
3 S.O.R. Þ b = m satisfy the equation
3. PbS + O ® PbO + SO2
(S= -2) 2 2 (S= +4) m3 – m · m2 + 3m – 2 = 0 Þ m = 2/3
9. (2x + 1) (2x + 3) (2x + 5) .... (2x + 99)
Change in O.S. = +4 – (–2) = [6]
éæ 1 öæ 3 öæ 5ö æ 99 ö ù
4. Na2B4O7 + 2HCl + 5H2O ® 2NaCl + 4H3BO3 Þ 250 ê ç x + 2 ÷ç x + 2 ÷ç x + 2 ÷ .... ç x + 2 ÷ ú
ëè øè øè ø è øû
Only [2] moles of HCl are required
Coefficient of x is equal to – (sum of roots)
49
é1 3 5 99 ù
= 250 ê + + + .... + ú = 2 ´ 2500
49
ë2 2 2 2û
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JEE (MAIN) MOCK TEST # 03
Lim ç
æ 3x -1 - 1 ö ( x - 1)( x - 2 )( x - 3) x7
÷
11. x ®1
è x - 1 ø sin ( ( x - 1)( x - 2 )( x - 3) ) ( x - 2 )( x - 3) \ f(0) = Þ C = 0 Þ f(x) =
(2x 7 + x 2 + 1)
Þ
ln 3 1
= ln 3 1
( )( ) 2
- - \ f(1) =
1 2 4
12. we have x5 + ex/5 = 1 + e1/5 2013
1 x/5 1
17. I= ò
1
(x - 1)(x - 2)(x - 3).........(x - 2013)dx
Þ ƒ'(x) = 5x4 + e = 5 + e1/5
5 5 b b
Using ò f(x)dx = ò f(a + b - x)dx
1 1/5 a a
Þ x = 1, ƒ(x) = 5 + e
5 2013
1 ò 1
(2013 - x)(2012 - x)......(1 - x) = -I
use g'(y) =
ƒ '(x) Þ 2I = 0 \ I = 0
1 1 18. Reqd. area = shaded region
Þ g'(1 + e1/ 5 ) = =
ƒ '(1) e1/ 5
5+ y y = x+1
5
sin(p[1 - p sin 2 tan(sin x)) (0,1)
13. Lim
x ®0 x2
y = 1–x
sin(p sin 2 t an(sin x)) æ p sin 2 tan(sin x) ö
= x®0 p sin 2 tan(sin x) ´ ç tan 2 sin x ÷
Lim x
è ø (–1,0) 0 (1,0)
æ tan 2 (sin x) ö æ sin 2 x ö 1
× ç sin 2 x ÷ ç x 2 ÷ = p
è øè ø = ò (x 2 - x1 )dy
0
f(1 + h) - f(1) 1
7
f'(1+) = lim
ò éë(1 - y ) - (y - 1)ùû dy = 6
14. 2
h ®0 h = sq. unit
0
1 + h - 1 - [[1 + h] - (1 + h)]
= lim =0 H
h ®0 h 19. cot a = d and H cot b = d
3
h(1 - 1 - h)
= lim =0 H H
h ®0 h or = tan a and = tan b
3d d
f(1 - h) - f(1)
f'(1–) = lim =0 H H 2H
h ®0 -h -
1 d 3d 3
tan(b - a) = =
- h ([1 - h] - (1 - h) ) 2 H2
= lim =0 1+ 2 H
h ®0 -h 3d q 3
a b
h (0 - 1 + h )
2
H 4H
= lim =1 Þ 1 + = d = 20m
h®0 -h 3d 2 3d
= f'(1+) = 0, f'(1–1) = + 1 Þ H 2 - 4dH + 3d 2 = 0
\ f(x) is not diff. at x = 1
5 7 Þ H2 - 80H + 3 (400) = 0
+ 8
5x8 + 7x 6 6
15.
ò æ 1 1 ö2 = ò æ x 1 x 1 ö2 dx Þ H = 20 or 60m
x14 ç 2 + 7 + 5 ÷ ç2 + 7 + 5 ÷
è x x ø è x x ø 20. Given S.D. = 2 Þ S.D.2 = 4
1 1 æ n ö
n
2
ç å xi ÷
Put 2 + 7 + 5 = t
x x å
i =1
x 2
i
è i=1 ø = 4
æ x7 ö = –
Þ f(x) = ç (2x 7 + x 2 + 1) ÷ + C n n2
è ø
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JEE (MAIN) MOCK TEST # 03
100 (20)2
Þ - 2 =4
n n
Þ 4n2 – 100n + 400 = 0
Þ n2 – 25n + 100 = 0
Þ n = 5 or 20
SECTION-II
1.
1 × 1 × 1 ×10 ×10 × 8 × 1
Only 5 at unit place and digits other than
2 & 7 at ten's place = 800 ways
1 1
4. I = ò xdx - ò [2x]dx
-1 -1
1
ïì ïü
-1/ 2
æ x2 ö 0 1/ 2 1
= ç 2 ÷ - í ò [2x]dx + ò [2x]dx + ò [2x]dx + ò [2x]dx ý
è ø-1 îï -1 -1/ 2 0 1/ 2 þï
ìï-1/ 2 0 1/ 2 1
üï
=0 í ò
- ( -2)dx + ò ( -1)dx + ò 0 dx + ò 1 dx ý
îï -1 -1/ 2 0 1/ 2 þï
=1
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