All India Aakash Test Series For JEE (Advanced) - 2023: TEST - 4A (Paper-1) - Code-A
All India Aakash Test Series For JEE (Advanced) - 2023: TEST - 4A (Paper-1) - Code-A
All India Aakash Test Series For JEE (Advanced) - 2023: TEST - 4A (Paper-1) - Code-A
ANSWERS
PHYSICS CHEMISTRY MATHEMATICS
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All India Aakash Test Series for JEE (Advanced)-2023 Test - 4A (Paper-1) (Code-A)_(Hints & Solutions)
– 30 + WA→B = 20 1
Wgas = P V + Kx 2
2
WA→B = 50 J
2. Answer (A)
( )
= 105 20 10 10−6 +
1
2
200 (0.1)2
a
Hint : tan = = 20 + 1
g
(12 − 8) 4 1 = 21 joules
Sol. : tan = = =
(12 + 8) 20 5 PiVi = nRTi
a 1
= 2 R 300
g 5 Vi = m3
105
g
a= = 2 m/s2
5 25
2 300
3. Answer (C) Vi = 3 106 cm3
5
Hint : Use Doppler effect 10
Sol. :
= 50000 cm3
V = 20 × 10 cm3
= 200 cm3
Vf = Vi + V = 50200 cm3
SD cos SD v
8. Answer (01.50)
and, = cos = 0
v0 v v Solution for Q. Nos. 7 and 8
v2 mv 2
n = n0 Hint : F =
v 2 – v 02 r
T (r )
3402 Sol. : dl =
n' = 1200 YA
3402 − 1702 dx
= 1600 Hz
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
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Test - 4A (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2023
dl =
(
1 L − r dr
2 2 2
) 13. Answer (B, D)
2 Y Hint : V AB = V A − V B
3V 3
2L3 Sol. : Urel = V + V cos60° = = 10
L = 2 2
3Y
= 15 m/s
1 10 ( 400 ) ( 0.5 )
4 2 3
= = 0.33 mm 30
3 2 1011 t = =2s
15
32 l 2 s = 10 × 2 = 20 m
Stress = = 1.5 × 108 Pa
8 14. Answer (A, B, C)
9. Answer (08.00) Hint : Use constraint relation
10. Answer (04.00) Sol. : T1 – mg = maA
Sol. of Q. 9 and Q. 10 m 2m
T1 = 4 ( g − aA )
1 3m
Hint : dK = (dm ) v y2
2 mg – T2 = maB and T1 = 2T2
Sol. : y = (A sinKx) (sint) 5g 16mg 8mg
aA = , T1 = , T2 =
2 11 11 11
1 dy
dK = (dm)
2 dt 3g
aB =
11
Kv =
K ( x ) dx = m2 A2 15. Answer (A, C)
l 8
Fext
Hint : acm =
N=8 mass
m2 A2 Sol. : Block moves downward and rightward.
Total energy is always constant = . Friction acts rightward on the wedge.
4
16. Answer (C, D)
11. Answer (A, B)
Hint : At pure rolling slipping stops.
Hint : Use principle of homogeneity
2 v 7 v
Sol. : [a] = [PV2] = ML5T–2 Sol. : mv 0R + mR 2 0 = mR 2
5 2R 5 R
[b] = [V] = L3
6v 0 7v 6v
= v= 0
PV = ML2T–2K–1
R = 5 5 7
T 6v 0
v0 − 7
12. Answer (C, D) =t v0
0 t0 =
1 2 g 7g
Hint : v = u + at, s = ut + at
2 17. Answer (50)
8−0 Hint : KE = Wext
Sol. : a1 = = 0.8 m/s2
10
( )
2
Sol. : l1 = 32 + 3 3 =6m
1
s1 = 0.8 (10 ) = 40 m
2
2 l 2 = 32 + 42 = 5 m
82 − 0 KE = F S = 50 × 1 = 50 J
a3 = = 0.5 m/s2
2 64
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All India Aakash Test Series for JEE (Advanced)-2023 Test - 4A (Paper-1) (Code-A)_(Hints & Solutions)
S H
Sol : lnK = −
R RT
H
= 2062.65 H = 17.12 kJ
R
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Test - 4A (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2023
26. Answer (12.00) 31. Answer (A, B, D)
27. Answer (03.00)
Hint : As, ng = 0 KP = KC
Hint and Sol. for Q.Nos. 24 and 25
Hint : Sol. : 2XY(g) X2 (g) + Y2 (g)
t=0 2 – –
x x
t = eq. 2–x
2 2
(x/2)2
= 0.16
Sol. : Energy of stabilisation by (2 − x)2
12
1C–H= = 3 kJ/mole x 2
4 =
2(2 − x) 5
28. Answer (09.00)
29. Answer (02.00) 8
x=
Hint and Sol. for Q.Nos. 28 and 29 9
[H+ ]2 [S2− ]
Hint : K1K 2 = = 10−21
[H2S]
(0.2)2 [S2− ]
Sol. : I. = 10−21
0.1
(0.001)2 [S2− ]
II. = 10−21 [S2− ] = 10 −16
0.1
G = 0
II. Mixing of two ideal gas at constant T
q = 0, U = 0, w = 0
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All India Aakash Test Series for JEE (Advanced)-2023 Test - 4A (Paper-1) (Code-A)_(Hints & Solutions)
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Test - 4A (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2023
18
y=3–x
Here (y – 46)2 – 9 (y2 – 20y + 244) ≥ 0
Let centre of circle be (x1, 3 – x1)
Then equation of circle passing though points M –8y2 + 88y – 80 ≥ 0
and N is y2 – 11y + 10 ≤ 0
minimum value of y = 1
( x − x1)2 + ( y − 3 + x1)2 = 2( x12 + 1) …(1)
And again
If this circle touches the x axis then
y2 + (2x – 20)y + (9x2 – 92x + 244) = 0
| 3 − x1 | =
2
2( x12 + 1)
y = (20 − 2x ) 4( x −10) − 4(9 x − 92 x + 244)
2 2
x12 + 6x1 − 7 = 0 18
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All India Aakash Test Series for JEE (Advanced)-2023 Test - 4A (Paper-1) (Code-A)_(Hints & Solutions)
27 3k 2
162 + 6k + − + 288 = 0
4 4
k2 – 8k – 609 = 0
k = 29 as k N.
45. Answer (33.00)
Hint : Use of Multinomial theorem 49. Answer (A, B)
Sol. : Number of permutations = Coefficients of x2 y2
Hint : Equation of normal to ellipse + =1
x2 a 2
b2
x3 in 3! (1 + x ) 1 + x +
3
2! a2 x b2 y
at point (x1, y1) is − = a2 − b2
x1 y1
= coefficients of x3 in
Sol. : Equation of normal to ellipse at point with
x2 eccentric angle is :
3!(1+ 3 x + 3 x 2 + x 3 ) 1+ x +
2
ax by
− = a2 − b2 …(1)
cos sin
3
= 6 + 3 + 1 = 33 And normal at point (–a sin , b cos ) is
2
ax by
46. Answer (09.00) + = b2 − a2 ...(2)
sin cos
Hint : Use of coefficients in multinomial theorem a
Slope of normal (1) = m1 = tan
Sol. : = Coefficient of x4 in b
a
2 Slope of normal (2) = m2 = − cot
x2 x2 x3 b
4! (1 + x ) 1 + x +
4
1 + x + +
2 2! 3!
m1 − m2
tan =
= Coefficient of x4 in 4! (1 + 4x + 6x2 + 4x3 + x4) 1 + m1m2
x4 x2 x3 a a
1 + 2 x +
2
+ 2x + x 3 1 + x + + tan + cot
4 2 6 b b
tan =
a2
= 1422 1− 2
b
Sum of digits of = 9.
1 1
47. Answer (07.00) 1 − e2 sin cos
tan =
48. Answer (06.00) 1
1−
1 − e2
Hint and Sol. for Q.Nos. 47 and 48
Hint : Sketch graphs and count number of prints 2cot e2 1 − e2 −e2 e2
= = or
of intersection. sin2 e2 − 1 1 − e2 1 − e2
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
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Test - 4A (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2023
50. Answer (A, B, D) 7!
Sol. : The words start with T is .
Hint : Let common ratio of G.P. be r then a = a, b 2!
= ar and c = ar2. Probability that words start with T
Sol. : a + b + c = 70 7!
1
a (1 + r + r2) = 70 …(i) = 2! =
8! 4
and 10b = 4a + 4c 2! 2!
5ar = 2a + 2ar2 Probability that word start with vowel
2r2 – 5r + 2 = 0 7! 7!
+
1 2! 2! 2! 3
r = 2 or …(ii) = =
2 8! 8
2! 2!
1
From (i) and (ii) : r = 2 then a = 10 and r = then
2 Probability that word start and end with I
a = 40. 6!
(a, b, c) = (10, 20, 40) or (40, 20, 10) 1
= 2! =
8! 28
51. Answer (A, C, D)
2! 2!
Hint : L.H.L. = lim f ( x ) = lim f (a − h)
− x→0
x→a Probability that word start with K and end with C
and R.H.L. = lim f ( x ) = lim f (a + h) is
+ h→ 0
x→a
6!
Sol. : lim f ( x ) = lim f (1 − h) = lim (1 − h)2 − 2 = −1
− h→ 0 h→ 0 2! 2! 1 1
x →1
= = =
8! 8 7 56
and lim f ( x ) = lim f (1+ h) = lim (1+ h) +1 = 2
+
x →1 h→ 0 h→ 0 2! 2!
z = x + iy then x [ −2, 2]
(x – 3)2 + y2 + (x – 7)2 + y2 = 40
For range of f(x) we have to get range of
x2 + y2 – 10x + 9 = 0
Centre of circle is (5, 0) and radius 4 units. y = x +2 + 2− x
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All India Aakash Test Series for JEE (Advanced)-2023 Test - 4A (Paper-1) (Code-A)_(Hints & Solutions)
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