PEA Association Pvt. Ltd.

Thapathali, Kathmandu, Tel: 5345730, 5357187

2080-6-27 Hints & Solution
Section – I ρl ρl
36.(d) R= =
1.(d) 2.(a) 3.(b) 4.(c) 5.(d) 6.(a) A πr2
7.(b) 8.(c) 9.(b) 10.(d) 11.(a) 12.(c) ρ × 2l
13.(a) 14.(c) 15.(c) 16.(c) 17.(d) 18.(a) R' = = 8R
π(r/2)2
19.(d) 20.(a) 21.(b) 22.(b) 23.(b) 24.(d) 37.(d) M=m×L
25.(c) 26.(c) if cut in 2 parts then
27.(b) P .B
A P = 0 then A
P ⊥BP, A
P ×C P =0 L M
M' = m × =
P ||C
P so B P ⊥C P 2 2
then A
1 2 1 A + δmin
= tan2θ1 = tan230° =   =
H1 sin
28.(b) 2
H2  3 3 38.(c) µ=
60°
1 sin
29.(c) E = mv2 2
2
1 1 3 60° + δmin
E' = m (2v)2 – mv2 = 3E or, = sin
2 2 2 2
h 60 + δmin
30.(c) sin45° = or, 60 =
l 2
h or, δmin = 60°
or, l = = 2h h h
sin45° 39.(a) λ= =
2
1 MC rms 2 KE p 2mE
31.(d) P= =
3 V 3 V 6.62 × 10–34
=
2 2 × 9.1 × 10–31 × 100 × 1.6 × 10–19
= E
3 = 1.22 × 10–10 m = 1.22 Å
2πx 40.(b)
32.(d) φ=
λ 41.(c) –3 < x < –1
2π × 0.4 1 –3 + 2 < x + 2 < –1 + 2
or, λ = = m –1 < x + 2 < 1
1.6π 2
|x + 2| < 1
v
f = = 330 × 2 = 660 Hz x
λ 42.(d) y = ⇒ Domain = ℜ – {0}
|x|
33.(c) 3f0c = 2f00 43.(a) A = {a, b, c} ⇒ n(A) = 3
3v 2v B = {d, e} ⇒ n(B) = 2
or, =
4lc 2l0 No. of functions = n(B)n(A) = 23 = 8
lc 3 a 3
or,
l0 4
= 44.(a) Here, 15 = ⇒ 15 =
1–r 1–r
34.(a) First case 15 – 15r = 3
Q1Q2 15r = 12
F=K 2 4
r
r=
2×6 5
12 = K 2 .... (i) 45.(b) Total number (n) = 7 + 2 = 9
r
nd
2 case Number of circular arrangements
–2 × 2 = (n – 1)! = 8!
F' = K ..... (ii) 46.(d) Free term,
r2
2n 2n
F' –4 Calculate r = = =n
= 1+1 2
F 12 2n
Ans: C(2n, n) = Cn
or, F' = –4N ∞
n 2 3 4 5
σ 47.(c) Σ n! = 2! + 3! + 4! + 5! + ....
35.(b) F = .Q n=2
ε0 1 1 1 1
If one of plate is removed then = + + + + .....
1! 2! 3! 4!
σ F 1 1 1
F' =
2ε0
Q=
2 ( )
= 1 + + + + ..... – 1 = e – 1
1! 2! 3!

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 PEA Association Pvt. Ltd. Thapathali, Kathmandu, Tel: 5345730, 5357187
2080-6-27 Hints & Solution
b c 60.(a) 2sin2θ + 3cos2θ
48.(b) If both roots are equal, then =
d e = 2 – 2cos2θ + 3cos2θ
⇒ be = dc = 2 + cos2θ
1 – 2i 1 – 2i Since, least value of cos2θ = 0
49.(a) Absolute value of
2+i
= | |
2+i ‡ least value of 2 + cos2θ
12 + (–2)2 =2+0=2
=
22 + 12
Section – II
5
= =1 P | |vP2 – vP1| |vP2 + (v
|∆v P1)|
5 61.(b) = =
t t t
50.(d) The determinant of odd ordered skew-
symmetric matrix is zero. 82 + 62
= = 1 m/s2
10
lim a + xn – a 1
51.(c)
x→0 xn
= u2
2 a 62.(d) Rmax = 2 + 4 = 6 =
g
Put a = 1, n = 1
or, u = 60
lim 1+x–1 1
= 2
x→0 x 2 t=
ucos45°
52.(a)
1 2
Z ∴ y = usin45t – gt
2
Q(0, 2, 3) P(1, 2, 3) 2
– g 
2 1 2
= usin45 ×
O Y ucos45° 2  60 cos45°
4 × 10 4
X =2– = m
1 3
2 × 60 ×
Using distance formula = 1 2
d f(x) 63.(a) t∝( h– 0)
53.(c) (e ) = ef(x).f '(x)
dx
h
d ex2 x2 d x2 –0
(e ) = ee . (e ) t' 2
dx dx So, =
t h
x2 ex
2
= 2xe .e t
54.(b) f(x) = 3x2 – 6x + 1 on [–1, 2] or, t' =
2
f '(x) = 6x – 6
3
For stationary point 6x – 6 = 0 64.(b) PV = mrT1 = mrT2
4
x=1
4T1 4 × 333 M1/M2=T2/T1 where
(3sinθ – sin3θ) or, T2 = = = 444 K
55.(d) Isin θ dθ = I
3
4
dθ 3 3
M1=m then m2=M-1/4M
= 171°C or M2=3/4M
= –3 cosθ +
1 cos3θ
K1Adθ K2Adθ
4 3 
+c and T is absolute
65.(b) Q= × t1 = × t2
l l
1 temperature
= [cosθ – 9cosθ] + c K1 t2 35 7
12 or, = = = Pressure and volume of
K2 t1 20 4
56.(b) log2y = 3 ⇒ y = 23 the gas remain constant.
f
57.(d) For the vertical line, equating coeff. of y with 66.(b) f, f – 3, f – 6 .....
2
[because vessel is open
zero. to atmosphere, pressure
f
–3 + k = 0
2
= f + (26 – 1) (–3) will be the same as
⇒ k=3 atmospheric pressure]
f
58.(b) Area of circle = πr2 or,
2
= f – 75
= π.52 = 25π sq. unit or, f = 150 Hz
59.(d) P.Pi )Pi + (p
(p P.Pj )Pj + (p
P.k
P)kP Here f18 = f + (18 – 1) (–3)
P P
p1 i + p2 j + p3k P = 150 – 51 = 99 Hz
= Pp

2
 PEA Association Pvt. Ltd. Thapathali, Kathmandu, Tel: 5345730, 5357187
2080-6-27 Hints & Solution
67.(a) u+v=x V0 200 2
or, v = x – u 72.(b) Vrms = = = 200 V
2 2
v Vrms
and m = Irms = = 200 × ωc
u Xc
or, x – u = mu = 200 × 100 × 10–6
x = 2 × 10–2 A = 20 × 10–3 A
or, u = &
m+1 = 20 mA
x x(m + 4 – 1) mx 73.(a) For Balmer series
v=x– = =
m+1 m+1 m+1
=R 2– 
1 1 1
Now = +
1 1 1 λB 2 ∞
f u v 4 4
R= = .... (i)
1 1 1 λB 3646
or, = +
f u v nd
2 case
1 m+1 m+1 For 1st member,
or, = +
f x mx 1 1 1
or,
1 m(m + 1) + (m + 1) (m + 1)2
f
=
mx
=
mx
λB' [
=R 2– 2
2 3 ]
36 36
mx or, λB' = = × 3646
∴ f= 5R 5 × 4
(m + 1)2
= 6563 Å
2Dλ
68.(b) 10β = 74.(b) 75.(a) 76.(b) 77.(b) 78.(c) 79.(b)
d
80.(a) 81.(a)
Dλ 2Dλ
or, 10 = 82.(b) Let φ = 6x2 + 5xy – 21y2 + 13x + 38y – 5
d d
∂φ
10 2 = 12x + 5y + 13
or, = ∂x
10–3 d
∂φ
2 × 10–3 = 5x – 42y + 38
or, d = = 2 × 10–4 m = 0.2 mm ∂y
10
For point of intersection
C1 C 2
69.(c) ∆E = (V – V2)2 12x + 5y + 13 = 0
2(C1 + C2) 1 5x – 42y + 38 = 0
3 × 10–6 × 5 × 10–6 –32 17
= (300 – 50)2 Solving x = ,y=
2 × 8 × 10–6 23 23
= 0.0586 J 83.(c) Given, y2 + 8x – 2y + 17 = 0
20 This equation can be reduced in (y – k)2 = 4a(x – h)
70.(a) l= = 20m
1 So, length of latus rectum |coeff. of x| = 8
R1 = x, R2 = (20 – x) 84.(d) ∆ = s(s – a) (s – b) (s – c),
R1 R2 a+b+c
Req = Where s =
R 1 + R2 2
x(20 – x)
or, 1.8 = = 14.1.6.7 = 14 3 sq. unit
20
1 1
or, 36 = 20x – x2 85.(a) tan–11 + tan–1 + tan–1
2 3
or, x2 – 20x + 36 = 0
x = 2m, 18m  2+3 
1 1
= + tan–1 
71.(c) B = BQ – BP π
µ0IQ µ0IP 4 1 – 1.1
=
a
–
a  2 3
2π 2π π π π π
2 2
= + tan–1 (1) = + =
µ0 4 4 4 2
= (5 – 2.5)
πa 86.(b) For orthogonals,
µ0 Pa .b
P=0
= 2.5
π×5 (Pi – 2Pj + 4kP).(2Pi + 7Pj + pk P) = 0
µ0 2 – 14 + 4p = 0
= 4p = 12 ⇒ p = 3
2π

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 PEA Association Pvt. Ltd. Thapathali, Kathmandu, Tel: 5345730, 5357187
2080-6-27 Hints & Solution
87.(b) The dc'sof the given lines 1 1 1 1
cos45°, cos60°, cos60°
1 1 1
[
= 2 1 – 1 + – + – + ....
2 3 4 5 ]
= 21 – 1 – + – + .... 
, , 1 1 1
2 2 2
The projection of the line joining points (–1, 2,
 ( 2 3 4 ) 
= 2[lne – ln(1 + 1)]
3) and (–1, 4, 0) is
e
=
1
2
1 1
(–1 + 1) + (4 – 2) + (0 – 3)
2 2
= 2ln() 2
3 1 92.(b) Given, cos105° + sin105°
=0+1– =– = cos(90° + 15°) + sin105°
2 2
1 = –sin15° + sin105°
∴ Projection is +ve = 105° + 15° 105° – 15°
2
88.(d) Since, this curve is parametric for of circle x2 +
= 2cos( 2 ) ( sin
2 )
y2 = 16 = 2cos60° sin45°
So, its area = π.42 1 1
= 16π sq. unit = 2. .
2 2
3 2b2
89.(b) Given, e = and =8 1
5 a =
2
b2
⇒ =4 1 π
a ∴ cosθ = ⇒ cosθ = cos
2 4
⇒ b = 4a
2

⇒ a2(e2 – 1) = 4a π
⇒ θ = 2nπ ±
4
9
(
⇒ a –1 + )
5
=4
93.(d) (
tan–1
6 + 5x
)
6 – 5x
= tan–16 + tan–15x
⇒ a=5 ∴ b2 = 20
d  6 + 5x  5
tan (
6 – 5x)
x2 y2 ∴ –1
=0+
Required hyperbola: 2 – 2 = 1
a b dx  1 + 25x 2

x2 y2 5
⇒ =
–
25 20
=1 1 + 25x2
lim yn ∞
⇒ 4x2 – 5y2 = 100 94.(b) form
(1 + i)2 1 + 2i + i2 2i y → ∞ ey ∞ 
90.(a) = = Using L. Hospital rule
i(2i – 1) i(2i – 1) i(2i – 1)
2 2i + 1 lim nyn–1
=
2i – 1
×( )2i + 1 y → ∞ ey
lim n(n – 1) yn–2
4i + 1
=
–4 – 1 y→∞ ey
4 1 Using successively n times,
=– i– lim n! n!
5 5 = =0
4 y → ∞ ey ∞
∴ Imaginary part = – 95.(b) (n + 1)! = 6(n – 1)!
5
1 1 1 (n + 1)n (n – 1)! = 6(n – 1)!
91.(a) + + + ..... n(n + 1) = 6
1.3 2.5 3.7
2 2 2 Only n = 2 satisfies
= + + + ...... 96.(b) Let α and β are two roots of the equation x2 +
2.3 4.5 6.7
1 1 1 px + q = 0, then, α + β = –p, αβ = q
=2 [ + +]
2.3 4.5 6.7
+ .... ∴ (α + β)2 = (α – β)2 + 4αβ
p2 = 1 + 4q
1 1 1 1 1 1
= 2[ – + – + – + ....] 97.(b) 98.(b) 99.(b) 100.(a)
2 3 4 5 6 7

…The End…