Download Solution of Practice Paper NEET 1020

Solution of Practice Paper NEET 2013
𝐖𝐨𝐫𝐤
1. (a) Power =
[T] =
𝐓𝐢𝐦𝐞
,
[P] =
[𝐌𝐋𝐓 −𝟐 ]
[𝐋]
𝐜
𝟏
𝟑 𝐗 𝟏𝟎𝟖
=
[𝐓]
= [ML0T-2]
𝟏.𝟓 𝐗 𝟏𝟎𝟖
= [ML2T-3],
Surface tension =
𝐄𝐧𝐞𝐫𝐠𝐲
Planck’s constant =
2. (c) Given: 1 = 1.5 X 108 m s-1,
1 =
[𝐌𝐋𝟐 𝐓 −𝟐 ]
= 2,
,
[h] =
𝐅𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝐲
𝐅𝐨𝐫𝐜𝐞
,
𝐥𝐞𝐧𝐠𝐭𝐡
[𝐌𝐋𝟐 𝐓 −𝟐 ]
[𝐓 −𝟏 ]
= [ML2T-1]
2 = 2.0 X 108 m s-1,
Refractive index for medium M1 is
Refractive index for medium M2 is
2 =
𝐜
𝟐
𝟑 𝐗 𝟏𝟎𝟖
=
=
𝟐𝟎.𝐗 𝟏𝟎𝟖
𝟑
𝟐
If i is angle of incidence and C is critical angle then for total internal reflection
sin i  sin C but sin C = 2/1,
or
sin i 
3. (a) For first ball, u = 0, a = g = 10 m s-2, t = 18 s,
For second ball, u = , a = g = 10 m s-2,
As
S2 = S1,
𝐕𝐬
4. (a) As
𝐕𝐩
=
𝐍𝐬
𝐍𝐩
,
𝟏
𝐪𝟏 𝐪𝟐
𝟑/𝟐
t = 18 – 6 = 12 s,
Vs =
𝐍𝐬
𝐍𝐩
Vp =
𝟒𝟐𝟎𝟎
𝟐𝟏𝟎𝟎
⃗,
⃗ +𝐁
⃗ +𝐂=𝟎
𝐀
̂)
𝐂 = - (3𝐢̂ + 4𝐤
, =
𝟗 𝐗 𝟏𝟎𝟗 (𝟏 𝐗 𝟏𝟎−𝟔 ) (𝟐 𝐗 𝟏𝟎−𝟔 )
(𝟏𝟎 𝐗 𝟏𝟎−𝟐 )𝟐
𝟏
𝟐
at2,
S1 = 0 +
S2 = ut +
=
X 120 = 240 V,
𝟑
i  sin-1 ( )
or
𝟐
From, S = ut +
As
𝐈𝐬
𝐈𝐩
𝟏
𝟏𝟐
𝐍𝐩
𝐍𝐬
𝟐
X 10 (18)2 = 1620 m
at2 =  X 12 +
𝟐
𝟏𝟔𝟐𝟎−𝟕𝟐𝟎
=
𝟒
𝟏
𝟏
𝟐
X 10 (12)2 = 12 + 720
= 75 m s-1
,
𝐍𝐩
Is =
Ip =
𝐍𝐬
𝟐𝟏𝟎𝟎
𝟒𝟐𝟎𝟎
X 10 = 5 A
̂ + 2𝐢̂ - 𝐣̂ + 3𝐤
̂ +𝐂=𝟎
⃗
𝐢̂ + 𝐣̂ + 𝐤
6. (b) Here, AB = BC = AC = 10 cm = 10 X 10-2 m,
𝟒𝟎 𝐀𝐂 𝟐

𝟏
12 + 720 = 1620,
5. (a) A body is in equilibrium, if,
̂ +𝐂=𝟎
⃗,
3𝐢̂ + 4𝐤
F1 =
𝟐
Force on charge 2 C at C due to charge 1 C at A is
= 1.8 N
Force on charge 2 C at C due to charge 1 C at B is
Similarly,
F2 =
𝟏
𝐪𝟏 𝐪𝟐
𝟒𝟎
𝐁𝐂 𝟐
Resultant of F1 and F2
, =
𝟗 𝐗 𝟏𝟎𝟗 (𝟏 𝐗 𝟏𝟎−𝟔 ) (𝟐 𝐗 𝟏𝟎−𝟔 )
(𝟏𝟎 𝐗 𝟏𝟎−𝟐 )𝟐
= 1.8 N
F = √𝐅𝟏𝟐 + 𝐅𝟐𝟐 + 𝟐𝐅𝟏 𝐅𝟐 𝐜𝐨𝐬 
= √(𝟏. 𝟖)𝟐 + (𝟏. 𝟖)𝟐 + 𝟐 (𝟏. 𝟖)(𝟏. 𝟖) 𝐜𝐨𝐬 𝟔𝟎𝟎 ,
7. (b) In case of projectile motion,
Range, R =
= √(𝟏. 𝟖)𝟐 + (𝟏. 𝟖)𝟐 + 𝟐 (𝟏. 𝟖)𝟐 𝐗
𝐮𝟐 𝐬𝐢𝐧 𝟐 
,
𝐠
Maximum height,
𝟏
𝟐
= 1.8 √𝟑 N
H=
𝐮𝟐 𝐬𝐢𝐧𝟐 
𝟐𝐠
where u is the initial velocity of projectile and  is the angle of projection,
(Range)2 = 48 (maximum height)2,
As per question,
𝐮𝟐 𝐬𝐢𝐧 𝟐 
𝐠
= 4√3 (
𝐮𝟐 𝐬𝐢𝐧𝟐 
𝟐𝐠
𝟐 𝐬𝐢𝐧  𝐜𝐨𝐬 
),
𝟒√𝟑
8. (b) Apparent coefficient of liquid in copper vessel
=
(
𝐮𝟐 𝐬𝐢𝐧 𝟐 
𝐠
𝐬𝐢𝐧𝟐 
𝟐
,
a = C,
)2 = 48 (
tan  =
𝐮𝟐 𝐬𝐢𝐧𝟐 
𝟒
𝟒√𝟑
𝟐𝐠
=
𝟏
)2
,
√𝟑
or
 = 300
Volume expansion coefficient of copper g = 3A
(Volume expansion coefficient is three times that of linear expansion coefficient)
Coefficient of real expansion of liquid
r = a + g,
Silver vessel r = S + 3s
…..(ii),
where s = linear expansion coefficient of silver.
From eqs. (i) and (ii), we get,
C + 3A = S + 3s,

9. (d) x1 = 4 sin (10t + ),
𝟔
A1 = 4 units, 1 = 10 units,
In SHM,
Energy, E =
For 1st particle,
E1 =
𝟏
𝟐
𝟏
𝟐
m2 A2,
m1 12A12,
As per question, m1 = m2, E1 = E2,
s =
r = C + 3A
…(i),
𝐂+𝟑𝐀−𝐒
𝟑
and x2 = 5cos (t),
A1 = 5 units, 2 =  units
where the symbols have their usual meanings
For 2nd particle,
12A12 = 22A22,
22 = 12 (
E2 =
𝐀𝟏
𝐀𝟐
)2,
𝟏
𝟐
m2 22A22
𝟒
2 = (10)2 ( )2 = 64,
𝟓
 = 8 units.
10. (c) Resistance of each piece =
RS = (
RP =
𝐑
𝟐𝟎
𝐑
) X 10 = ,
𝟏𝟎
=
𝐑
𝟐𝟎𝟎
𝟐𝟎
, When 10 such piece are connected in series, its effective resistance is
When remaining 10 piece are connected in parallel, its effective resistance is
𝟐
𝐑
( )
𝟐𝟎
𝐑
,
When these two combinations are connected in series, its effective resistance is
𝐑
Reff = RS + RP =
𝟐
+
𝐑
𝟐𝟎𝟎
=
𝟏𝟎𝟏
𝟐𝟎𝟎
R
11. (a) Gravitational potential due to the shell at any point inside it = -
𝐆𝐌
𝐚
Gravitational potential due to the particle at the centre at a point P distant
=-
𝐆𝐌
𝐚/𝟐
=-
𝟐𝐆𝐌
𝐚
,
Net gravitational potential at P,
=-
𝐆𝐌
𝐚
-
𝟐𝐆𝐌
𝐚
-6
12. (c) m = 1 mg = 1 X 10 kg, From Einstein’s mass energy equivalence relation,
𝑎
2
=-
from the centre
𝟑𝐆𝐌
𝐚
Energy released, E = mc2
= 3 X 108 m s-1,
where, c = speed of light in vacuum,
E = 1 X 10-6 X (3 X 108)2,
= 1 X 10-6 X 9 X 1016 = 9 X 1010 J
⃗ X𝐁
⃗ is perpendicular to the plane of 𝐀
⃗ and 𝐁
⃗ . Also, 𝐀
⃗ -𝐁
⃗ in the plane of 𝐀
⃗ and 𝐁
⃗ . So, 𝐀
⃗ X𝐁
⃗ is
13. (b) 𝐀
⃗ -𝐁
⃗.
perpendicular to 𝐀
Or
=

𝟐
radian
14. (a) Let T be the tension of a rope. When monkey climbs up with acceleration a
or
T – m1g = m1a
When another monkey climbs down with same acceleration a,
or
m2g – T = m2a
Adding (i) and (ii), we get,
or
(1 -
or
𝟐
𝟐
𝟑
𝟑
(m2 – m1)g = (m1 + m2) a,
(1 - )g = [ + 1] a,
15. (b) Distance travelled by the body,
g=
𝟑
𝟓
𝟑
a
or
a=
𝐦𝟐
)g=[
𝐦𝟏
𝐦𝟐
+ 1]a
𝐠
𝟓
= Area under velocity-time graph
= Area of trapezium OEDC,
̂,
16. (a) Here, 𝐫 = 𝐢̂ + 2𝐣̂- 𝐤
𝟏
or
𝐦𝟏
=
𝟏
𝟐
[10 + 40] X 10 = 250 m
̂,
⃗ = 3𝐢̂ + 4𝐣̂ - 2𝐤
𝐩
⃗,
Angular momentum, 𝐋 = 𝐫 X 𝐩
̂ (4 – 6),
𝐋 = 𝐢̂ (- 4 – (- 4) ) + 𝐣̂ (- 3 – (- 2)) + 𝐤
̂ (4 – 6),
= 𝐢̂ (- 4 + 4) + 𝐣̂ (- 3 + 2) + 𝐤
𝑖̂ 𝑗̂ 𝑘̂
𝐋 = |1 2 − 1|,
3 4 −2
̂
̂
= 0𝐢̂ - 𝐣̂ - 2𝐤 = - 𝐣̂ - 2𝐤
Since the angular momentum is in yz plane, i.e.., perpendicular to x-axis.
𝟏
X 10 X (10)2 – 0 = 500 J
17. (b) By work-energy theorem,
W = K =
18. (b) Loss in energy = mg (h – h’),
= 0.1 X 10 X (10 – 5.4) = 4.6 J
4.6 J = msT,
= 0.1 X 460 X T,
19. (c) Capacitance of parallel plate capacitor,
Capacitance of IInd half,
C2 = K
𝟎 𝐀/𝟐
=
20. (c) Current, I =
𝐏
𝐕
=
𝟏𝟐𝟓 𝐕
𝟐
𝟐
= 800 A,
𝐤𝐠
𝐂
𝐂
𝐬
) (800 ) (60 s),
21. (b) Here, 
⃗ = 3𝐢̂ + 2𝐣̂ m s-1,
̂ T,
⃗ = 2𝐣̂ + 3𝐤
𝐁
(
Capacitance of 1st half,
= 4 F,
𝟑
Mass of chlorine liberated, m = zIt,
= (0.367 X 10-6
𝟎 𝐀
𝐝
𝐊 𝟎 𝐀
So, net capacitance C = C1 + C2 = 2 + 6,
𝟏𝟎𝟎 𝐗 𝟏𝟎𝟑 𝐖
T = 0.10C
or
C=
𝐝
𝟐
𝟐
)=
X 4 F = 6 F,
⃗ = 𝐅/m =
𝐚
𝐦
𝟎 𝐀/𝟐
𝐝
=
𝟎 𝐀
𝟐𝐝
=
𝟒𝐅
𝟐
=2F
C1 and C2 are connected in parallel,
=8F
According to the Faradays’ first law of electrolysis
m = (0.367 X 10-6 kg/C) (800 A) (60 s)
= 17.61 X 10-3 kg
Specific charge of proton = e/m = 0.96 X 108 C kg-1
̂ )]
⃗ ),
𝐅 e (
⃗ X𝐁
= e [(3𝐣̂ + 2𝐣̂) X (2𝐣̂ + 3𝐤
Force on a proton in a uniform magnetic field is
̂ - 9𝐣̂ + 6𝐢̂),
̂ ),
̂ ) N,
= e (6𝐤
= e (6𝐢̂ - 9𝐣̂ + 6𝐤
= e (6𝐢̂ - 9𝐣̂ + 6𝐤
̂)
𝐞 (𝟔𝐢̂− 𝟗𝐣̂+ 𝟔𝐤
C1 =
̂ ) m s-2,
, = 0.96 X 108 (6𝐢̂ - 9𝐣̂ + 6𝐤
Acceleration of the proton,
̂ ) m s-2
= 2.88 X 108 (2𝐢̂ - 3𝐣̂ + 2𝐤
22. When the force acting on a body is directed towards a fixed point, then it changes only the direction of
motion of the body without changing its speed. So the body will describe a circular motion.
23. (c) Here, T = 7270C = 1000 K,
According to Stefan’s law,
t = 0.3 min = 0.3 X 60s = 18 s, A = 0.1 m2,
 = 5.67 X 10-8 W m-2 K-4
Heat radiated by a perfectly black body is
H = AT4t
= (5.67 X 10-8 W m-2 K-4) (0.1 m2) (1000 K)4 (18 s),
= 102060 J
24. (c) The equivalent circuit of the given network is as shown in figure.
It is a balanced Wheatstone’s bridge. Hence no current flows through
resistance of arm BD. The equivalent resistance between A and C is
𝟏
𝐑 𝐞𝐪
𝟏
=
(𝟏𝟎+𝟐𝟎)
Current, I =
𝟏
+
=
(𝟓+𝟏𝟎)
𝐕
𝐑 𝐞𝐪
𝟓𝐕
=
𝟏𝟎 
𝟏
𝟑𝟎
+
𝟏
𝟏𝟓
=
𝟑
𝟑𝟎
=
𝟏
𝟏𝟎
,
or
Req = 10 
= 0.5 A
25. (c) A diode is said to be forward biased if p – type semiconductor of p-n junction is at higher potential and ntype semiconductor of p-n junction is at lower potential.
A diode is said to be reverse biased if p-type
semiconductor of p-n junction is at lower potential and n-type semiconductor of p-n junction is at higher
potential.
Hence, in figure (i) the diode is forward biased and in figure (ii) the diode is reverse biased.
u1 = 72 km h-1 = 72 X
26. (b) m1 = 400 kg,
u2 = 9 km h-1 = 9 X
𝟓
𝟏𝟖
𝟓
𝟏𝟖
m s-1 = 2.5 m s-1,
m s-1 = 20 m s-1,
m2 = 4000 kg,
1 = - 18 km h-1 = - 18 X
𝟓
m s-1 = - 5 m s-1,
𝟏𝟖
According to law of conservation of momentum
m1u1 + m2u2 = m11 + m22,
400 X 20 + 4000 X 2.5 = 400 X (- 5) + 40002,
8000 + 10000 = - 2000 + 4000 2
𝟐𝟎𝟎𝟎𝟎
2 =
𝟒𝟎𝟎𝟎
= 5 m s-1 = 5 X
,
𝟏𝟖
𝟓
2 = ?
km h-1 = 18 km h-1.
27. (b) When the key between terminals 1 and 2 is plugged in, R is in the circuit.
V1 = IR = 1R = Kl1
(I = 1 A),
are in the circuit.
V2 = I (R + X) = 1 (R + X) = Kl2,
𝐀+ 𝐦
𝐬𝐢𝐧(
)
𝟐
,
𝐀
𝐬𝐢𝐧
𝟐
28. (c) As  =
cos
𝐀
𝟐

=
or
𝟐
𝐀
𝟐
When the key between terminals 1 and 3 is plugged in, R and X, both
V2 - V1 = R + X – R = K(l2 – l1)
As per question, m = A,

= cos-1 ( ),

=
𝐬𝐢𝐧 𝐀
=
𝐀
𝐬𝐢𝐧
𝟐
𝐀
𝟐
𝟐 𝐬𝐢𝐧 𝐜𝐨𝐬
𝐀
𝟐
𝐀
𝐬𝐢𝐧
𝟐
= 2 cos
A = 2 cos-1 ( )
or
𝟐
=
𝐀+𝐀
)
𝟐
𝐀 ,
𝐬𝐢𝐧
𝟐
𝐬𝐢𝐧(
𝟐
29. (d) The minimum force required to start pushing a body up a rough inclined
plane is
F1 = mg sin  +  mg cos ,
Minimum force needed to prevent the body from sliding down the inclined
plane is
𝐅𝟏
𝐅𝟐
=
F2 = mg sin  -  mg cos 
𝐬𝐢𝐧  +  𝐜𝐨𝐬 
𝐬𝐢𝐧  −  𝐜𝐨𝐬 
𝟏
30. For Lyman series,
𝟏
𝐋
=R[
𝟏
𝟏𝟐
-
𝟏
]
𝟐𝟐
=
𝐋
𝐭𝐚𝐧  + 
=
𝟐 + 
𝐭𝐚𝐧  −  𝟐 − 
𝟏
𝟏
=R[
𝐧𝟐𝟏
𝟏
-
𝐧𝟐𝟐
],
𝟏
= R [1 - ],
𝐋
𝟒
For first line, n1 = 2, n2 = 3,
Dividing (i) by (ii), we get,
31. (d) The time period of satellite
As g =
𝐆𝐌𝐞
𝐑𝟐𝐞
or T2 =
(24 X 60 X 60)2 =
𝟒𝐫 𝟑
𝐠 𝐑𝟐𝐞
,
𝟒 𝐗 (𝟑.𝟏𝟒)𝟐 𝐫 𝟑
𝟗.𝟖 𝐑𝟐𝐞
=3
For first line n1 = 1, n2 = 2
=
𝟑
…(i)
𝟒
𝟏
𝐁
𝐁
𝐋
=R[
=
𝟑
𝟒
𝟏
𝟐𝟐
RX
-
𝟏
For Balmer series,
𝟏
𝟓𝐑
𝟏
𝟏
𝟏
𝟒
𝟗
𝐁
] = R [ - ],
𝟑𝟐
𝟑𝟔 𝟐𝟕
=
𝟓
T = 2 √𝐫 𝟑 /𝐆𝐌𝐞 ,
,
=
B =
𝐁
𝟓
𝟑𝟔
𝟐𝟕
𝟓
=R[
R
L =
𝟏
𝐧𝟐𝟏
𝟏
𝐧𝟐𝟐
],
….(ii)
𝟐𝟕
𝟓

Squaring both sides, we get
Substituting the given values in above equation, we get
or r = 6.6 Re
-
T2 =
𝟒𝟐 𝐫 𝟑
𝐆𝐌𝐞
𝐀
𝟐
32. (c)
33. (b) Here, I1 = I2 = 5 A, r = 0.5 m, Force per unit length between the two wires is
f=
𝟎 𝐈𝟏 𝐈 𝟐
𝟐
𝐫
=
𝟐 𝐗 𝟏𝟎−𝟕 𝐗 𝟓 𝐗 𝟓
𝟎.𝟓
= 10-5 N m-1, The currents are in the opposite direction, so force will be
repulsive.
34. (b) Time period of simple pendulum (in stationary lift)
𝐓′
T’ = 2 √𝐥/𝐠′ or
g’ = g +
35. (a)
𝐠
𝟑
=
𝟒
𝟑
𝐓
= √𝐠/𝐠′,
𝐓′
or
36. (d) I =
𝐓
=
𝐪
𝟐/
=
𝐪
𝟐
,
𝐓
37. (a) Here, Phase difference,  = 1320 ,
Frequency,  = 105 Hz,
𝟐

𝟐
X x =
38. (a)
𝟑
𝐠
or
𝐓′
𝐓
= √𝟑/𝟒
=

or
𝟐𝟐
0
𝟏𝟖𝟎𝟎
X 1320 =
Phase difference =
T’ =
𝟑
𝟐

 rad,
√𝟑
T
𝟐
B=
𝟎 𝐈
𝟐𝐑
=
𝟎
𝟐𝐑
X
𝐪 (𝟐)
𝟐
=
𝟎 𝐪
𝟐𝐑
Path difference, x = 11 m
X Path difference,
 =
𝟐

X x
Phase velocity,  =  = (105 Hz) (3m) = 315 m s-1
X 11 m = 3 m,
𝟐𝟐
𝟑
𝐠
= √𝟒
Magnetic induction at the centre of ring
0
=
In moving lift
Lift is accelerating upwards with an acceleration g/3.
g,
𝐪
T = 2√𝐥/𝐠,
39. (a)
40. (a) Age of pottery is determined by the ratio isotope of carbon.
41. (a)
42. (c) Here, Work function, 0 = 5 eV,
𝟏𝟐𝟒𝟎𝟎 𝐞𝐕 Å
Take hc = 12400 eV Å,
E=
Kmax = h - 0,
= 6.2 eV – 5 eV = 1.2 eV,
𝟐𝟎𝟎𝟎 Å
Energy of incident photon, E = h = hc/
= 6.2 eV,
According to Einstein’s photoelectric equation
As Kmax = eVs = 1.2 eV, Vs = 1.2 V
43. (d) Electric field of an electromagnetic wave in free space is given by
E = 10 cos (107t + kx) 𝐣̂ V m-1,
which is acting along y direction. As E is varying with x and t, hence propagation of electromagnetic wave
takes place along-x-axis. Thus statement (4) is wrong. Comparing the relation, E = 10 cos (107t + kx) with
standard equation of electromagnetic wave
we get, E0 = 10 V m-1.
 = 60 = 60 X
K=
𝟐

=
𝟐
𝟔𝟎
=
𝟐𝟐
𝟕
𝟏
𝟑𝟎
𝟐
X 19 X 1010 (

(t + x) = E0 cos (
𝟐

= 107 or
 188.4 m,
Thus statement (1) is correct.
= 0.033 rad m-1
Thus statement (2) is wrong.
[Young’s modulus Y =
𝟏
𝟐
Thus statement (3) is correct.
44. (d) Potential energy stored per unit volume
=
E = E0 cos
𝐒𝐭𝐫𝐞𝐬𝐬
𝐒𝐭𝐫𝐚𝐢𝐧
𝟏 𝐗 𝟏𝟎−𝟑 2
𝟓
]
),
u=
u=
𝟏
𝟐
𝟏
𝟐
X Stress X Strain =
𝟏
𝟐
𝟐
t+
47. (b)
48. (c)
49. (b)
50. (c)

= 107
X Y X (Strain)2
= 3.8 X 109 X 10-6 = 3.8 X 103 J m-3
46. (c)
x)


𝟐 𝐗 (𝟑 𝐗 𝟏𝟎𝟖 )
X 19 X 1010 (l/l)2,
45. (a)
𝟐