Solution
NEET MOCK TEST-3
Entrance Exam - NEET-UG
PHYSICS
1.
(c) 6 m
Explanation:
Given, t1 = 2 s and t2 = 1 s
Now, s = ut + (1/2)at2
∴ s1 = 0 + (1/2) × 2 × 4 = 4 m
and s2 = (1/2) × 4 × 1 = 2 m
∴ s1 + s 2 = 6 m
2.
(b) 4v
Explanation:
We know, for the fluid flowing through the nonuniform pipe the velocity of a fluid is
inversely proportional to the area of cross-section.
Hence, according to the problem given, if v1, v2 are the velocities at A and B and a1, a2
are the area of cross-sections at A and B, then
v2 a1
=
v1 a2
Here, incompressible fluid flows steadily through a cylindrical pipe which has radius 2R
at point A and radius R at point B farther along the flow direction, hence
2
π(2R) (v)
v2 =
a1 v1
=
a2 2
πR
∴ v2= 4v
3. (a) V is always negative.
Explanation:
In elliptical orbits, linear speed changes from position to position. Hence T is not
conserved.
V is always negative (attraction).
Angular momentum is conserved, both in magnitude and direction.
The total energy is always negative.
4.
(d) none of these
Explanation:
The magnitude of components of a vector depends on orientation of vector with the
components and it may be equal to or less than or greater than the magnitude of the
vector itself.
5.
(c)
3
m
As
Explanation:
The first part of equation is useless and E y ∝ Jx BZ (Given)
∴ Constant of proportionality
, Ey 3
c m
K = = =
BZ Jx Jx As
[As E
= c (speed of light) and J = I
]
B
Area
6.
(b) X experiences extension and Y compression
Explanation:
X experiences extension and Y compression
7.
(d) 3v
4a
Explanation:
Angular momentum about A will be conserved, i.e.,
Li = Lf
or mva = Iω
2
m(2a)
or mva = 3
⋅ ω
3v
ω =
4a
8.
(c) mgy 0
Explanation:
By law of conservation of mechanical energy
Δk = −ΔU
⇒ kf − ki = Ui − Uf ⇒ kf = mgy − mg [y − y0 ] [∵ k
i = 0, Ui = mgy and U
f
= mg (y − y0 ) ]
⇒ kf = mgy0
9.
–
(c) √3v 0
Explanation:
2 2
As v = u + 2as
∴ u ...(i)
2
∝ s
For given condition:
...(ii)
′
2
u ∝ 3s
From equations (i) and (ii),
′
–
or u ( ∵ u = v0 )
2
u ′
2
= 3 = √3v0
u
10. (a) 1 : 9
Explanation:
As we know that,
P= 4σ
r
or P ∝ 1
r
...(i)
P1 r2 3
∴ = =
P2 r1 1
2 2
A1 4πr r1
1
∵ = 2
= ( )
A2 4πr r2
2
=( ...(By using eq. (i))
2
1 1
) =
3 9
, 11.
(d) J G
6
Explanation:
Work = (Uf - Ui)
=− 5G
6
− (−G) =
G
6
J
12. (a) R
r
Explanation:
=ω
aR 2 R
R × × r
ar 2
ω r
2
=
Tr R
×
2 r
TR
= R
r
Since, Tr = TR
13.
(c) [M0L3T-1]
Explanation:
[M0L3T-1]
14.
(b) 2.3
Explanation:
2.3
15. (a) 5mr2
Explanation:
Since, the disc is massless, the total moment of inertia of the system is due to the 5
masses placed on the ring.
2 2
Hence, I = mr × 5 = 5mr
16. (a) MgL/18
Explanation:
Mass of hanging portion is M / 3 (one-third) and centre of mass c, is at a distance h =
L
6
below the table top. Therefore, the required work done is,
W = mgh = ( M L M gL
) (g) ( ) =
3 6 18
17. (a) 18 km/h
Explanation:
Total distance travelled = s
Total time taken,
s s s
t= 10
3
+
3
20
+
3
60
= 10s
180
=
18
s
vav = s
t
=
s
s
= 18 km/h
18
NEET MOCK TEST-3
Entrance Exam - NEET-UG
PHYSICS
1.
(c) 6 m
Explanation:
Given, t1 = 2 s and t2 = 1 s
Now, s = ut + (1/2)at2
∴ s1 = 0 + (1/2) × 2 × 4 = 4 m
and s2 = (1/2) × 4 × 1 = 2 m
∴ s1 + s 2 = 6 m
2.
(b) 4v
Explanation:
We know, for the fluid flowing through the nonuniform pipe the velocity of a fluid is
inversely proportional to the area of cross-section.
Hence, according to the problem given, if v1, v2 are the velocities at A and B and a1, a2
are the area of cross-sections at A and B, then
v2 a1
=
v1 a2
Here, incompressible fluid flows steadily through a cylindrical pipe which has radius 2R
at point A and radius R at point B farther along the flow direction, hence
2
π(2R) (v)
v2 =
a1 v1
=
a2 2
πR
∴ v2= 4v
3. (a) V is always negative.
Explanation:
In elliptical orbits, linear speed changes from position to position. Hence T is not
conserved.
V is always negative (attraction).
Angular momentum is conserved, both in magnitude and direction.
The total energy is always negative.
4.
(d) none of these
Explanation:
The magnitude of components of a vector depends on orientation of vector with the
components and it may be equal to or less than or greater than the magnitude of the
vector itself.
5.
(c)
3
m
As
Explanation:
The first part of equation is useless and E y ∝ Jx BZ (Given)
∴ Constant of proportionality
, Ey 3
c m
K = = =
BZ Jx Jx As
[As E
= c (speed of light) and J = I
]
B
Area
6.
(b) X experiences extension and Y compression
Explanation:
X experiences extension and Y compression
7.
(d) 3v
4a
Explanation:
Angular momentum about A will be conserved, i.e.,
Li = Lf
or mva = Iω
2
m(2a)
or mva = 3
⋅ ω
3v
ω =
4a
8.
(c) mgy 0
Explanation:
By law of conservation of mechanical energy
Δk = −ΔU
⇒ kf − ki = Ui − Uf ⇒ kf = mgy − mg [y − y0 ] [∵ k
i = 0, Ui = mgy and U
f
= mg (y − y0 ) ]
⇒ kf = mgy0
9.
–
(c) √3v 0
Explanation:
2 2
As v = u + 2as
∴ u ...(i)
2
∝ s
For given condition:
...(ii)
′
2
u ∝ 3s
From equations (i) and (ii),
′
–
or u ( ∵ u = v0 )
2
u ′
2
= 3 = √3v0
u
10. (a) 1 : 9
Explanation:
As we know that,
P= 4σ
r
or P ∝ 1
r
...(i)
P1 r2 3
∴ = =
P2 r1 1
2 2
A1 4πr r1
1
∵ = 2
= ( )
A2 4πr r2
2
=( ...(By using eq. (i))
2
1 1
) =
3 9
, 11.
(d) J G
6
Explanation:
Work = (Uf - Ui)
=− 5G
6
− (−G) =
G
6
J
12. (a) R
r
Explanation:
=ω
aR 2 R
R × × r
ar 2
ω r
2
=
Tr R
×
2 r
TR
= R
r
Since, Tr = TR
13.
(c) [M0L3T-1]
Explanation:
[M0L3T-1]
14.
(b) 2.3
Explanation:
2.3
15. (a) 5mr2
Explanation:
Since, the disc is massless, the total moment of inertia of the system is due to the 5
masses placed on the ring.
2 2
Hence, I = mr × 5 = 5mr
16. (a) MgL/18
Explanation:
Mass of hanging portion is M / 3 (one-third) and centre of mass c, is at a distance h =
L
6
below the table top. Therefore, the required work done is,
W = mgh = ( M L M gL
) (g) ( ) =
3 6 18
17. (a) 18 km/h
Explanation:
Total distance travelled = s
Total time taken,
s s s
t= 10
3
+
3
20
+
3
60
= 10s
180
=
18
s
vav = s
t
=
s
s
= 18 km/h
18
