Solution
NEET MOCK TEST-4
Entrance Exam - NEET-UG
PHYSICS
1. (a) 7 × 10-2 N/m
Explanation:
In MKS system its value is 7 × 10-2 N/m
2.
(b) Bernoulli's principle
Explanation:
According to the equation of continuity, av = a constant. It means, as area increases,
velocity decreases. Thus, as blood flows from narrow arteries to wider ones, velocity
decreases. According to Bernoulli’s theorem, P +
1
2
2
ρv = a constant. It means, as velocity
decreases, pressure increases. Thus, when arteries become narrow, blood pressure
increases.
3.
(b) 2 × 10 N/m 11 2
Explanation:
11 2
2 × 10 N/m
4. (a) F
Explanation:
The centripetal force acting on satellite is provided by the earth's gravitational force,
therefore the net force acting on satellite is the centripetal force = gravitational force =
F
2
5. (a) V
T
2
t
2
Explanation:
Work done on the body is equal to gain in the kinetic energy.
Acceleration of the body, a = V
T
Velocity acquired in timer, v = at = V
T
t
Now, KE acquired ∝ v 2
Work done ∝
2 2
V t
∴
2
T
6.
–
(d) 1 : √2
Explanation:
According to the conditions,
For A to B
S= 1
2
gt
2
For A to C
2S = gt2 1
2
Dividing (i) by (ii) we get
t 1
′
=
t √2
, 7.
(b) (0, 0)
Explanation:
∑ mi xi
xcom =
∑ mi
1 1 1 1
1× +1×(− )+1×(− )+1×( )
2 2 2 2
=
4
xcom = 0
Similarly y
∑ mi y
i
cm
=
∑ yi
1 1 1 1
1× +1× +1×(− )+1×(− )
2 2 2 2
ycm =
4
ycm = 0
Hence coordinate of com
(xcm, ycm) = (0, 0)
8.
(d) 6 m s-1
Explanation:
Radius of circle r =4 m
acceleration g=9 m/sec2
for minimum velocity at the highest point
2
v
m = mg
r
2
v = rg
−
−− −−−−
v = √rg = √4 × 9 = 6 m/sec
9.
(d) Angle of twist
Explanation:
Angle of twist
10.
(c) 1.6 × 10-4 J
Explanation:
3 3
∵ R = nr
3
3 −2
3
∴ n= R
r
3
= (6) = 216
= (
0.6×10
−3
)
1×10
2 2/3 1/3
Now, W = 4πr T n (n - 1)
= 4× 22
7
× 10
−6
× 7 × 10
−2
× 36 × (6 - 1)
≈ 1.6 × 10-4 J
11.
(c) A
, Explanation:
Note that A is stretched the least and recovers the original length after the removal of
the deforming force.
12.
(b) i, ii, iii
Explanation:
All given heavenly bodies move around sun in elliptical orbits and follow Kepler’s law.
13.
(c) 7.5 R
Explanation:
Let the body of mass M covers a distance of x before the collision, then
Mx = 5M(9R - x)
Mx = 45MR - 5Mx
6Mx = 45MR
x= 45M R
6M
= 7.5R
14.
(c) C
Explanation:
Because the slope is the highest at C, v = ds
dt
is maximum.
15.
(b) L
6
Explanation:
L
6
16.
(b) independent of r
Explanation:
Centripetal force,
2
mv 1
∝
r r
or
2
mv k
=
r r
−−
or v = √ k
m
= constant
Clearly, speed v is independent of r.
17.
(b) 21.6
Explanation:
1 2 -3 -1
[V] = [M L T A ]
2 −3 −1
1
n2 = n 1(
M1
M2
) (
L1
) (
T1
) (
A1
)
L T A
2 2 2
NEET MOCK TEST-4
Entrance Exam - NEET-UG
PHYSICS
1. (a) 7 × 10-2 N/m
Explanation:
In MKS system its value is 7 × 10-2 N/m
2.
(b) Bernoulli's principle
Explanation:
According to the equation of continuity, av = a constant. It means, as area increases,
velocity decreases. Thus, as blood flows from narrow arteries to wider ones, velocity
decreases. According to Bernoulli’s theorem, P +
1
2
2
ρv = a constant. It means, as velocity
decreases, pressure increases. Thus, when arteries become narrow, blood pressure
increases.
3.
(b) 2 × 10 N/m 11 2
Explanation:
11 2
2 × 10 N/m
4. (a) F
Explanation:
The centripetal force acting on satellite is provided by the earth's gravitational force,
therefore the net force acting on satellite is the centripetal force = gravitational force =
F
2
5. (a) V
T
2
t
2
Explanation:
Work done on the body is equal to gain in the kinetic energy.
Acceleration of the body, a = V
T
Velocity acquired in timer, v = at = V
T
t
Now, KE acquired ∝ v 2
Work done ∝
2 2
V t
∴
2
T
6.
–
(d) 1 : √2
Explanation:
According to the conditions,
For A to B
S= 1
2
gt
2
For A to C
2S = gt2 1
2
Dividing (i) by (ii) we get
t 1
′
=
t √2
, 7.
(b) (0, 0)
Explanation:
∑ mi xi
xcom =
∑ mi
1 1 1 1
1× +1×(− )+1×(− )+1×( )
2 2 2 2
=
4
xcom = 0
Similarly y
∑ mi y
i
cm
=
∑ yi
1 1 1 1
1× +1× +1×(− )+1×(− )
2 2 2 2
ycm =
4
ycm = 0
Hence coordinate of com
(xcm, ycm) = (0, 0)
8.
(d) 6 m s-1
Explanation:
Radius of circle r =4 m
acceleration g=9 m/sec2
for minimum velocity at the highest point
2
v
m = mg
r
2
v = rg
−
−− −−−−
v = √rg = √4 × 9 = 6 m/sec
9.
(d) Angle of twist
Explanation:
Angle of twist
10.
(c) 1.6 × 10-4 J
Explanation:
3 3
∵ R = nr
3
3 −2
3
∴ n= R
r
3
= (6) = 216
= (
0.6×10
−3
)
1×10
2 2/3 1/3
Now, W = 4πr T n (n - 1)
= 4× 22
7
× 10
−6
× 7 × 10
−2
× 36 × (6 - 1)
≈ 1.6 × 10-4 J
11.
(c) A
, Explanation:
Note that A is stretched the least and recovers the original length after the removal of
the deforming force.
12.
(b) i, ii, iii
Explanation:
All given heavenly bodies move around sun in elliptical orbits and follow Kepler’s law.
13.
(c) 7.5 R
Explanation:
Let the body of mass M covers a distance of x before the collision, then
Mx = 5M(9R - x)
Mx = 45MR - 5Mx
6Mx = 45MR
x= 45M R
6M
= 7.5R
14.
(c) C
Explanation:
Because the slope is the highest at C, v = ds
dt
is maximum.
15.
(b) L
6
Explanation:
L
6
16.
(b) independent of r
Explanation:
Centripetal force,
2
mv 1
∝
r r
or
2
mv k
=
r r
−−
or v = √ k
m
= constant
Clearly, speed v is independent of r.
17.
(b) 21.6
Explanation:
1 2 -3 -1
[V] = [M L T A ]
2 −3 −1
1
n2 = n 1(
M1
M2
) (
L1
) (
T1
) (
A1
)
L T A
2 2 2
