33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Motion in a Plane
4. (d) : Let the two vectors be Aand B.
Then, magnitude of sum of Aand B,
2 2
| A+ B| = A + B + 2AB
cosθ
and magnitude of difference of Aand
B
2 2
| A− B|= A + B − 2AB
cos θ,
| A+ B| = | A− B|( given)
or A 2 + B 2 + 2AB cosθ
= A 2 + B 2 − 2AB
cosθ
⇒ 4 AB cosq = 0
Q 4 AB ≠ 0, \ cosq = 0 or q = 90°
5. (b) : Let q be angle between A
and B
2 2
| A+ B| = | A− B|, then | A+ B| = | A−
B|
A 2 + B 2 + 2ABcosq = A 2 + B 2 – 2ABcosq
or 4ABcosq = 0 or cosq = 0 or q = 90°
6. (c) : A + B = A + B if A B
. θ = 0°
7. (a) : Let q be angle between A
and B
.
Given: A= | A|
= 3 units
B= | B|
= 4units
C= | C|
= 5units
| A+ B| = | C |
A 2 + 2ABcosq + B 2 = C 2
9 + 2ABcosq + 16 = 25 or 2ABcosq = 0
or cosq = 0 \ q = 90º.
8. (b) : x = 5t – 2t 2 , y = 10t
dx
= 5− 4t, dy = 10 ∴ vx
= 5− 4t,
vy
= 10
dt dt
dv dv
x
y
=− 4, = 0 ∴ ax
=− 4,
ay
= 0
dt dt
^ ^ ^
Acceleration, a= a i+ a j=−4
i
x
y
\ The acceleration of the particle at t = 2 s is –4 m s –2 .
^
^
9. (a) : Here, R= 4sin( 2πt) i+
4cos( 2πt)
j
The velocity of the particle is
dR d
v = = [ sin( t) ^
^
4 2π i+
4cos( 2π
t)]
j
dt dt
= 8πcos( 2π ) ^
^
t i−8πsin( 2πt)
j
Its magnitude is
2 2
| v| = ( 8πcos( 2πt)) + ( −8πsin( 2πt))
2 2 2 2
= 64π cos ( 2πt) + 64π sin ( 2πt
)
2 2 2
= 64π [cos ( 2πt) + sin ( 2πt
)]
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2 2 2
= 64π ( As sin θ+ cos θ = 1)
= 8p m/s
10. (d) : At time t = 0, the position vector of the particle
is r ^
1 = 2i + 3
^. j
21
At time t = 5 s, the position vector of the particle is
^
r2 = 13 i + 14
^. j
Displacement from r
r
1 to 2 is
∆r r r ^ ^ ^ ^ ^ ^
= 2 − 1 = ( 13 i + 14 j) − ( 2i + 3j)
= 11i + 11 j
\ Average velocity,
^ ^
r i j
vav = ∆
i j
t
= 11 + 11 11 ^ ^
= ( + )
∆ 5−
0 5
11. (d) :
Velocity towards east direction, v
^
1 = 30 i ms /
Velocity towards north direction, v
2 = 40
^m/s j
Change in velocity, ∆v v v ^ ^
= 2 − 1 = ( 40 j−30
i)
∴ | | = | ^ ^
∆v 40 j− 30 i | = 50 ms /
Change in velocity
Averageacceleration, a av =
Time interval
v v v
aav = 2 − 1 ∆
=
∆t
∆t
| ∆v
| 50 ms / 2
| aav
| = = = 5 ms /
∆t
10 s
x
12. (b) : x= asinωt
or = sinωt
a
y
y= acosωt
or = cosωt
a
Squaring and adding, we get
2 2
x y
2 2
+ = 1 ( Q cos ωt
+ sin ωt
= 1)
2 2
a a
or x 2 + y 2 = a 2
This is the equation of a circle. Hence particle follows a
circular path.
13. (b) : Let q be the angle
which the particle makes
with an x-axis.
From figure,
3
tanθ= = 3
3
−1
or, θ= tan ( 3)
= 60°
14. (a) : v 1 = 50 km/hr due north
v 2 = 50 km/hr due west W
−v 1 = 50 km/hr due south
Magnitude of change in velocity
= | v − v | = | v + ( − v )| = v + ( −v
)
2 1 2 1 2 2 1 2
N
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