33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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22 NEET-AIPMT Chapterwise Topicwise Solutions Physics
2 2
= ( 50) + ( 50) = 70. 7 km / hr
1 km
21. (a) : v Resultant = = 4 km/hr
v = 70.7 km/hr along south-west direction
1/
4 hr
15. (d) : v 2 – u 2 = 2ax
2 2
∴ vRiver
= 5 − 4 = 3 km/hr
For case I : v 2 = u 2 – 2( g sin q 1 ) x 1
22. (a) : Let v be the velocity of river water. As shown in
[ a = –g sin q]
figure,
2
u
x1
=
2g
sinθ
[ v = 0]
sin30°= v 1
05 .
0.5 m/s
For case II :
or, v = 0.5 sin30°
v 2 = u 2 – (2g sin q 2 ) x 2
2
q
= 0.5 × (1/2) = 0.25 m/s
u
x2
=
2g
sinθ
g cos q 23. (d) : Let particle B move upwards with velocity v,
2
g
then
x1
sinθ2
sin30°
∴ = =
x
° = 1
2 sinθ1
sin60
tan 60°= v ; v = 3× 10=
173 . m/s.
3
10
^ ^ ^ ^
16. (b) :
Here, u= 2i + 3j, a= 03 . i + 02 . j,
t = 10 s
As v = u + at
24. (b) : Let the velocity of river be v R and velocity of
boat is v B .
^ ^ ^ ^
2 2
∴ v = ( 2i + 3j) + (. 03i + 02 . j)( 10 )
∴ Resultantvelocity = vB + vR + 2vBvRcosθ
^ ^ ^ ^ ^ ^
= 2 i + 3j+ 3i + 2j= 5i + 5j
( 10) = v 2 B + vR 2 + 2vBvR
cos90°
2 2
| v | = () 5 + () 5 = 5 2 units
( 10) = () 8 2 + vR
2 or ( 10) 2 = () 8
2 + vR
2
^ ^
2
17. (b) : Here, Initialvelocity, u= 3i + 4 j
vR
= 100 − 64 or vR
= 6 km / hr
^ ^
25. (a) : The equation of trajectory is
Acceleration, a= 04 . i + 03 . j ; time, t = 10 s
Let v 2
gx
be velocity of a particle after 10 s.
y= xtanθ
−
2 2
2u
cos θ
Using, v = u+
at
where q is the angle of projection and u is the velocity
^ ^ ^ ^
∴ v = ( 3i + 4 j) + (. 04i + 03 . j)( 10)
with which projectile is projected.
^ ^ ^ ^ ^ ^
For equal trajectories and for same angles of projection,
= 3i + 4 j+ 4i + 3j= 7i + 7 j
g
Speed of the particle after 10 s = | v |
u 2 = constant
As perquestion, 98 .
= g ′
2 2
2 2
= () 7 + () 7 = 7 2 units
5 3
where g′ is acceleration due to gravity on the planet.
18. (c) : Vertical acceleration in both the cases is g,
98 . × 9
whereas horizontal velocity is constant.
′ = =
−
g
35 . ms 2
25
v
19. (b) :
r = 10 m/s
26. (a) : At point B, X
f
component of velocity
remains unchanged while Y
v s = 20 m/s
component reverses its
q
direction.
\ The velocity of the projectile
^ ^
10 1
at point B is 2i− 3j
m/s.
cosφ= = or φ= 60 °
20 2
27. (b) : Horizontal range, R u 2
sin2θ
=
⇒ q = 180° – (90° + 60°) = 30°
g
a a
where u is the velocity of projection and q is the angle of
20. (d) : t = =
v ′ 2
v − v1 2 projection.
sin
Maximumheight, H u 2 2
θ
=
2g
g sin q
u a
v
According to question R = H