33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Telegram @unacademyplusdiscounts
44 NEET-AIPMT Chapterwise Topicwise Solutions Physics
8. (c) : Here, m = 10 g = 10 –2 kg, R = 6.4 cm = 6.4 × 10 –2 m,
K f = 8 × 10 –4 J, K i = 0, a t = ?
Using work energy theorem,
Work done by all the forces = Change in KE
W tangential force + W centripetal force = K f – K i
K
−4
f
8×
10
⇒ at
= =
4πRm
− −
4 × 22
2 2
× 64 . × 10 × 10
7
= 0.099 ≈ 0.1 m s –2
9. (b) : Work done = change in kinetic energy of the
body
1
2 2
W = × 001 . [( 1000) − ( 500) ] = 3750 joule
2
10. (c) : Here ^ ^
r i j r ^ ^
1 = ( − 2 + 5 ) m, 2 = ( 4 j+
3k)
m
^ ^
F = ( 4i + 3j)
N, W = ?
Work done by force F in moving from r
r
1 to 2,
^ ^ ^ ^ ^ ^
W = F⋅( r2 −r1) ⇒ W = ( 4i+ 3j) ⋅ ( 4j+ 3k+ 2i−5j)
^ ^ ^ ^ ^
= ( 4i+ 3j) ⋅( 2i− j+
3k ) = 8 + (–3) = 5 J
11. (c) : Here, F ^ ^
= ( 3 i + j)
N
Initial position, r ^ ^
1 = ( 2 i + k)
m
Final position, r ^ ^ ^
2 = ( 4i + 3j−
k)
m
Displacement, r = r −r
2 1
^ ^ ^ ^ ^ ^ ^ ^
r = ( 4i + 3j−k) m − ( 2i+ k)
m = 2i+ 3j−2k
m
^ ^ ^ ^ ^
Work done, W = F⋅ r = ( 3i + j) ⋅ ( 2i+ 3j−
2 k)
= 6 + 3 = 9 J
12. (a) : Distance (s) = 10 m; Force (F) = 5 N and work
done (W) = 25 J
Work done (W) = Fs cosq
\ 25 = 5 × 10 cosq = 50 cosq
or cosq = 25/50 = 0.5 or q = 60°
^ ^ ^
13. (a) : Force F = ( − 2i+ 15j+
6 k)
N, and distance,
d = 10 j^ m
Work done, W = F⋅d
= ( − 2 ^ i + 15 ^ j + 6k ^ ) ⋅(
10
^ j )
= 150 Nm = 150 J
14. (a)
15. (d) : Let the speed of the
third fragment of mass 3m be v′.
From law of conservation
of linear momentum,
2 v
3mv′ = 2mv
⇒ v′ = ...(i)
3
\ Energy released during the process is,
K.E. = ⎛ ( )
⎝ ⎜ ⎞
2 1 2 1
⎠
⎟ + ′
2 2 3 2 2
mv mv = mv + m v 2
1
(
2 3 )
29
(Using eqn. (i))
2 mv 4 2
= mv + = mv .
3 3
16. (b) :
2
Let v ′ be velocity of third piece of mass 2m.
Initial momentum, p i = 0 (As the body is at rest)
Final momentum, ^ ^
p mvi mv j mv
f = + + 2 ′
According to law of conservation of momentum
^ ^
pi
= p
f or, 0= mv i+ mvj+ 2mv′
or, v ^ v ^
v′ =− i − j
2 2
The magnitude of v′ is
′ = ⎛ 2
⎞
−
⎝
⎜
⎠
⎟ + ⎛ 2
v v⎞
v
v
−
⎝
⎜
⎠
⎟ =
2 2 2
Total kinetic energy generated due to explosion
1 1 1
= mv 2 + mv 2 + ( ′
2 2 2 2 mv )
2
2
1 2 1 2 1 ⎛ ⎞
= mv + mv + (
2 2 2 2 m)
⎝
⎜
v
⎠
⎟
2
2
2 mv 3 2
= mv + = mv
2 2
17. (d) : Velocity of water is v, mass flowing per unit
length is m.
\ Mass flowing per second = mv
\ Rate of kinetic energy or K.E. per second
1 2 1 3
= ( mv) v = mv .
2 2
18. (c) : mv = m
Mv ′ ⇒ v ′ = ⎛ ⎝ ⎜ ⎞
M ⎠
⎟ v
1 1
Total K.E. of the bullet and gun = mv
2 + Mv′
2
2 2
Total K.E. = 1 2
2 1 2
mv
2
+ M m 2
⋅ 2
M v
Total K.E. = 1 2 ⎧ m ⎫
mv ⎨1
+ ⎬⎭
2 ⎩ M
⎧1
⎫ ⎧ ⎫
or = ⎨ × 02⎬
⎨1 +
02 . 2
. ⎬v
= 105 . × 1000 J
⎩2
⎭ ⎩ 4 ⎭
⇒ v 2 4
= × 105 . × 1000
= 100
2 \ v = 100 m s –1
01 . × 42 .
19. (b) : According to law of conservation of linear
momentum,
30 × 0 = 18 × 6 + 12 × v
⇒ –108 = 12v ⇒ v = – 9 m/s.
Negative sign indicates that both fragments move in
opposite directions.