33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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Work, Energy and PowerTelegram @unacademyplusdiscounts43ANSWER KEY1. (a) 2. (a) 3. (b) 4. (b) 5. (d) 6. (a) 7. (c) 8. (c) 9. (b) 10. (c)11. (c) 12. (a) 13. (a) 14. (a) 15. (d) 16. (b) 17. (d) 18. (c) 19. (b) 20. (a)21. (c) 22. (a) 23. (a) 24. (b) 25. (d) 26. (a) 27. (b) 28. (c) 29. (c) 30. (d)31. (d) 32. (d) 33. (b) 34. (c) 35. (a) 36. (b) 37. (b) 38. (a) 39. (d) 40. (d)41. (b) 42. (c) 43. (a) 44. (b) 45. (b) 46. (a) 47. (a) 48. (a) 49. (d) 50. (a)51. (d) 52. (a) 53. (d) 54. (c) 55. (b) 56. (c) 57. (c) 58. (d) 59. (b) 60. (b)61. (d) 62. (d) 63. (c) 64. (b) 65. (b) 66. (c) 67. (b) 68. (c) 69. (b) 70. (c)71. (d) 72. (c) 73. (a) 74. (d) 75. (b) 76. (a) 77. (d) 78. (a) 79. (d) 80. (c)81. (a)Hints & Explanations1. (a) : ^ ^Given, r = cosωtx+sinωt y∴ dr^^v = =− ωsinωtx+ωcosωtydt dva = =−ω 2 ^ωtx− ω 2 ^cos sinωty=−ω2 rdtSince position vector () r is directed away from the origin,so, acceleration ( −ω 2 r ) is directed towards the origin.Also, ^ ^ ^ ^r⋅ v = (cosωtx+ sin ωty) ⋅( − ωsinωtx+ωcos ωty)= –w sinwt coswt + w sinwt coswt = 0⇒ r ⊥v2. (a) : Two vectors A and B are orthogonal to eachother, if their scalar product is zero ie .. A ⋅ B=0 ^ ^Here, A= cosωt i+sinωt jand B ωt^ ωt^= cos i + sin j2 2 ⎛⎞∴ A⋅ B= (cosωti ^ + sin ωtj^ ) ⋅ωt t+⎝⎜cos ^ ω i sin^ j⎠⎟2 2ωtωt= cosωtcos + sinωtsin2 2 = ⎛⎝⎜ − ⎞cos ωω tt⎠⎟2But A ⋅ B= 0 (as A and B are orthogonal to each other)⎛ ⎞∴ cos −⎝⎜ωω tt⎠⎟ = 02⎛cos ω ω t ⎞t cosπ ωtω t−π⎝⎜⎠⎟ = or − =2 2 2 2ωtπ π= or t =2 2 ω3. (b) : ^ ^ ^ ^ ^ ^a= 2i+ 3j+ 8k, b = 4 j− 4 i+α k a⋅ b= 0if a ⊥ b( 2 ^ i+ 3 ^ j+ 8k ^ ) ⋅− ( 4 ^ i+ 4 ^ j+ α k^ ) = 0or, –8 + 12 + 8a = 0 ⇒ 4 + 8a = 0⇒ a = –1/2. 4. (b) : Given:( F1+ F2) ⊥( F1−F2) ∴ ( F1+ F2) ⋅( F1− F2)= 0 F −F −F ⋅ F + F ⋅ F = 0 ⇒ F = F1 2 2 2 1 2 2 1 1 2 2 2i.e. F 1 , F 2 are equal to each other in magnitude.5. (d) : Position vector of the particle^^r = ( acos ωt) i+( asin ωt)jvelocity vector drv = = ( − a sin t) ^^ω ω i+( aωcos ωt)jdt= ω[( − sin ω ) ^^a t i+( acos ωt)]j ^ ^ ^ ^v⋅ r = ωa[ − sinωti+ cos ωt j)][ ⋅ acosωti+asin ωt j]= w[–a 2 sin wt cos wt + a 2 cos wt sin wt] = 0Therefore velocity vector is perpendicular to the positionvector.6. (a) : ^ ^ ^A= i+ j+ k B ^ ^ ^3 4 5 and = 3i+ 4 j−5k ^ ^ ^ ^ ^ ^A⋅Bcos θ= A B = ( 3i+ 4 j+ 5k) ⋅ ( 3i+ 4 j−5k)2 2 2 2 2 2[ ( 3) + ( 4) + ( 5) ] × [ ( 3) + ( 4) + ( 5) ]9= + 16 − 25= 0 or θ = 90°.507. (c) : Here, m = 1 g = 10 –3 kg,h = 1 km = 1000 m, v = 50 m s –1 , g = 10 m s –2 .(i) The work done by the gravitational force= mgh = 10 –3 × 10 × 1000 = 10 J(ii) The total work done by gravitational force and theresistive force of air is equal to change in kinetic energyof rain drop.\ W g + W r = 1 2mv − 02or10 + W r = 1 −3× 10 × 50 × 502W r = –8.75 J
Telegram @unacademyplusdiscounts44 NEET-AIPMT Chapterwise Topicwise Solutions Physics8. (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 KEW tangential force + W centripetal force = K f – K iK−4f8×10⇒ at= =4πRm− −4 × 222 2× 64 . × 10 × 107= 0.099 ≈ 0.1 m s –29. (b) : Work done = change in kinetic energy of thebody12 2W = × 001 . [( 1000) − ( 500) ] = 3750 joule210. (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 r1 to 2, ^ ^ ^ ^ ^ ^W = F⋅( r2 −r1) ⇒ W = ( 4i+ 3j) ⋅ ( 4j+ 3k+ 2i−5j)^ ^ ^ ^ ^= ( 4i+ 3j) ⋅( 2i− j+3k ) = 8 + (–3) = 5 J11. (c) : Here, F ^ ^= ( 3 i + j)NInitial position, r ^ ^1 = ( 2 i + k)mFinal position, r ^ ^ ^2 = ( 4i + 3j−k)mDisplacement, r = r −r2 1 ^ ^ ^ ^ ^ ^ ^ ^r = ( 4i + 3j−k) m − ( 2i+ k)m = 2i+ 3j−2km ^ ^ ^ ^ ^Work done, W = F⋅ r = ( 3i + j) ⋅ ( 2i+ 3j−2 k)= 6 + 3 = 9 J12. (a) : Distance (s) = 10 m; Force (F) = 5 N and workdone (W) = 25 JWork done (W) = Fs cosq\ 25 = 5 × 10 cosq = 50 cosqor 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 J14. (a)15. (d) : Let the speed of thethird fragment of mass 3m be v′.From law of conservationof linear momentum,2 v3mv′ = 2mv⇒ v′ = ...(i)3\ Energy released during the process is,K.E. = ⎛ ( )⎝ ⎜ ⎞2 1 2 1⎠⎟ + ′2 2 3 2 2mv mv = mv + m v 21(2 3 )29(Using eqn. (i))2 mv 4 2= mv + = mv .3 316. (b) :2Let 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 mvf = + + 2 ′According to law of conservation of momentum ^ ^pi= pf or, 0= mv i+ mvj+ 2mv′or, v ^ v ^v′ =− i − j2 2The magnitude of v′ is′ = ⎛ 2⎞−⎝⎜⎠⎟ + ⎛ 2v v⎞vv−⎝⎜⎠⎟ =2 2 2Total kinetic energy generated due to explosion1 1 1= mv 2 + mv 2 + ( ′2 2 2 2 mv )221 2 1 2 1 ⎛ ⎞= mv + mv + (2 2 2 2 m)⎝⎜v⎠⎟222 mv 3 2= mv + = mv2 217. (d) : Velocity of water is v, mass flowing per unitlength is m.\ Mass flowing per second = mv\ Rate of kinetic energy or K.E. per second1 2 1 3= ( mv) v = mv .2 218. (c) : mv = mMv ′ ⇒ v ′ = ⎛ ⎝ ⎜ ⎞M ⎠⎟ v1 1Total K.E. of the bullet and gun = mv2 + Mv′22 2Total K.E. = 1 22 1 2mv2+ M m 2⋅ 2M vTotal 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= 1002 \ v = 100 m s –101 . × 42 .19. (b) : According to law of conservation of linearmomentum,30 × 0 = 18 × 6 + 12 × v⇒ –108 = 12v ⇒ v = – 9 m/s.Negative sign indicates that both fragments move inopposite directions.
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Work, Energy and Power
Telegram @unacademyplusdiscounts
43
ANSWER KEY
1. (a) 2. (a) 3. (b) 4. (b) 5. (d) 6. (a) 7. (c) 8. (c) 9. (b) 10. (c)
11. (c) 12. (a) 13. (a) 14. (a) 15. (d) 16. (b) 17. (d) 18. (c) 19. (b) 20. (a)
21. (c) 22. (a) 23. (a) 24. (b) 25. (d) 26. (a) 27. (b) 28. (c) 29. (c) 30. (d)
31. (d) 32. (d) 33. (b) 34. (c) 35. (a) 36. (b) 37. (b) 38. (a) 39. (d) 40. (d)
41. (b) 42. (c) 43. (a) 44. (b) 45. (b) 46. (a) 47. (a) 48. (a) 49. (d) 50. (a)
51. (d) 52. (a) 53. (d) 54. (c) 55. (b) 56. (c) 57. (c) 58. (d) 59. (b) 60. (b)
61. (d) 62. (d) 63. (c) 64. (b) 65. (b) 66. (c) 67. (b) 68. (c) 69. (b) 70. (c)
71. (d) 72. (c) 73. (a) 74. (d) 75. (b) 76. (a) 77. (d) 78. (a) 79. (d) 80. (c)
81. (a)
Hints & Explanations
1. (a) : ^ ^
Given, r = cosωtx+
sinωt y
∴
dr
^
^
v = =− ωsinωtx+
ωcosωty
dt
dv
a = =−ω 2 ^
ωtx− ω 2 ^
cos sinωty=−ω
2 r
dt
Since position vector () r is directed away from the origin,
so, acceleration ( −ω 2 r ) is directed towards the origin.
Also,
^ ^ ^ ^
r⋅ v = (cosωtx+ sin ωty) ⋅( − ωsinωtx+
ωcos ωty)
= –w sinwt coswt + w sinwt coswt = 0
⇒ r
⊥v
2. (a) : Two vectors A
and B
are orthogonal to each
other, if their scalar product is zero ie .. A ⋅ B
=0
^ ^
Here, A= cosωt i+
sinωt j
and B
ωt
^ ωt
^
= cos i + sin j
2 2
⎛
⎞
∴ A⋅ B= (cosωti ^ + sin ωtj
^ ) ⋅
ωt t
+
⎝
⎜cos ^ ω i sin
^ j
⎠
⎟
2 2
ωt
ωt
= cosωt
cos + sinωt
sin
2 2 = ⎛
⎝
⎜ − ⎞
cos ω
ω t
t
⎠
⎟
2
But A
⋅ B
= 0 (as A
and B
are orthogonal to each other)
⎛ ⎞
∴ cos −
⎝
⎜ω
ω t
t
⎠
⎟ = 0
2
⎛
cos ω ω t ⎞
t cos
π ωt
ω t
−
π
⎝
⎜
⎠
⎟ = or − =
2 2 2 2
ωt
π π
= or t =
2 2 ω
3. (b) : ^ ^ ^ ^ ^ ^
a= 2i+ 3j+ 8k, b = 4 j− 4 i+
α k
a⋅ b= 0if a ⊥ b
( 2 ^ i+ 3 ^ j+ 8k ^ ) ⋅− ( 4 ^ i+ 4 ^ j+ α k
^ ) = 0
or, –8 + 12 + 8a = 0 ⇒ 4 + 8a = 0
⇒ a = –1/2.
4. (b) : Given:( F1+ F2) ⊥( F1−
F2)
∴ ( F1+ F2) ⋅( F1− F2)
= 0
F −F −F ⋅ F + F ⋅ F = 0 ⇒ F = F
1 2 2 2 1 2 2 1 1 2 2 2
i.e. F 1 , F 2 are equal to each other in magnitude.
5. (d) : Position vector of the particle
^
^
r = ( acos ωt) i+
( asin ωt)
j
velocity vector
dr
v = = ( − a sin t) ^
^
ω ω i+
( aωcos ωt)
j
dt
= ω[( − sin ω ) ^
^
a t i+
( acos ωt)]
j
^ ^ ^ ^
v⋅ r = ωa[ − sinωti+ cos ωt j)][ ⋅ acosωti+
asin ωt j]
= w[–a 2 sin wt cos wt + a 2 cos wt sin wt] = 0
Therefore velocity vector is perpendicular to the position
vector.
6. (a) : ^ ^ ^
A= i+ j+ k B ^ ^ ^
3 4 5 and = 3i+ 4 j−5k
^ ^ ^ ^ ^ ^
A⋅
B
cos θ=
A B = ( 3i+ 4 j+ 5k) ⋅ ( 3i+ 4 j−
5k)
2 2 2 2 2 2
[ ( 3) + ( 4) + ( 5) ] × [ ( 3) + ( 4) + ( 5) ]
9
= + 16 − 25
= 0 or θ = 90°
.
50
7. (c) : Here, m = 1 g = 10 –3 kg,
h = 1 km = 1000 m, v = 50 m s –1 , g = 10 m s –2 .
(i) The work done by the gravitational force
= mgh = 10 –3 × 10 × 1000 = 10 J
(ii) The total work done by gravitational force and the
resistive force of air is equal to change in kinetic energy
of rain drop.
\ W g + W r = 1 2
mv − 0
2
or
10 + W r = 1 −3
× 10 × 50 × 50
2
W r = –8.75 J
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