PHYSICS-11th (PQRS) TEST-3.pdf

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PHYSICS REVIEW TEST-3 Take g = 10 m/s2 where ever required in this paper. Q.1 Answer following short questions (Show all important steps involved): (i) If the extension of the spring in figure 1 is x1 and in figure 2 is x2, the systems being in constant acceleration, then find the ratio x1/x2. Pulley and strings are ideal. [2] (ii) Consider the two configurations shown in equilibrium. Find ratio of TA/TB. (Ignore the mass of the rope and the pulley) [2] (iii) Four identical 2 kg blocks are in equilibrium as shown. All surfaces are rough. Aforce of 10 N is applied to block C parallel to incline as shown. Rank the NORMAL on the BOTTOM surface of each block? [2] (iv) Find the acceleration of the blocks. [2] (i) (ii) (v) Four identical balls of mass m are arranged as shown. The mass of the block in figure (3) is 4 m and having acceleration a, the angle of the incline is =30° in figure (2) and all the surfaces are frictionless. Rank the TENSIONS in the labelled ropes (from lower to higher)? Consider wedge and table to be fixed. [2] (1) (2) (3) [Sol: (i) x1 = 32 50 K ; x2 = K ; (ii) TA = Mg ...(1) 2TB = Mg ...(2)  TA TB = 2 Ans. (iii) N1 = 20, N2 = 40, N3 = 20 + 10sin N4 = 20cos  N4 < N1 < N3 < N2 Ans. 4F (iv) (i) a = M  ; (ii) a = 4F  Mg  M (v) TB = mg ...(1) TA = 2mg ...(2) TC = mgsin  TC = mg/2 ...(3) Ans. ] Q.2 Answer following short questions: (i) Two blocks of masses M1 and M2 are connected to each other through a light spring as shown in the figure. If we push mass M1 with a force F and cause acceleration a1 in it, what will be the acceleration of M2? [3] (Assume friction less surface everywhere). (ii) A particle of mass 10 kg is acted upon by a force F along the line of motion which varies as shown in the figure. The initial velocity of the particle is 10 ms–1. Find the maximum velocity attained by the particle before it comes to instantaneous rest. [3] (iii) For the given system of blocks, massless and frictionless pulley, and ideal strings. (a) Find the minimum coefficient of friction (min) for which the system stays in equilibrium. (b) If  = min/2 find the acceleration of the system, tensions in string A and B. [1+2] (iv) If Block B comes to rest instantaneously after striking the ground, find the time taken by the block A to strike the pulley. (Assume friction less surface everywhere). [3] (v) Two blocks A & B are kept on a partically rough surface. Two forces F1 & F2 act as shown on blocks A & B respectively. Find the acceleration of the system & friction force on block A.(with direction) [2+1] (vi) An aeroplane A is flying horizontally due east at a speed of 400 km/hr. Passengers in A, observe another aeroplane B moving perpendicular to direction of motion at A. Aeroplane B is actually moving in a direction 30° north of east in the same horizontal plane as shown in the figure. Determine the velocity of B. [3] [Sol.(i) F – kx = M1a1 kx

XI(PQRS) PHYSICS REVIEW TEST-3
Take g = 10 m/s2 where ever required in this paper.
Q.1 Answer following short questions (Show all important steps involved):
(i) If the extension of the spring in figure 1 is x1 and in figure 2 is x2, the
systems beinginconstantacceleration,thenfindtheratiox1/x2.Pulley
and strings are ideal. [2]
(ii) Considerthetwoconfigurations showninequilibrium.FindratioofTA/TB.(Ignorethemassoftherope
andthe pulley) [2]
(iii) Fouridentical2kgblocks areinequilibrium asshown.All surfaces are
rough.Aforceof10NisappliedtoblockCparalleltoinclineasshown.
Rank the NORMAL on the BOTTOM surface of each block?
[2]
(iv) Find the accelerationof the blocks. [2]
(i) (ii)
(v) Four identical balls of mass m are arranged as shown. The mass of the block in figure (3) is 4 m and
having accelerationa,the angleoftheincline is =30° in figure(2) and all thesurfaces are frictionless.
Rank the TENSIONS in the labelled ropes (from lower to higher)? Consider wedge and table to be
fixed. [2]
(1) (2) (3)
[Sol:
(i) x1 =
K
32
; x2 =
K
50
; 25
16
2
1

x
x
(ii)
TA =Mg ...(1)
2TB = Mg ...(2)

B
A
T
T
= 2 Ans.
(iii)
N1 = 20, N2 = 40, N3 = 20 + 10sin N4 = 20cos
 N4 < N1 < N3 < N2 Ans.
(iv) (i) a =
M
F
4
 ; (ii) a =
M
Mg
F 
4

(v) TB = mg ...(1)
TA =2mg ...(2)
TC =mgsin TD = 4mg/5 ...(5)
 TC = mg/2 ...(3)
TC < TD < TB < TA Ans. ]
Q.2 Answer following short questions:
(i) Two blocks of masses M1 and M2 are connected to each other through a light
spring as shown in the figure. If we push mass M1 with a force F and
cause acceleration a1 in it, what will be the acceleration of M2? [3]
(Assumefrictionless surfaceeverywhere).
(ii) A particle of mass 10 kg is acted upon by a force F along the line of
motion whichvaries as shown inthe figure. The initial velocityof the
particle is 10ms–1.Findthe maximumvelocityattainedbythe
particle before itcomes to instantaneous rest. [3]
(iii) Forthegivensystem ofblocks,masslessand frictionless pulley,
andidealstrings.
(a) Findtheminimumcoefficientoffriction(min)forwhichthesystem
staysinequilibrium.
(b) If = min/2 find theacceleration of the system, tensions in string
A and B.
[1+2]
(iv) If BlockB comes torest instantaneouslyafterstrikingthe ground,
find the time taken bythe blockAto strikethe pulley.
(Assumefrictionless surfaceeverywhere).
[3]
(v) Two blocksA& B are kept on a particallyrough surface.
Two forces F1 & F2 act as shown on blocksA& B
respectively.Findthe accelerationofthesystem & friction
force onblockA.(with direction)
[2+1]
(vi) An aeroplaneAis flying horizontally due east at a speed of 400 km/hr.
Passengers inA, observe another aeroplane B moving perpendicular to
directionofmotionatA.Aeroplane Bis actuallymovinginadirection30°
northofeastin thesamehorizontal planeas shownin
thefigure.DeterminethevelocityofB. [3]
[Sol.(i) F – kx = M1a1 kx = M2a2
kx = F – M1a1 ...(1) F – M1a1 = M2a2
 a2 =
2
1
1
M
a
M
F 
Ans.
(ii) change in velocityis area under a-t graph. a is +ve till t = 10s.
areatill 10s
½ × 10 × 2 = 10
Vmax – 10 = 10 Vmax = 20m/s.
(iii) T1 is tensioninstringA
T2 is tensioninstringB
(a) T2 = minMg
T2 = T1 + g
T1 = g
 T = 2g
 2g = min Mg
min =
M
2
(b)  =
2
min

=
M
1
fs = Mg = g
T2 – g = Ma .......... (1)
T1 + g – T2 = a
g – T1 = a
2g – T2 = 2a  T2 = 2g – 2a .............(2)
Eq. (1) and (2) implies
a =
M
g

2
=
1
2 


g
(iv) a = 0, f = 60 N leftward
(v) V =  
5
.
0
2
2 




 g
= 5 m/s
0.5 = ½ × g/2 × t2  t =
5
1
s
T =
5
1
+
5
5
.
0
=
10
5
3
s
(vi) j
V
i
V
V B
B
B
ˆ
2
ˆ
2
3



i
V
V A
A
ˆ


A
B
V /

= j
V
i
V
V
V B
A
B
A
B
ˆ
2
ˆ
2
3
/ 











 VB =
3
2 A
V
=
3
800
 B
V

= j
i ˆ
3
400
ˆ
400  ]
Q.3nl Krrishissavingachildinabuildingcaught byfire.Heandchild aretrappedand heties aropeanddrops
it out of window. Unfortunately, the tensilestrength of the rope is 900 N and Krrish & child's combine
mass is 100 kilograms. He therefore decides to let the rope slide through his hands, pulling just hard
enough on it so it doesn't break. What will be his speed at the bottom ofthe rope as hehits the ground?
Assumetheropeis 12 meterslongand thatKrrish'svertical speed is zeroas heleaves thewindowat the
top of the rope. [3]
[Sol. a =
100
900
1000 
= 1 m/s2
v2 = 2 × 1 × 12
v = 24 m/s Ans. ]
Q.4nl Asmall cubical block is placedona triangular blockM sothat theytouch each
other along a smooth inclined contact plane as shown. The inclined surface
makesananglewiththehorizontal.AhorizontalforceFistobeappliedonthe
block m so that the two bodies move without slipping against each other.
Assumingthefloorto besmoothalso, determinethe
(a) normal force with which m andM press against each other and
(b) the magnitude of external force F. Express your answers in terms of m, M,  and g. [3+3]
[Sol. F = (M + m) a ....(1)
mg sin  + ma cos  = F cos  ....(2)
cos  F = mg sin  +
M
m
F
cos
m


F = 










)
M
m
(
cos
m
cos
)
M
m
(
= mgsin 
F = 


cos
M
sin
mg
)
M
m
(
= 
 tan
)
M
m
(
M
mg
N sin =
m
M
F
M

 N = 





 m
M
F
M
cosec  ]
Q.5nl Thesystemshowninthefigureisinequilibrium.Findtheinitialaccelerationof
A, B and C just after the spring-2 is cut. [2+2+2]
[Sol.
3mg = KX1 ....(1)
2mg + KX1 = KX2
 2mg + 3mg = KX2  5 mg = KX2 ....(2)
KX3 = KX2 + mg  KX3 = 5mg + mg = 6 mg ....(3)
when spring2is cut
KX3 – mg = ma3  a3 = 5g 
KX1 + 2mg = 2ma2  a2 =
2
mg
5

acceleration of 3 m will be zero. ]
Q.6nl Atruck undergoes constantacceleration, a=5 m/s2. Onthebackoftruck atwo-sided,frictionless ramp
is fixed, with each side of the ramp at an angle  from the horizontal.Two masses, m1 and m2, sit on
either sideof the ramp as shown, connected byamassless string wound over a massless pulley.
(a) Draw free bodydiagram (FBD) of m1 & m2 in ground frame.
(b) Calculate the ratio of masses, m2/m1, forwhich the masses donot slide on theramp when released.
[3+3]
[Sol. (a)
(b) m2 g sin = m2a cos + T
m2g sin  = T + m2 a cos  ....(1)
m1g sin  + m1a cos  = T ....(2)
From eg. (1) & (2)
m2g sin  = m2 a cos  + m1 g sin  + m1 a cos 
m2[g sin  – a cos ] = m1 [g sin  + a cos ]

1
2
m
m
= 





cos
a
sin
g
cos
a
sin
g
=
5
11
Ans. ]
Q.7 A time - dependent force, F = 10t, starts to act on a block of mass 5 kg at
t =0 as shown in the figure.
(i) Whendoes it lose contact with the ground?
(ii) Draw the a – t graph for 0-10 seconds.
(iii) Find the speed of the block at t = 10 sec. [2 + 3 + 1]
[Sol: (i) t = 5 sec.
(ii)
5
50
10 
t
(iii) v = 25m/s ]
Q.8nl A block withmass m is pushed alonga horizontal floor bya force P that makes an angle withthe
horizontal as shown.Thecoefficientsofkineticandstaticfriction betweentheblockandthefloorarek
and s, respectively.
(a) Findthemaximumforcethatcanbeapplied withoutmovingtheblock.
(b) The applied force Pis larger than the maximum force calculated in (a).As a
consequence, theblock will start to move. Find the accelerationof the
block.
[3+3]
[Sol.
(a) Max. force that can be applied
P cos  = sN = s (mg + P sin )
P [ Cos  – s sin ] = s mg ; P =





sin
cos
mg
s
s
Ans.
(b) Now blockstarts movingso frictionactingon the blockis kinetic
P cos  – k (P sin  + mg) = ma ; a =
m
)
mg
sin
P
(
cos
P k 




]
Q.9nl Thesystemshownin the figureis connected byflexible inextensible cord.The
coefficient of friction between block C & the rigid surface is 0.3. The system
starts from rest when height of blockAabove ground is d.
Pulleyandstrings are ideal.
(a) Find velocityof C just beforeAis going to touch ground in terms of d and the
othergivenvalues.
(b) Find acceleration of B after A touches the ground and comes to rest
instantaneously.
(c) Find theinitialdistance'd'betweenAand theground, so thatthe system comes
to rest when bodyB just touchesA. [3+2+2]
[Sol. fmax = 0.3 × 5 mg
= 1.5 mg
commonacceleration a =
m
8
mg
5
.
1
mg
3 
=
8
15
m
8
mg
5
.
1
 m/s2
After verticallya distance'd'
v2 = 0 + 2 ×
8
15
× d ( v2 = u2 + 2as)
v2 =
4
15
d ....(1)
After this 2 m comes to rest and system retards
Retardation =
m
mg
mg
6
5
.
1 
=
6
5
m/s2
Now After travelling1.8 m, masses cometo rest
 O = v2 – 2 ×
6
5
× 1.8  v2 = 3
Putting thevalue of v2 from eq. (1)
4
15
d = 3  d =
5
4
m Ans. ]
Q.10327kin/sss An aircraft is flying horizontallywith a constant velocity= 200 m/s, at a height = 1 km above the
ground.At the moment shown, a bombis released from the aircraft and the cannon-gun below fires a
shell withinitialspeed 200m/s, atsomeangle.
(a) Forwhat value of ''will theprojectile shelldestroythe bombinmid-air?
(b) If the valueof  is 53°, findtheminimumdistancebetweenthebomb and theshell as theyflypast each
other. Take sin 53° = 4/5. [3+4]
[Sol.(i) Supposeshelldestroythe bomb attime 't'then forhorizontal motion
t(200 + 200cos) = 3 × 1000
 t(1 + cos) = 5 3 ...(1)
Forverticalmotion
2
1
gt2 + (200sin)t –
2
1
gt2 = 1000
 (sin)t =5 ...(2)
from (1) and (2)



cos
1
sin
=
3
1
On solving,  = 60° Ans.
(ii) A
v = 200 î
B
v = –200cos53 î + 200sin53 j
ˆ = –200 ×
5
3
î + 200 ×
5
4
j
ˆ = –120 î + 160 j
ˆ
B
/
A
v = A
v – B
v = (200 + 120) î – 160 j
ˆ
tan =
320
160
 =
2
1
AB = 2 km
BP=minimum distance=ABsin(30° –)
 BP = 2[sin30°cos – cos30°sin] = 2 








5
1
2
3
5
2
2
1
=
5
3
2 
Ans. ]

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PHYSICS-11th (PQRS) TEST-3.pdf

  • 1. XI(PQRS) PHYSICS REVIEW TEST-3 Take g = 10 m/s2 where ever required in this paper. Q.1 Answer following short questions (Show all important steps involved): (i) If the extension of the spring in figure 1 is x1 and in figure 2 is x2, the systems beinginconstantacceleration,thenfindtheratiox1/x2.Pulley and strings are ideal. [2] (ii) Considerthetwoconfigurations showninequilibrium.FindratioofTA/TB.(Ignorethemassoftherope andthe pulley) [2] (iii) Fouridentical2kgblocks areinequilibrium asshown.All surfaces are rough.Aforceof10NisappliedtoblockCparalleltoinclineasshown. Rank the NORMAL on the BOTTOM surface of each block? [2] (iv) Find the accelerationof the blocks. [2] (i) (ii) (v) Four identical balls of mass m are arranged as shown. The mass of the block in figure (3) is 4 m and having accelerationa,the angleoftheincline is =30° in figure(2) and all thesurfaces are frictionless. Rank the TENSIONS in the labelled ropes (from lower to higher)? Consider wedge and table to be fixed. [2] (1) (2) (3) [Sol: (i) x1 = K 32 ; x2 = K 50 ; 25 16 2 1  x x
  • 2. (ii) TA =Mg ...(1) 2TB = Mg ...(2)  B A T T = 2 Ans. (iii) N1 = 20, N2 = 40, N3 = 20 + 10sin N4 = 20cos  N4 < N1 < N3 < N2 Ans. (iv) (i) a = M F 4  ; (ii) a = M Mg F  4  (v) TB = mg ...(1) TA =2mg ...(2) TC =mgsin TD = 4mg/5 ...(5)  TC = mg/2 ...(3) TC < TD < TB < TA Ans. ] Q.2 Answer following short questions: (i) Two blocks of masses M1 and M2 are connected to each other through a light spring as shown in the figure. If we push mass M1 with a force F and cause acceleration a1 in it, what will be the acceleration of M2? [3] (Assumefrictionless surfaceeverywhere). (ii) A particle of mass 10 kg is acted upon by a force F along the line of motion whichvaries as shown inthe figure. The initial velocityof the particle is 10ms–1.Findthe maximumvelocityattainedbythe particle before itcomes to instantaneous rest. [3] (iii) Forthegivensystem ofblocks,masslessand frictionless pulley, andidealstrings. (a) Findtheminimumcoefficientoffriction(min)forwhichthesystem staysinequilibrium. (b) If = min/2 find theacceleration of the system, tensions in string A and B. [1+2]
  • 3. (iv) If BlockB comes torest instantaneouslyafterstrikingthe ground, find the time taken bythe blockAto strikethe pulley. (Assumefrictionless surfaceeverywhere). [3] (v) Two blocksA& B are kept on a particallyrough surface. Two forces F1 & F2 act as shown on blocksA& B respectively.Findthe accelerationofthesystem & friction force onblockA.(with direction) [2+1] (vi) An aeroplaneAis flying horizontally due east at a speed of 400 km/hr. Passengers inA, observe another aeroplane B moving perpendicular to directionofmotionatA.Aeroplane Bis actuallymovinginadirection30° northofeastin thesamehorizontal planeas shownin thefigure.DeterminethevelocityofB. [3] [Sol.(i) F – kx = M1a1 kx = M2a2 kx = F – M1a1 ...(1) F – M1a1 = M2a2  a2 = 2 1 1 M a M F  Ans. (ii) change in velocityis area under a-t graph. a is +ve till t = 10s. areatill 10s ½ × 10 × 2 = 10 Vmax – 10 = 10 Vmax = 20m/s. (iii) T1 is tensioninstringA T2 is tensioninstringB (a) T2 = minMg T2 = T1 + g T1 = g  T = 2g  2g = min Mg min = M 2 (b)  = 2 min  = M 1 fs = Mg = g T2 – g = Ma .......... (1) T1 + g – T2 = a g – T1 = a 2g – T2 = 2a  T2 = 2g – 2a .............(2) Eq. (1) and (2) implies a = M g  2 = 1 2    g
  • 4. (iv) a = 0, f = 60 N leftward (v) V =   5 . 0 2 2       g = 5 m/s 0.5 = ½ × g/2 × t2  t = 5 1 s T = 5 1 + 5 5 . 0 = 10 5 3 s (vi) j V i V V B B B ˆ 2 ˆ 2 3    i V V A A ˆ   A B V /  = j V i V V V B A B A B ˆ 2 ˆ 2 3 /              VB = 3 2 A V = 3 800  B V  = j i ˆ 3 400 ˆ 400  ] Q.3nl Krrishissavingachildinabuildingcaught byfire.Heandchild aretrappedand heties aropeanddrops it out of window. Unfortunately, the tensilestrength of the rope is 900 N and Krrish & child's combine mass is 100 kilograms. He therefore decides to let the rope slide through his hands, pulling just hard enough on it so it doesn't break. What will be his speed at the bottom ofthe rope as hehits the ground? Assumetheropeis 12 meterslongand thatKrrish'svertical speed is zeroas heleaves thewindowat the top of the rope. [3] [Sol. a = 100 900 1000  = 1 m/s2 v2 = 2 × 1 × 12 v = 24 m/s Ans. ] Q.4nl Asmall cubical block is placedona triangular blockM sothat theytouch each other along a smooth inclined contact plane as shown. The inclined surface makesananglewiththehorizontal.AhorizontalforceFistobeappliedonthe block m so that the two bodies move without slipping against each other. Assumingthefloorto besmoothalso, determinethe (a) normal force with which m andM press against each other and (b) the magnitude of external force F. Express your answers in terms of m, M,  and g. [3+3] [Sol. F = (M + m) a ....(1) mg sin  + ma cos  = F cos  ....(2) cos  F = mg sin  + M m F cos m   F =            ) M m ( cos m cos ) M m ( = mgsin 
  • 5. F =    cos M sin mg ) M m ( =   tan ) M m ( M mg N sin = m M F M   N =        m M F M cosec  ] Q.5nl Thesystemshowninthefigureisinequilibrium.Findtheinitialaccelerationof A, B and C just after the spring-2 is cut. [2+2+2] [Sol. 3mg = KX1 ....(1) 2mg + KX1 = KX2  2mg + 3mg = KX2  5 mg = KX2 ....(2) KX3 = KX2 + mg  KX3 = 5mg + mg = 6 mg ....(3) when spring2is cut KX3 – mg = ma3  a3 = 5g  KX1 + 2mg = 2ma2  a2 = 2 mg 5  acceleration of 3 m will be zero. ] Q.6nl Atruck undergoes constantacceleration, a=5 m/s2. Onthebackoftruck atwo-sided,frictionless ramp is fixed, with each side of the ramp at an angle  from the horizontal.Two masses, m1 and m2, sit on either sideof the ramp as shown, connected byamassless string wound over a massless pulley. (a) Draw free bodydiagram (FBD) of m1 & m2 in ground frame. (b) Calculate the ratio of masses, m2/m1, forwhich the masses donot slide on theramp when released. [3+3]
  • 6. [Sol. (a) (b) m2 g sin = m2a cos + T m2g sin  = T + m2 a cos  ....(1) m1g sin  + m1a cos  = T ....(2) From eg. (1) & (2) m2g sin  = m2 a cos  + m1 g sin  + m1 a cos  m2[g sin  – a cos ] = m1 [g sin  + a cos ]  1 2 m m =       cos a sin g cos a sin g = 5 11 Ans. ] Q.7 A time - dependent force, F = 10t, starts to act on a block of mass 5 kg at t =0 as shown in the figure. (i) Whendoes it lose contact with the ground? (ii) Draw the a – t graph for 0-10 seconds. (iii) Find the speed of the block at t = 10 sec. [2 + 3 + 1] [Sol: (i) t = 5 sec. (ii) 5 50 10  t (iii) v = 25m/s ] Q.8nl A block withmass m is pushed alonga horizontal floor bya force P that makes an angle withthe horizontal as shown.Thecoefficientsofkineticandstaticfriction betweentheblockandthefloorarek and s, respectively. (a) Findthemaximumforcethatcanbeapplied withoutmovingtheblock. (b) The applied force Pis larger than the maximum force calculated in (a).As a consequence, theblock will start to move. Find the accelerationof the block. [3+3] [Sol. (a) Max. force that can be applied P cos  = sN = s (mg + P sin ) P [ Cos  – s sin ] = s mg ; P =      sin cos mg s s Ans.
  • 7. (b) Now blockstarts movingso frictionactingon the blockis kinetic P cos  – k (P sin  + mg) = ma ; a = m ) mg sin P ( cos P k      ] Q.9nl Thesystemshownin the figureis connected byflexible inextensible cord.The coefficient of friction between block C & the rigid surface is 0.3. The system starts from rest when height of blockAabove ground is d. Pulleyandstrings are ideal. (a) Find velocityof C just beforeAis going to touch ground in terms of d and the othergivenvalues. (b) Find acceleration of B after A touches the ground and comes to rest instantaneously. (c) Find theinitialdistance'd'betweenAand theground, so thatthe system comes to rest when bodyB just touchesA. [3+2+2] [Sol. fmax = 0.3 × 5 mg = 1.5 mg commonacceleration a = m 8 mg 5 . 1 mg 3  = 8 15 m 8 mg 5 . 1  m/s2 After verticallya distance'd' v2 = 0 + 2 × 8 15 × d ( v2 = u2 + 2as) v2 = 4 15 d ....(1) After this 2 m comes to rest and system retards Retardation = m mg mg 6 5 . 1  = 6 5 m/s2 Now After travelling1.8 m, masses cometo rest  O = v2 – 2 × 6 5 × 1.8  v2 = 3 Putting thevalue of v2 from eq. (1) 4 15 d = 3  d = 5 4 m Ans. ]
  • 8. Q.10327kin/sss An aircraft is flying horizontallywith a constant velocity= 200 m/s, at a height = 1 km above the ground.At the moment shown, a bombis released from the aircraft and the cannon-gun below fires a shell withinitialspeed 200m/s, atsomeangle. (a) Forwhat value of ''will theprojectile shelldestroythe bombinmid-air? (b) If the valueof  is 53°, findtheminimumdistancebetweenthebomb and theshell as theyflypast each other. Take sin 53° = 4/5. [3+4] [Sol.(i) Supposeshelldestroythe bomb attime 't'then forhorizontal motion t(200 + 200cos) = 3 × 1000  t(1 + cos) = 5 3 ...(1) Forverticalmotion 2 1 gt2 + (200sin)t – 2 1 gt2 = 1000  (sin)t =5 ...(2) from (1) and (2)    cos 1 sin = 3 1 On solving,  = 60° Ans. (ii) A v = 200 î B v = –200cos53 î + 200sin53 j ˆ = –200 × 5 3 î + 200 × 5 4 j ˆ = –120 î + 160 j ˆ B / A v = A v – B v = (200 + 120) î – 160 j ˆ tan = 320 160  = 2 1 AB = 2 km BP=minimum distance=ABsin(30° –)  BP = 2[sin30°cos – cos30°sin] = 2          5 1 2 3 5 2 2 1 = 5 3 2  Ans. ]