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12th Grade > Physics

MOTION IN TWO DIMENSION MCQs

Total Questions : 31 | Page 1 of 4 pages
Question 1. For a given velocity, a projectile has the same range R for two angles of projection. If t1 and t2 are the times of flight in the two cases then   
  1.    t1t2 α R2
  2.    t1t2 α R
  3.    t1t2 α 1R
  4.    t1t2 α 1R2
 Discuss Question
Answer: Option B. -> t1t2 α R
:
B
For same range angles of projection should be θ and 90 - θ
So, time of flights t1 = 2usinθg and
t2 = 2usin(90θ)g = 2ucosθg
By multiplying = t1t2 = 4u2sinθcosθg2
t1t2 = 2g (u2sin2θ)g = 2Rg t1t2 α R
Question 2. A body of mass m is thrown upwards at an angle θ with the horizontal with speed v.  While rising up the speed of the mass after t seconds will be 
  1.    √(vcosθ)2+(vsinθ)2
  2.    √(vcosθ)2−(vsinθ)2−gt
  3.    √v2+g2t2−(2vsinθ)gt
  4.    √v2+g2t2−(2vcosθ)gt
 Discuss Question
Answer: Option C. -> √v2+g2t2−(2vsinθ)gt
:
C
Instantaeous speed of rising mass after t sec will be vt = vx2+vy2
where vx = v sin θ = Horizontal component of velocity
vy = v sin θ - gt = Vertical component of velocity
vt = (vcosθ)2+(vsinθgt)2
Question 3. An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation of 6 km towards a point directly above the target on the earth's surface. At an appropriate time, the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be falling
  1.    On a parabolic path as seen by pilot in the plane
  2.    Vertically along a straight path as seen by an observer on the ground near the target
  3.    On a parabolic path as seen by an observer on the ground near the target
  4.    On a zig-zag path as seen by pilot in the plane
 Discuss Question
Answer: Option C. -> On a parabolic path as seen by an observer on the ground near the target
:
C
The pilot will see the ball falling in straight line because the reference frame is moving with the same horizontal velocity but the observer at rest will see the ball falling in parabolic path.
Question 4. A man sitting in a bus travelling in a direction from west to east with a speed of 40 km/h observes that the rain-drops are falling vertically down. To the another man standing on ground the rain will appear  
  1.    To fall vertically down
  2.    To fall at an angle going from west to east
  3.    To fall at an angle going from east to west       
  4.    The information given is insufficient to decide the direction of rain. 
 Discuss Question
Answer: Option B. -> To fall at an angle going from west to east
:
B
A man is sitting in a bus and travelling from west to east, and the rain drops are appears falling vertically down.
A Man Sitting In A Bus Travelling In A Direction From West T...
vm = velocity of man
vr = Actual velocity of rain which is falling at an angle θ with vertical
vrm = velocity of rain w.r.t to moving man
If the another man observe the rain then he will find that actually rain falling with velocity v, at an angle going from west to east.
Question 5. A string of length L is fixed at one end and carries a mass M at the other end.  The string makes 2π revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is 
  1.    ML    
  2.    2 ML
  3.    4 ML
  4.    16 ML
 Discuss Question
Answer: Option D. -> 16 ML
:
D
A String Of Length L Is Fixed At One End And Carries A Mass ...
T sin θ = M ω2 R
T sin θ = M ω2 L sin θ
From (i) and (ii)
T = M ω2 L
= M 4π2n2L
= M 4 π2(2π2)L
= 16 ML
Question 6. A car is moving on a circular path and takes a turn. If R1 and R2  be the reactions on the inner and outer wheels respectively, then   
  1.    R1 = R2
  2.    R1 < R2
  3.    R1 > R2
  4.    R1 ≥ R2
 Discuss Question
Answer: Option B. -> R1 < R2
:
B
Reaction on inner wheel R1 = 12M[g - v2hra]
Reaction on outer wheel R2 = 12M[g + v2hra]
where, r = radius of circular path, 2a = distance between two wheels and h = height of centre of gravity of car
Question 7. A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle θ1. In the next 2 sec, it rotates through an additional angle θ2 . The ratio of angles is 
  1.    1
  2.    2
  3.    3 
  4.    5
 Discuss Question
Answer: Option C. ->
:
C
Using relation θ = ω0t + 12 α t2
θ1 = 12 (α) 22 = 2 α ..(i) (As ω0=0,t=2sec)
Now using same question for t = 4 sec , ω0 = 0
θ1 + θ2 = 12 (4)2 = 8 α ..(ii)
From (i) and (ii), θ1 = 2 α and θ2 = 6α θ1θ2 = 3
Question 8. A particle P is moving in a circle of radius 'a' with a uniform speed v . C is the centre of the circle and AB is a diameter. When passing through B the angular velocity of P about A  and B  are in the ratio  
  1.    1 : 1
  2.    1 : 2
  3.    2 : 1 
  4.    4 : 1
 Discuss Question
Answer: Option B. -> 1 : 2
:
B
A Particle P Is Moving In A Circle Of Radius 'a' With A Unif...
Angular velocity of particle P about point A,
ωA = vrAB = v2r
Angular velocity of particle P about point C,
ωC = vrBC = vr
Ratio ωAωC = v2rvr = 12
Question 9. Three identical particles are joined together by a thread as shown in figure.  All the three particles are moving in a horizontal plane in circular path with ‘O’ as center.  If the velocity of the outermost particle is v0, then the ratio of tensions in the three sections of the string is Three Identical Particles Are Joined Together By A Thread As...
  1.    3 : 5 : 7 
  2.    3 : 4 : 5
  3.    7 : 11 : 6   
  4.    3 : 5 : 6
 Discuss Question
Answer: Option D. -> 3 : 5 : 6
:
D
Three Identical Particles Are Joined Together By A Thread As...
Let ω is the angular speed of revoultion
T3 =m ω2 3l
T2 -T3 =mω2 2l T2 = mω25l
T1 - T2 = mω2l T1 = mω26l
T3:T2:T1 = 3 : 5 : 6
Question 10. The resultant of P and Q  is perpendicular to P. What is the angle between  P and Q.
  1.    cos−1(PQ)
  2.    cos−1(−PQ)
  3.    sin−1(PQ)
  4.    sin−1(−PQ)
 Discuss Question
Answer: Option B. -> cos−1(−PQ)
:
B
The Resultant Of →P And →Q  Is Perpendicular To →P. ...
tan90 = QsinθP+Qcosθ P + Q cos θ = 0
cos θ = PQ θ = cos1(PQ)

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