1) Three equal masses P, Q and R are pulled with a constant force F. They are connected to each other with strings. The ratio of the tension between PQ and QR is:
(answer should come 2:1)

2) The pulleys and strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle Ф should be (answer should come 45 degree)

3) A string of negligible mass going over a clamped pulley of mass 'm' supports a block of mass M as shown in figure. The force on the pulley by the clamp is given by
{answer should come \sqrt{(M+m)^{2} + M^{2}} g

4) When a bus suddenly takes a turn, the passengers are thrown outwards because of----------

5) A body of mass m stands on one end of a wooden plank of length L and mass M. The plank is floating on water. If the boy walks from one end of the plank to the other end at a constant peed, the resulting displacement of the plank is given by (answer should come  \frac{mL}{(M+m)} )

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2014-12-15T18:43:53+05:30

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1) Tension in string between P and Q = T1
         tension in string between Q and R = T2
     let all of them move with a uniform acceleration of  a.
          let mass of P , Q and R be  m.
           F - T1 = m a     for  P
            T1 - T2 = m a    for Q
              T2 = m a   for R
        => T1 = ma + ma = 2 ma 
             F = 3 ma
          ratio = 2

2)  
      T1 = m g
       T1 Sin Ф + T2 Cos Ф = √2 m g
            T2 = m g 
         Sin Ф + Cos Ф = √2 
               square it,      1 + sin 2Ф = 2
              Sin 2Ф = 1     => Ф = 45 deg

3) 
            
 please see diagram for free body diagram of pulley.  
        there are vertical forces M g (same as tension T) and m g  weight of pulley.  there is a horizontal force   Mg = Tension.
          resultant force =  √ [ (Mg+mg)² + (Mg)² ] 
            same as your expected answer

4)   inertia of motion.    There is not enough centripetal force acting on them. So they continue to move in a straight line.

5) 
      Center of mass of  body m and plank of mass M is at   [ M * L/2 + m * 0 ] / (M+m) , from the end where the body m is present.
              that is = CM at   ML/2(M+m)
    We assume there is no friction between water and plank.
         When the body goes to the other end, the center of mass will be at ML/2(M+m) from that end.
        Since there is no external force on the two body  system together, the CM stays at the same point.
        Hence the plank moves by L - ML/2(M+m)  - ML/2(M+m) = L * m /(M+m)

             


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