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YSA Model Paper 

Subject: Physics                                                                                            Class: IX

Answer Key

Section – C
 
Q.3      a) From book page no. 18. under the headings of random error and                    
systematic error.
            b) Final velocity,       vf = 0ms-1
                        Height,            h = 1.8m.
            Acceleration due to gravity, g = 10ms-2
                        Initial Velocity, Vi = ?
            Sol.
                        2aS = vf2 – vi2
                        2gs = vf2 – vi2
                        2(-10ms-2) (1.8m) = (0)2 (vi)2
                        36m2S-2 = - vi2
            Taking square root on both sides.
                        Vi = 6mS-1
Q.4      a) Book Page No. 47, under the heading motion of bodies by string.      Case – I
            b) Distance towards North   = 800m
                -       --         --         West   = 1300m
                    -       --         -- in 3rd direction = R = ?
 
            From fig.          
R2 = (OA)2 + (OB)2
                  = (800)2 + (1300)2
                  = 640000 + 1690000
                  = 2330000
        
               R = 1526.4m.
            Distance covered in 3rd direction is 152.6 m.
Q.5      a) Weight of Painter, W1 = 700N.
             --- --- --  --Platform, W2 = 200N.
            Let T1 and T2 are tensions in the strings
            According to first condition of equilibrium.
                                    Up ward force = Down ward force
                                    T1 + T2 = W1 + W2
                                    T1 + T2 = 700N + 200N
                                    T1 + T2 = 900N
            Calculating torques at Pt. D.
            Clockwise torque = Anti clockwise torque
            T1 * CD                   W1 * AD + W2 * BD
            T1 * 3m           =          700N * 2m + 200N * 1.5m.
            3T1m               =          1400Nm + 300Nm
            3T1m               =          1700Nm
            T                    =         
            T                    =          566.66
                                    =          566.67N
 
            As T1 + T      =          900N
                        T        =          900 – 566.7
                                    =          333.3N
Q.5      b) Page No. 92, heading 7.8 from 2nd paragraph to end of topic.
 
Q.6      a) Page No. 149. Definition under the heading of conduction, convection         and 
radiation of Pascal’s law and hydraulic Press.
            b) Page No. 127 of the text book.
            c) Unlike other materials the behaviour of water is unusual. By increasing or decreasing the temperature 
of ∆ C0 , the water expands this          unusual behaviour is called anomalous expansion of water.
 
 
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