Page 6 - ECAT Past Paper 2006
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ECAT PAST PAPER-2006                                                 BY www.studyplex.net



                       C) 4 (Correct)                                    D) 5
                       Explanation:
                              A typical four petrol engine also undergoes four successive processes in each cycle.
               Q.20  Speed of sound in Vacuum is ---- m/s.
                       A) 332                                            B) 340
                       C) 0 (Correct)                                    D) 350
                       Explanation:
                              Speed of Sound for:
                               Air: 332 m/s
                               Vacuum: 0 m/s
                               Oxygen: 315 m/s
               Q.21  When effective value of current is 10.What is its peak value?
                       A) 12                                             B) 13
                       C) 10                                             D) 14.1 (Correct)
                       Explanation:
                              As we know that:
                              I rms=Io/√2
                              Irms=10
                              so; 10*√2=Io
                              Io=1.41*10
                              Io=14.1 A (Answer)
                                                                                     st
               Q.22  A body moves with the uniform acceleration with a distance x for 1  second and y for
                       next 4 seconds then what is relation b/w x and y?
                       A) y=24x                                          B) y=16x (Correct)
                       C) y=12x                                          D) y=8x
                       Explanation:
                              As we know that:
                                         2
                              S= vit+ ½ at
                              as body movies with uniform acceleration so vi=0
                              so S= ½ at
                                       2
                                                 2
                              Let it be x and y ½ at  for some interval .
                                                  2
                                                               2
                              For 1  second: x= ½ at   x= ½ a(1) x= ½ a ---(1)
                                   st
                              For 4  second: y= ½ at y= ½ a(4)  y=16 (½ a)—(2)
                                   th
                                                              2
                                                  2
                              As we are to find the relation b/w x and y so:
                              From (2)  y= 16x (Answer)
               Q.23  An electron volt is equal to:
                                 -12
                                                                                  -12
                       A) 0.62*10 J                                      B) 1.6*10 J
                                 12
                       C) 0.62*10 J                                      D) 1.6*10  J (Correct)
                                                                                  -19
                       -Explanation:
                                           18
                              1 J=6.25*10  eV
                                            -19
                              1eV= 1.6*10  J
                                                                                                      -1
               Q.24  A wire 3.0 m long, of uniform cross section area 2.0 mm  has a conductance 1.25 ohm ,
                                                                           2
                       What is resistivity of material of the wire.
                       A) 5.3*10  ohm-m (Correct)                        B) 2.15 ohm-m
                                -7
                               -7
                                                                                  -7
                       C) 8.3*10  ohm-m                                  D) 1.9*10  ohm-m
                       Explanation:
                              Given data:
                              L=3.0 m
                                                 2
                              A=2 mm  or 2*10 m
                                      2
                                              -6
                                                                   -1
                              G= 1.25 so R=1/G R=1/1.25 R=8*10  ohm
                              As we know that:
                              p= RA/L
                                     -1
                                            -6
                              p= 8*10 *2*10 /3
                                      -7
                              p=5.3*10  ohm-m (Answer)
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