Sunday, 23 August 2020

Numericals of chapter -2(motion)

1. a car covers a distance of 160 km between two cities in four hour what is the average speed of the car ?
 sol. d=160 km 
 t= 4h 
 average speed= ? 
we know that s= d/t 
so, average speed = total distance travelled/total time taken 
 =160 km / 4 h 
 =40 km/h 
2. A train travels a distance of 300 Kilometre with an average speed of 60 kilometre per hour .how much time does it take to cover the distance? 
sol.d=300 km
S= 60km / h
t=?
We know , s=d/t
So, t=d/s
  t=300km/60kmh‐¹
t=5h.
3.A boy travels with an average speed of 10 metre per second for 20 minutes. how much distance does he travel?
Sol. S=10 m/s
t= 20 minutes
  = 20×1 minute  
  =20×60 seconds   (1 minute=60 seconds)
  =1200 seconds
But , s=d/t
       d=s×t
d=10 m/s ×1200 s
d=12000m
d=12× 1000m
d=12× 1km    (1km=1000m)
d=12 km.
4. A boy walks a distance of 30 metre in 1 minute and another 30 metre in 1.5 minutes .describe the type of motion of the boy and find his average speed in metre per second?
Sol. d¹  = 30 m
 t¹ = 1 minute
     =60 sec
d² = 30 m
t² = 1.5 minute
    =1.5 ×60 sec
   = 90sec
 S¹ =d¹/t¹
      =30 m/60 sec
      =0.5 m/s
S² = d²/t²
     = 30 m/ 90 sec
     =0.3333 m/s
 S¹  is not equal to s².
Type of the motion of the boy  is non-uniform motion.
Average speed= total distance/ total time
 =(d¹ + d²) / (t¹ + t² )
 =30 m + 30 m / 60 s + 90 s
 = 60 m / 150 s
 = 0.4 m/s
5. A cyclist Travels a distance of 1 kilometre in the first hour, 0.5 km in the second hour and 0.3 km in the third hour. find the average speed of the cyclist in a) kilometre per hour  b) metre per second

Ans.d¹ = 1km
             = 1000m
t¹ = 1h
d² = 0.5 km
     =0.5× 1000 m
     = 500 m
t² = 1 h
d³= 0.3 km
    = 0.3 × 1000 m
    = 300 m
t³= 1h
Average speed = d¹ + d² + d³ / t¹ + t²+ t ³
        =1000 m +500 m + 300m /1 h + 1h + 1h
        =1800 m/ 3h
        =1800 m/ 180×60 s
        =1 m / 6 s
        =0.167 m/s.


Average speed=1.8 km/3 h
          =0.6 km/h.
6. A car travels with a speed of 30 km per hour for 30 minutes and then with speed 40 km per hour for 1 hour .find 
a)the total distance travelled by the car 
b) the total time of travel and 
c) the average speed of car
Sol.  S¹= 30 km/h
         t= 30 minutes
          = 30× 1/60 h
          = 0.5 h
d¹ =s¹ × t¹
     = 30 km/h × 0.5 h
     = 15 km
s² = 40 km/ h
t² = 1h
d² = s² ×t²
     =40 km/h × 1h
     = 40 km
Total distance travelled= d¹ + d²
                   =15 km + 40 km
                   =55 km
Total time of travel = t¹ + t²
              =0.5 h + 1.0 h
              = 1.5 h
The average speed= Total distance travelled/ total time taken
=55 km/ 1.5h
=36.67 km /h .
7. On earth the weight of a body of mass 1.0 kg is is 10 Newton . what will be the weight of a boy of mass 37 kg in a) kgf    b) N ?
Sol. 1.0 kg =10 N

W = m×g
     =37 kg × g
     =37 kgf
Weight of boy of mass 37 kg
 =37 kg
= 37×1kg
=37× 10N   (since 1kg=10 N)
=370N
8. The weight of a body of mass 6.0 kg on Moon is 10 Newton. if a boy of mass 30 kg goes from Earth to the Moon surface ,what will be his  a) mass  b) weight
Sol.mass is constant .
So mass of the boy remains same .
Mass on moon=30 kg.
Weight of 6.0 kg on moon=10N
So, weight of 1kg on moon=10/6 N
Weight of 30kg boy = 30×1 kg
                                   =30×10/6 N  (since 1kg=10/6 N
                                   =5×10N
                                   =50N.





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