Copy link:
Copy

A PARTICLE IS THROWN WITH SPEED OF 20 m/s IN VERTICAL UPWARD DIRECTION.THE MAXIMUM HEIGHT THE PARTICLE COULD REACH WILL BE(ASSUME ACCELERATION DUE TO GRAVITY IS 10 m/s2 )
40m
20m
10m
NONE OF ABOVE

A PARTICLE IS THROWN IN VERTICAL UPWARD DIRECTION WITH SPEED OF 20m/s FROM A TOWER OF HEIGHT 100m.AFTER HOW MUCH TIME THE PARTICLE WILL HIT THE GROUND APPROXIMATLY ?(ASSUME ACCELERATION DUE TO GRAVITY IS 10m/s2)
1 sec
4 sec
7sec
10sec

A PARTICLE IS DROPPED FROM A HIGH TOWER. THE DISTANCE IT WOULD COVER IN 3RD SECOND WILL BE( TAKE ACCELERATION DUE TO GRAVITY IS 10m/s2)
20m
25m
30m
35m

TWO PARTICLE,ONE IS RELEASED FROM THE TOP OF TOWER WHILE OTHER IS THROWN FROM BASE OF TOWER IN VERTICAL UPWARD DIRECTION WITH SPEED OF 20m/s/ IF THEY MEET IN MID JOURNEY AFTER 2 SEC.THEN HIEGHT OF TOWER WILL BE (ASSUME ACCELERATION DUE TO GRAVITY IS 10m/s2)
20m
40m
60m
80m

A BALOON START RISING FROM GROUND WITH AN ACCELERATION OF 2m/s2 , AFTER 10 SECOND OF RISE A PARTICLE IS DROPPED FROM THE BALOON WITH RESPECT TO BALLON.THE VELOCITY OF PARTICLE JUST AFTER RELEASE WILL BE (ASSUME ACCELERATION DUE TO GRAVITY IS 10m/s2)
0 m/s
10m/s
20m/s
40m/s

A PARTICLE IS THROWN IN UPWARD DIRECTION WITH SPEED OF 25m/s FROM THE BASE OF POLE OF CERTAIN HEIGHT. IT REACHES THE LEVEL OF TOP OF POLE TWICE IN DURATION OF 1 SEC.THE HEGHT OF TOWER WILL BE(ASSUME ACCELERATION DUE TO GRAVITY IS 10m/s2)
20m
30m
40m
50m

A PARTICLE IS THROWN FROM GROUND WITH VELOCITY OF U=(30i+40j)m/s IN xy PLANE. THE ANGLE OF PROJECTION OF PARTICLE WILL BE?
370
53O
600
750

IF A PARTICLE IS THROWN FROM THE GROUND WITH SPEED OF 20m/s AT AN ANGLE OF 60O TO THE HORIZONTAL THEN MINIMUM SPEED OF PARTICLE DURING ITS TRAJECTORY WILL BE(ASSUME ACCELERATION DUE TO GRAVITY IS 10m/s2)
20m/s
10m/s
5m/s
2m/s

Copy link:
Copy

Authored by:

Tests Created: 1

  • No Follower
  • Send message
Discussion
Your Facebook Friends on WizIQ