(i) The diameter of neck and bottom of a bottle are 2cm and 10cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2kgf, what force is exerted on the bottom of the bottle?
(ii) A ball is thrown vertically upwards. It returns 6s later. Calculate : (i) the greatest height reached by the ball, and (ii) the initial velocity of the ball. (Take g = 10metre per second square).
Answers
Answered by
1
i) The same force (1.2 kgf) would be exerted at the bottom of the bottle, as the force exerted in a fluid gets transmitted undiminished in all directions.
(ii)Given, u=?
v=0
g=-10m/s^2
returns after 6 seconds
s=?
If the ball returns after 6 s, then the time taken by it to reach the height is 6/2= 3s
So, using
first eq of motion, v=u+gt
0=u+(-10)*3 (here, g=-10 m/s^2 as the ball is thrown against gravity)
-u=-30 m/s
u=30m/s (initial velo)
Now using third eq
v^2=u^2+2gs
0=(30)^2+ 2 (-10)s
0=900-20s
20s=900
s=900/20=45 m
(ii)Given, u=?
v=0
g=-10m/s^2
returns after 6 seconds
s=?
If the ball returns after 6 s, then the time taken by it to reach the height is 6/2= 3s
So, using
first eq of motion, v=u+gt
0=u+(-10)*3 (here, g=-10 m/s^2 as the ball is thrown against gravity)
-u=-30 m/s
u=30m/s (initial velo)
Now using third eq
v^2=u^2+2gs
0=(30)^2+ 2 (-10)s
0=900-20s
20s=900
s=900/20=45 m
Similar questions