Solution:- (i) Velocity of packet:-
Given data:-
1):- vi = 2 m/s ( initial velocity of packet will become same as velocity of helicopter moving). ,
2):- t = 2s , ( time as given)
3):- g = a = -10 m/s (Because as we know that value of "a" while coming downward is (-ve) and value of "a" while going upward is (+ve)).
Must remember this:-
Soln:-
vf=vi+at ............(first equation of motion).
vf= 2+(-10)(2) = 2 + (-20)
vf = -18 m/s.
(-ve) sign shows that packet is coming downward and here we used simple formula because time and initial velocity is given .
(ii) how far packet is from helicopter after 2 mintues:-
(a):- So first we calculate the total distance covered by helicopter,
Distance coverd by helicopter: d= v*t
d=2*2= 4m . (Here we simply use velocity-distance formula because helicopter is going in (x-axis) horizontal direction . so as we also know that in x-asis , acceleration is zero.)
(b):- Distance coverd by packet : S =(vf2 - vi2 )/2g
S = ((-18)2 - (2)2)/2(10)
S = ( 324 - 4)/20 = 320/20
S = 16m.
Here we used 3rd equation of motion (2gS =vf2 - vi2 ) , because packet is going downward and we have to add value g also.
Total Distance between packet and helicopter:-
So total distance = D = d+S
D= 4+16
D = 20m. Answer .
3
) Different radioactive materials have ......................... ,
A ) Different half life , B ) same total life ,
C ) Both of these , D
) None of these ,
Correct
Answer : - C ) Both of these ,
4
) From Laplace correction , the
propagation of sound in gas is ........................ ,
A ) adiabatic process , B ) Isothermic process ,
C ) Both of these , D ) None of these ,
Correct Answer : - A ) adiabatic process ,
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