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b ) F=-k/2x^2
ma=-k/2x^2
vdv/dx = -k/2mx^2
integreate both sides with th elimits 0-v and 1-.25

we get v=root3
now v=dx/dt
root3 dt=dx
integrate with limit 0-t for time and 1- 0.25 for x
we get t =root3/4 + c
c=pie/3

it depends on the frame u choose
if the observer is in the lift too
then what ever the accelration is the tension remains the same
but when the observer is not in lift then ans is as given by INDIAN ARMY

well same here
have tests in skul and bt tests tooo
me in lot of trouble da
sorrry

plzz do post the diagram
from next time
absence of the diagram created a lot of confusion here

oh man
got my mistakle
i thought it was atwood machine where the pulley is aateched to the spring
u should have posted the diagram

here is the solution
at the maximum extension
Kx=3mg---(1)

when the string is cut obviously since it is freely falling
m has accelration g
for 2m
kx -2mg = 2ma
3mg-2mg=2ma
a =g/2

plzz from next post the diagram

just apply projectile formula
T=2vsin@/g
@=90
sin90=1
3 = 2v/g
v =3g/2

edited

@sonu
as for ur doubt
as soon as u cut the string
the tension becmoes 0
now the only force causing the extension is due to tension
now tension is 0
so the spring keeps oscilating whereas the bodies fall of with acclerations g and g

every freely falling body has accelration g
since no other external force acts on it
both will have accelration g,g

a big floooppp

cooooooool congrats champs
seeems many toppers r here

paper-1 92
paper-2 98

congrats everyone

impossible
if speed is changing
then the magnitude of the velocity changes so velocity changes tooo

in the variation o\f masses problem in momentum

F ext +/- F thrust = Ma

how do u detremine the direction of thrust force and the sign

consider O the last electron is in the 2p orbital || | |
since 2p is close to the nuclues its tuff remove it
so this is ruled out

for O2 the last electrons are in 3p orbital || | |
FOR O2 - the configuraton is || || |
for O2 + the configuration is | | |

now calculate bond order for
we come to know that 02 - wil have the least bond order
so i think it will have least ionisation enthalpy

correct me if i'm wrong