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Apologies fr not going thru anchits ans...it seemed too scary to me....here goes my attempt...

m assuming the particle to execute SHM ( though i dun have enough reasons to come to tat conclusion...it makes things easier :D )

The particle starts from rest at x=0 and again comes to rest at x=1 ....so surely...x=0 and x=1 are the extreme positions ,

also amplitude=(1-0)/2=1/2

time taken to complete half oscillation = 1sec...so T= 2 sec

now w =pi rad/sec (w=omega)

so the eq of SHM is x=1/2 sin (pi t+c)

so max acceleration can be found which will be greater than or equal to 4 (w^2A)

obviously Z cant hold the same sign throughout...

so ans =a,c

Thanx a lot !

lol...thanq

Let the number of F atoms be f, and the number of Cl atoms be c,

Energy required for the removal of $1e^{-}$ from 1 F atom = $27.91X10^{-22}KJ$

Energy required for the removal of $1e^{-}$ from f F atoms = $27.91X10^{-22}KJ$ $\times f$

lly,

Energy required for the removal of $1e^{-}$ from c Cl atoms = $20.77X10^{-22}KJ \times c$

Energy released on addition of $1e^{-}$ to f F atoms = $-5.53\times 10^{-22}KJ \times f$

Energy released on addition of $1e^{-}$ to c Cl atoms = $-5.78\times 10^{-22}KJ \times c$

$27.91X10^{-22}KJ$ $\times f$ $20.77X10^{-22}KJ \times c$ =284KJ

$-5.53\times 10^{-22}KJ \times f$ $-5.78\times 10^{-22}KJ \times c$ =68.8KJ

Solve for f and c,

$\frac{c}{c+f} \times 100$ gives the percentage composition of Cl

$\frac{f}{c+f} \times 100$ gives the percentage composition of F

Area under the curve f(x) - Area under the curve g(x) = Area enclosed by the closed curve ...

$Put\ x=k^6\\\ \\<br/>I=\int{k^2{\left(2+k^{3\right)}^2 6k^5dk\\\ \\$

$=\int{6k^{13}+24k^{10}+24k^{7}}$

$=\frac{6}{14} k^{14}+\frac{24}{11}k^{11}+\frac{24}{8}k^8$

$sub, k=x^{\frac{1}{6}}$

$I=\frac{3}{7}x^{\frac{7}{3}}+\frac{24}{11}x^{\frac{11}{6}}+3x^{\frac{4}{3}}$ +C