The longest wavelength in Balmer series of hydrogen spectrum will be

Options

A. 6557 Å
B. 1216 Å
C. 4800 Å
D. 4800 Å

Explanation

Side of step ladder AB and AC is 1.6 m
`therefore AB=1.6m and AC=1.6m`
Suppose `angleABC=angleACB=angleADE=angleAED=theta`
Normal force on point B and C is `N_(B) and N_(C)` length of rope `DE=0.5m`
Weight suspended `W=40kgf`
`=40xx9.8`
`=392N`
`AD=BD=AE=EC=(1.6)/(2)=0.8m`
[`because` D and E is the midpoint of AB and AC]
`BF=BD+DF`
`=0.8+0.4`
`=1.2m`
`AF=AB-BF=1.6-1.2=0.4m`
Suppose A is midpoint of rope DE
`therefore DA.=(0.5)/(2)=0.25m`
`DF.=F.A.`
`=(1)/(2)DA.`
`=(1)/(2)xx0.25`
`therefore F.A.=0.125m`
For equilibrium of base of ladder
`N_(B)+N_(C)=W`
`therefore N_(B)+N_(C)=392N....(1)`
For side AB, couple about point A
`N_(B)xxBC.=WxxF.A.+TxxA A.`
but `BC=ABcostheta`
and `A A.ADsintheta`
`therefore N_(B)xxABcostheta=Wxx0.125+TxxADsintheta....(2)`
but in `DeltaDFF.`
`costheta=(DF.)/(DF)=(0.125)/(0.4)=0.3125`
`therefore theta=72.8^(@)`
`therefore costheta=cos72.8^(@)=0.3125`
`sintheta=sin72.8^(@)=0.9553`
Putting value in eqn. (2)
`N_(B)xx1.6xx0.3125=392xx0.125+Txx0.8xx0.9553`
`therefore 0.5N_(B)=49+0.76424T....(3)`
Now couple (moment of force) about point A of side AC
`N_(C)xxC C.=TxxA A`.
but `C C.=ACcosthetaandA A.=AEsintheta`
`therefore N_(C)xxACcostheta=TxxAEsintheta`
`therefore N_(C)xx1.6xx0.3125=Txx0.8xx0.9553`
`0.5N_(C)=0.76424T....(4)`
putting value of eqn. (4) in eqn. (3)
`0.5N_(B)=49+0.5N_(C)`
`therefore 0.5(N_(B)-N_(C))=49`
`therefore N_(B)-N_(C)=(49)/(0.5)=98.....(5)`
Now solving eqn. (1) and eqn. (4)
`N_(B)+N_(C)=392`
`N_(B)-N_(C)=98`
Sum these
`2N_(B)=490`
`therefore N_(B)=245N`
and from eqn. (1)
`N_(C)=392-N_(B)`
`=392-245`
`=147N`
Now from eqn. (4)
`0.5xx147=0.76424T`
`therefore T=(0.5xx147)/(0.76424)`
`therefore T=96.1739N`
`therefore T~~96.2N`