Two uniform spheres A (Hollow) and B (solid) of same radius R (

Options

A. `(r)/(NR)^(3//2)`
B. `(NR/r)^(1//2)`
C. `(NRr)^(3//2)`
D. `(NRr)^(1//2)`

Answer

B

Explanation

As each are containing n lamps , hence Resisitance of each arec=nr.Number of `arcs=N//n` Total resisitacne s is given by
`(1)/(s)=sigma(1)/(nr)=(N)/(n)((1)/(nr))rArrS=(n^(2)r)/(N)`
`therefore Total resistance =r+S=R+(n^(2)r)N`
If E is the emf of the machine current entering the arcs is `(E//R+S)` and in each arc is `(E//R+S)N`. Hence ,current passing through each lamp
`I=(nE)/(N(r+n^(2))(r//N)=(E)/(N)[(R)/(n+(nr))/(N)]^(-1)`
Now heat produced per second in the lamps is `H=N,I^(2)`
Since, light emitted is proportional of `H^(2)` therefore lighty produced is maximum when `h^(2)` and hence H is maximum or `[(R)/(n)+(nr)/(N)]` is minimum.Hence we can write
`R/n+(nr)/(N)=[((R)/(n))^(1)/(2)-((nr)/(N))^(1/2)]^(-2)+2((R)/(n))+2((Rr)/(N))^(1)/(2)`
This is minimum when `sqrt(R//n)-sqrt(nr//N)=0`
or very small or very small or n is closely equal to `(NR//R)^(3)/(2)`