n identical reistance are taken in which `(n )/(2)` resistors are joined in series in the left gap and the remaining `(n)/(2)` resistances are joined in parallel in the right gap of metre bridge. Balancing length in cm is

Options

A. `100,(n^(2))/(n^(2)+4)`
B. `100. (n^(2))/(n^(2)+1)`
C. `400. (1)/(n^(2)+4)`
D. `400. (1)/(n^(2)+1)`

Answer

A

Explanation

Meter bridge is shown in the figure below

When `.(n)/(2)` resistances are joined in series in left gap each of resistance `R_(1)` then equivalent resistance in left gap.
`R= (R _(1))/(2) + (R_(1))/(2) + (R_(1))/(2) + ..... (n)/(2) ` times `= (R_(1)n)/( 2)`
When `(n)/(2)` resistors are joined in parallel in right gap then equivalent resistance in right gap
`(1)/(s ) = (1)/(R_(1)) + (1)/(R_(1)) + (1)/(R_(1)) + ..... (n )/(2) ` times `= (n )/( 2R_(1))`
`rArr s= (2R_(1))/( n )`
If l be the balacing length in the meter bridge wire then
`(R )/(s) = (1)/(100-1) rArr ((R_(1)n)/(2 ))/((2R_(1))/(n )) = (l )/(100-l)`
`(n^(2))/(4) = (l)/(100-1) rArr l = (100n^(2))/( n^(2) + 4)`