In a metre bridge experiemnt, null point is obtained at `20cm` from one end of the wire when resistance `X` is balanced against another resistnace `Y`. If `XltY`, then the new position of the null point from the same end,if one decides to balance a resistance of `4X`s against `Y` will be at

Options

A. 50cm
B. 80cm
C. 40cm
D. 70cm

Explanation

Meter bridge is an arrangement which works on wheatstone’s principle, so the balancing condition
where `I_2=100-l_1`
Case `1R = X, S = Y, l_11 = 20cm , l2 = 100 – 20 = 80 cm ` `therefore X/Y=20/80`
Case II Let the position of null put be obtained at a distance l from same end.
`therefore R = 4X, S = Y , l_1 = l, l_2 = 100 – l`
So, from equation (i), we get :
`(4x)/Y=1/(100-1) , X/Y=1/(4(100-l))`.....(ii)
Therefore, for equation (i) and (ii), we get :
`1/(4(100-1))=20/80 implies 1/(4(100-1))=1/4 or 1=100-1 or 2 times l=100`
Hence I=50cm
`I=E/(R+r)`
`I=E/R` =constant
where R= external resistance