In a potentionmeter experiment, when three cells A, B and C are connected in series, the balancing length is found to be 740 cm. If A and B are connected in series, balancing length is 440 cm and for B and C connected in series, it is 540 cm. The emf of `E_(A),E_(B)` and `E_(C)` are respectively (in volts)

Options

A. 1, 1.2 and 1.5
B. 1, 2 and 3
C. 1.5, 2 and 3
D. 1.5, 2.5 and 3.5

Answer

A

Explanation

Let `epsilon_(A),epsilon_(B)andepsilon_(C)` be the emf of three cells A,B and C respectively.
As per question, `epsilon_(A)+epsilon_(B)+epsilon_(C)=kl_(1)=kxx740" "...(i)`
`epsilon_(A)+epsilon_(B)=kl_(2)=kxx440" "...(ii)`
`epsilon_(B)+epsilon_(C)=kl_(3)=kxx540" "...(iii)`
Inserting the value of `epsilon_(A)+epsilon_(B)` from (ii) into (i), we get `epsilon_(C)=300k`
Inserting the value of `epsilon_(B)+epsilon_(C)` from (iii) into (i), we get `epsilon_(A)=200k`
Inserting this value of `epsilon_(A)` into (ii), we get `epsilon_(B)=240k`
`therefore epsilon_(A):epsilon_(B):epsilon_(C)=200k:240k:300k`
`=10:12:15=1:1.2:1.5`
`implies epsilon_(A)=1V,epsilon_(B)=1.2V,epsilon_(C)=1.5V`