Correct Answer - B::C
Given, for first student, `R_(2)=10Omega,R_(1)=5Omega,R_(3)=5Omega`
For second student, `R_(1)=500Omega,R_(3)=5Omega`.
Now, according to whatstone bridge rule,
`(R_(2))/(R)=(R_(1))/(R_(3))impliesR=R_(3)xx(R_(2))/(R_(1))`
Now putting all the values in Eq. (i) we get `R=10Omega` for both students. Thus, we can analyse that the wheatstone bridge is mot sensitive and accurate if resistances are of same value.
Thus, the errors of measurement of the two students depend on the accuracy and sensitivity of the bridge, which inturn depends on the accuracy with which `R_(2) and R_(1)` can be measured. when `R_(2) and R_(1)` are larger, the currents through the arms of bridge is very weak. this can make the determination of null point accurately more difficult.