A 100 V voltmeter of internal resistance 20 `kOmega` in series with a high resistance R is connected to a 110 V line. The voltmeter reads 5 V, the value of R is

Explanation

As shown in fig., the voltmeter is in series with R so
`V=V_(1)+V_(R)`, i.e., `110=5+V_(R)`
i.e., `V_(R)=110-5=105V`
And as in series potential divides in proportion to resistance,
`(V_(R))/(V_(1))=R/(R_(1)),` i.e., `105/5=R/(20kOmega)` or
`R=420kOmega`