If we use 100 Omega and 200 Omega in pla...

If we use `100 Omega` and `200 Omega` in place of `R` and `X` we get null point deflection, `l = 33cm`. If we interchange the resistors, the null point length is found to be `67 cm` Find end corrections `prop and beta`.

`alpha=1cm`, `beta=1cm`

`alpha=2cm`, `beta=1cm`

`alpha=1cm`, `beta=2cm`

None of these

Text Solution

Verified by Experts

The correct Answer is:

`alpha=(R_(2)l_(1)-R_(1)l_(2))/(R_(1)-R_(2))`
`=((200)(33)-(100)(67))/(100-200)=1cm`
`beta=(R_(1)l_(1)-R_(2)l_(2))/(R_(1)-R_(2))-100`
`=((100)-(200)(67))/(100-200)-100=1cm`

Topper's Solved these Questions

EXPERIMENTAL PHYSICS
NARAYNA|Exercise Level-VI(Single answer)|25 Videos
ELECTROSTATICS AND GAUSS LAW
NARAYNA|Exercise Intergers type question|11 Videos
MAGNETISM
NARAYNA|Exercise LEVEL-II (H.W)|24 Videos

With resistances P and Q in the left gap and right gap respectively, of a meter bridge. The null point is obtained at 60 cm from left end. When Q is increased by 10Omega the null point is obtained at 40 cm from left end. The values of P and Q are

`12Omega, 8Omega`

`8Omega, 12Omega`

`8Omega, 8Omega`

`12Omega, 12 Omega`

The length of a potentiometer wire is 600 cm and it carries a current of 40 mA . For a cell of emf 2V and internal resistance 10Omega , the null point is found to be at 500 cm . On connecting a voltmeter acros the cell, the balancing length is decreased by 10 cm The resistance of the voltmeter is

`500Omega`

`290Omega`

`490Omega`

`20Omega`

The length of a potentiometer wire is 600 cm and it carries a current of 40 mA . For a cell of emf 2V and internal resistance 10Omega , the null point is found to be ast 500 cm . On connecting a voltmeter acros the cell, the balancing length is decreased by 10 cm The voltmeter reading will be

`1.96V`

`1.8V`

`1.64V`

`0.96V`