The potentiometer wire of length 100cm and resistance 9- is joined to a cell of e-m-f 10V and internal resistance 1- Another cell of e-m-f 5V and internal resistance 2- is connected and shown in Fig- The galvanometer G shows no deflection when the length AC is

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Magnetic Field Due to a Circular Arc at the Centre

The potentiom...

Question

The potentiometer wire of length $100 c m$ and resistance $9 Ω$ is joined to a cell of e.m.f $10 V$ and internal resistance $1 Ω$ . Another cell of e.m.f $5 V$ and internal resistance $2 Ω$ is connected and shown in Fig. The galvanometer $G$ shows no deflection when the length AC is

52.52 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

53.56 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

54.2 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

55.55 c m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $55.55 c m$
First of all we should find potential gradient
Current in potentiometer $= \frac{10}{9 + 1} = 1 A$
Potential difference across potentiometer $= i R = 1 \times 9 = 9 V$
potential $C_{1}$ radient $= k = \frac{V}{L} = \frac{9}{100} \times 100 9 \frac{V}{m}$
$E M F m = K l (l = b a l a n c i n g l e n g t h)$
$5 = K l$
$l = \frac{5}{9} m \frac{5}{9} \times 100 l = \frac{500}{9}$
$A C = 55.55 c m$

Suggest Corrections

Similar questions

Q. In the fig. the potentiometer wire

A B

of length

L

& resistance

9 r

is joined to the cell

D

of e.m.f.

ε

and internal resistance

r

. The cell

C s

e.m.f. is

ε / 2

and its internal resistance is

2 r

. The galvanometer

G

will show no deflection then find length

A J

Q. A potentiometer wire

A B

having length

L

and resistance

12 r

is joined to a cell

D

of emf

ϵ

and internal resistance

r

. A cell

C

having emf

ϵ / 2

and internal resistance

3 r

is connected. The length

A J

at which the galvanometer as shown in fig. shows no deflection is

Q. In the figure, the potentiometer wire

A B

of length

L

and resistnace

9 r

is joined to the cell

D

of e.m.f and internal resistance

r

. The cell

C

's emf is

ε / 2

and its internal resistance is

2 r

.
The galvanometer

G

will show no deflection when the length

A J

Q. In the figure, the potentiometer wire of length 1000 cm and resistance 9 ohm is joined to a cell of emf 2V and internal resistance of 1 ohm. The galvanometer G will show null deflection when the length AC is

Q. In the potentiometer circuit shown in figure , the internal resistance of the 12V battery is

1 Ω

and length of the wire AB is 100 cm. when CB is 60 cm. the galvanometer show no deflection. the emf of cell E is (the resistance of wire AB is

3 Ω

)

Join BYJU'S Learning Program

The potentiometer wire of length 100cm and resistance 9Ω is joined to a cell of e.m.f 10V and internal resistance 1Ω. Another cell of e.m.f 5V and internal resistance 2Ω is connected and shown in Fig. The galvanometer G shows no deflection when the length AC is

The correct option is D 55.55cmFirst of all we should find potential gradientCurrent in potentiometer =109+1=1APotential difference across potentiometer =iR=1×9=9Vpotential C1 radient =k=VL=9100×1009VmEMFm=Kl(l=balancing length)5=Kll=59m59×100l=5009AC=55.55cm

The potentiometer wire of length $100 c m$ and resistance $9 Ω$ is joined to a cell of e.m.f $10 V$ and internal resistance $1 Ω$ . Another cell of e.m.f $5 V$ and internal resistance $2 Ω$ is connected and shown in Fig. The galvanometer $G$ shows no deflection when the length AC is