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

Byju's Answer

Standard XI

Physics

Meter Bridge

In fig., AB i...

Question

In fig., $A B$ is a potentiometer wire of length $10 m$ and resistance $2 Ω$ . With key $(K)$ open the balancing length is $5.5 m$ . However, on closing key $(K)$ the balancing length reduces to $5 m$ . The internal resistance of the cell $(E_{1})$ is –

0.01 Ω

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

0.1 Ω

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

0.2 Ω

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

1 Ω

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

Open in App

Solution

The correct option is B $0.1 Ω$
When $K$ is open we are balancing the emf $E_{1}$ , hence $E_{1} = \frac{V}{L} \times 5.5 = \frac{2}{10} \times 5.5....... (1)$
When $K$ is closed we are balancing the potential across the discharging battery which is $\frac{r}{r + r_{1}} E_{1}$ where $r_{1}$ is the internal resistance of the battery.
Hence,
$\frac{E_{1}}{r + r_{1}} = \frac{2}{10} \times 5..... (2)$
On dividing $(1)$ and $(2)$ , we get
$r_{1} + 1 = 1.1 \Rightarrow r_{1} = 0.1 Ω$

Suggest Corrections

Similar questions

Q. In fig.,

A B

is a potentiometer wire of length

10 m

and resistance

2 Ω

. With key

(K)

open the balancing length is

5.5 m

. However, on closing key

(K)

the balancing length reduces to

5 m

. The internal resistance of the cell

(E_{1})

is –

Q. In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of

2 m

, when the cell is shunted by a

5 Ω

resistance and at a length of

3 m

, when the cell is shunted by a

10 Ω

resistance. The internal resistance of the cell is

Q. The internal resistance of a cell is to be determined using a potentiometer. In an experiment, an external resistance of

R

=

60 Ω

is used across the given cell. When the key

K_{1}

is closed and

K_{2}

is open, the balance point is found at a length of

72

cm

. The balance length on the potentiometer is found at

60

cm

when both the keys are closed. Calculate the internal resistance of the cell.

For the arrangement of the potentiometer shown in the figure, the balance point is obtained at a distance 75cm from A when the key k is open. The second balance point is obtained at 60 cm from A when the key k is closed. Find the internal resistance (in $Ω$ ) of the battery $E_{1}$ . ___

Q. A cell balances against a length of

200 c m

on a potentiometer wire, when it is shunted by a resistance of

8 Ω

. The balancing length reduces by

40 c m

, when it is shunted by a resistance of

4 Ω

. Calculate the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.

Join BYJU'S Learning Program

In fig., AB is a potentiometer wire of length 10 m and resistance 2 Ω. With key (K) open the balancing length is 5.5 m. However, on closing key (K) the balancing length reduces to 5 m. The internal resistance of the cell (E1) is –

In fig., $A B$ is a potentiometer wire of length $10 m$ and resistance $2 Ω$ . With key $(K)$ open the balancing length is $5.5 m$ . However, on closing key $(K)$ the balancing length reduces to $5 m$ . The internal resistance of the cell $(E_{1})$ is –