Question

in potentiometer of the cell emf 1.5v balances at the length of 270cm.if another cell balances at 360cm for the same current it's emf will be..​

Answer 1

Appropriate Question:

• In a potentiometer a cell of emf 1.5V gives a balanced point at 270 cm length of the wire. If the cell is replaced by another cell then the balance point shifts to 360 cm then the emf of second cell is?

Given that:

• Emf of a cell = 1.5V
• Balanced point = 270 cm

When the cell is replaced by another cell,

• Let the emf another cell be x.
• New balanced point = 360 cm

Formula used:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \mathtt{\dfrac{E_2}{E_1} = \dfrac{l_2}{l_1}}}$

Where,

• E₁ = Emf of a cell = 1.5V
• E₂ = Emf another cell = x
• l₁ = Balanced point = 270 cm
• l₂ = New balanced point = 360 cm

Finding the emf another cell:

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \: \: \: {\mathtt{\dfrac{x}{1.5} = \dfrac{360}{270}}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \: \: \: {\mathtt{x = \dfrac{36 \times 1.5}{27}}}\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \: \: \: {\mathtt{x = \dfrac{ \cancel{36} \: \: ^4 \times 1.5}{\cancel{27} \: \: ^3}}}\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \: \: \: {\mathtt{x = \dfrac{4 \times 1.5}{3}}}\\ \\ \: \: \: \: \: \: \rightarrow \: \: \: {\mathtt{x = \dfrac{\cancel{6} \: \:^2 }{\cancel{3}\: \:^1}}}\\ \\ \rightarrow \: \: \: {\mathtt{x =2}}$

Hence,

• The emf of second cell is 2V.

Answer 2

Given :-

In potentiometer of the of emf 1.5V gives a balanced point at 270 cm length of the wire.

To Find :-

If the cell is replaced by another cell then the balance point shifts to 360 cm then the emf of second cell is?

Solution :-

Let the emf be a

emf is proportinal to length

a/1.5 = 360/270

a/15 × 10 = 36/27

10a/15 = 4/3

2a/3 = 4/3

a = 4/3 × 3/2

a = 12/6

a = 2