Question

In the diagram, balance length for E' is lo. If cross sectional area of wire AB is doubled then what will be balancing length.

(1) 1.5 lo
(2) 2 lo
(3) 2.5 lo
(4) 0.5 lo​

Answer 1

Answer:

The balancing length is $1.5l_{0}$

Explanation:

Resistance is a unit of measurement for the restriction of current flow in an electrical circuit.

The balance length E' for is $l_{0}$.

The cable is 10 meters long.

Two resistance 1Ω and 9Ω are connected in series with E.

The cross-sectional area of the wire AB is doubled.

To Find- The balancing length.

Now, we know that the resistance is given by

R = ρl/A

where ρ is the resistivity, l is the length and A is the Area.

The resistance of the wire AB, $R_{p} =$ 10 Ω.

When the resistance is doubled, then

A' = 2A

Therefore, Half of the previous resistance is the new resistance.

$R_{p} '$ = $\frac{R_{p} }{2}$

$R_{p} ' = \frac{10}{2} = 5$

Now, for the battery E',

$E ' =\frac{ER_{p} }{R_{p} +r_{s} +R_{S} } *\frac{l}{L}$

Putting all the values, we get

$E ' =\frac{E(5) }{5 +1+10} *\frac{L_{0} }{10}$

$E' = \frac{E*l_{0} }{20}$   Equation (1)

When the area becomes doubled, then

$E' = \frac{E*L'}{30}$  Equation (2)

Equating (1) and (2), we get

$\frac{E*l_{0} }{20} = \frac{E*L'}{30}\\l' = \frac{3l_{0} }{2}$

Therefore, The balancing length is $1.5l_{0}$

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