In an aluminium (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are $2.7 \times 10^{- 8} Ω m$ and $1.0 \times 10^{- 7} Ω m$ , respectively. The electrical resistance between the two faces P and Q of the composite bar is [2015]

Question

Smart Achievers · Accepted Answer

(2)

As resistance of a wire, $R = \int ρ \frac{l}{A}$

For iron,

$R_{Fe} = \frac{ρ_{Fe} \times l_{Fe}}{A_{Fe}} = \frac{10^{- 7} \times 50 \times 10^{- 3}}{4 \times 10^{- 6}}$ $= \frac{25}{2} \times 10^{- 4} Ω$

For aluminium,

$R_{Al} = \frac{ρ_{Al} \times l_{Al}}{A_{Al}} = \frac{2.7 \times 10^{- 8} \times 50 \times 10^{- 3}}{(49 - 4) \times 10^{- 6}}$ $= \frac{2.7 \times 50}{45} \times 10^{- 5}$

$= 0.3 \times 10^{- 4} Ω$

As the potential difference across both resistors is the same, they are in parallel combination.

Therefore,

$R_{P Q} = \frac{R_{Fe} \times R_{Al}}{R_{Fe} + R_{Al}}$ $= \frac{12.5 \times 10^{- 4} \times 0.3 \times 10^{- 4}}{12.8 \times 10^{- 4}}$

$= \frac{1875}{64} μ Ω$

Solution

1 $\frac{2475}{64} μ Ω$

2 $\frac{1875}{64} μ Ω$

3 $\frac{1875}{49} μ Ω$

4 $\frac{2475}{132} μ Ω$

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Solution

1 247564 μΩ

2 187564 μΩ

3 187549 μΩ

4 2475132 μΩ

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1 $\frac{2475}{64} μ Ω$

2 $\frac{1875}{64} μ Ω$

3 $\frac{1875}{49} μ Ω$

4 $\frac{2475}{132} μ Ω$