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Byju's Answer

Standard XIII

Physics

Wheatstone Bridge

In the metre ...

Question

In the metre bridge arrangement, an unknown resistance $R_{1}$ is connected in series with resistance of $10 Ω$ . When the combination is connecetd to left gap of the meter bridge and $R_{2}$ is connected to right gap, balance point from left end is at $50 cm$ . When $10 Ω$ resistor is removed, the balance point shifts to $40 cm$ . The value of $R_{1}$ is

20 Ω

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10 Ω

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60 Ω

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40 Ω

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Solution

The correct option is A $20 Ω$
$\frac{R_{1} + 10}{R_{2}} = \frac{50}{50}$

$\Rightarrow R_{1} + 10 = R_{2} . . . (1)$

Now, when $10 Ω$ is removed, then

$\frac{R_{1}}{R_{2}} = \frac{40}{60}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{2}{3}$

$\Rightarrow R_{2} = \frac{3}{2} R_{1} . . . (2)$

Put $(2)$ in $(1)$ , we get

$R_{1} + 10 = \frac{3}{2} R_{1}$

$\Rightarrow 2 R_{1} + 20 = 3 R_{1}$

$∴ R_{1} = 20 Ω$

Hence, the correct answer is (A).

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The correct option is A 20 ΩR1+10R2=5050 ⇒R1+10=R2 ...(1) Now, when 10 Ω is removed, then R1R2=4060 ⇒R1R2=23 ⇒R2=32R1 ...(2) Put (2) in (1), we get R1+10=32R1 ⇒2R1+20=3R1 ∴R1=20 Ω Hence, the correct answer is (A).