Question

In a metre bridge experiment , resistance box with 2 Ω is connected in the left gap and the unknown resistance S in the right gap. If the balancing length be 40 cm, value of S will be *
(a) 2 Ω
(b) 3 Ω
(c) 4 Ω
(d) 5 Ω​

Answer 1

$\large{\underline{\boxed{\boxed{\sf Let's \: Understand \: Concept \: F1^{st}}}}}$

Here, we have given that in a metre bridge experiment resistance box with 2Ω is connected in the left group and the unknown resistance 'S' in the right gap and the balancing length is 40cm then what will be value of S. Thus, by using balanced wheatstone bridge condition we, will obtain the value of unknown resistance 'S'.

$\huge{\underline{\boxed{\sf AnSwer}}}$

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$\large{\underline{\sf Given:-}}$

▶Left gap resistance = 2Ω

▶ Balancing Length of Bridge Wire = 40cm

$\large{\underline{\sf Find:-}}$

⇨Right gap resistance

$\large{\underline{\sf Formula \: Used:-}}$

$\huge{\underline{\boxed{\sf \dfrac{P}{S} = \dfrac{l}{100-l}}}}$

$\large{\underline{\sf Solution:-}}$

Here, using balanced Wheatstone bridge condition:-

$\mapsto\sf \dfrac{P}{S} = \dfrac{l}{100-l}$

$\purple{\sf where} \small{\begin{cases}\pink{\sf P = 2 \Omega} \\ \green{\sf l = 40cm} \end{cases}}$

$\dashrightarrow\sf \dfrac{P}{S} = \dfrac{l}{100-l}$

$\dashrightarrow\sf \dfrac{2}{S} = \dfrac{40}{100-40}$

$\dashrightarrow\sf \dfrac{2}{S} = \dfrac{40}{60}$

$\dashrightarrow\sf \dfrac{2}{S} = \dfrac{4}{6}$

$\dashrightarrow\sf \dfrac{2}{S} = \dfrac{2}{3}$

$\dashrightarrow\sf S \times 2 = 3 \times 2 \qquad \big \lgroup Cross-multiplication \big\rgroup$

$\dashrightarrow\sf 2S= 6$

$\dashrightarrow\sf S= \dfrac{6}{2}$

$\dashrightarrow\sf S= 3 \Omega$

$\therefore\sf S= 3 \Omega$

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Hence, value of unknown resistance in the right gap 'S' is 3Ω.

Therefore, Option (B) 3Ω is right answer.

Answer 2

Given:

In a metre bridge experiment , resistance box with 2 Ω is connected in the left gap and the unknown resistance S in the right gap. Balancing length is 40 cm.

To find:

Value of S ?

Calculation:

• We know that for a BALANCED WHEATSTONE BRIDGE:

$\boxed{ \dfrac{R}{S} = \dfrac{L}{100 - L} }$

• Here, R is known resistance, S is unknown resistance and L is balancing length.

$\implies \dfrac{2}{S} = \dfrac{40}{100 - 40}$

$\implies \dfrac{2}{S} = \dfrac{40}{60}$

$\implies \dfrac{2}{S} = \dfrac{2}{3}$

$\implies 2S = 6$

$\implies S = 3 \: ohm$

Value of S is 3 ohm (i.e. OPTION b) ✔️