
In a meter bridge, the balance length from left end (standard resistance of $1Ω$ is in the right gap) is found to be 20 cm, the length of resistance wire in left gap) is found to be ${20\,cm}$ the length of resistance wire in left gap is $\dfrac{1}{2}$ m and radius is ${2\,mm}$ its specific resistance is
A. $\pi \times {10^{ - 6}}{ohm\,m}$
B. $2\pi \times {10^{ - 6}}{ohm\,m}$
C. $\dfrac{\pi }{2} \times {10^{ - 6}}{ohm\,m}$
D. $3\pi \times {10^{ - 6}}{ohm\,m}$
Answer
561k+ views
Hint:In order to solve the above question, we need to understand the concept of Meter Bridge. We will use the formula of balancing length to calculate the specific resistance.
Complete step by step answer:
From given data:
$R=1Ω$, $X={20\,cm}$ and the remaining length is ${80\,cm}$.Let $s$ be the value of specific resistance.
$X = \dfrac{{sl}}{A}$…….. (1)
$\dfrac{X}{R} = \dfrac{{20}}{{80}} \\$
Substituting the value of (1) in above equation
$
\dfrac{X}{R} = \dfrac{1}{4} \\
\Rightarrow\dfrac{1}{4} = \dfrac{{sl}}{{RA}} \\
\Rightarrow\dfrac{1}{4} = \dfrac{{s \times \dfrac{1}{2}}}{{\pi \times {{(2 \times {{10}^{ - 3}})}^2}}} \\
\Rightarrow\dfrac{1}{4} = \dfrac{s}{{2\pi \times {{(2 \times {{10}^{ - 3}})}^2}}} \\
\therefore s = 2\pi \times {10^{ - 6}}\Omega m$
Hence the correct option is B.
Additional information:
Meter Bridge instrument consists of a one meter standard resistance wire of uniform cross-sectional area fixed between two ends of metallic strip. There are two gaps, one on the right side and the other on the left side, we have two resistors, one of which is known resistance and the other is unknown resistance which are attached. When the jockey is moved on the wire until the galvanometer shows null reading. Certain positions on the wire show null deflection then reading is obtained.
Note:The instrument meter bridge also called by another name that is sliding Wire Bridge and it is a device or an instrument which works on the principle of a Wheatstone bridge. Usually we find the unknown resistance of a conductor by using the Wheatstone bridge.
Complete step by step answer:
From given data:
$R=1Ω$, $X={20\,cm}$ and the remaining length is ${80\,cm}$.Let $s$ be the value of specific resistance.
$X = \dfrac{{sl}}{A}$…….. (1)
$\dfrac{X}{R} = \dfrac{{20}}{{80}} \\$
Substituting the value of (1) in above equation
$
\dfrac{X}{R} = \dfrac{1}{4} \\
\Rightarrow\dfrac{1}{4} = \dfrac{{sl}}{{RA}} \\
\Rightarrow\dfrac{1}{4} = \dfrac{{s \times \dfrac{1}{2}}}{{\pi \times {{(2 \times {{10}^{ - 3}})}^2}}} \\
\Rightarrow\dfrac{1}{4} = \dfrac{s}{{2\pi \times {{(2 \times {{10}^{ - 3}})}^2}}} \\
\therefore s = 2\pi \times {10^{ - 6}}\Omega m$
Hence the correct option is B.
Additional information:
Meter Bridge instrument consists of a one meter standard resistance wire of uniform cross-sectional area fixed between two ends of metallic strip. There are two gaps, one on the right side and the other on the left side, we have two resistors, one of which is known resistance and the other is unknown resistance which are attached. When the jockey is moved on the wire until the galvanometer shows null reading. Certain positions on the wire show null deflection then reading is obtained.
Note:The instrument meter bridge also called by another name that is sliding Wire Bridge and it is a device or an instrument which works on the principle of a Wheatstone bridge. Usually we find the unknown resistance of a conductor by using the Wheatstone bridge.
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