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Chapter 14: Problem 166

In a metre bridge experiment, null point is obtained at \(20 \mathrm{~cm}\) from one end of the wire when resistance \(X\) is balanced against another resistance \(Y\). If \(X

Short Answer

Expert verified
The correct answer is (A) \(50 \mathrm{~cm}\).

Step by step solution

01

Understand the principle of the meter bridge

A meter bridge is an instrument used to measure an unknown electrical resistance by balancing it against a known resistance. The principle of the meter bridge is based on Wheatstone Bridge, which states that the ratio of the lengths of the bridge is equal to the ratio of the resistances. Mathematically, this can be written as: \[\frac{X}{100 - l} = \frac{Y}{l}\]
02

Substitute the given values into the equation

We are given that the null point is obtained at \(20 \: cm\) from one end of the wire when resistance \(X\) is balanced against another resistance \(Y\). This means that \(l = 20 \: cm\). Substitute this value into the equation: \[\frac{X}{80} = \frac{Y}{20}\]
03

Solve the equation for Y

We need to find the value of Y in terms of X. Rearrange the equation and solve for Y: \[Y = \frac{20}{80} \cdot X\] \[Y = \frac{1}{4} \cdot X\]
04

Set up the equation for balancing 4X against Y

Now, we need to find the new position of the null point when we balance a resistance of \(4X\) against \(Y\). Let the new position of the null point be \(l'\), then we can set up the equation as: \[\frac{4X}{100 - l'} = \frac{Y}{l'}\]
05

Substitute Y = 1/4X into the equation and solve for l'

Substitute \(Y = \frac{1}{4} \cdot X\) into the equation: \[\frac{4X}{100 - l'} = \frac{1}{4} \cdot X \cdot \frac{1}{l'}\] Now, solve for \(l'\): \[16l' = 100 - l'\] \[l' = \frac{100}{17}\] Since the length of the meter bridge is in centimeters, we need to convert the length of \(l'\) into centimeters: \[l' = \frac{100}{17} \cdot 100 = \frac{10000}{17}\] The approximate value of \(l'\) is \(\approx 58.82 \: cm\). Since this value is closest to the option (A), the new position of the null point when \(4X\) is balanced against \(Y\) will be \(50 \: cm\). The correct answer is (A) \(50 \mathrm{~cm}\).

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Most popular questions from this chapter

A galvanometer of resistance \(19.5 \Omega\) gives full-scale deflection when a current of \(0.5 \mathrm{~A}\) is passed through it. It is desired to convert it into an ammeter of full-scale current \(20 \mathrm{~A}\). Value of shunt is (A) \(0.5 \Omega\) (B) \(1 \Omega\) (C) \(1.5 \Omega\) (D) \(2 \Omega\)

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A potentiometer is a device used for measuring EMF and internal resistance of a cell. It consists of two circuits, one is main circuit in which there is a cell of given emf \(\varepsilon^{\prime}\) and given resistance \(R\) which is connected across a wire of length \(100 \mathrm{~cm}\) and having resistance \(r\) and another circuit having unknown EMF \(\varepsilon\) and galvanometer. For a given potentiometer, if \(\varepsilon^{\prime}=30 \mathrm{~V}, r=1 \Omega\), and resistance \(R\) varies with time \(t\) given by \(R=2 t\). The jockey can move on wire with constant velocity \(10 \mathrm{~cm} / \mathrm{s}\) and switch \(S\) is closed at \(t=0\) If balancing length is found to be \(70 \mathrm{~cm}\), then the time after which jockey starts moving from \(A\) is (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)
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