JEE Main 2021

LEVELJEE Main

Animated Solution for Physics - Current Electricity: Consider a 72 cm long wire $A B$ as shown in the figure. The galvanometer jockey is placed at $P$ on $A B$ at a distance $x$ cm from $A$ . The galvanometer shows zero deflection. The value of $x$ , to the nearest integer, is ......... .

Enter Numerical Value:

Visualized Solution

Meter Bridge Setup

Meter bridge setup with wire $A B$ of length $72 cm$ .

Balanced Condition

Balanced Wheatstone Bridge Condition:
$\frac{R _{1}}{R _{2}} = \frac{R _{A P}}{R _{P B}}$

Substitution

$\frac{12}{6} = \frac{x}{72 - x}$

Simplification

$2 = \frac{x}{72 - x}$

Cross Multiplication

$144 - 2 x = x$

Solving for $x$

$3 x = 144$
$x = 48 cm$

Final Answer

Final Answer: $48$

Variations

Consider variations:
1. Non-uniform wire.
2. Shunting one of the resistors.

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The Sigma Insight: Electrical Instruments

Solution Diagram

Analyzing the Setup

Imagine you are in the lab, looking at a classic meter bridge setup. We have a uniform wire

A B

that is

72 cm

long.

Connected to the upper arms of this circuit are two resistors: a

12 Ω

resistor and a

6 Ω

resistor. A galvanometer is connected from the junction of these two resistors to a jockey that slides along the wire

A B

The problem states a crucial piece of information: the galvanometer shows zero deflection when the jockey is at point

P

, which is at a distance

x

from point

A

The Master Equation

When the galvanometer shows zero deflection, it means no current is flowing through that middle branch. This is the hallmark of a balanced Wheatstone bridge.

In a balanced state, the ratio of the resistances in the upper arms must perfectly match the ratio of the resistances in the lower arms (the two segments of the wire).

Since the wire is uniform, its resistance is directly proportional to its length (

R \propto L

). Therefore, the resistance of segment

A P

is proportional to

x

, and the resistance of segment

P B

is proportional to the remaining length,

(72 - x)

We can set up our master equation:

\frac{12}{6} = \frac{x}{72 - x}

Final Calculation

Now, it's just a matter of simple algebra. Let's simplify the left side of our equation. Twelve divided by six is exactly two.

2 = \frac{x}{72 - x}

Next, we cross-multiply to get rid of the fraction. Multiplying

2

(72 - x)

gives us:

144 - 2 x = x

Let's group the

x

terms together by adding

2 x

to both sides:

144 = 3 x

Finally, dividing by

3

, we find the value of

x

x = 48 cm

The jockey must be placed exactly

48 cm

from point

A

to achieve the balance point. The nearest integer is 48.

Common Traps & FAQs

Why is the resistance of the wire proportional to its length?

For a wire of uniform cross-section and material, resistance

R = ρ \frac{L}{A}

. Since resistivity

ρ

and area

A

are constant, the resistance

R

is directly proportional to the length

L

What happens if the battery and galvanometer are swapped?

The balance point remains completely unchanged. This is a fundamental and symmetrical property of the Wheatstone bridge.

Why do we completely ignore the resistance of the galvanometer?

Because no current flows through it in the balanced condition. Since current is zero, there is no potential drop across it, and its resistance doesn't affect the voltage distribution in the rest of the circuit.

What if the wire gets heated up during the experiment?

Heating changes the resistance. However, if the wire is uniform, the resistance of both segments changes by the exact same factor, so the ratio might still hold. Practically, though, uneven heating can cause slight errors.

Can we use this exact setup to measure very high resistances?

No, a meter bridge is most accurate for moderate resistances. For very high or very low resistances, the sensitivity of the bridge decreases significantly, making it hard to find an accurate null point.

Similar Questions

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A potentiometer wire $A B$ having length $L$ and resistance $12 r$ is joined to a cell $D$ of EMF $ε$ and internal resistance $r$ . A cell $C$ having emf $\frac{ε}{2}$ and internal resistance $3 r$ is connected. The length $A J$ at which the galvanometer as shown in figure shows no deflection is

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(A)

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A thin uniform wire $A B$ of length $1 m$ , an unknown resistance $X$ and a resistance of $12 Ω$ are connected by thick conducting strips, as shown in the figure. A battery and galvanometer (with a sliding jockey connected to it are also available). Connections are to be made to measure the unknown resistance $X$ using the principle of Wheatstone bridge. Answer the following questions. (a) Are there positive and negative terminals on the galvanometer? (b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points. (c) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of $60 cm$ from $A$ . Obtain the value of the resistance $X$ .

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During an experiment with a meter bridge, the galvanometer shows a null point when the jockey is pressed at $40.0 cm$ using a standard resistance of $90 Ω$ , as shown in the scale used in the meter bridge is $1 mm$ . The unknown resistance is

(A)

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In the circuit shown, a four-wire potentiometer is made of a $400 cm$ long wire, which extends between $A$ and $B$ . The resistance per unit length of the potentiometer wire is $r = 0.01 Ω/ cm$ . If an ideal voltmeter is connected as shown with jockey $J$ at $50 cm$ from end $A$ , the expected reading of the voltmeter will be

The length of a potentiometer wire is $1200 cm$ and it carries a current of $60 mA$ . For a cell of emf $5 V$ and internal resistance of $20 Ω$ , the null point on it is found to be at $1000 cm$ . The resistance of whole wire is

(A)

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(B)

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In the given circuit of potentiometer, the potential difference $E$ across $A B$ (10 m length) is larger than $E_{1}$ and $E_{2}$ as well. For key $K_{1}$ (closed), the jockey is adjusted to touch the wire at point $J_{1}$ , so that there is no deflection in the galvanometer. Now, the first battery ( $E_{1}$ ) is replaced by second battery ( $E_{2}$ ) for working by making $K_{1}$ open and $K_{2}$ closed. The galvanometer gives then null deflection at $J_{2}$ . The value of $\frac{E _{1}}{E _{2}}$ is $\frac{a}{b}$ , where $a = \dots\dots$ .

JEE Main 2011

LEVELJEE Main

A meter bridge is set-up as shown in figure, to determine an unknown resistance $X$ using a standard $10 Ω$ resistor. The galvanometer shows null point when tapping-key is at $52 cm$ mark. The end-corrections are $1 cm$ and $2 cm$ respectively for the ends $A$ and $B$ . The determined value of $X$ is

(A)

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(B)

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(D)

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Enter Numerical Value:

Visualized Solution

Meter Bridge Setup

Balanced Condition

Substitution

Simplification

Cross Multiplication

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Final Answer

Variations

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

Solving for $x$