JEE Meter Bridge Cheatsheet

Cheatsheet Content

### Meter Bridge Cheatsheet #### Principle & Diagram - **Principle:** Based on Wheatstone Bridge. Null deflection in galvanometer indicates balanced bridge. - **Diagram (Conceptual):** ``` A---[R_KNOWN]----B | | [G] [Rx_UNKNOWN] | | C--------------D (1m wire) ^ Balance Point J ``` (A-C is 0cm, D is 100cm. J is the balance point.) #### Key Formulas - **Condition for Balance:** $$\frac{R_{KNOWN}}{l_1} = \frac{R_X}{l_2}$$ - Where $l_1$ is length from one end to balance point, $l_2$ is length from other end to balance point. - Typically, $l_1 = l_{cm}$ and $l_2 = (100 - l_{cm})$ cm. - **Unknown Resistance ($R_X$):** $$\mathbf{R_X = R_{KNOWN} \left( \frac{100 - l}{l} \right)}$$ - $l$ is the balance length from the end connected to $R_{KNOWN}$. - **Resistivity ($\rho$):** $$\rho = \frac{R_X \cdot A}{L} = \frac{R_X \cdot \pi r^2}{L}$$ - $A$: cross-sectional area of the wire, $r$: radius, $L$: length of the wire whose resistance is $R_X$. - **End Corrections:** Add $x$ and $y$ to lengths $l_1$ and $l_2$ respectively for non-ideal contacts. - Corrected balance condition: $$\frac{R_{KNOWN}}{l_1 + x} = \frac{R_X}{l_2 + y}$$ - **Galvanometer Internal Resistance:** Does NOT affect balance point if ideal (infinite resistance at null). If non-ideal, it only affects sensitivity, not the balance condition. #### Units & Conversions - Resistance: Ohms ($\Omega$) - Length: centimeters (cm) for $l_1, l_2$; meters (m) for $L$ in resistivity. - Area: $m^2$ or $cm^2$. Ensure consistency. ($1 m = 100 cm$, $1 cm^2 = 10^{-4} m^2$) #### Approximations & Tips - **Neglecting Wire Resistance:** The resistance of the connecting wires is usually assumed to be negligible compared to $R_{KNOWN}$ and $R_X$. - **Accuracy:** Max accuracy when balance point is near the center (50 cm), as percentage error is minimized. ### Quick Solution Template (6 Steps) 1. **Draw & Label:** Sketch the meter bridge, label $R_{KNOWN}$, $R_X$, and balance lengths $l$ and $(100-l)$. 2. **Identify Knowns/Unknowns:** List given values (resistances, lengths) and what needs to be found. 3. **Write Balance Condition:** Apply the Wheatstone bridge principle: $\frac{R_{KNOWN}}{l} = \frac{R_X}{(100-l)}$. 4. **Substitute Values:** Plug in the given numerical values into the equation. 5. **Solve:** Algebraically solve for the unknown quantity. 6. **Check Units:** Ensure all units are consistent and the final answer has appropriate units. ### Fully Worked Solved Examples #### Example 1: Finding Unknown Resistance (Conceptual) **Question:** In a meter bridge, the balance point is found to be at 40 cm from the zero end when a resistor of 10 $\Omega$ is placed in the left gap. Determine the value of the unknown resistance in the right gap. **Solution:** 1. **Draw & Label:** $R_{KNOWN} = 10 \Omega$ (left gap), $R_X$ (right gap). Balance length $l = 40$ cm from $R_{KNOWN}$ side. So $l_1 = 40$ cm, $l_2 = (100 - 40) = 60$ cm. 2. **Identify Knowns/Unknowns:** $R_{KNOWN} = 10 \Omega$, $l_1 = 40$ cm, $l_2 = 60$ cm. Find $R_X$. 3. **Write Balance Condition:** $\frac{R_{KNOWN}}{l_1} = \frac{R_X}{l_2}$ 4. **Substitute Values:** $\frac{10 \Omega}{40 \text{ cm}} = \frac{R_X}{60 \text{ cm}}$ 5. **Solve:** $R_X = 10 \Omega \times \frac{60 \text{ cm}}{40 \text{ cm}} = 10 \Omega \times 1.5 = 15 \Omega$. 6. **Check Units:** Resistance in $\Omega$. **Answer:** The unknown resistance $R_X = 15 \Omega$. #### Example 2: Resistivity Calculation **Question:** A resistance of 100 $\Omega$ is connected in the left gap of a meter bridge. The balance point is found to be at 30 cm from the left end. The unknown resistance is a wire of length 150 cm and radius 0.1 mm. Calculate the resistivity of the material of the wire. (Assume galvanometer has negligible internal resistance). **Solution:** 1. **Draw & Label:** $R_{KNOWN} = 100 \Omega$ (left gap), $R_X$ (right gap). Balance length $l = 30$ cm from $R_{KNOWN}$ side. So $l_1 = 30$ cm, $l_2 = (100 - 30) = 70$ cm. 2. **Identify Knowns/Unknowns:** $R_{KNOWN} = 100 \Omega$, $l_1 = 30$ cm, $l_2 = 70$ cm. Wire length $L = 150$ cm $= 1.5$ m. Wire radius $r = 0.1$ mm $= 0.1 \times 10^{-3}$ m $= 10^{-4}$ m. Find $\rho$. 3. **Write Balance Condition:** First find $R_X$: $\frac{R_{KNOWN}}{l_1} = \frac{R_X}{l_2}$ 4. **Substitute Values (for $R_X$):** $\frac{100 \Omega}{30 \text{ cm}} = \frac{R_X}{70 \text{ cm}}$ 5. **Solve (for $R_X$):** $R_X = 100 \Omega \times \frac{70 \text{ cm}}{30 \text{ cm}} = \frac{700}{3} \Omega \approx 233.33 \Omega$. Now calculate resistivity $\rho = \frac{R_X \cdot A}{L} = \frac{R_X \cdot \pi r^2}{L}$. $A = \pi (10^{-4} \text{ m})^2 = \pi \times 10^{-8} \text{ m}^2$. $\rho = \frac{(700/3 \Omega) \times (\pi \times 10^{-8} \text{ m}^2)}{1.5 \text{ m}} = \frac{700 \pi}{4.5} \times 10^{-8} \Omega \text{ m} \approx 488.69 \times 10^{-8} \Omega \text{ m} \approx 4.89 \times 10^{-6} \Omega \text{ m}$. 6. **Check Units:** Resistivity in $\Omega \cdot m$. **Answer:** The resistivity of the material of the wire is approximately $4.89 \times 10^{-6} \Omega \cdot m$. ### JEE-Style PYQs with Short Solutions 1. **JEE Main 2021 (Feb/Shift 1) - Unknown Resistance:** **Q:** In a meter bridge experiment, the null point is obtained at 20 cm from the left end when the resistance $S$ is in the right gap and a resistance $R$ is in the left gap. If $S$ is shunted by a resistance of 10 $\Omega$, the null point is found at 30 cm. Find $R$. **Sol:** Initial: $R/20 = S/80 \Rightarrow 4R=S$. Final: $R/30 = (S \parallel 10)/70 \Rightarrow 7R/3 = (10S)/(10+S)$. Substitute $S=4R$: $7R/3 = (40R)/(10+4R) \Rightarrow 7(10+4R) = 120 \Rightarrow 70+28R=120 \Rightarrow 28R=50 \Rightarrow R \approx 1.78 \Omega$. **Difficulty:** ★★★ 2. **JEE Main 2019 (Jan/Shift 2) - End Correction:** **Q:** A meter bridge is set up with resistance $R_1$ in the left gap and $R_2$ in the right gap. The null point is obtained at 40 cm. When $R_1$ is shunted by 10 $\Omega$, the null point shifts to 50 cm. The end corrections are $x$ and $y$ for left and right ends respectively. If $x=1$ cm, find $y$. **Sol:** Initial: $R_1/(40+x) = R_2/(60+y)$. Final: $(R_1 \parallel 10)/(50+x) = R_2/(50+y)$. With $x=1$ cm, initial: $R_1/41 = R_2/(60+y)$. Final: $(10R_1/(10+R_1))/51 = R_2/(50+y)$. Complex algebraic solution. (Often requires specific values for $R_1, R_2$ for a direct solution.) For such problems, if exact $R_1, R_2$ are not given, it implies a ratio. **Difficulty:** ★★★ (Conceptual, requires careful setup) 3. **JEE Main 2017 (Offline) - Balance Point Shift:** **Q:** In a meter bridge, the balancing length from the left end is 20 cm when a standard resistance of 1 $\Omega$ is in the right gap. The unknown resistance is in the left gap. If the unknown resistance is interchanged with the standard resistance, what will be the new balancing length? **Sol:** Initial: $R_X/20 = 1/80 \Rightarrow R_X = 0.25 \Omega$. Interchanged: $1/l' = R_X/(100-l')$. Substitute $R_X$: $1/l' = 0.25/(100-l') \Rightarrow 100-l' = 0.25l' \Rightarrow 100 = 1.25l' \Rightarrow l' = 80$ cm. **Difficulty:** ★★ 4. **JEE Main 2018 (Online/Shift 1) - Series/Parallel:** **Q:** In a meter bridge, with $R$ in the left gap and $S$ in the right gap, the null point is at 40 cm. If a resistance of 10 $\Omega$ is connected in series with $R$, the null point shifts to 50 cm. Find $R$ and $S$. **Sol:** Initial: $R/40 = S/60 \Rightarrow 3R=2S$. Final: $(R+10)/50 = S/50 \Rightarrow R+10=S$. Substitute $S=R+10$ into $3R=2S$: $3R=2(R+10) \Rightarrow 3R=2R+20 \Rightarrow R=20 \Omega$. Then $S=20+10=30 \Omega$. **Difficulty:** ★★ 5. **JEE Main 2023 (April/Shift 1) - Power Dissipation (Indirect):** **Q:** In a meter bridge experiment, a resistance $R$ is connected in the left gap and $S$ in the right gap. The null point is at 60 cm. If $R$ and $S$ are interchanged, the null point shifts by $x$ cm. Find $x$. **Sol:** Initial: $R/60 = S/40 \Rightarrow 2R=3S$. Interchanged: $S/l' = R/(100-l')$. Substitute $S = 2R/3$: $(2R/3)/l' = R/(100-l') \Rightarrow 2/3l' = 1/(100-l') \Rightarrow 200-2l' = 3l' \Rightarrow 200=5l' \Rightarrow l'=40$ cm. Shift $x = |60-40| = 20$ cm. **Difficulty:** ★★ ### Practice Problems (with Answers) 1. A meter bridge is balanced at 30 cm from the left end. If the left gap has a 6 $\Omega$ resistor, what is the unknown resistance in the right gap? **Answer:** $14 \Omega$ 2. The balance point in a meter bridge is found at 50 cm. If a 20 $\Omega$ resistor is in the right gap, what is the resistance in the left gap? **Answer:** $20 \Omega$ 3. In a meter bridge, a wire of length 200 cm and diameter 0.4 mm is in the right gap, and a 5 $\Omega$ resistor is in the left gap. The balance point is at 40 cm. Calculate the resistivity of the wire material. **Answer:** $1.88 \times 10^{-7} \Omega \cdot m$ 4. If the balance point of a meter bridge with an unknown resistor in the left gap and 10 $\Omega$ in the right gap is at 60 cm, what is the unknown resistance? **Answer:** $15 \Omega$ 5. A meter bridge is balanced at 45 cm. If the known resistance is $R_1$ and the unknown is $R_X$, and $R_1$ is doubled, how does the balance point change? **Answer:** Shifts towards the $R_X$ side. New balance point $l' = 60$ cm. ### Common Mistakes & Time-Saving Tips - **Common Mistakes:** - Incorrectly identifying $l_1$ and $l_2$ (always relative to the resistor in that gap). - Forgetting to convert units (e.g., mm to m for resistivity calculations). - Algebraic errors when solving for unknowns, particularly with shunt/series combinations. - Ignoring end corrections if explicitly mentioned. - **Time-Saving Tips:** - Always draw the circuit and label clearly; it avoids confusion. - For balance point shifts, write down the two balance conditions and solve them as simultaneous equations. - Remember $\frac{R_1}{R_2} = \frac{l_1}{l_2}$. If $R_1$ increases, $l_1$ must also increase relative to $l_2$ to maintain balance, and vice-versa. - If the balance point is far from 50 cm, end corrections are more likely to be significant (though usually ignored unless specified). ### One-Line Memory Aids 1. **"Ratio of R's equals ratio of L's."** (Direct reminder of balance condition) 2. **"Null point means no current through galvanometer."** (Core principle) 3. **"Right resistance, right length; Left resistance, left length."** (Helps map $R_X$ to $(100-l)$ or $l$ correctly based on setup)