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Question

A meter bridge with two resistances $R_1$ and $R_2$ as shown in figure was balanced (null point) at 40 cm from the point $P$. The null point changed to 50 cm from the point $P$, when $16 \ \Omega$ resistance is connected in parallel to $R_2$. The values of resistances $R_1$ and $R_2$ are _________.

The correct answer is
$R_2 = 16 \ \Omega, R_1 = \frac{16}{3} \ \Omega$

To find the values of resistances \( R_1 \) and \( R_2 \), we will use the principle of a meter bridge, which is a practical application of Wheatstone bridge. The principle states that at balance (null point), the ratio of resistances is equal to the ratio of the lengths of the bridge wire:

Initially, when the null point is at 40 cm from point \( P \):

\[ \frac{R_1}{R_2} = \frac{40}{60} \]

When a 16 Ω resistor is connected in parallel with \( R_2 \), the effective resistance becomes:

\[ R_{2 \text{ eff}} = \frac{R_2 \cdot 16}{R_2 + 16} \]

Now, the null point changes to 50 cm from \( P \):

\[ \frac{R_1}{R_{2 \text{ eff}}} = \frac{50}{50} = 1 \]

Using the equation, we get:

\[ R_1 = R_{2 \text{ eff}} \] \[ R_1 = \frac{R_2 \cdot 16}{R_2 + 16} \]

Substitute the value of \( R_1 \) from the first condition:

\[ \frac{R_1}{R_2} = \frac{2}{3} \implies R_1 = \frac{2}{3} R_2 \]

Equate both expressions for \( R_1 \):

\[ \frac{2}{3} R_2 = \frac{R_2 \cdot 16}{R_2 + 16} \]

Cross-multiplying gives:

\[ 2R_2 (R_2 + 16) = 3 \cdot 16 \cdot R_2 \]

Solving it:

\[ 2R_2^2 + 32R_2 = 48R_2 \] \[ 2R_2^2 = 16R_2 \] \[ R_2^2 = 8R_2 \] \[ R_2 = 8 \, \text{or} \, R_2 = 0 \]

Since a zero resistance is not practical, \( R_2 = 8 \, \Omega \). Plugging back to find \( R_1 \):

\[ R_1 = \frac{2}{3} R_2 = \frac{2}{3} \times 8 = \frac{16}{3} \, \Omega \]

Therefore, the resistances are \( R_2 = 8 \, \Omega \) and \( R_1 = \frac{16}{3} \, \Omega \).

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Important Questions from Electricity and Magnetism

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    then the refractive index of the medium is ________.
    (All given measurement are in SI units)
  4. Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by $15 \text{ cm}$ length of wire $Q$ is________.

    $(\mu_o = 4\pi \times 10^{-7} \text{ T.m/A})$

  5. For the two cells having same EMF $E$ and internal resistance $r$, the current passing through the external resistor $6\text{ }\Omega$ is same when both the cells are connected either in parallel or in series. The value of internal resistance $r$ is ________$\Omega$.
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