In a meter bridge, initially the balance point is found at 50 cm. When the resistance in the left gap is increased by 5 ohms, the balance point shifts to 60 cm. Calculate the original resistances in t

In a meter bridge, the balance condition is given by the ratio of resistances being equal to the ratio of the corresponding lengths:

Since the total length of the wire is $100 \text{ cm}$, if $L_1$ is the length from one end, then $L_2 = (100 - L_1) \text{ cm}$.
So, the formula becomes:

Initially, the balance point is at $L_1 = 50 \text{ cm}$.
Substituting this into the formula:

This implies that $R_1 = R_2$. Let's call this original resistance $R$. So, $R_1 = R$ and $R_2 = R$.

The balance point shifts from $50 \text{ cm}$ to $60 \text{ cm}$. This means $L_1$ increased.
For $L_1$ to increase, the ratio $\frac{R_1}{R_2}$ must increase.
If $R_1$ is increased, then $\frac{R_1 + 5}{R_2}$ would be greater than $\frac{R_1}{R_2}$, causing $L_1$ to increase.
If $R_2$ is increased, then $\frac{R_1}{R_2 + 5}$ would be less than $\frac{R_1}{R_2}$, causing $L_1$ to decrease.
Since the balance point shifted from $50 \text{ cm}$ to $60 \text{ cm}$ (an increase in $L_1$), it means the resistance $R_1$ (the one corresponding to $L_1$) was increased by $5\Omega$.

The new resistance in the first gap is $R_1' = R + 5$.
The resistance in the second gap remains $R_2' = R$.
The new balance point is $L_1' = 60 \text{ cm}$.
Substituting these values into the meter bridge formula:

Simplify the ratio $\frac{60}{40}$:

Cross-multiply:

Subtract $2R$ from both sides:

The original resistances were $R_1 = R$ and $R_2 = R$.
Therefore, the original resistances are $10 \, \Omega$ and $10 \, \Omega$.

The original resistances are $\text{10 } \Omega \text{ and 10 } \Omega$.