The sensitivity of the Wheatstone bridge is defined as the change in the output (balancing) voltage $E$ per unit change in the measured resistance $R$ (i.e., $\frac{dE}{dR}$).

1. Define Parameters and Output Voltage

The bridge output voltage $E$ is the potential difference between the midpoints of the two arms, assuming the meter is disconnected (open circuit voltage $V_{oc}$).

Let $V_L$ be the voltage at the left midpoint (between $R$ and $S$), and $V_R$ be the voltage at the right midpoint (between $P$ and $Q$). $E = V_L - V_R$.

The source voltage is $V = 10 \text{ V}$. The bottom rail is the reference (ground).

• $R = 50 \ \Omega$
• $S = 5 \text{ k}\Omega = 5000 \ \Omega$
• $P = 10 \ \Omega$
• $Q = 1 \text{ k}\Omega = 1000 \ \Omega$

The output voltage $E$ is:

$$E = V \left( \frac{S}{R + S} - \frac{Q}{P + Q} \right)$$

In this problem, $R$ is the variable resistor, and the sensitivity is calculated with respect to $R$.

$$E = V \left[ \frac{S}{R + S} - \frac{Q}{P + Q} \right]$$

2. Calculate the Sensitivity ($\frac{dE}{dR}$)

Since only $R$ is the variable, the term $\frac{Q}{P + Q}$ is a constant. We differentiate $E$ with respect to $R$:

$$\frac{dE}{dR} = \frac{d}{dR} \left( V \frac{S}{R + S} \right) - \frac{d}{dR} \left( V \frac{Q}{P + Q} \right)$$

The second term is zero. We use the chain rule on the first term:

$$\frac{d}{dR} \left( \frac{S}{R + S} \right) = S \cdot (-1) (R + S)^{-2} \cdot 1 = - \frac{S}{(R + S)^2}$$

$$\frac{dE}{dR} = - V \frac{S}{(R + S)^2}$$

3. Substitute Numerical Values

The initial state values are used for calculation:

• $V = 10 \text{ V}$
• $R = 50 \ \Omega$
• $S = 5000 \ \Omega$

$$\frac{dE}{dR} = - 10 \text{ V} \cdot \frac{5000 \ \Omega}{(50 \ \Omega + 5000 \ \Omega)^2}$$ $$\frac{dE}{dR} = - 10 \cdot \frac{5000}{(5050)^2}$$ $$\frac{dE}{dR} = - \frac{50000}{25,502,500}$$ $$\frac{dE}{dR} \approx -0.0019606 \text{ V}/\Omega$$

The question asks for the sensitivity in terms of magnitude ($\text{mV}/\Omega$).

$$\left| \frac{dE}{dR} \right| = 0.0019606 \text{ V}/\Omega$$

Convert to $\text{mV}/\Omega$ ($1 \text{ V} = 1000 \text{ mV}$):

$$\text{Sensitivity} = 0.0019606 \times 1000 \text{ mV}/\Omega \approx 1.9606 \text{ mV}/\Omega$$

4. Final Formatting

Rounding off to two decimal places:

$$\text{Sensitivity} = 1.96 \text{ mV}/\Omega$$

This result falls within the constraint range [1.94, 2].

The sensitivity of the bridge is 1.96 $\text{mV}/\Omega$.