1. **Problem Statement:** We have a tapered cylindrical conductor of length $\ell$ with radius varying linearly from $a$ at the left end to $b$ at the right end. It is connected across a battery of emf $V$ and has resistivity $\rho$ and resistance $R$. We want to find the electric field $E$ at a point $P$ located at a distance $x$ from the left end. 2. **Understanding the Geometry and Resistance:** The radius at distance $x$ is given by linear interpolation: $$r(x) = a + \frac{b - a}{\ell} x$$ The cross-sectional area at $x$ is: $$A(x) = \pi r(x)^2 = \pi \left(a + \frac{b - a}{\ell} x\right)^2$$ 3. **Differential Resistance:** A small element of length $dx$ at position $x$ has resistance: $$dR = \frac{\rho \, dx}{A(x)} = \frac{\rho \, dx}{\pi \left(a + \frac{b - a}{\ell} x\right)^2}$$ 4. **Total Resistance:** Integrate $dR$ from $x=0$ to $x=\ell$ to get total resistance $R$: $$R = \int_0^{\ell} \frac{\rho}{\pi \left(a + \frac{b - a}{\ell} x\right)^2} dx$$ Let $u = a + \frac{b - a}{\ell} x$, then $du = \frac{b - a}{\ell} dx$ or $dx = \frac{\ell}{b - a} du$. Limits change from $x=0 \to u=a$ and $x=\ell \to u=b$. So, $$R = \int_a^b \frac{\rho}{\pi u^2} \cdot \frac{\ell}{b - a} du = \frac{\rho \ell}{\pi (b - a)} \int_a^b u^{-2} du$$ Calculate the integral: $$\int_a^b u^{-2} du = \left[-\frac{1}{u}\right]_a^b = \frac{1}{a} - \frac{1}{b}$$ Therefore, $$R = \frac{\rho \ell}{\pi (b - a)} \left(\frac{1}{a} - \frac{1}{b}\right) = \frac{\rho \ell}{\pi a b}$$ (This is a known formula for tapered conductor resistance.) 5. **Current and Electric Field:** The current $I$ through the conductor is: $$I = \frac{V}{R}$$ The electric field $E$ at position $x$ is related to the voltage drop per unit length: $$E = \frac{dV}{dx}$$ Since $dV = I dR = I \frac{\rho dx}{A(x)}$, we have: $$E = \frac{dV}{dx} = I \frac{\rho}{A(x)} = \frac{V}{R} \cdot \frac{\rho}{\pi \left(a + \frac{b - a}{\ell} x\right)^2}$$ 6. **Substitute $R$ and simplify:** Recall: $$R = \frac{\rho \ell}{\pi a b}$$ So, $$E = V \cdot \frac{1}{R} \cdot \frac{\rho}{\pi \left(a + \frac{b - a}{\ell} x\right)^2} = V \cdot \frac{\pi a b}{\rho \ell} \cdot \frac{\rho}{\pi \left(a + \frac{b - a}{\ell} x\right)^2} = \frac{V a b}{\ell \left(a + \frac{b - a}{\ell} x\right)^2}$$ Multiply numerator and denominator inside the square by $\ell$: $$E = \frac{V a b \ell^2}{\ell \left(\ell a + (b - a) x\right)^2} = \frac{V a b \ell}{\left(\ell a + (b - a) x\right)^2}$$ 7. **Final Expression:** The electric field at point $P$ is: $$\boxed{E = \frac{V a b \ell}{\left(\ell a + (b - a) x\right)^2}}$$ **Note:** None of the given options exactly match this expression, so the correct choice is (4) None of these. --- **Summary:** The electric field at distance $x$ along the tapered conductor is $$E = \frac{V a b \ell}{(\ell a + (b - a) x)^2}$$ where $a$ and $b$ are the radii at the ends, $\ell$ is the length, and $V$ is the emf.