Identify the relationship between Y (electric potential) and X (distance from the surface of an isolated charged sphere) and select the correct graph.

Here are the solutions to the questions you've provided:

Question 48:
Y: Electric potential
X: Distance from the surface of an isolated charged sphere

• Step 1: Recall the formula for electric potential. For an isolated charged sphere with charge $Q$ and radius $R$, the electric potential $V$ at a distance $r$ from its center (for $r \ge R$) is given by:

  $$V = \frac{kQ}{r}$$

  where $k$ is Coulomb's constant.

• Step 2: Relate distance from the surface to distance from the center. Let $X$ be the distance from the surface of the sphere. The distance from the center of the sphere is $r = R + X$.

• Step 3: Substitute into the potential formula.

  $$Y = V = \frac{kQ}{R+X}$$

• Step 4: Analyze the relationship. This equation shows that the electric potential $Y$ is inversely proportional to $(R+X)$. As $X$ increases, $Y$ decreases, approaching zero asymptotically. This is a characteristic inverse curve.

• Step 5: Match with the graphs. Graph A shows a curve that starts at a high positive value (when $X=0$, $Y=kQ/R$) and decreases as $X$ increases, approaching zero. This matches the derived relationship.

• Answer for 48: $\boxed{\text{A}}$

Question 49:
Y: The radius of the orbit of an electron moving at constant speed in a perpendicular uniform magnetic field
X: The momentum of the electron

• Step 1: Identify the forces acting on the electron. When an electron with charge $q$ and mass $m$ moves with speed $v$ in a uniform magnetic field $B$ perpendicular to its velocity, the magnetic force $F_B$ provides the centripetal force $F_c$.

• Step 2: Write the formulas for magnetic and centripetal forces. Magnetic force: $F_B = qvB$ Centripetal force: $F_c = \frac{mv^2}{r}$ where $r$ is the radius of the orbit.

• Step 3: Equate the forces and solve for $r$.

  $$qvB = \frac{mv^2}{r}$$ $$r = \frac{mv}{qB}$$

• Step 4: Relate to momentum. Momentum $p$ is defined as $p = mv$. So, $Y = r = \frac{p}{qB}$.

• Step 5: Analyze the relationship. Since $q$ and $B$ are constants (charge of electron and uniform magnetic field), the radius $Y$ is directly proportional to the momentum $X$. This is a linear relationship passing through the origin ($Y=0$ when $X=0$).

• Step 6: Match with the graphs. Graph B shows a straight line passing through the origin with a positive slope, indicating a direct proportionality.

• Answer for 49: $\boxed{\text{B}}$

Question 50:
Y: Reactance of an inductor
X: Frequency

• Step 1: Recall the formula for inductive reactance. The inductive reactance $X_L$ of an inductor with inductance $L$ at a frequency $f$ is given by:

  $$X_L = 2\pi f L$$

• Step 2: Relate to Y and X. Here, $Y = X_L$ and $X = f$. So, $Y = 2\pi L X$.

• Step 3: Analyze the relationship. Since $2\pi L$ is a constant (for a given inductor), the inductive reactance $Y$ is directly proportional to the frequency $X$. This is a linear relationship passing through the origin ($Y=0$ when $X=0$).

• Step 4: Match with the graphs. Graph B shows a straight line passing through the origin with a positive slope, indicating a direct proportionality.

• Answer for 50: $\boxed{\text{B}}$