A pendulum bob weighs 80 N in air. When immersed in water, it weighs 70 N. When immersed in liquid Y, it weighs 65 N. What is the relative density of liquid Y?

When the pendulum bob is totally immersed in water, the buoyant force ($F_{B, \text{water}}$) is equal to the weight of the water displaced.

$$F_{B, \text{water}} = 10 \text{ N}$$

When the pendulum bob is totally immersed in liquid Y, the buoyant force ($F_{B, \text{Y}}$) is equal to the weight of liquid Y displaced.

$$F_{B, \text{Y}} = 15 \text{ N}$$

The buoyant force is given by the formula $F_B = \rho_{\text{fluid}} V_{\text{displaced}} g$, where $\rho_{\text{fluid}}$ is the density of the fluid, $V_{\text{displaced}}$ is the volume of the fluid displaced, and $g$ is the acceleration due to gravity.
Since the bob is totally immersed, the volume of fluid displaced is equal to the volume of the bob ($V_{\text{bob}}$).
So, for water:

$$F_{B, \text{water}} = \rho_{\text{water}} V_{\text{bob}} g = 10 \text{ N} \quad (1)$$

And for liquid Y:

$$F_{B, \text{Y}} = \rho_Y V_{\text{bob}} g = 15 \text{ N} \quad (2)$$

The relative density (or specific gravity) of liquid Y is defined as the ratio of its density to the density of water:

$$\text{Relative Density of Y} = \frac{\rho_Y}{\rho_{\text{water}}}$$

To find this ratio, we can divide equation (2) by equation (1):

$$\frac{F_{B, \text{Y}}}{F_{B, \text{water}}} = \frac{\rho_Y V_{\text{bob}} g}{\rho_{\text{water}} V_{\text{bob}} g}$$

The terms $V_{\text{bob}}$ and $g$ cancel out, leaving:

$$\frac{F_{B, \text{Y}}}{F_{B, \text{water}}} = \frac{\rho_Y}{\rho_{\text{water}}}$$

Substitute the given buoyant forces:

$$\text{Relative Density of Y} = \frac{15 \text{ N}}{10 \text{ N}}$$ $$\text{Relative Density of Y} = 1.5$$

The final answer is $\text{B) } 1.50$.