Among the following compounds, geometrical isomerism is exhibited by:

To determine which compound exhibits geometrical isomerism, we first recall the necessary condition for its existence. For a molecule containing a carbon-carbon double bond to show geometrical (cis-trans or E/Z) isomerism, each carbon of the double bond must be attached to two different substituents. If either carbon is bonded to two identical groups, geometrical isomerism is not possible.

In 4-chloromethylenecyclohexane, the exocyclic double-bonded carbon is attached to two identical hydrogen atoms $$\left(=CH_2\right)$$. Since one carbon of the double bond has identical substituents, the compound cannot exhibit geometrical isomerism.

In 3-methyl-1-chloromethylenecyclohexane, the exocyclic carbon of the double bond is attached to one hydrogen atom and one chlorine atom $$\left(=CHCl\right)$$, which are different substituents. The ring carbon involved in the double bond is connected to two non-equivalent paths around the ring because of the methyl substituent at the $$3$$-position. As both double-bonded carbons have two different groups attached, this compound can exhibit geometrical isomerism.

In 5-methyl-1-chloromethylenecyclohexane, the arrangement is symmetrical, making the two paths around the ring equivalent. Consequently, the ring carbon of the double bond is effectively attached to identical groups, so the compound does not exhibit geometrical isomerism.

Similarly, in 1-chloromethylenecyclohexane, the ring carbon involved in the double bond is attached to two identical sides of the unsubstituted cyclohexane ring. Therefore, it cannot exhibit geometrical isomerism.

Hence, among the given compounds, only 3-methyl-1-chloromethylenecyclohexane (Option B) satisfies the conditions required for geometrical isomerism and can exist as E/Z isomers.