The verified correct answer isD (Five minutes). In Kubernetes, node health is continuously monitored. When a node stops reporting status (heartbeats from the kubelet) or is otherwise considered unreachable, the Node controller updates the Node’s Ready condition toUnknown(or it can becomeFalse). From that point, Kubernetes has to balance two risks: acting too quickly might cause unnecessary disruption (e.g., transient network hiccups), but acting too slowly prolongs outage for workloads that were running on the failed node.

The “default eviction timeout” refers to the control plane behavior that determines how long Kubernetes waits before evicting Pods from a node that appears unhealthy/unreachable. After this timeout elapses, Kubernetes begins eviction of Pods so controllers (like Deployments) can recreate them on healthy nodes, restoring the desired replica count and availability.

This is tightly connected to high availability and self-healing: Kubernetes does not “move” Pods from a dead node; it replaces them. The eviction timeout gives the cluster time to confirm the node is truly unavailable, avoiding flapping in unstable networks. Once eviction begins, replacement Pods can be scheduled elsewhere (assuming capacity exists), which is the normal recovery path for stateless workloads.

It’s also worth noting that graceful operational handling can be influenced by PodDisruptionBudgets (for voluntary disruptions) and by workload design (replicas across nodes/zones). But the question is testing the default timer value, which isfive minutesin this context.

Therefore, among the choices provided, the correct answer isD.

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