Constructors, Classes, and Inheritance

The protected access specifier works as private for outside world, but works as public for derived classes in Java.

If the first statement of a constructor has the form this(parameter list), then the constructor calls another constructor of the same class. For example:
class Emp
{
    int m;
    double n;
    public Emp(int x)
    {
            this(789.89,x);
     }
}
Here, this(789.89,x) statement calls the constructor Emp(double,int)

The protected access specifier is very useful in multilevel inheritance because the protected member of super class A can be accessed by the subclass C in the following way.
A(Super Class)$\rightarrow$ B(Subclass of A) $\rightarrow$C(Subclass of B)

This is a correct answer. String is not a wrapper class in java.

This is a wrong answer. Interfaces cannot be marked as final.
This is the correct answer. Final keyword can be used with classes, methods and data members.

This is the correct answer. All data members inside an interface need to be final. Once the data members are final, their values cannot be changed. If we do not make the data members final, it will result in compile time error.

This is the correct answer. As class B is extending from class A, hence class A is the superclass and class B is the subclass.

If there are more than one constructor in a class, then its termed as constructor overloading.

A constructor can be without a parameter also. The constructor without any parameter is called a default constructor.

The given structure is a representation of simple inheritance.

The given structure is a representation of multilevel inheritance.

Multiple inheritance can be implemented in Java using interfaces. One or more classes can implement the defined interface.

A final class cannot be extended. If we try the compiler will give an error.

The implementation in Java for the given structure is as follows:
class A
{
}
class B
{
}
class C implements A,B
{
}