To answer this question, we need to understand the concept of method overriding in Java.

In the given code, there are two classes: Dog and Alsation. The Dog class has a static method called bark() that prints "woof ". The Alsation class extends the Dog class and also has a static method called bark(). However, the Alsation class does not provide an implementation for the bark() method.

In the Bark class's main() method, two objects are created: woofer of type Dog and nipper of type Alsation. When the bark() method is called on woofer, it will call the bark() method of the Dog class, which prints "woof ". When the bark() method is called on nipper, it will call the bark() method of the Alsation class. Since there is no implementation of the bark() method in the Alsation class, it will use the implementation of the bark() method from the superclass (i.e., Dog class).

Therefore, the output will be "woof woof ".

Option B) woof woof - This option is correct because it matches the output explained above.

The correct answer is B) woof woof.

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• Wrong explanation

To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) The class compiles and runs, but does not print anything. - This option is incorrect. The code contains an assertion statement that will throw an AssertionError if the condition specified in the assertion is false. Since the condition i != j evaluates to false when i is equal to j, an AssertionError will be thrown.

Option B) The number 1 gets printed with AssertionError - This option is correct. The code contains an assertion statement assert i != j : i;. When i is equal to j (as is the case when i is 1 and j is also 1), the assertion condition evaluates to false, and an AssertionError is thrown. The value of i (which is 1) is included in the AssertionError message.

Option C) The number 2 gets printed with AssertionError - This option is incorrect. The code does not reach the point where i is equal to 2, so this option is not possible.

Option D) The number 3 gets printed with AssertionError - This option is incorrect. The code does not reach the point where i is equal to 3, so this option is not possible.

Option E) The program generates a compilation error - This option is incorrect. The code does not contain any syntax errors, so it will compile successfully.

The correct answer is Option B) The number 1 gets printed with AssertionError. This option is correct because the assertion condition i != j evaluates to false when i is equal to j, and an AssertionError is thrown with the value of i (which is 1) included in the message.

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• Wrong explanation

To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) compile time error at lines 1,2,3,4 - This option is incorrect. It suggests that there are compile time errors at all four lines, but this is not the case. There are no compile time errors at lines 1 and 2.

Option B) compile time error at line 3 - This option is incorrect. There is no compile time error at line 3. The "protected" access modifier is valid for interface methods.

Option C) compile time error at line 1 - This option is incorrect. There is no compile time error at line 1. The default access modifier is assumed if no access modifier is specified, making the method accessible to any class within the same package.

Option D) compile time error at lines 3,4 - This option is correct. There is a compile time error at line 4. The "private" access modifier is not allowed for interface methods. Additionally, there is no compile time error at line 3.

Option E) None of the above - This option is incorrect. As explained above, option D is the correct answer.

Therefore, the correct answer is D) compile time error at lines 3,4. This option is correct because the "private" access modifier is not allowed for interface methods.

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• Wrong explanation

To determine the number of objects eligible for garbage collection after line 6, we need to analyze the code.

Let's break down the code and track the objects:

1. C c1 = new C(); - This creates an object c1.
2. C c2 = m1(c1); - This invokes the method m1 with c1 as a parameter. Inside the method, a new object ob1 is created and returned. This new object is assigned to c2. So, we have two objects c1 and ob1.
3. C c3 = new C(); - This creates a new object c3.
4. c2 = c3; - This assigns the object c3 to c2. Now, c2 refers to c3, and the object previously referred to by c2 (which was ob1) becomes eligible for garbage collection.
5. anothermethod(); - This method call doesn't affect object eligibility for garbage collection.

After line 6, we have two objects (c1 and c3) that are still referenced, and one object (ob1) that is eligible for garbage collection since it is no longer referenced.

Therefore, the correct answer is B) 2 objects.

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• Wrong explanation

To answer this question, we need to understand the concepts of constructors and inheritance in Java.

In the given code, there are two classes: base and Super. The Super class extends the base class.

The base class has two constructors:

1. The default constructor base() which prints "base".
2. The parameterized constructor base(int i1).

The Super class has one constructor Super(). Inside this constructor, the super() statement is used to invoke the default constructor of the base class.

In the main method, an object b1 is created using the new keyword. The object is of type base and is assigned as a reference to the Super class.

Now, let's go through each option to determine the correct answer:

Option A) Compile time error - This option is correct. When the Super class extends the base class, it inherits the default constructor of the base class. However, when a class has a parameterized constructor and the default constructor is not explicitly defined, the default constructor is no longer available. In this case, the base class does not have an explicit default constructor, so the super() statement in the Super class constructor will result in a compile-time error.

Option B) Prints base and super - This option is incorrect because the code will not compile, and hence, nothing will be printed.

Option C) Prints super and base - This option is incorrect because the code will not compile, and hence, nothing will be printed.

Option D) None of the above - This option is incorrect because the code will not compile, and hence, nothing will be printed.

Therefore, the correct answer is A) Compile time error.

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• Wrong explanation

To answer this question, let's discuss the "switch" selection structure.

The "switch" selection structure is used to evaluate an expression and execute different code blocks based on the value of that expression. It consists of multiple "case" statements and an optional "default" case.

In the "switch" structure, each "case" statement checks if the value of the expression matches a specific value. If a match is found, the code block associated with that "case" statement is executed. If none of the "case" statements match the value of the expression, the code block associated with the "default" case is executed.

Now, let's analyze the given statement: "The 'switch' selection structure must end with the default case."

This statement is incorrect. The "switch" selection structure does not necessarily have to end with the "default" case. The "default" case is optional and can be placed anywhere within the "switch" structure. It is used to provide a default action when none of the "case" statements match the value of the expression.

Therefore, the correct answer is B) False.

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• Wrong explanation

To answer this question, you need to understand the concept of the modulus operator (%) in Java.

The modulus operator (%) in Java calculates the remainder of dividing one number by another. It can be used with variables of integer type, but it can also be used with variables of floating-point types, such as float or double.

Therefore, the statement that the modulus operator (%) in Java can be used only with variables of integer type is incorrect.

Option A) True - This option is incorrect because the modulus operator (%) in Java can be used with variables of floating-point types as well. Option B) False - This option is correct because the modulus operator (%) in Java can be used with variables of both integer and floating-point types.

The correct answer is B) False. This option is correct because the modulus operator (%) in Java can be used with variables of integer and floating-point types.

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• Wrong explanation