1) If there is one class template which has one static member variable that static variable will belong to

Answer : a) Every instance of class template will share the same copy of static variable. 
Every class instantiated from a template class is like a ordinary and separate class. So by definition of static member variable, every such class should have its own copy of static variable so answer is a .

2) What is template specialization ?

Answer : Template specialization is a way of informing the compiler to use that version of the template for specific data types in case of template class and for specific arguments in case of function template.
For e.g say we define template class as
template class container{}
So a particular container class will be generated depending on the template argument which is T


container d;

the above statement will generate container class from the template class.

If we prefer to optimize the implementation by making all containers of pointers share a common implementation, we could specialize the above template for pointer template argument as below


template class container{}
In this all the below statements generate classes out of the specialized template class
container
container

This is called template specialization.

3)How we will overload operator*+=such that
obj1 *+= obj2;
implies that
obj1=obj1*(obj1+obj2);
four choice were there last option was d) it is not possible I checked that option.


Answer : Its not possible to define new operator using operator overloading, only existing operators can be overloaded. There is no such ternary operator as *+= in C++, so its not possible to overload that operator.

4)IN C++ what does the operator overloading means.
a) Giving new meaning to existing C++ operators
b) defining functionality of existingC++ operator for user define
objects.
c)defining new operators.
d) don't remember.


Answer : b. operator overloading doesn't allow changing the meaning of existing operators, also we can't change the meaning of the operator in case of built in data types. Only operations involving user defined data types can be overloaded.

5)
What is '>>' in C++
a) right shift operator and insertion operator depend upon the use
b) right shift operator and extractionoperator depend upon the
context use
c) right shift operator and insertion/extractionoperator depend upon
the use


Answer : In C++ operator">>" is still a right shift operator as it was in C. Howeve the operator was been overloaded in almost all of library classes as an extraction operator. So its not a built in extraction operator but still a right shift operator.

6) class A
{
int a ,b;
A() : a(0)
{b=0;}
};
if you create obj of this class as A obj;
a) b will be initialized before a
b) a will be initialized before b
c) both will be initialized together
d) none of these.

Answer : a) exec

The order of initialization is not as per the order of member initializer list rather the member variables are initialized int the order of declaration in the class. So the variable a will be initialized first followed by b. So the answer is b.