Voltmeters confusion Voltmeters must be connected in parallel with the resistor. It must have  a very high resistance so that it does not alter the amount of current through the resistor. But what will happen if its resistance is not that high compared to the resistance of resistor. Will the voltmeter still show same reading? How does a voltmeter measure the voltage?

Suppose two  resistance of 1000ohm and 2000ohm are placed in series with 50V mains supply. What wll be the reading on a voltmeter of internal resistance 2000ohm when placed across the 1000ohm resistor? What is the fractional change in voltage when voltmeter is connected?

 A: voltmeters are of two types: active and passive. the active voltmeters contain buffer amplifiers which allow the voltage measurement to be made with almost zero current draw. all digital voltmeters are of this active design. 
Older, analog meters (typically called volt-ohm-meters or "VOM's") are passive, in that the voltage is measured with a moving-coil meter movement which requires a finite amount of current draw to occur through the meter in order to make a voltage measurement. 
The internal DC resistance of such a meter is in the range of 1000 ohms to 50,000 ohms, which means if, for example, the 1000 ohm meter is measuring the voltage across a 1000 ohm resistor, then the same amount of current flowing through the resistor will also be flowing through the meter, which will cut the voltage across the resistor in half and thereby yield a false low reading on the meter.
a hybrid design using an analog meter movement but containing a primitive vacuum-tube buffer amp was also available for some years, before the digital models were invented. These were called vacuum-tube voltmeters or "VTVM's". 
A: I will not solve this homework type of problem for you but will highlight:
(1) The voltmeter will display the voltage between its ports.
(2) If you estimate that the voltmeter affects the circuit, you first need to solve the circuit (find voltages at points of interest) by replacing the voltmeter with its internal resistance.
A: A voltmeter is designed to measure the potential difference across its terminals $V_{\rm voltmeter}$.  
Suppose you want to measure the potential difference across a resistor  $V_{\rm resistor}$ with resistance $R$ which has a current $I$ passing through it.  $V_{\rm resistor}=IR$  
As soon as you connect the voltmeter in parallel across the resistor you have altered the original circuit and changed the current passing through the resistor this means that the potential difference across the resistor (which is the reading on the voltmeter) has changed.
The amount by which the potential difference across the resistor has changed depends on the relative values of the resistance of the resistor, the resistance of the voltmeter and the values of the circuit elements in the rest of the circuit.
What is happening is that the current through the resistor is not the same before and after the voltmeter is added to the circuit.  
In your example because the resistance of the resistor is comparable with the resistance of the voltmeter there will be a significant change between the potential difference across the resistor before and after the voltmeter is placed across the resistor.  
With the voltmeter in place you basically have added an extra resistor (the resistance of the voltmeter) to the circuit.  
A: The voltmeter must have an input resistance sufficiently greater than the resistance across which it is connected so that it does not materially affect the voltage across that resistance. Without the voltmeter in the circuit the voltage will divide as 33.33 volts across the 2000 ohm resistor and 16.67 volts across the 1000 ohm resistor (between a and b) (Confirm for yourself). See Fig 1 below. 
Now put in the meter. Fig 2. Lets assume the internal meter resistance of 2000 ohms includes the meter movement resistance. Calculate the equivalent parallel resistance of the meter and the 1000 ohm circuit resistor across a and b. Put that in series with the 2000 ohm circuit resistor and calculate the voltage between a and b.  From that you can calculate the fractional change. Not surprisingly you will find the voltage measured by the meter much different than what the actual voltage would have been without the meter. 
Hope this helps

