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We can approach the solution to this problem by first considering how resistance applied in the series affects the full-scale deflection of a voltmeter. If you are having troubles recalling the relation, let us tell you that the full-scale deflection is directly proportional to the resistance. Using this information, we can easily solve the question, as you’ll see.

As told in the hint section, we will approach the solution to the question considering how resistance applied in the series affects the full-scale deflection of a voltmeter.

We have now established the fact that the full-scale deflection of the voltmeter is directly proportional to the resistance applied in series. Hence, all we need to do to solve this question is to find some basic ratios and place the numerical values in the right place and manner.

In the question, it has been given to us that for a $3V$ full deflection, grading is $6000\dfrac{\Omega }{V}$.

We have to find the resistance that should be connected in series to the voltmeter, so that it reads $12V$ at full-scale deflection.

The ratio of the new full-deflection to the original full-deflection is:

$\dfrac{{12V}}{{3V}}\, = \,4$

Since the full-scale deflection varies linearly to the resistance applied, the ratio of the new net resistance should be 4 times that of the old resistance.

We can write it in equation as:

$\dfrac{{{R_{new}}}}{{{R_{old}}}}\, = \,4$

We already know the value of old resistance, which is $6000\Omega $

Putting this value in the relation, we get:

$\begin{gathered}

{R_{new}}\, = \,4\, \times \,6000\Omega \\

{R_{new}}\, = \,24000\Omega \\

\end{gathered} $

Since ${R_{new}}$ is the net new resistance, it also took in the old resistance, so the extra resistance that needs to be added can be found out as:

$\begin{gathered}

{R_{new}}\, = \,{R_{old}}\, + \,r \\

24000\Omega \, = \,6000\Omega \, + \,r \\

r\, = \,18000\Omega \\

or \\

r\, = \,1.8\, \times \,{10^4}\Omega \\

\end{gathered} $

**Complete step by step answer:**As told in the hint section, we will approach the solution to the question considering how resistance applied in the series affects the full-scale deflection of a voltmeter.

We have now established the fact that the full-scale deflection of the voltmeter is directly proportional to the resistance applied in series. Hence, all we need to do to solve this question is to find some basic ratios and place the numerical values in the right place and manner.

In the question, it has been given to us that for a $3V$ full deflection, grading is $6000\dfrac{\Omega }{V}$.

We have to find the resistance that should be connected in series to the voltmeter, so that it reads $12V$ at full-scale deflection.

The ratio of the new full-deflection to the original full-deflection is:

$\dfrac{{12V}}{{3V}}\, = \,4$

Since the full-scale deflection varies linearly to the resistance applied, the ratio of the new net resistance should be 4 times that of the old resistance.

We can write it in equation as:

$\dfrac{{{R_{new}}}}{{{R_{old}}}}\, = \,4$

We already know the value of old resistance, which is $6000\Omega $

Putting this value in the relation, we get:

$\begin{gathered}

{R_{new}}\, = \,4\, \times \,6000\Omega \\

{R_{new}}\, = \,24000\Omega \\

\end{gathered} $

Since ${R_{new}}$ is the net new resistance, it also took in the old resistance, so the extra resistance that needs to be added can be found out as:

$\begin{gathered}

{R_{new}}\, = \,{R_{old}}\, + \,r \\

24000\Omega \, = \,6000\Omega \, + \,r \\

r\, = \,18000\Omega \\

or \\

r\, = \,1.8\, \times \,{10^4}\Omega \\

\end{gathered} $

**So the correct answer is option (A).****Note:**Many students forget that the relation of full deflection is directly proportional to the resistance and take it as inversely proportional, which is wrong. Many students forget to find out the extra resistance that needs to be added and only solve up to the new net resistance, which causes them to lose marks.Recently Updated Pages

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