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$A)\text{ }1:1$

$B)\text{ }1:2$

$C)\text{ 2}:1$

$D)\text{ }1:4$

Answer

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The bulbs can be considered to be resistors. Since they are in series, the same current will flow through them. We will use the formula for the power dissipated by a resistor in terms of the current through it and its resistance.

The power $P$ dissipated by a resistor of resistance $R$ is given by

$P={{I}^{2}}R$ --(1)

where $I$ is the current passing through the resistor.

Now, let us analyze the question.

Let the resistances of the two bulbs be ${{R}_{1}}$ and ${{R}_{2}}$ respectively.

According to the question,

${{R}_{1}}:{{R}_{2}}=1:2$

$\therefore \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{1}{2}$ --(2)

Let the current flowing through each of them be $I$.

Let the power dissipated by the two bulbs be ${{P}_{1}}$ and ${{P}_{2}}$ respectively.

We have to find out ${{P}_{1}}:{{P}_{2}}$.

Now using (1), we get

${{P}_{1}}={{I}^{2}}{{R}_{1}}$ --(3)

${{P}_{2}}={{I}^{2}}{{R}_{2}}$ --(4)

Therefore, dividing (3) by (4), we get

$\dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{{{I}^{2}}{{R}_{1}}}{{{I}^{2}}{{R}_{2}}}=\dfrac{{{R}_{1}}}{{{R}_{2}}}$

Using (2) in the above equation, we get

$\dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{1}{2}$

$\therefore {{P}_{1}}:{{P}_{2}}=1:2$

Hence, we have got the required ratio of the powers dissipated by the two bulbs.