Four bulbs of $25 W, 40 W, 60 W$ and $100 W$ are connected in series and their combination is connected across a main power supply. Match the following list I with the potential difference across each bulb in List II
List I List II
(P) $25 W$ (1) Highest potential differences
(Q) $100 W$ (2) Second highest potential difference
(R) $60 W$ (3) Third highest potential difference
(S) $40 W$ (4) Lowest potential difference

P- 2; Q- 3; R- 1; S- 4

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P- 1; Q- 2; R- 4; S- 3

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P- 1; Q- 4; R- 3; S- 2

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P- 2; Q- 4; R- 1; S- 2

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Solution

The correct option is C P- 1; Q- 4; R- 3; S- 2
As $P = \frac{V^{2}}{R}$
$R_{1} = \frac{V^{2}}{25}$
$R_{2} = \frac{V^{2}}{40}$
$R_{3} = \frac{V^{2}}{60}$
$R_{4} = \frac{V^{2}}{100}$
Potential difference across any bulb = IR
Highest potential will be across 25 W bulb.
Similarly, we can arrange.

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Q. Four bulbs having power rating

25 W, 40 W, 60 W a n d 100 W

are connected in series and this combination is connected across the a voltage

V

then match the following:

Match the following:

Column -I	Column - II
a. 25 W bulb	(i) The highest potential difference
b.100 W bulb	(ii) The second highest potentisl difference
c.60 W bulb	(iii) The third highest potential difference
d.40 W bulb	(iv) The lowest potential difference

Two electric bulbs one of 200V-40W and other of 200V-100W are connected in series to a 200 volt line, then

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Q. Two bulbs

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and

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and rated voltage

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are connected in series across a potential difference of

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. Which bulb will work at above its rated voltage?

$\begin{matrix} L i s t - I & L i s t - I I P . & I f_{α} = \frac{π}{7} t h e n \frac{1}{c o s α} + \frac{2 c o s α}{c o s 2 α} = & 1. & 2 Q . & u n d e r s e t x \to \infty L t [(x - 1) (x - 2) (x + 3) (x + 5) (x + 10)]^{\frac{1}{5}} - x = & 2. & 3 R . & \int_{- 2}^{2} \frac{3 x^{2} d x}{1 + e^{x}} = & 3. & 4 S . & L e t f (x) = x^{3} + x^{2} + 2 x - 1. T h e m i n i m u m i n t e g r a l v a l u e o f x i f x s a t i s f i e s f (f (x)) > f (2 x + 1) i s & 4. & 8 \end{matrix}$

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List I	List II
(P) $25 W$	(1) Highest potential differences
(Q) $100 W$	(2) Second highest potential difference
(R) $60 W$	(3) Third highest potential difference
(S) $40 W$	(4) Lowest potential difference

The correct option is C P- 1; Q- 4; R- 3; S- 2As P=V2RR1=V225R2=V240R3=V260R4=V2100Potential difference across any bulb = IRHighest potential will be across 25 W bulb.Similarly, we can arrange.

The correct option is C P- 1; Q- 4; R- 3; S- 2
As $P = \frac{V^{2}}{R}$
$R_{1} = \frac{V^{2}}{25}$
$R_{2} = \frac{V^{2}}{40}$
$R_{3} = \frac{V^{2}}{60}$
$R_{4} = \frac{V^{2}}{100}$
Potential difference across any bulb = IR
Highest potential will be across 25 W bulb.
Similarly, we can arrange.