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Salsk061 [2.6K]
3 years ago
10

Which statement is true regarding copper's ability to conduct electricity?

Physics
2 answers:
zhenek [66]3 years ago
7 0
<span>Copper is a good conductor of electricity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms.</span>
Rasek [7]3 years ago
5 0

Answer:

(i) Copper is a good conductor of electricity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms.

Explanation:

Copper atoms has loosely held electrons in their outer shell. These electrons are not permanently associated with the copper atoms, instead They make a cloud around the outside of these atoms and are able to move freely and quickly through the solid.  

Option (ii) cannot be selected because copper is a good conductor. For the conductivity, free movement of electrons is a condition which is unsatisfied in the (iii). In (iv), it is given that copper has 1 electron in the outermost shell which can flow to the other copper atoms which is not true because in copper many atoms can flow to other atoms.

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Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

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and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

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Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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