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Chapter 7: Problem 1059

Mercury thermometers can be used to measure temperatures up to (A) \(100^{\circ} \mathrm{C}\) (B) \(212^{\circ} \mathrm{C}\) (C) \(360^{\circ} \mathrm{C}\) (D) \(500^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The correct option is (C) \(360^{\circ} \mathrm{C}\), as it is the closest option to the boiling point of Mercury. This is the highest temperature a Mercury thermometer can measure.

Step by step solution

01

Recall the boiling point of Mercury

The boiling point of Mercury is \(356.73^{\circ} \mathrm{C}\), which is the temperature at which Mercury transitions from a liquid to a gas. This means that a Mercury thermometer would not be able to measure temperatures above this point.
02

Find the closest option

Now we have to check each of the available options and find the one that is closest to the boiling point of Mercury, which is \(356.73^{\circ} \mathrm{C}\).
03

Option A - \(100^{\circ} \mathrm{C}\)

This option is too low, as the boiling point of Mercury is greater than \(100^{\circ} \mathrm{C}\).
04

Option B - \(212^{\circ} \mathrm{C}\)

This option is also too low, as the boiling point of Mercury is greater than \(212^{\circ} \mathrm{C}\).
05

Option C - \(360^{\circ} \mathrm{C}\)

This option is the closest to the actual boiling point of Mercury, which is \(356.73^{\circ} \mathrm{C}\). It's slightly above the boiling point, making this option as the most accurate one.
06

Option D - \(500^{\circ} \mathrm{C}\)

This option is too high, as the boiling point of Mercury is less than \(500^{\circ} \mathrm{C}\).
07

Conclusion

The correct option is (C) \(360^{\circ} \mathrm{C}\), as it is the closest option to the boiling point of Mercury. This is the highest temperature a Mercury thermometer can measure.

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The surface tension of a liquid is \(5 \mathrm{~N} / \mathrm{m}\). If a thin film of the area \(0.02 \mathrm{~m}^{2}\) is formed on a loop, then its surface energy will be (A) \(5 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(2 \times 10^{-1} \mathrm{~J}\) (D) \(5 \times 10^{-1} \mathrm{~J}\)

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