Units And Measurements CBSE Questions & Answers
Units And Measurements
This is Physics Class 11 Units and Measurements CBSE Questions & Answers. There are 15 questions in this test with each question having around four answer choices.
Questions & Answers
1
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area of the sheet to correct significant figures.
- A8.722 \({{\rm{m}}^{\rm{2}}}\)
- B8.7221 \({{\rm{m}}^{\rm{2}}}\)
- C8.8 \({{\rm{m}}^{\rm{2}}}\)
- D8.72 \({{\rm{m}}^{\rm{2}}}\)Correct
2
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the volume of the sheet to correct significant figures.
- A0.0855 \({{\rm{m}}^{\rm{3}}}\)Correct
- B0.0755 \({{\rm{m}}^{\rm{3}}}\)
- C0.08552 \({{\rm{m}}^{\rm{3}}}\)
- D0.1855 \({{\rm{m}}^{\rm{3}}}\)
3
A physical quantity P is related to four observables a, b, c and d as follows : \(P = {{{a^3}{b^2}} \over {\sqrt c d}}\) The percentage errors of measurement in a, b, c and d are 1\(\% \), 3\(\% \), 4\(\% \) and 2\(\% \),respectively. What is the percentage error in the quantity P?
- A13\(\% \)Correct
- B12\(\% \)
- C11\(\% \)
- D15\(\% \)
4
A physical quantity P is related to four observables a, b, c and d as follows: \(P = {{{a^3}{b^2}} \over {\sqrt c d}}\) The percentage errors of measurement in a, b, c and d are 1\(\% \), 3\(\% \), 4\(\% \) and 2\(\% \), respectively. If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?
- A3.71
- B3.76
- C3.8Correct
- D4.0
5
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion. Choose the correct formula
- Ay = a sin 2 \(\pi \) t/TCorrect
- By = a sin 2 \(\pi \) t
- Cy = a sin vt
- Dy = (a/T) sin t/a
6
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by \({\mathop A\limits^0}:{\rm{ 1 }}{\mathop A\limits^0}{\rm{ }} = {\rm{ 1}}{0^{ - {\rm{1}}0}}\) m. The size of a hydrogen atom is about \(0.{\rm{5 }}{\mathop A\limits^0}\) . What is the total atomic volume in \({{\rm{m}}^{\rm{3}}}\) of a mole of hydrogen atoms ?
- A\( \cong 2.9 \times {10^{ - 6}}{m^3}\)
- B\( \cong 3.3 \times {10^{ - 7}}{m^3}\)
- C\( \cong 3 \times {10^{ - 7}}{m^3}\)Correct
- D\( \cong 3.2 \times {10^{ - 6}}{m^3}\)
7
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å).
- A\( \cong {10^3}\)
- B\( \cong {10^4}\)Correct
- C\( \cong {10^2}\)
- D\( \cong {10^5}\)
8
The principle of parallax is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit \( \approx {\rm{ 3 }} \times {\rm{ 1}}{0^{{\rm{11}}}}{\rm{m}}\) . However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1”(second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?
- A\( \cong 3 \times {10^{12}}m\)
- B\( \cong 3 \times {10^{16}}m\)Correct
- C\( \cong 3 \times {10^{13}}m\)
- D\( \cong 3 \times {10^{14}}m\)
9
The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?
- A1.34 parsec; 2.44” (second of arc)
- B1.32 parsec; 2.64” (second of arc)Correct
- C1.35 parsec; 2.54” (second of arc)
- D1.22 parsec; 2.24” (second of arc)
10
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding \({\text{1}}{0^{\text{7}}}\) K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun = 2.0 × \({\rm{1}}{0^{{\rm{3}}0}}\) kg, radius of the Sun = 7.0 × \({\rm{1}}{0^{\rm{8}}}\) m.
- A\(1.6 \times {10^3}kg{m^{ - 3}}\)
- B\(1.5 \times {10^3}kg{m^{ - 3}}\)
- C\(1.3 \times {10^3}kg{m^{ - 3}}\)
- D\(1.4 \times {10^3}kg{m^{ - 3}}\)Correct
11
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72” of arc. Calculate the diameter of Jupiter.
- A\(1.632 \times {10^5}km\)
- B\(1.429 \times {10^5}km\)Correct
- C\(1.321 \times {10^5}km\)
- D\(1.523 \times {10^5}km\)
12
A man walking briskly in rain with speed v must slant his umbrella forward making an angle \(\theta \) with the vertical. A student derives the following relation between \(\theta \) and v : tan \(\theta \) = v and checks that the relation has a correct limit: as \({\rm{v }} \to {\rm{ }}0,{\rm{ }}\theta {\rm{ }} \to 0\), as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Which of the following is correct?
- Aderivation is wrong; correct formula is tan \(\theta \) = v/v’ where v’ is speed of rainCorrect
- Bderivation is wrong; correct formula is sin \(\theta \) = v/v’ where v’ is speed of rain
- Cderivation is wrong; correct formula is cos \(\theta \) = v/v’ where v’ is speed of rain
- Dderivation is correct; correct formula is tan \(\theta \) = v
13
It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?
- AAccuracy of 2 parts in \({\rm{1}}{0^{{\rm{11}}}}{\rm{to 1}}{0^{{\rm{12}}}}\)
- BAccuracy of 10 parts in \({\rm{1}}{0^{{\rm{11}}}}{\rm{to 1}}{0^{{\rm{12}}}}\)
- CAccuracy of 1 part in \({\rm{1}}{0^{{\rm{11}}}}{\rm{to 1}}{0^{{\rm{12}}}}\)Correct
- DAccuracy of 5 parts in \({\rm{1}}{0^{{\rm{11}}}}{\rm{to 1}}{0^{{\rm{12}}}}\)
14
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 \({\mathop A\limits^0}\). (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m3. Are the two densities of the same order of magnitude? If so, why?
- A\( \cong 0.75 \times {10^3}kg{m^{ - 3}}\) . In the solid phase atoms are tightly packed, so atomic mass density is close to the mass density of the solid
- B\( \cong 0.6 \times {10^3}kg{m^{ - 3}}\) . In the solid phase atoms are tightly packed, so atomic mass density is close to the mass density of the solidCorrect
- C\( \cong 0.7 \times {10^7}kg{m^{ - 3}}\). In the solid phase atoms are tightly packed, so atomic mass density is close to the mass density of the solid
- D\( \cong 0.78 \times {10^3}kg{m^{ - 3}}\). In the solid phase atoms are tightly packed, so atomic mass density is close to the mass density of the solid
15
The significant digits in 0.000532 are<
- A5, 3, 2Correct
- B2, 3
- C5, 3
- D0,5,3,2