Question:

Fill in the blanks (a) 1 kgm2/s2=………gcm2/s2 (b) 1 m = ………. ly (c) 3.0 m s−2 = ………km h−2 (d) G = 6.67 × 10−11 N m2 (kg)−2 = ………. (cm)3 s−2 g−1

Answer:

(a) 1000 gcm2/s2 (b) 9.4605284 × 10^15 ly (c) 10800 km h−2 (d) 6.67 × 10−8 (cm)3 s−2 g−1

Question:

The unit length convenient on the atomic scale is known as an angstrom and is denoted by A˚:(1A˚=10^−10m). The size of a hydrogen atom is about 0.5 A˚. What is the total atomic volume in m3 of a mole of hydrogen atoms?

Answer:

Step 1: Calculate the volume of a single hydrogen atom. Volume of a single hydrogen atom = 4/3πr3 = 4/3π(0.5 A˚ )3 = 2.09 x 10-29 m3

Step 2: Calculate the number of hydrogen atoms in a mole. Number of hydrogen atoms in a mole = 6.022 x 1023 atoms

Step 3: Calculate the total atomic volume in m3 of a mole of hydrogen atoms. Total atomic volume in m3 of a mole of hydrogen atoms = Volume of a single hydrogen atom x Number of hydrogen atoms in a mole = 2.09 x 10-29 m3 x 6.022 x 1023 atoms = 1.25 x 10-5 m3

Question:

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0​ of a particle in terms of its speed v and the speed of light, c. (This relation first arose asa consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : m=mo​​/(1−v^2)^1/2 Guess where to put the missing c.

Answer:

The correct equation is: m=mo​​/(1−v^2/c^2)^1/2

Question:

Write the dimensional formula of Force. A MLT B MLT^−2 C MLT^2 D None of the above

Answer:

Answer: B MLT^−2

Question:

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 and volume of the sheet to correct significant figures.

Answer:

Area = (4.234 m) x (1.005 m) = 4.25 m^2

Volume = (4.234 m) x (1.005 m) x (0.0201 m) = 0.085 m^3

Question:

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.

Answer:

Step 1: Convert the angular diameter of Jupiter from arc to radians.

1 arc = (π/180) radians

Therefore, 35.72 arc = (35.72 × π)/180 radians

= 0.6224 radians

Step 2: Calculate the diameter of Jupiter using the formula:

Diameter of Jupiter (D) = 2 × 824.7 million kilometers × tan (0.6224 radians)

= 143,948.3 kilometers

Question:

It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1s ?

Answer:

1. This implies that the standard cesium clock is highly accurate in measuring a time-interval of 1s, since the difference between two cesium clocks running for 100 years is only 0.02s.

Question:

1 parsec = __________ metres

Answer:

1 parsec = 3.08567758 x 10^16 metres

Question:

Explain this statement clearly: “To call a dimensional quantity ’large’ or ‘small’ is meaningless without specifying a standard for comparison.” In view of this, which of the following statements are complete : (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light.

Answer:

(a) atoms are very small objects - Complete (b) a jet plane moves with great speed - Complete (c) the mass of Jupiter is very large - Complete (d) the air inside this room contains a large number of molecules - Complete (e) a proton is much more massive than an electron - Complete (f) the speed of sound is much smaller than the speed of light - Complete

Question:

A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J=1 kgm^2s^−2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α^−1β^−2γ^2 in terms of the new units.

Answer:

1. A calorie is a unit of heat or energy and it equals about 4.2 J.

2. 1 J=1 kgm^2s^−2.

3. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s.

4. A calorie has a magnitude 4.2 α^−1β^−2γ^2 in terms of the new units.

Question:

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10^7K, and its outer surface at a temperature of about 6000K. At these high temperatures, no substances remain 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 is correct from the following data: mass of the Sun =2.0×10^30kg, radius of the Sun 7.0×10^8m.

Answer:

1. The temperature of the Sun’s inner core is greater than 10^7K, so no substances remain in a solid or liquid phase.

2. The mass density of the Sun can be calculated using the equation Density = Mass/Volume.

3. The mass of the Sun is 2.0×10^30kg and the radius of the Sun is 7.0×10^8m.

4. The volume of the Sun can be calculated using the equation Volume = 4/3πr^3, where r is the radius of the Sun.

5. Substituting the values into the equation, we get Volume = 1.41×10^27m^3.

6. The mass density of the Sun can be calculated by substituting the mass and volume into the equation Density = Mass/Volume.

7. The mass density of the Sun is 1.42×10^3kg/m^3, which is in the range of densities of gases.

8. This is correct, as the mass density of the Sun is within the range of densities of gases.

Question:

Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity): (a) The total mass of rain-bearing clouds over India during the Monsoon. (b) The mass of an elephant (c) The wind speed during a storm (d) The number of strands of hair on your head (e) The number of air molecules in your classroom

Answer:

(a) The total mass of rain-bearing clouds over India during the Monsoon: Estimate: One way to estimate the total mass of rain-bearing clouds over India during the Monsoon is to calculate the average rainfall in India during the Monsoon season and then multiply it by the approximate surface area of India. This will give an estimate of the total mass of the rain-bearing clouds.

(b) The mass of an elephant: Estimate: One way to estimate the mass of an elephant is to use a scale to weigh a similar-sized animal, such as a cow, and then multiply that weight by the approximate size ratio between a cow and an elephant. This will give an estimate of the mass of an elephant.

(c) The wind speed during a storm: Estimate: One way to estimate the wind speed during a storm is to measure the speed of the wind at the start of the storm and then use that as an upper bound for the wind speed during the storm.

(d) The number of strands of hair on your head: Estimate: One way to estimate the number of strands of hair on your head is to count the number of hairs on a small area of your head and then multiply that number by the approximate size of your head. This will give an estimate of the number of strands of hair on your head.

(e) The number of air molecules in your classroom: Estimate: One way to estimate the number of air molecules in your classroom is to calculate the approximate volume of the classroom and then multiply that by the average number of air molecules per unit volume. This will give an estimate of the number of air molecules in your classroom.

Question:

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the sun and the earth in terms of the new unit if light takes 8 min and 20 sec to cover this distance? A 300 B 400 C 500 D 600

Answer:

Answer: B

Step 1: Calculate the speed of light in terms of the new unit. Speed of light = 1

Step 2: Calculate the distance between the sun and the earth. Distance = (8 min and 20 sec) x (1 unit/min) = 8.33 units

Step 3: Compare the calculated distance to the answer choices. The closest answer is 400, so the answer is B.

Question:

Answer the following: (a) You are given a thread and a metre scale. How will you estimate the diameter of the thread? (b) A screw gauge has a pitch of 1.0mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Answer:

(a) To estimate the diameter of the thread, wrap the thread around the metre scale and measure the length of the thread required to make one complete round. Divide this length by 3.14 to get the approximate diameter of the thread.

(b) No, it is not possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale. This is because the pitch of the screw gauge remains the same, and so the accuracy is limited by the pitch of the screw gauge.

(c) A set of 100 measurements of the diameter is expected to yield a more reliable estimate than a set of 5 measurements only because a larger sample size will provide a better representation of the true value of the mean diameter. The larger sample size will also reduce the chance of any outliers skewing the results.

Question:

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion : (a) y=asin 2π^t/T (b) y = a sin υ t
(c) y = (a/T) sin t/a (d) y = (a/√2) (sin 2π / T + cos 2π / T) (a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Answer:

1. Identify the dimensional formula of each variable: a. a = [L] b. v = [L/T] c. T = [T]

2. Calculate the dimensional formula of each formula: a. y = a sin 2π^t/T → [L] b. y = a sin υ t → [L/T] c. y = (a/T) sin t/a → [L/T] d. y = (a/√2) (sin 2π / T + cos 2π / T) → [L]

3. Rule out the wrong formula: Formula (b) is wrong as it does not have the same dimensional formula as the other formulas.

Question:

Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the trains motion, but the distant objects (hill tops, the Moon, the stars etc.)seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer:

1. When a train is moving at a fast speed, the objects that are close to the window of the train appear to move in a direction opposite to the motion of the train.

2. This is because the motion of the train is faster than the motion of the nearby objects.

3. On the other hand, the objects that are far away (such as hills, the Moon, and stars) appear to be stationary.

4. In fact, since the person is aware that they are moving, it appears that these distant objects are also moving with them.

5. This is because the motion of the train is too slow to be noticed when compared to the motion of these distant objects.

Question:

A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 ms^−1).

Answer:

1. Convert 77.0 seconds into milliseconds: 77.0 s = 77,000 ms

2. Calculate the distance of the enemy submarine using the speed of sound in water: Distance = Speed x Time = 1450 ms^−1 x 77,000 ms = 112,550,000 ms = 112,550 m

Question:

The principle of ‘parallax’ in section 2.3.1 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 ≈3×10^11m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order 1"(second) of an 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 ?

Answer:

A parsec is equal to 3.26 light years or 3.086 x 10^16 metres.

Question:

Precise measurements of physical quantities are a need of science. For example, to as certain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer:

1. Precise measurements of physical quantities are essential for science, as they allow us to accurately measure and study phenomena. For example, to measure the speed of an aircraft, we must be able to measure its position at two different points in time with a high degree of accuracy. This was the impetus behind the development of radar during World War II.

2. In modern science, precise measurements of length, time, and mass are needed in a variety of fields. For example, in astronomy, precise measurements of the position of stars and planets are needed in order to accurately measure their motion. In particle physics, precise measurements of the mass of particles are needed to study their interactions. In engineering, precise measurements of lengths and angles are needed to design and construct structures.

3. The precision needed for these measurements varies depending on the application. For example, in astronomy, the precision needed to measure the position of stars and planets may be as small as a few millimeters, while in particle physics, the precision needed to measure the mass of particles may be as small as a few micrograms. In engineering, the precision needed to measure lengths and angles may be as small as a few micrometers.

Question:

Which of the following is the most precise device for measuring length? A A vernier callipers with 20 divisions on the sliding scale. B A screw gauge of pitch 1 mm and 100 divisions on the circular scale. C An optical instrument that can measure length to within a wavelength of light. D All instruments have same precision.

Answer:

Answer: C An optical instrument that can measure length to within a wavelength of light.

Question:

Fill in the blanks (a) The volume of a cube of side 1cm is equal to ________ m^3 (b) The surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to ________ (mm)^2 (c) A vehicle moving with a speed of 18km/h covers ________ metres in one second. (d) The relative density of lead is 11.3. Its density is ________ g/cm^3 or ________kg/m^3

Answer:

(a) 1 x 10^-6 m^3 (b) 125.66 (mm)^2 (c) 5 m (d) 11.3 g/cm^3, 11300 kg/m^3

Question:

A physical quantity P is related to four observables a, b, c and d as follows : P a^3b^2 / √cd 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 ? 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 ?

Answer:

Answer:

1. The percentage error in the quantity P is calculated as the square root of the sum of the squares of the percentage errors of the observables a, b, c and d, which is 5.07%.
2. The value of P should be rounded off to 3.76.

Question:

Estimate the average mass density of a sodium atom assuming its size to be about 2.5. (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 kgm3. Are the two densities of the same order of magnitude? If so, why?

Answer:

Step 1: Calculate the volume of a sodium atom. Volume = 4/3πr3 Volume = 4/3π (2.5)3 Volume = 65.44 cm3

Step 2: Calculate the mass of a sodium atom. Mass = Avogadro’s number × Atomic mass of sodium Mass = 6.02 × 1023 × 22.99 Mass = 1.37 × 1025 g

Step 3: Calculate the average mass density of a sodium atom. Density = Mass/Volume Density = 1.37 × 1025 g/65.44 cm3 Density = 2.09 × 1022 g/cm3

Step 4: Compare the density of a sodium atom with the density of sodium in its crystalline phase. The density of sodium in its crystalline phase is 970 kg/m3, which is much higher than the average mass density of a sodium atom (2.09 × 1022 g/cm3). Therefore, the two densities are not of the same order of magnitude.

Step 5: Explain why the two densities are not of the same order of magnitude. The two densities are not of the same order of magnitude because the density of a sodium atom is much lower than the density of sodium in its crystalline phase. This is because the density of a sodium atom is calculated on the basis of its individual atoms, whereas the density of sodium in its crystalline phase is calculated on the basis of the entire crystal structure. The crystal structure of sodium is much more dense than individual sodium atoms, resulting in a much higher density.

Question:

The unit of length convenient on the nuclear scale is a fermi : 1 f = 10−15 m. Nuclear sizes obey roughly the following empirical relation : r=r0​A^1/3 where, r is the radius of the nucleus, A its mass number, and ro​ is a constant equal to about 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus.

Answer:

Step 1: The formula for nuclear radius is r = r0A^1/3.

Step 2: The formula implies that the volume of a nucleus is proportional to A^1/3.

Step 3: The mass of a nucleus is proportional to A.

Step 4: Therefore, the mass density of a nucleus is proportional to A^-2/3.

Step 5: Since A^-2/3 is a small number, the mass density of a nucleus is nearly constant for different nuclei.

Step 6: The mass number of sodium nucleus is 23.

Step 7: The mass density of sodium nucleus can be estimated as follows: r0A^-2/3 = 1.2 * 10−15 * 23^-2/3 ≈ 2.1 * 10^14 kg/m^3.

Question:

The farthest objects in our universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

Answer:

1. Calculate the speed of light: Speed of light = 299,792,458 m/s

2. Calculate the time taken for light to travel from the quasar to the Earth: Time taken = 3.0 billion years = 3,000,000,000 years

3. Calculate the distance in km of the quasar from the Earth: Distance = Speed of light x Time taken Distance = 299,792,458 m/s x 3,000,000,000 years Distance = 899,377,374,000,000,000 km

Question:

In what range do you expect the density of the Sun to be, in the range of densities of solids and liquids or gases? Briefly explain with the help of the following data : mass of the Sun = 2.0×10^30kg, radius of the Sun =7.0×10^8m

Answer:

Answer: The density of the Sun can be expected to be in the range of gases. This is because the mass of the Sun (2.0×10^30kg) is much greater than the mass of a solid or liquid of the same size (7.0×10^8m). The mass-to-volume ratio of the Sun is much greater than that of solids and liquids, so its density must be much higher than that of either a solid or liquid.

Question:

A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair? A 0.035mm B 0.007mm C 3.5mm D None of the above

Answer:

Answer: B 0.007mm

Question:

It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely covers the disk of the Sun. From this fact .determine the approximate diameter of the moon.

Answer:

1. A total solar eclipse occurs when the Moon passes between the Sun and the Earth, and the Moon’s shadow falls on Earth.

2. The Moon’s shadow is approximately the same size as the Moon itself.

3. Therefore, the approximate diameter of the Moon can be determined by measuring the diameter of the Sun during a total solar eclipse.

4. The diameter of the Sun is approximately 864,400 miles.

5. Therefore, the approximate diameter of the Moon is 864,400 miles.

Question:

A man walking briskly in rain with speed v must slant his umbrella forward making an angle with the vertical. A student derives the following relation between θ and v:tanθ=v and checks that the relation has a correct limit: as v→0,θ→0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.

Answer:

No, this relation cannot be correct. The correct relation should be tanθ=v/g, where g is the acceleration due to gravity. This relation is based on the fact that the man must tilt his umbrella forward to keep the rain off his face, and the force of gravity is necessary to balance the force of the rain.

Question:

A LASER is a source of very intense, monochromatic, and unidirectional beam oflight. These properties of a laser light can be exploited to measure long distances.The distance of the Moon from the Earth has been already determined very preciselyusing a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moons surface. How much is the radius of the lunar orbit around the Earth ?

Answer:

1. Determine the speed of the laser light: Speed = Distance/Time = (Distance of the Moon from the Earth)/(2.56 seconds)

2. Calculate the circumference of the lunar orbit: Circumference = 2πr, where r is the radius of the lunar orbit.

3. Solve for the radius of the lunar orbit: r = Circumference/(2π) = (Speed x Time)/(2π) = (Distance of the Moon from the Earth x 2.56 seconds)/(2π)

Question:

The photograph of a house occupies an area of 1.75 cm^2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55m^2. What is the linear magnification of the projector-screen arrangement.

Answer:

1. Calculate the area of the photograph on the 35 mm slide: 1.75 cm^2

2. Calculate the area of the photograph on the screen: 1.55 m^2

3. Calculate the linear magnification of the projector-screen arrangement:

Linear magnification = (Area of photograph on screen / Area of photograph on 35 mm slide)^1/2

Linear magnification = (1.55 m^2 / 1.75 cm^2)^1/2

Linear magnification = 27.3

Question:

State the number of significant figures in the following: (a)0.007m^2 (b)2.64×10^24kg (c)0.2370gcm^−3 (d)6.320J (e)6.032Nm^−2 (f)0.0006032m^2

Answer:

(a) 3 significant figures (b) 8 significant figures (c) 4 significant figures (d) 4 significant figures (e) 5 significant figures (f) 5 significant figures

Question:

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 ?

Answer:

1. The nearest star to our solar system is 4.29 light years away. To convert this distance to parsecs, use the equation 1 parsec = 3.26 light years. Therefore, 4.29 light years is equal to 1.32 parsecs.

2. Alpha Centauri would show a parallax of 0.75 arcseconds when viewed from two locations of the Earth six months apart in its orbit around the Sun. This can be calculated using the equation parallax = 1/distance (in parsecs). Therefore, 0.75 arcseconds = 1/1.32 parsecs.

Question:

A famous relation in physics relates the moving mass m0​ of a particle in terms of its speed v and the speed of light c. (This relation first arose as a consequence of the special theory of relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes m=m0​​/(1−v^2)^1/2. Guess where to put the missing c.

Answer:

The correct relation is m=m0​/(1−v^2/c^2)^1/2. The missing c should be placed in the denominator, so that the denominator is (1−v^2/c^2).