GSEB Class 11 Physics Solutions Chapter 2 Units and Measurements

Question 1. Fill in the blanks:
(a) The volume of a cube of side 1 cm is equal to ................. m³.
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ................. (mm)².
(c) A vehicle moving with a speed of 18 kmh⁻¹ covers ................. m in 1s.
(d) The relative density of lead is 11.3. Its density is ................. g cm⁻³ or ................. kg m⁻³.
Answer:
(a) The volume of a cube of side 1 cm is given by, \( V = (1 \text{ cm})^3 \) or \( V = (10^{-2} \text{m})^3 = 10^{-6} \text{m}^3 \).
(b) The surface area of a solid cylinder of radius r and height h is given by:
\( A = \text{Area of two caps} + \text{curved surface area} \)
\( = 2\pi r^2 + 2\pi rh \)
Here \( r = 2 \text{ cm} = 20 \text{ mm} \), \( h = 10 \text{ cm} = 100 \text{ mm} \).
\( \therefore A = 2 \times \frac{22}{7} \times 20 (20 + 100) \text{ (mm)}^2 \)
\( = 15099 \text{ mm}^2 \)
\( = 1.5099 \times 10^4 \text{ mm}^2 \)
\( = 1.5 \times 10^4 \text{ mm}^2 \)
(c) Here \( v = 18 \text{ kmh}^{-1} = \frac{18 \times 1000 \text{m}}{3600 \text{s}} = 5 \text{ ms}^{-1} \)
\( t = 1 \text{s} \)
\( \therefore x = ut = 5 \times 1 = 5 \text{m} \)
(d) Relative density of lead = 11.3,
density of water = 1 gm cm⁻³
We know that relative density of lead
\( = \frac{\text{density of lead}}{\text{density of water}} \)
\( \therefore \text{density of lead} = \text{relative density of lead} \times \text{density of water} \)
\( = 11.3 \times 1 \text{ gm cm}^{-3} \)
\( = 11.3 \text{ gm cm}^{-3} \)
Also in S.I. system density of water \( = 10^3 \text{ kg m}^{-3} \)
\( \therefore \text{density of lead} = 11.3 \times 10^3 \text{ kgm}^{-3} \)
\( = 1.13 \times 10^4 \text{kgm}^{-3} \)
(a) \( 10^{-6} \)
(b) \( 1.5 \times 10^4 \)
(c) \( 5 \text{m} \)
(d) 11.3, \( 1.13 \times 10^4 \)
In simple words: We convert all measurements to standard units. For volume, a 1 cm cube becomes \( 10^{-6} \text{m}^3 \). For surface area, we calculate it in \( \text{mm}^2 \) using the given radius and height. For distance, we change \( \text{kmh}^{-1} \) to \( \text{ms}^{-1} \) and multiply by time. For density, we multiply the relative density by the density of water in both g \( \text{cm}^{-3} \) and kg \( \text{m}^{-3} \) units.

Exam Tip: Remember to always convert all units to a consistent system (like SI or CGS) before performing calculations. Be careful with exponents and unit conversions, especially for powers of ten.

Question 2. Fill in the blanks by suitable conversion of units.
(a) \( 1 \text{ kg m}^2 \text{s}^{-2} = \) ................. g cm² s⁻²
(b) \( 1 \text{ m} = \) ................. ly (light year)
(c) \( 3.0 \text{ ms}^{-2} = \) ................. kmh⁻²
(d) \( G = 6.67 \times 10^{-11} \text{ Nm}^2 \text{ kg}^{-2} = \) ................. cm³ s⁻² g⁻¹.
Answer:
(a) \( 1 \text{ kg m}^2 \text{s}^{-2} = (10^3 \text{g}) (10^2 \text{cm})^2 \text{s}^{-2} \)
\( = 10^3 \times 10^4 \text{g cm}^2 \text{s}^{-2} \)
\( = 10^7 \text{g cm}^2 \text{s}^{-2} \)
(b) \( 1 \text{ light year (ly)} = 9.46 \times 10^{15} \text{m} \)
\( \therefore 1 \text{m} = \frac{1}{9.46 \times 10^{15}} \text{ly} = 1.057 \times 10^{-16} \text{ly} \)
(c) \( 3.0 \text{ ms}^{-2} = 3 \times 10^{-3} \text{ km} \times (\frac{1}{60 \times 60} \text{ h}^{-1})^{-2} \)
\( = 3 \times 10^{-3} \times (3600)^2 \text{ kmh}^{-2} \)
\( = 3.888 \times 10^4 \text{ kmh}^{-2} \)
\( = 3.9 \times 10^4 \text{ kmh}^{-2} \)
(d) \( G = 6.67 \times 10^{11} \text{ Nm}^2 \text{kg}^{-2} \)
\( = 6.67 \times 10^{-11} (10^5 \text{ dyne}) (10^2 \text{ cm})^2 (10^3 \text{ g})^{-2} \)
\( = 6.67 \times 10^{-11} \times 10^5 \times 10^4 \times 10^{-6} \text{ dyne cm}^2 \text{ g}^{-2} \)
\( = 6.67 \times 10^{-8} (\text{g cm s}^{-2}) \text{ cm}^2 \text{ g}^{-2} \)
\( = 6.67 \times 10^{-8} \text{ cm}^3 \text{ g}^{-1} \text{ s}^{-2} \).
In simple words: To convert units, we replace each base unit with its equivalent in the new system. For example, kilograms become grams, meters become centimeters, and seconds stay seconds unless converted to hours. We must carefully apply the powers to these conversions.

Exam Tip: When converting units, treat the units algebraically. Multiply by conversion factors that are equal to 1, ensuring the original units cancel out and the desired units remain. Pay close attention to exponents.

Question 3. A calorie is a unit of heat or energy and it equals about 4.2 J, where 1J = 1 kg m²s⁻². Suppose we employ a system of units in which the unit of mass equals \( \alpha \)kg, the unit of length \( \beta \)m, the unit of time is \( \gamma \)s. Show that a calorie has a magnitude. \( 4.2 \alpha^{-1} \beta^{-2} \gamma^2 \) in terms of new units?
Answer:
We know that \( n_1 u_1 = n_2 u_2 \)
\( n_2 = n_1 \frac{u_1}{u_2} \)
\( = n_1 \frac{[M_1^a L_1^b T_1^c]}{[M_2^a L_2^b T_2^c]} \)
\( = n_1 [\frac{M_1}{M_2}]^a [\frac{L_1}{L_2}]^b [\frac{T_1}{T_2}]^c \)
1 cal \( = 4.2 \text{ J} = 4.2 \text{ kg m}^2 \text{s}^{-2} \),
\( \therefore a = 1, b = 2, c = -2 \)
SI system:
\( n_1 = 4.2 \)
\( M_1 = 1 \text{ kg} \)
\( L_1 = 1 \text{ m} \)
\( T_1 = 1 \text{ s} \)
New system:
\( n_2 = ? \)
\( M_2 = \alpha \text{ kg} \)
\( L_2 = \beta \text{ m} \)
\( T_2 = \gamma \text{ s} \)
\( \therefore n_2 = 4.2 [\frac{1 \text{kg}}{\alpha \text{kg}}]^1 [\frac{1 \text{m}}{\beta \text{m}}]^2 [\frac{1 \text{s}}{\gamma \text{s}}]^{-2} \)
\( n_2 = 4.2 \alpha^{-1} \beta^{-2} \gamma^2 \)
\( \therefore 1 \text{ cal} = 4.2 \alpha^{-1} \beta^{-2} \gamma^2 \) in new system
Hence proved.
In simple words: We want to change the units of a calorie into a new system. We use a formula that relates the numerical value of a quantity in one unit system to another. By plugging in the exponents for mass, length, and time from the Joule's definition and the new unit definitions, we can show that 1 calorie is \( 4.2 \alpha^{-1} \beta^{-2} \gamma^2 \) in the new units.

Exam Tip: For unit conversion problems, always write down the dimensions of the quantity in both old and new systems. Use the conversion factor formula \( n_2 = n_1 (\frac{M_1}{M_2})^a (\frac{L_1}{L_2})^b (\frac{T_1}{T_2})^c \) carefully, ensuring each term is raised to its correct dimensional power.

Question 4. Explain the statement clearly: "To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison.” In view of this, reframe the following statements wherever necessary:
1. Atoms are very small objects.
2. A jet plane moves with great speed.
3. The mass of Jupiter is very large.
4. The air inside this room contains a large number of molecules.
5. A proton is much more massive than an electron.
6. The speed of sound is much smaller than the speed of light.
Answer:
The statement is true. Physical quantities are called large or small in comparison to some standard (or the unit) of measurement. For example, the distance between two cities on Earth is measured in kilometers, but the distance between stars or intergalactic distances are measured in Parsec. The latter standard Parsec is \( 3.08 \times 10^{16} \text{m} \) or \( 3.08 \times 10^{13} \text{ km} \), which is larger than meters or kilometers. Therefore, interstellar or inter-galactic distances are certainly larger than the distances between two cities on Earth.
1. The size of an atom is smaller than the sharp tip of a pin.
2. A jet plane moves faster than a superfast train.
3. The mass of Jupiter is very large as compared to the mass of Earth.
4. The air inside this room contains more number of molecules than in one mole of air.
5. The statement is already correct.
6. The statement is already correct.
In simple words: The idea of "large" or "small" only makes sense when you compare it to something else. Without a comparison, these words have no real meaning. For instance, an atom is small compared to a pin's tip, and a jet plane is fast compared to a superfast train. Some statements are already clear, like a proton being much heavier than an electron.

Exam Tip: When evaluating comparative statements in physics, always identify the reference point or "standard for comparison." A quantity's size or magnitude is only meaningful when measured against another related quantity or a defined unit.

Question 5. A new unit of length is chosen such that the speed of light in vaccum is unity. What is the distance between the Sun and the Earth in terms of the new unit of light takes 8 min and 20s to cover this distance?
Answer:
We are given that velocity of light in vacuum, \( c = 1 \) new unit of length \( \text{s}^{-1} \).
Time taken by light of sun to reach the Earth, \( t = 8 \text{ min } 20 \text{s} \)
\( = 8 \times 60 + 20 = 500 \text{s} \)
\( \therefore \) distance between the Sun and the Earth,
\( x = c \times t = 1 \text{ new unit of length s}^{-1} \times 500 \text{s} \)
\( = 500 \) new units of length
Aliter: Distance between Sun and Earth
\( = \text{speed of light in vacuum} \times \text{time taken by light to travel from Sun to Earth} \).
\( = 3 \times 10^8 \text{ms}^{-1} \times 500 \text{s} = 3 \times 10^8 \text{ m} \)
In the new system, the speed of light in vacuum is unity
\( \therefore \) the new unit of length is \( 3 \times 10^8 \text{ m} \)
\( \therefore \) distance between Sun and Earth \( = 500 \) new units.
In simple words: If we use a new unit where the speed of light is 1, and we know light takes 8 minutes and 20 seconds to travel from the Sun to the Earth, we simply multiply this time by the speed of light (which is 1 in our new units). This tells us the distance in these new units.

Exam Tip: In problems involving new unit systems, always clearly define the new unit in terms of standard units. Then, apply basic kinematic equations, ensuring all values are expressed consistently in the chosen unit system before calculations.

Question 6. Which of the following is the most precise device for measuring length:
1. A vernier callipers with 20 divisions on the sliding scale.
2. A screw gauge of pitch 1 mm and 100 divisions on the circular scale.
3. An optical instrument that can measure length to within a wavelength of light?
Answer:
The most precise device is that whose least count is minimum. Now:
1. Least count of vernier callipers
\( = 1 \text{ MSD} - 1 \text{ VSD} \).
\( = \frac{1}{20} \text{mm} = \frac{1}{200} \text{ cm} \)
\( = 0.005 \text{ cm} \)
2. Least count of screw gauge \( = \frac{\text{pitch}}{\text{no. of divisions on circular scale}} \)
\( = \frac{1}{100} \text{ mm} = \frac{1}{1000} \text{ cm} \)
\( = 0.001 \text{ cm} \)
3. Wavelength of light, \( \lambda \approx 10^{-5} \text{ cm} = 0.00001 \text{ cm} \)
\( \therefore \) Least count of optical instrument \( = 0.00001 \text{ cm} \)
Thus, clearly the optical instrument is the most precise.
In simple words: The best tool for measuring length is the one that can measure the smallest differences, called the least count. Comparing vernier calipers, a screw gauge, and an optical instrument, the optical instrument has the smallest least count because it measures down to the wavelength of light, making it the most precise option.

Exam Tip: Precision in measurement is determined by the least count of the instrument. The smaller the least count, the more precise the measurement. Always calculate and compare the least counts of different instruments to determine which is most precise.

Question 7. 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 3.5mm. What is the estimate on the thickness of hair?
Answer:
Estimate on the thickness of hair is given by:
Magnification \( = \frac{\text{Observed width}}{\text{Real width}} \)
Real width \( = \frac{\text{Observed width}}{\text{magnification}} \)
\( = \frac{3.5 \text{mm}}{100} = 0.035 \text{ mm} \)
\( = 3.5 \times 10^{-2} \text{ mm} \).
In simple words: To find the real thickness of the hair, we divide the width seen through the microscope by the microscope's magnification. The student saw the hair as 3.5mm wide under 100x magnification, so the actual thickness is 0.035mm.

Exam Tip: Remember that magnification makes objects appear larger. To find the actual size of an object viewed through a magnifying instrument, divide the observed size by the magnification factor.

Question 8. 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 diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurement of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Answer:
(a) The diameter of a thread is so small that it cannot be measured directly using a meter scale. We wind the thread closely on a pencil or a cylindrical glass rod so that the turns touch one another and are tight. Count the number of turns (n) and find the average length (l) of the coiled thread using the meter scale. Divide this average length by the number of turns to get the diameter, i.e.,
Diameter \( = \frac{\text{Average length}}{\text{Number of turns}} = \frac{l}{n} \)
(b) Yes, the accuracy of the gauge increases by increasing the number of divisions of the circular scale because the least count is given by:
Least count \( = \frac{\text{pitch}}{\text{no. of divisions on the circular scale}} \)
Practically, it may not be possible to take the reading precisely due to low resolution of the human eye.
(c) By taking a large number of observations (100) and then taking their mean for the result, the random error reduces significantly compared to taking only five observations and then taking their mean. Therefore, a set of 100 measurements of the diameter is expected to yield a more reliable estimate than a set of 5 measurements only.
In simple words: (a) To measure a thin thread's diameter with a meter scale, wrap it many times around a pencil, measure the total length, and divide by the number of turns. (b) Increasing divisions on a screw gauge improves accuracy by reducing the least count, but human eyesight limits how much we can improve. (c) Taking many measurements and averaging them helps reduce random errors, making the result more dependable than just a few measurements.

Exam Tip: When dealing with small measurements, indirect methods like the winding method for thread diameter are often necessary. Remember that while instrumental precision can be enhanced, practical limitations like human vision can affect the overall accuracy of a measurement. Averaging multiple readings is a key strategy to minimize random errors.

Question 9. The photograph of a house occupies an area of 1.75 cm² on a 35mm slide. The slide is projected on to a screen and the area of the house on the screen is 1.55m². What is the linear magnification of the projector-screen arrangement?
Answer:
Here, size of an object \( = \text{Area of object} \)
\( = 1.75 \text{ cm}^2 = 1.75 \times 10^{-4} \text{ m}^2 \)
Size of the image \( = \text{Area of the image} \)
\( = 1.55 \text{ m}^2 \)
Areal magnification \( = \frac{\text{size of image}}{\text{size of the object}} \)
\( = \frac{1.55 \text{m}^2}{1.75 \times 10^{-4} \text{m}^2} = 8857.1 \)
\( \therefore \) Linear magnification \( = \sqrt{8857.1} = 94.1 \).
In simple words: First, we convert the slide's area to square meters. Then, we find the areal magnification by dividing the screen area by the slide's area. Since linear magnification is the square root of areal magnification, we take the square root of the result to get the linear magnification.

Exam Tip: Remember the relationship between linear and areal (or volumetric) magnification. Areal magnification is the square of linear magnification (\( M_{area} = M_{linear}^2 \)), and volumetric magnification is the cube of linear magnification (\( M_{vol} = M_{linear}^3 \)). Always ensure units are consistent before calculating ratios.

Question 10. State the number of significant figures in the following:
1. \( 0.007 \text{m}^2 \)
2. \( 2.64 \times 10^{24} \text{ kg} \)
3. \( 0.2370 \text{ g cm}^{-3} \)
4. \( 6.320 \text{ J} \)
5. \( 6.032 \text{ Nm}^{-2} \)
6. \( 0.0006032 \text{ m}^2 \)
Answer:
The number of significant figures is as given below:
1. 1
2. 3
3. 4
4. 4
5. 4
6. 4
In simple words: We need to count the important digits in each number. Leading zeros (zeros before the first non-zero digit) are not significant. Trailing zeros (zeros at the end) are significant only if the number contains a decimal point. Non-zero digits are always significant. Digits between non-zero digits are also significant.

Exam Tip: To count significant figures: (1) All non-zero digits are significant. (2) Zeros between non-zero digits are significant. (3) Leading zeros (before non-zero digits) are not significant. (4) Trailing zeros are significant only if there is a decimal point. Numbers in scientific notation follow these rules for the mantissa.

Question 11. The length, breadth and thickness of a rectangular sheet of metal are 4.234m, 1.005m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures?
Answer:
Here, length, \( l = 4.234 \text{m} \)
breadth, \( b = 1.005 \text{m} \)
thickness, \( h = 0.0201 \text{m} = 2.01 \text{ cm} \)
Area of the sheet \( = 2 (lb + bh + hl) \).
\( = 2(4.234 \times 1.005 + 1.005 \times 0.0201 + 0.0201 \times 4.234) \)
\( = 8.7209468 \text{m}^2 \)
As the least number of significant figures in thickness is 3,
\( \therefore \) Area \( = 8.72 \text{ m}^2 \)
Volume \( = l \times b \times h \)
\( = 4.234 \times 1.005 \times 0.0201 \text{m}^3 \)
\( = 0.0855 \text{m}^3 \).
In simple words: First, ensure all measurements are in the same unit, converting centimeters to meters. Then, calculate the total surface area and volume using the given formulas. For the final answer, round these calculations to the smallest number of significant figures present in the original measurements, which is three, based on the thickness.

Exam Tip: When performing calculations with measured values, the final result should be rounded to the same number of significant figures as the measurement with the fewest significant figures. For addition/subtraction, round to the least number of decimal places.

Question 12. The mass of a box measured by a grocer's balance is 2.3 kg. Two gold pieces of masses 20.15g and 20.17g are added to the box. What is (a) the total mass of the box (b) the difference in the mass of the pieces to correct significant figures?
Answer:
Here, mass of the box, \( m = 2.300 \text{ kg} \)
Mass of one gold piece, \( m_1 = 20.15 \text{g} = 0.02015 \text{ kg} \)
Mass of second gold piece, \( m_2 = 20.17 \text{g} = 0.02017 \text{ kg} \)
(a) Total mass \( = m + m_1 + m_2 \)
\( = (2.300 + 0.02015 + 0.02017) \text{ kg} \)
\( = 2.34032 \text{ kg} \)
As the least number of significant figures in the mass of box is 2, so maximum number of significant figures in the result can be 2
\( \therefore \) Total mass \( = 2.3 \text{ kg} \)
(b) Difference in masses \( = m_2 - m_1 \)
\( = 20.17 - 20.15 \)
\( = 0.02 \text{g} \)
Since there are two significant figures in the difference (0.02g), the difference in masses to the correct significant figures is \( 0.02 \text{g} \).
In simple words: First, convert all masses to the same unit (kilograms). Then, to find the total mass, add all the masses together. The final answer for total mass should be rounded to match the precision of the least precise measurement. For the difference in masses, simply subtract the smaller gold piece mass from the larger one, keeping the correct number of significant figures.

Exam Tip: When adding or subtracting measurements, the result should have no more decimal places than the measurement with the fewest decimal places. When multiplying or dividing, the result should have no more significant figures than the measurement with the fewest significant figures.

Question 13. A physical quantity P is related to four observables a, b, c, and d as follows: \( P = \frac{a^3 b^2}{\sqrt{cd}} \) 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:
\( P = \frac{a^3 b^2}{\sqrt{c} d} \)
\( \therefore \) Relative error in P is given by:
\( \frac{\Delta P}{P} = 3\frac{\Delta a}{a} + 2\frac{\Delta b}{b} + \frac{1}{2} \frac{\Delta c}{c} + \frac{\Delta d}{d} \)
\( \therefore \) Percentage error in P is given by:
\( \frac{\Delta P}{P} \times 100 = (3 \frac{\Delta a}{a} \times 100) + (2 \frac{\Delta b}{b} \times 100) + (\frac{1}{2} \frac{\Delta c}{c} \times 100) + (\frac{\Delta d}{d} \times 100) \)
Here,
\( \frac{\Delta a}{a} \times 100 = 1\% \), \( \frac{\Delta b}{b} \times 100 = 3\% \)
\( \frac{\Delta c}{c} \times 100 = 4\% \), \( \frac{\Delta d}{d} \times 100 = 2\% \)
\( \therefore \frac{\Delta P}{P} \times 100 = 3 \times 1\% + 2 \times 3\% + \frac{1}{2} \times 4\% + 2\% \)
\( = 3 + 6 + 2 + 2 = 13\% \)
The calculation of error clearly shows that the number of significant figures is 2, so the result of P may be rounded off to two significant digits i.e. \( P = 3.763 = 3.8 \).
In simple words: To find the percentage error in P, we sum the percentage errors of a, b, c, and d, multiplying each by its power in the formula. For example, the error in 'a' is multiplied by 3. Then, we round the calculated value of P to the number of significant figures determined by the least precise measurement in the input.

Exam Tip: When calculating percentage error for a quantity derived from multiple measurements, remember that powers become multipliers for percentage errors, and roots become fractional multipliers. For rounding, the final result should reflect the lowest number of significant figures from the input values.

Question 14. A book with many printing errors contains four different formulae for the displacement y of a particle under going a certain periodic motion.
(a) \( y = a \sin \frac{2\pi t}{T} \)
(b) \( y = a \sin vt \)
(c) \( y = (\frac{a}{T}) \sin (\frac{t}{a}) \)
(d) \( y = (\frac{a}{\sqrt{2}}) (\sin \frac{2\pi t}{T} + \cos \frac{2\pi t}{T}) \)
(a = maximum displacement of the particle, v = speed of the particle, T = time period of motion). Rule out the wrong formulae on dimensional grounds.
Answer:
Let us use the principle of homogeneity of dimensions. If in a given relation, the dimensions of each term are the same, then it will be correct; otherwise, it is wrong. The argument of a trigonometric function, i.e., an angle, is dimensionless. Now here in each case, dimensions of L.H.S. is [L] and dimensions of R.H.S. are:
(a) \( = [L] \) (angle \( \frac{2\pi t}{T} \) is dimensionless)
(b) \( = [L]\sin[(LT^{-1})(T)] = [L]\sin[L] \) (angle is not dimensionless here)
(c) \( = \frac{[L]}{[T]} \sin \frac{[T]}{[L]} = [LT^{-1}] \) (angle is not dimensionless here)
(d) \( = [L] [\sin \frac{T}{T} + \cos \frac{T}{T}] = [L] \)
\( \therefore \) formulae (b) and (c) are wrong.
In simple words: We check each formula using dimensional analysis. For a formula to be correct, the dimensions on both sides must match. Also, the input to trigonometric functions (like sin or cos) must always be dimensionless. If these rules are broken, the formula is incorrect. Formulas (b) and (c) fail these dimensional rules.

Exam Tip: The Principle of Homogeneity states that a physical equation is dimensionally correct if the dimensions of the terms on both sides of the equation are the same. Also, arguments of trigonometric, logarithmic, and exponential functions must always be dimensionless.

Question 15. A famous relation in physics relates 'moving mass' m to the 'rest mass' \( m_0 \) of a particle in terms of its speed v and speed of light c. (This relation first arose as a 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 \( m = \frac{m_0}{(1-v^2)^{1/2}} \). Guess where to put the missing c?
Answer:
According to the principle of homogeneity of dimensions, powers of M, L, T on either side of the formulae must be equal. For this, on R.H.S., the denominator \( (1 - v^2)^{1/2} \) must be dimensionless. 1 is dimensionless. So factor at \( v^2 \) must be dimensionless i.e. there must be \( v^2/c^2 \). Hence the correct formula should be,
\( m = \frac{m_0}{\left(1-\frac{v^2}{c^2}\right)^{1 / 2}} \).
In simple words: For a physics formula to be correct, all parts of it must have consistent units. In this case, the part inside the square root in the denominator must be unitless. Since \( v^2 \) has units of speed squared, it needs to be divided by another speed squared to make it unitless. That missing speed is the speed of light, \( c \), so the expression becomes \( v^2/c^2 \).

Exam Tip: Always use the principle of dimensional homogeneity to verify physical equations. Every term in an equation must have the same dimensions. This is particularly useful for identifying missing constants or correcting the form of expressions involving physical quantities.

Question 16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å. \( 1 \text{ Å} = 10^{-10} \text{m} \). The size of the hydrogen atom is about \( 0.5 \text{Å} \). What is the total atomic volume in \( \text{m}^3 \) of a mole of hydrogen atoms?
Answer:
Here \( r = 0.5 \text{Å} \)
\( = 0.5 \times 10^{-10} \text{ m} \)
\( V_1 = \text{Volume of each hydrogen atom} \)
\( = \frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.14 \times (0.5 \times 10^{-10})^3 \)
\( = 5.236 \times 10^{-31} \text{m}^3 \)
According to Avogadro's hypothesis, one mole of hydrogen contains:
\( N = 6.023 \times 10^{23} \) atoms
Atomic volume of 1 mole of hydrogen atoms,
\( V = NV_1 \)
or \( V = 6.023 \times 10^{23} \times 5.236 \times 10^{-31} \)
\( = 3.154 \times 10^{-7} \text{m}^3 \)
\( = 3 \times 10^{-7} \text{m}^3 \).
In simple words: First, convert the hydrogen atom's radius from angstroms to meters. Then, use the formula for the volume of a sphere to find the volume of a single hydrogen atom. Finally, multiply this single atom's volume by Avogadro's number (the number of atoms in one mole) to find the total atomic volume for a mole of hydrogen atoms.

Exam Tip: Remember to convert all length units to SI units (meters) before calculating volume. Avogadro's number is a key constant for converting between properties of single atoms/molecules and a mole of that substance.

Question 17. One mole of an ideal gas at standard temperature and pressure occupies 22.4L (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 \( 1 \text{Å} \)). Why is this ratio so large?
Answer:
d = diameter of hydrogen molecule \( = 1 \text{Å} \)
Molar volume of one mole of hydrogen
\( = 22.4 \text{L} = 22.4 \times 10^{-3} \text{m}^3 \)
r = radius of one molecule of hydrogen
\( = \frac{d}{2} = 0.5 \text{Å} \)
\( = 0.5 \times 10^{-10} \text{m} \)
Volume of one molecule of hydrogen
\( = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.5 \times 10^{-10})^3 \)
\( = 5.236 \times 10^{-31} \text{m}^3 \)
1 mole has \( 6.023 \times 10^{23} \) atoms or molecules of H\( _2 \)
\( \therefore \) Atomic volume of one mole of hydrogen
\( = 6.023 \times 10^{23} \times 5.236 \times 10^{-31} \text{m}^3 \)
\( = 3.154 \times 10^{-7} \text{m}^3 \)
\( \therefore \frac{\text{Molar volume}}{\text{Atomic volume}} = \frac{22.4 \times 10^{-3} \text{m}^{3}}{3.154 \times 10^{-7} \text{m}^{3}} \)
\( = 7.1 \times 10^4 \)
\( = 7 \times 10^4 \)
The large value of the ratio shows that the intermolecular separation in a gas is much larger than the size of a molecule.
In simple words: First, convert the molar volume from liters to cubic meters and the hydrogen molecule's diameter to meters to find its radius. Calculate the volume of one hydrogen molecule. Multiply this by Avogadro's number to get the total atomic volume for a mole of hydrogen. Finally, divide the molar volume by the atomic volume. This ratio is very large because the particles in a gas are very far apart compared to their actual size.

Exam Tip: Pay close attention to unit conversions (liters to \( \text{m}^3 \), angstroms to meters). The large ratio between molar volume and atomic volume for a gas signifies that gases are mostly empty space, with molecules far apart compared to their size.

Question 18. 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 train's 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:
The line joining the object to the eye is called the line of sight. When a train moves rapidly, the line of sight of a nearby tree changes its direction of motion rapidly, meaning near objects make a greater angle change than distant objects. Therefore, the trees appear to run in the opposite direction.
On the other hand, the angular change, i.e., the line of sight of far-off objects (hill tops, the moon, the stars etc.), changes its direction extremely slowly, and hence the relative shift in their position is negligible. Therefore, they appear to be stationary, or they appear to move with the observer in the train.
In simple words: When you're on a fast train, things close by, like trees, seem to zoom past in the opposite direction because their angle of view changes quickly. But things far away, like hills or the moon, hardly change their angle of view, so they appear to stay still or even move along with the train.

Exam Tip: This phenomenon relates to angular velocity. Nearby objects have a larger angular velocity relative to an observer in motion, making them appear to move quickly. Distant objects have a much smaller angular velocity, causing them to appear stationary or move with the observer.

Question 19. 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 \( = 3 \times 10^{11} \text{m} \). 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?
Answer:
We know that parallax of \( \theta \) of a star is the angle made by the semi-major axis of the Earth's orbit \( l \) to the direction of the star as shown in figure.
From parallax method, we know that \( S = \frac{b}{\theta} \)

Here \( \theta' = \angle AOB = 2\angle COB = 2\theta \), where \( \theta \) is the parallax angle.
\( \theta \) = position star
\( b = \text{AB} = \text{base line} \)
\( = \text{Diameter of Earth's orbit} = 3 \times 10^{11} \text{m} \)
\( \theta = 1 \text{ second of arc} = 4.85 \times 10^{-6} \text{rad} \)
\( S = \frac{b}{2\theta} = \frac{3 \times 10^{11}}{2 \times 4.85 \times 10^{-6}} = 3.09 \times 10^{16} \text{m} \)
Also \( 1 \text{ parsec} = 3.08 \times 10^{16} \text{m} \)
In simple words: Parallax uses the shift in a star's apparent position as Earth moves around the Sun to calculate its distance. We use Earth's orbital diameter as our baseline. A parsec is defined as the distance at which a star would show a parallax of one arcsecond. By converting this angle and using the baseline, we can calculate that one parsec is about \( 3.08 \times 10^{16} \) meters.

Exam Tip: Remember that parallax is an angle, and for astronomical distances, it is very small, measured in arcseconds. The relationship \( S = \frac{b}{\theta} \) (where \( \theta \) is in radians) is fundamental for calculating distances using parallax, and 1 parsec is defined based on a parallax angle of 1 arcsecond with a baseline of 1 astronomical unit (Earth-Sun distance).

Question 20. 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:
Distance \( = 4.29 \text{ light year} \)
\( = 4.29 \times 9.46 \times 10^{15} \text{m } \) (\( 1 \text{ ly} = 9.46 \times 10^{15} \text{m} \))
\( = \frac{4.29 \times 9.46 \times 10^{15}}{3.08 \times 10^{16}} \) (\( : 1 \text{ parsec} = 3.08 \times 10^{16} \text{m} \))
\( = 1.318 \text{ parsec} = 1.32 \text{ parsec} \)
Aliter: We know that \( 3.26 \text{ ly} = 1 \text{ parsec} \)
\( \therefore 4.29 \text{ light year} = \frac{4.29}{3.26} = 1.32 \text{ parsec} \)
Required parallax \( = 2\theta \). where \( \theta \) is the annual parallax. Since parsec is the distance corresponding to an annual parallax of one second of an arc so, \( \theta = 1.32 \) second of an arc.
Required Parallax \( = 2\theta \) where \( \theta \) is the annual parallax
Since parsec is the distance corresponding to an annual parallax of one second of an arc so, \( \theta = \frac{1''}{\text{distance in parsec}} \).
\( \theta = \frac{1''}{1.32} = 0.757 \text{ second of arc} \).
\( \therefore \) Required Parallax \( = 2 \times 0.757 = 1.514 \text{ second of arc} \).
In simple words: First, convert the star's distance from light-years to meters, then to parsecs, knowing that 1 parsec is approximately \( 3.08 \times 10^{16} \) meters. To find the parallax, we use the fact that parallax in arcseconds is the inverse of the distance in parsecs. Since the question asks for parallax from two Earth locations six months apart (which uses a baseline equal to Earth's orbit diameter), we use \( 2\theta \), where \( \theta \) is the annual parallax.

Exam Tip: Remember the direct conversion: \( 1 \text{ parsec} \approx 3.26 \text{ light-years} \). For parallax calculations, the angle \( \theta \) (in arcseconds) is related to distance \( D \) (in parsecs) by \( \theta = 1/D \). When viewing from two locations six months apart, the total parallax observed is twice the annual parallax.

Question 21. Precise measurements of physical quantities are a need of science. For example, to ascertain 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 Word War II. Think of different examples in modern science where precision measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed?
Answer:
1. Precision measurement of time and mass are needed for nuclear and atomic reactions, in nuclear weapons, in nuclear power plants etc. (order \( 10^{-9} \text{s} \) and \( 10^{-9} \text{ kg} \)).
2. Precise location, size and mass of tumor in body are very essential for radio and laser therapy. It is of the order of \( 10^{-9} \text{m} \).
3. The vast applications of laser depend on the interval of time and distance. The time involved is of the order of \( 10^{-9} \text{s} \).
4. The whole range of thin film technology, semiconductor devices depends on the development of precise thickness of the layer of material. The function layer also plays an important role. The common range of these thickness are micro (micrometer \( \mu \)). Apart from this molecular thick layers in photographic films, metallurgy, anti corrosive electroplating and many other devices are needed.
5. High technological satellite launching and putting in, orbits, super computers, ultra and copy, spectroscopy all need very precise measurements of mass, length and time from micro (\( 10^{-6} \)) to nano (\( 10^{-9} \)) or even upto fermi (\( 10^{-15} \text{m} \)) levels.
In simple words: Accurate measurements are crucial in many scientific areas. For nuclear reactions and power, very precise time and mass measurements are needed, often down to \( 10^{-9} \) units. In medical treatments like laser therapy, knowing the exact location and size of a tumor to \( 10^{-9} \) meters is vital. Lasers themselves rely on precise timing and distance measurements, again in the \( 10^{-9} \) second range. Advanced technologies like thin films and semiconductors require exact layer thicknesses, typically in micrometers or even finer. Finally, complex systems like satellite launches and supercomputers demand extremely precise measurements across vast scales.

Exam Tip: Be prepared to provide specific examples from various fields of science and technology where precision is paramount. For each example, try to quantify the level of precision (e.g., order of magnitude in meters, seconds, or kilograms) to demonstrate a deeper understanding.

Question 22. 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 our head.
(e) the number of air molecules in your classroom.
Answer:
(a) The mass of rainwater over India, which is the same as the mass of clouds, can be estimated. You multiply the average rainfall by the area of India and the density of water. So, \( 10 \, \text{cm} \times 3.3 \times 10^{12} \, \text{m}^2 \times 10^3 \, \text{kgm}^{-3} = 10 \times 10^{-2} \, \text{m} \times 3.3 \times 10^{15} \, \text{kgm}^{-1} = 3.3 \times 10^{14} \, \text{kg} \). This gives a very large estimate for the total mass.
(b) The mass of an elephant can be estimated using the lever principle. An adult elephant usually weighs around 3000 kg. Another way to figure this out is to put the elephant into a boat with a known bottom area. By measuring how deep the boat sinks with the elephant inside, we can find the volume of water that the boat and elephant move. Then, we take away the volume of water moved by the empty boat. This leaves us with the volume of water moved by just the elephant. If we multiply this volume by the density of water, we will get the elephant's mass.
(c) The wind speed during a storm can be measured by a few tools, such as balloons. Storm winds typically start at 80 km/h and can reach speeds of up to 300 km/h in very strong storms.
(d) The number of human hair strands on a head is calculated by dividing the total area of the head by the cross-sectional area of a single hair. If we assume hair thickness \( t = 5 \times 10^5 \, \text{m} \) (or \( 5 \times 10^3 \, \text{m} \)), and the average radius of a human head \( r = 8 \, \text{cm} \), then the head's area is \( \pi r^2 = \pi (8)^2 = 64\pi \, \text{cm}^2 \). The cross-sectional area of a hair is \( \pi \frac{t^2}{4} = \pi \times \frac{25}{4} \times 10^{-6} \, \text{cm}^2 \). By dividing \( \frac{64 \pi \times 4}{25 \pi \times 10^{-6}} \), we get approximately \( 10 \times 10^6 = 10^7 \) hairs. This suggests a very large number of hairs.
(e) To find the number of air molecules in a classroom, we use Avogadro's number. One mole of air at standard temperature and pressure takes up \( 22.4 \) liters, which is \( 22.4 \times 10^{-3} \, \text{m}^3 \). We also know that one mole contains Avogadro's number of atoms, \( N = 6.023 \times 10^{23} \). So, the number of air molecules in a classroom is given by \( \frac{\text{Volume of classroom}}{22.4 \times 10^{-3} \, \text{m}^3} \times N \). Using the formula \( \frac{6.023 \times 10^{23}}{22.4 \times 10^{-3} \, \text{m}^3} \times V = 0.2689 \times 10^{26} \times V \), and assuming a typical classroom size is \( 5 \, \text{m} \times 4 \, \text{m} \times 3 \, \text{m} = 60 \, \text{m}^3 \), we find the number of molecules to be \( 0.2689 \times 10^{26} \times 60 \approx 16 \times 10^{26} \) molecules, which simplifies to approximately \( 10^{27} \) molecules. This is an incredibly large number.
In simple words: Estimation involves using basic concepts and common knowledge to make a rough calculation. For clouds, multiply rainfall by area and water density. For an elephant, use a lever or water displacement. Wind speed is found with devices like balloons, typically ranging from 80 to 300 km/h. Hair count is total head area divided by hair cross-section. Air molecules in a classroom are found by multiplying Avogadro's number by the room's volume, divided by molar volume.

Exam Tip: When making estimates, clearly state any assumptions you make and always try to provide both a method and a numerical range or value.

Question 23. The sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding \( 10^7 \)K and its outer surface at a temperature of about 6000K. 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 Sun = \( 2.0 \times 10^{30} \) kg, radius of the Sun = \( 7.0 \times 10^8 \)m.
Answer: Given the Sun's characteristics, with its inner core hotter than \( 10^7 \, \text{K} \) and its outer layer around 6000 K, no material can stay solid or liquid. We would expect its mass density to be similar to solids and liquids, rather than gases. To confirm this, let's calculate the density using the given data: Mass of Sun \( M = 2.0 \times 10^{30} \, \text{kg} \), and radius of the Sun \( R = 7.0 \times 10^8 \, \text{m} \).
The mass density of the Sun is calculated as:
\( \text{Mass density of Sun} = \frac{\text{Mass of Sun}}{\text{Volume of Sun}} = \frac{M}{\frac{4}{3}\pi R^3} \)
\( = \frac{2 \times 10^{30}}{\frac{4}{3} \times 3.14 \times (7 \times 10^8)^3} \)
\( = \frac{3 \times 2 \times 10^{30}}{4 \times 3.14 \times 343 \times 10^{24}} \, \text{kgm}^{-3} \)
\( \approx 1.4 \times 10^3 \, \text{kgm}^{-3} \)
This calculated density falls within the range typical for solids and liquids, not gases. This happens because even though the Sun is plasma (ionized matter) at very high temperatures, its immense self-gravity causes a strong inward pull. This inward gravitational attraction on its outer layers by the inner layers leads to a very high density.
In simple words: Despite being extremely hot plasma, the Sun's huge gravitational force pulls everything inward, making its density very high, more like a solid or liquid than a gas. This is because its mass and radius result in a density around \( 1.4 \times 10^3 \, \text{kgm}^{-3} \).

Exam Tip: Remember that even extremely hot plasma under immense gravitational pressure can behave like a dense fluid, not an expanded gas, due to the powerful compressing forces.

Question 24. When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.
Answer: To find the diameter of Jupiter, we can use the angular diameter formula. The distance of Jupiter from Earth, \( d = 824.7 \) million kilometers, which is \( 824.7 \times 10^6 \, \text{km} \). The angular diameter, \( \theta = 35.72'' \), which converts to \( 35.72 \times 4.85 \times 10^{-6} \, \text{rad} \) (since \( 1'' = 4.85 \times 10^{-6} \, \text{rad} \)).
Using the relationship \( \theta = \frac{D}{d} \), where \( D \) is the diameter of Jupiter, we can calculate \( D \):
\( D = \theta \times d \)
\( = 35.72 \times 4.85 \times 10^{-6} \times 824.7 \times 10^6 \, \text{km} \)
\( = 142873 \, \text{km} \)
This can also be written as \( 142.873 \times 10^3 \, \text{km} \) or approximately \( 1.42873 \times 10^5 \, \text{km} \).
In simple words: We find Jupiter's diameter using its distance from Earth and its apparent size in the sky. By multiplying the angular diameter (converted to radians) by the distance, we calculate its actual physical diameter.

Exam Tip: For angular diameter calculations, ensure that the angular measurement is converted to radians before using the formula \( D = \theta \times d \).