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Question:
2.6 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 ? 2.7 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 3.5 mm. What is the estimate on the thickness of hair ? 2.8 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 1.0 mm 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 ? 2.9 The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2 . What is the linear magnification of the projector-screen arrangement. 2.10 State the number of significant figures in the following : (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm–3 (d) 6.320 J (e) 6.032 N m–2 (f) 0.0006032 m2 2.11 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. 2.12 The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ? 2.13 A physical quantity P is related to four observables a, b, c and d as follows : = ( ) 3 2 P a b / 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 ? 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 ? 2.14 A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion : (a) y = a sin 2π t/T (b) y = a sin vt (c) y = (a/T) sin t/a (d) y a t T t T = ( ) 2 (sin 2 / + cos 2 / ) π π (a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds. 2.15 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the 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 m 1 v 0 = − 2 1/2 . Guess where to put the missing c. 2.16 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ? 2.17 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). Why is this ratio so large ? 2.18 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). 2.19 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 × 1011m. 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 ? 2.20 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 ? 2.21 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 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. 2.22 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 your head (e) the number of air molecules in your classroom. 2.23 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 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 ×1030 kg, radius of the Sun = 7.0 × 108 m. 2.24 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:

2.6

(a) Least count of vernier calipers = 1 /20  = 0.05 cm

(b) Least count of screw gauge =1/100 = 0.01 cm

(c) Least count of an optical device =1/ 105 = 0.00001 cm

 Wavelength of light  is of order of 10–5 cm

 

Optical instrument has least count, Hence it is the most precise device for measuring length. Ans ©

 

2.7

Magnification = 100

Average width of the hair in the field of view of the microscope is 3.5 mm

Actual estimate  thickness of hair = average width on field of view / magnification

                                                                   =3.5 mm / 100  = 0.035 mm

Estimate on the thickness of hair = 0.035 mm 

 

2.8

  1. a) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a meter scale. The diameter of the thread is given by Length of the thread / Number of turns
  2. b) Increasing the number of divisions on the circular scale, does not increase the accuracy of the screw gauge

c)Performing a set of 100 measurements is more reliable than a set of 5, as it reduces the random errors encountered

 

2.9

The area of the house on the photograph (object) = 1.75 cm2

The area of the house on the screen (image) = 1.55 m2 = 1.55 * 10-4 cm2

Arial magnification = v/u = 155 * 10-4 / 1.75 = 8857

Linear magnification = square root of (8857) = 94.11

 

2.10

(a)

Value is 0.007 m2.

We never count zeros before non zero value in significant figure. So 7 is only 1 significant figure

 

(b)

The given that

Value is 2.64 × 1024 kg.

Powers of 10 is irrelevant for the determination of significant figures. Hence 2,6 and 4 are significant figures

Number of significant digit = 3  

(c)

Value is 0.2370 g cm–3.

Zeros after non zero number and decimal point are always significant. So all four numbers are significant

Number of significant digit = 4  

(d)

Value is 6.320 J.

Zeros after non zero number and decimal point are always significant. So all four numbers are significant

Number of significant digit = 4  

(e)

Value is  6.032 Nm–2.

All zeroes between two non-zero digits are always significant. So all four numbers are significant

Number of significant digit = 4  

(f)

Value is 0.0006032 m2.

We never count zeros before non zero value in significant figure and All zeroes between two non-zero digits are always significant. So that last four digits are significant.

Number of significant digit = 4

 

 

2.11 Length = 4.234 m

         Breadth = 1.005m

         Thickness = 2.01 cm = 2.01 * 10-2m

 

         Maximum significant figure for area and volume is 3. Hence, need to round appropriately

         Area = 2(length*breadth + breadth*Thickness + Thickness*length) as there are 6 sides and two sides of each measurement

                   = 2(4.3604739) = 8.7209478 m2 which when rounded to two decimals is 8.72

         Volume = length * breadth * thickness

                       = 4.234 * 1.005 * 2.01*10-2 = 0.0855 m3

2.12 Mass of the box at grocer balance = 2.300 Kg

         Mass of gold piece 1 = 20.15 gm = 0.02015 Kg

         Mass of gold piece 2 = 20.17 gm = 0.02017 Kg  

                             

 

         Total mass of the box = 2.300 + 0.02015 + 0.02017 = 2.34032 kg = 2.3 kg retaining the decimal places as in the grocer balance

         Difference in masses = 20.17 – 20.15 = 0.02 retaining the number as per the least decimal place number

 

2.13 P = a3b2 / (square root c) * d

         Hence, maximum fractional error in P is given by ΔP / P which is plus or minus [ 3 Δa/a + 2Δb/b + ½ Δc/c + Δd/d]

           = plus or minus [3(1/100) + 2(3/100) + ½(4/100) + 2/100]

           = plus or minus 0.13

 

            Percentage of error = ΔP/P * 100 = plus or minus 13%

            Given value of P = 3.763; Hence, rounding off the one decimal we get 3.8

2.14 The dimension of displacement y is length. The trigonometric functions and its arguments are dimensionless

         (a) 2πt / T = Dimensionless

         (b) vt = L/T * T = L

         © t /a = T

        (d) 2πt / T = Dimensionless

         Hence, b and c, are incorrect

2.15 The given formula would be dimensionally correct if the dimension on the LHS is equal to the dimension on the RHS. The LHS dimension is M. As m0 also has the same dimension, we need to consider. Hence, 1-v2 should be dimensionless. This is possible only if we have 1-v2/c2 Hence, the formula is m = m0 / (1 – v2 / c2) 1/2

 

2.16 Radius of hydrogen atom = 0.5 * 10-10m

         Volume of hydrogen atom = 4/3 π r3 = 4/3 * 22/7 * (0.5 * 10-10) 3  = 0.524 * 10-30 m3

         1 mole of hydrogen atom contains 6.023 * 1023 hydrogen atoms

         Hence, volume of 1 mole of hydrogen atom = 0.524 * 10-30    * 6.023 * 1023 = 3.16 * 10-7 m3

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