Units & Measurement

NCERT Solution

Part 2

Question 5: 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 s to cover this distance?

Answer: Given, speed of light = 1 unit per second

Time taken for light to travel between the Sun and the Earth = 8 min 20 s

= 8 × 60 + 20 s = 480 + 20 = 500 s

So, distance between the Sun and the Earth = 500 unit

Question 6: Which of the following is the most precise device for measuring length?

a vernier callipers with 20 divisions on the sliding scale
a screw gauge of pitch 1 mm and 100 divisions on the circular scale
an optical instrument that can measure length to within a wavelength of light

Answer: (c) an optical instrument that can measure length to within a wavelength of light

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 3.5 mm. What is the estimate on the thickness of hair?

Answer: Magnification = 100

With as seen in microscope = 3.5 mm

So, actual width = 3.5 ÷ 100 = 0.035 mm

Question 8: Answer the following:

(a)You are given a thread and a metre scale. How will you estimate the diameter of the thread?

Answer: Diameter of thread can be indirectly measured using a meter scale. For this, you need to wrap the thread around a cylindrical object, e.g. a pencil. Then measure the length of coil of thread on pencil. Divide this with number of turns of thread to get estimation of 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?

Answer: Theoretically, it is possible to increase the accuracy of a screw gauge by increasing the number of divisions on circular scale, because least count is given by pitch divided by number of divisions on circular scale. Increasing the number of divisions on circular scale will decrease the least count and as a result increase the accuracy of instrument. However, there is a limit to this method. More divisions would mean it will be difficult to observe them by human eye.

(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: We know that random errors are possible when taking measurement. A large number of observations ensure better accuracy because we then rely on mean of all observations. So, a set of 100 measurements will be more reliable than a set of 5 measurements.

Question 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?

Answer: Area on slide = 1.75 cm² = 1.75 × 10^-4 m²

Area on screen = 1.55 m²

Magnification in Area = 1.55 m² ÷ 1.75 × 10^-4 m²

= 8.857 × 10³

So, linear magnification `=sqrt(8.857×10^3)=94.1`

Question 10: State the number of significant figures in the following:

0.007 m²
2.64 × 10²⁴ kg
0.2370 g cm^–3
6.320 J
6.032 N m^–2
0.0006032 m²

Answer: (a) 1, (b) 3, (c) 4, (d) 4, (e) 4, (f) 4

Question 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.

Answer: Area = 4.234 × 1.005 × 2 = 8.510 m²

Volume = 4.234 × 1.005 × 0.0201 = 0.855 m³

Question 12: The mass of a box measured by a grocer’s balance is 2.30 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?

Answer: Total mass = 2.30 + .002015 + .002017 = 2.3 kg

Difference in masses of pieces = 20.17 – 20.15 = 0.02 g

Question 13: A physical quantity P is related to four observables a, b, c and d as follows:

`P=a^3b^2/(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? 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=(a^3b^2)/(sqrtcd)=a^3b^2c^(-1/2)d^(-1)`

So, maximum fractional error in measurement

`(ΔP)/P=3(Δa)/a+2(Δb)/b+1/2(Δc)/c+(Δd)/d`

Given, `(Δa)/a=1%`, `(Δb)/b=3%`, `(Δc)/c=4%` and `(Δd)/d=2%`

So, maximum fractional error in measurement

`(ΔP)/P=3×1%+2×3%+1/2×4%+2%`

= 3% + 6% + 2% + 2% = 13%

If P = 3.768 then ΔP = 13% of P

`=(13P)/(100)= (13×3.768)/(100)=0.489`

Since the error lies in the first decimal place, answer should be rounded off to first decimal place. So, after rounding off the value of P = 3.8