

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
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
