Class 9 Physics New book Chapter 1 Numerical
Introduction:
Chapter 1: Physical Quantities and Measurements (Numerical) – Introduction
In this chapter, we will solve Class 9 Physics New book Chapter 1 numerical. Students learn how to solve numerical, Class 9 Physics Chapter 1 Punjab board problems, based on basic concepts of physics like physical quantities, units, measurement, and significant figures. The focus is on applying formulas related to speed, distance, time, area, and volume. These numerical help to build a strong foundation for solving real-world problems using physics.
1.1 Calculate the number of seconds in a (a) day (b) week (c) month and state you answer using SI Prefixes.
Number of minutes in 1 hour = 60
Seconds in 1 hour = (60) (60) = 3600
Seconds in 1 day = (3600) (24) = 86,400
$$=(\frac{86400}{1000})\;(1000)s\\\\=(86.4)\;(1000)s\\\\=86.4\;Ks\\$$
Number of seconds in a week = (86,400) (7)
= 604,800
$$=(\frac{604800}{1000})\;(1000)s\\\\=(604.8)\;(1000)s\\\\=604.8\;Ks\\$$
Number of seconds in 1 month = (86,400) (30)
= 2,592,000
$$=(\frac{2592000}{1000000})\;(1000000)s\\\\=(2.592)\;(1000000)s\\\\=2.592\;Ms\\$$
==========================================
1.2 State the answers of problem 1.1 in scientific notation
Number of minutes in 1 hour = 60
seconds in 1 hour = (60) (60) = 3600
Its mean seconds in 1 day = (3600) (24) = 86,400 = $$8.64\;\times\;10^4$$
seconds in a week = (86,400) (7) = 604,800 = $$6.048\;\times\;10^5$$
seconds in 1 month = (86,400)(30) = 2,592,000 = $$2.592\;\times\;10^6$$
==========================================
1.3 Solve the following addition or subtraction. State your answer in scientific notation
$$Light\;year\;is\;a\;unit\;of\;dis\tan ce\;used\;in\;Astronomy.\\It\;is\;the\;dis\tan ce\;covered\;by\;light\;in\;one\;year.\\Taking\;tha\;speed\;of\;light\;as\;3.0\;\times10^8\;m\;s^{-1}\\calculate\;the\;dis\tan ce\\\\\\$$
$$(a)\;\;4\times10^{-4}kg\;+3\times10^{-5}kg\\\\=\;10^{-4}(4\;+\;3\times10^{-1}\;)\;kg\\\\=\;10^{-4}(4\;+\;\frac3{10}\;)\;kg\\\\=\;10^{-4}(4\;+\;0.3)\;kg\\\\=\;10^{-4}(4.3\;)\;kg\\\\=\;4.3\;\times10^{-4}\;kg\\\\\\(b)\;\;5.4\times10^{-6}\;m\;-\;3.2\times10^{-5}\;m\\\\=\;10^{-5}(5.4\;\times10^{-1}-3.2)m\\\\=\;10^{-5}(\frac{5.4}{10}-3.2)m\\\\=\;10^{-5}(0.54-3.2)m\\\\=\;10^{-5}(-2.66)m\\\\=\;-2.66\;\times10^{-5}m\\\\\\\\\\$$
==========================================
1.4 Solve the following multiplication or division. State your answer in scientific notation.
$$(a)\;\;\;(5\times10^4m)\;\times\;(3\times10^{-2}m)\\(b)\;\;\;\frac{6\times10^8kg}{3\times10^4\;m^3}\;\\$$
$$(a)\;\;\;(5\times10^4m)\;\times\;(3\times10^{-2}m)\\\\=\;\;5\times3\times10^4\times10^{-2}\times m\times m\\\\=\;15\times10^{4-2}\;\times m^2\\\\=\;15\times10^2\;m^2\\\\=\;1.5\;\times10^1\;\times10^2\;m^2\\\\=\;1.5\;\times10^{1+2}m^2\\\\=\;1.5\;\times10^3m^2\\\\\\(b)\;\;\;\frac{6\times10^8kg}{3\times10^4\;m^3}\;\\=\;2\times10^8\times10^{-4}\;kg\;m^{-3}\\\\=\;2\times{(10)}^{8-4}\;kg\;m^{-3}\\\\=\;2\times10^4\;kg\;m^{-3}\\\\$$
==========================================
1.5 Calculate the following. State your answer in scientific notation.
$$\frac{(3\times10^2kg)\times(4.0km)}{5\times10^2\;s^2}\\\\\frac{12}5kg\;km\;s^{-2}\\\\2.4\;kg\;km\;s^{-2}\\\\2.4\;\times\;1000\;kg\;m\;s^{-2}\\\\2.4\;\times10^3\;kg\;m\;s^{-2}\\\\$$
==========================================
1.6 State the number of significant digits in each measurement.
$$(a)\;0.0045m\;\;(b)\;2.047m\;\\(c)\;3.40m\;\;(d)\;3.420\;\times\;10^4m\\\\(a)\;0.0045m\\All\mathit\;non\mathit\;zero\mathit\;digits\mathit\;are\mathit\;significant\mathit.\\Zeros\;are\;not\;significant\;here.\;\\4\;and\;5\;are\;significant.\\Hence,\;there\;are\;only\;2\;significant\;figures.\\\\(b)\;2.047m\\All\mathit\;non\mathit\;zero\mathit\;digits\mathit\;are\mathit\;significant\mathit.\;\\Zero\;between\;non\;zeros\;are\;significant.\;\\All\;the\;digits\;are\;significant\;here.\\Hence,\;there\;are\;4\;significant\;figures.\\\\c)\;3.40m\\All\mathit\;non\mathit\;zero\mathit\;digits\mathit\;are\mathit\;significant\mathit.\\Zero\;after\;the\;decimical\;point\;is\;significant.\\Hence,\;there\;are\;3\;significant\;figures.\\\\(d)\;3.420\;\times\;10^4m\\All\mathit\;non\mathit\;zero\mathit\;digits\mathit\;are\mathit\;significant\mathit.\\Zero\;after\;the\;decimical\;point\;is\;significant.\\Hence,\;there\;are\;4\;significant\;figures.$$
==========================================
1.7 Write in scientific notation
$$(a)\;\;0.0035m\\\\=\;3.5\;\times\;10^{-3}\;m\\\\\\(b)\;\;206.4\;\times10^2\;m\\\\=\;2.064\;\times\;10^2\;\times\;10^2\;m\\\\=\;2.064\;\times\;10^{2+2}\;m\\\\=\;2.064\;\times\;10^4\;m\\\\$$
==========================================
1.8 Write using correct prefixes
$$\boldsymbol(\boldsymbol a\boldsymbol)\boldsymbol\;\boldsymbol\;\mathbf5\boldsymbol.\mathbf0\boldsymbol\;\boldsymbol\times\boldsymbol\;\mathbf{10}^{\mathbf4}\boldsymbol\;\boldsymbol c\boldsymbol m\\\\=\;5.0\;\times\;10^4\;\times\;10^{-2}m\\\\=\;5.0\;\times\;10^{4-2}m\\\\=5.0\;\times\;10^2m\\\\=\;\frac{5.0\;\times\;10^2km}{1000}\\\\=\;\frac{5.0\;\times\;100\;km}{1000}\\\\=\;\frac{500\;km}{1000}\\\\=\;0.5\;km\\\\\boldsymbol(\boldsymbol b\boldsymbol)\boldsymbol\;\boldsymbol\;\mathbf{580}\boldsymbol\;\boldsymbol\times\boldsymbol\;\mathbf{10}^{\mathbf2}\boldsymbol\;\boldsymbol g\\\\=\frac{\;580\;\times\;10^2\;kg}{1000}\\\\=\;\frac{\;580\;\times\;100\;kg}{1000}\\\\=\;58\;kg\\\\\boldsymbol(\boldsymbol c\boldsymbol)\boldsymbol\;\boldsymbol\;\mathbf{45}\boldsymbol\;\boldsymbol\times\boldsymbol\;\mathbf{10}^{\boldsymbol-\mathbf4}\boldsymbol\;\boldsymbol s\\\\=\;45\;\times\;10^{-4}\;\times\;1000\;\;ms\\\\=\;\frac{45\;\;\times\;1000\;\;ms}{10^4}\\\\=\;\frac{\;45000\;\;ms}{10000}\\\\=\;4.5\;ms\\\\$$
==========================================
1.9
$$Light\;year\;is\;a\;unit\;used\\in\;Astronomy.\;It\;is\;the\\dis\tan ce\;covered\;by\\light\;in\;one\;year.\\Taking\;the\;speed\;of\;light\\as\;3.0\times10^8\;m\;s^{-1},\\calculate\;the\;dis\tan ce\;$$
$$Number\;of\;seconds\;in\;one\;minute\;=\;60\\\\Number\;of\;seconds\;in\;one\;hour\;=\;60\;\times\;60\\\\Number\;of\;seconds\;in\;one\;day\;=\;60\;\times\;60\;\times\;24\\\\number\;of\;seconds\;in\;one\;year\;=\;60\times60\times24\times365\\\\Number\;of\;seconds\;in\;one\;year\;=\;31,536,000\\\\Dis\tan ce\;=\;31,536,000\;\times\;3.0\;\times\;10^8\;m\\=\;94,608,000\;\times\;10^8\;m\\=\;9.4608\;\times\;10^7\;\times\;10^8\;m\\=\;9.4608\;\times\;10^{7+8}\;m\\=\;9.4608\;\times\;10^{15}\;m$$
==========================================
1.10
$$Express\;the\;density\\of\;mercury\;given\;as\\13.6\;g\;c\;m^{-3}\;in\;kg\;m^{-3}\\\\\\$$
$$13.6\;g\;c\;m^{-3}\\\\=\;\frac{13.6\;\;kg\;c\;m^{-3}}{1000}\\\\=\;0.0136\;\;kg\;c\;m^{-3}\\\\\mathbf1\boldsymbol\;\boldsymbol c\boldsymbol m\boldsymbol\;\boldsymbol=\boldsymbol\;\mathbf{10}^{\boldsymbol-\mathbf2}\boldsymbol\;\boldsymbol m\\\\\mathbf1\boldsymbol\;\boldsymbol c\boldsymbol\;\boldsymbol m^{\mathbf3}\boldsymbol\;\boldsymbol=\boldsymbol\;\mathbf{10}^{\boldsymbol-\mathbf6}\boldsymbol\;\boldsymbol m^{\mathbf3}\\\\\;0.0136\;\;kg\;c\;m^{-3}\\\\=\frac{0.0136\;kg}{\;c\;m^3}\\\\=\;\frac{0.0136\;kg}{\;10^{-6}\;m^3}\\\\=0.0136\;\times10^6\;\;kg\;\;\;m^{-3}\\\\=\;1.36\;\times\;10^{-2\;}\times10^6\;\;kg\;\;\;m^{-3}\\\\=\;1.36\;\times10^{-2+6}\;\;kg\;\;\;m^{-3}\\\\=\;1.36\;\times10^4\;\;kg\;\;\;m^{-3}\\\\$$
==========================================
Check! Some important links below
