This page contains NCERT book Solutions class -9 Gravitation starting from page 143 at the end of the chapter. I have also been writing the notes for this chapter to learn more about notes you can follow this<\/p>\n
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Question 12<\/span><\/strong><\/p>\n Gravitational force on the surface of the moon is only 1\/6 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?<\/p>\n Answer<\/span><\/strong><\/p>\n Weight of any object on moon=$\\frac{1}{6} \\times\u00a0$ weight of object on earth<\/p>\n Also, Weight = mass $\\times$ acceleration<\/p>\n Acceleration due to gravity $g=9.8 ms^{2}$<\/p>\n So, weight of an object of mass 10 Kg on earth = $10 \\times 9.8=98N $<\/p>\n And weight of the same object on moon =\u00a0$\\frac{1}{6} \\times 98=16.3N$<\/p>\n Question 13<\/strong> Answer<\/strong><\/p>\n (i) According to question<\/p>\n Initial velocity of the ball (u)=$49ms^{-1}$<\/p>\n final velocity of the ball (v)=$0ms^{-1}$<\/p>\n Downward gravity (g) = $9.8ms^{-2}$<\/p>\n Upward\u00a0gravity (g) = $-9.8ms^{-2}$<\/p>\n Now we have to find the maximum height $h$ attained by the ball<\/p>\n From equation of motion we have<\/p>\n $v^{2}-u^{2}=2gh$<\/p>\n putting respective values in above equation we get<\/p>\n $(0)^{2}-(49)^{2}=2\\times (-9.8)\\times h$<\/p>\n $h=\\frac{49 \\times 49}{2 \\times 9.8}=122.5m$<\/p>\n (ii) let $t$ be the time taken by the ball to reach the height 122.5 m , then according to the equation of motion $v=u+gt$<\/p>\n we get,<\/p>\n $0=49+t \\times (-9.8)$<\/p>\n $9.8t=49$<\/p>\n or,<\/p>\n $t=\\frac{49}{9.8}=5s$<\/p>\n But, time of ascent=time of descent<\/p>\n Therefore , total time taken by ball to return back = 5+5 = 10 s<\/p>\n Question 14<\/strong><\/p>\n A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.<\/p>\n Answer<\/strong><\/p>\n According to question<\/p>\n initial velocity of stone =\u00a0$0 ms^{-1}$<\/p>\n acceleration due to gravity (g)=$9.8 ms^{-2}$<\/p>\n height (h) = $-19.6 m$<\/p>\n From equation of motion of object under gravity we have<\/p>\n $v^{2}-u^{2}=2gh$<\/p>\n putting respective values in above equation we get<\/p>\n $v^{2}-(0)^{2}=2 \\times 9.8 \\times 19.6$<\/p>\n $v^{2}=2 \\times 9.8 \\times 19.6$<\/p>\n $v^{2}=( 19.6)^{2}$<\/p>\n or<\/p>\n $v=19.6 ms^{-1}$<\/p>\n Hence velocity of \u00a0the stone just before touching the ground is $19.6 ms^{-1}$<\/p>\n Question 15<\/strong><\/p>\n A stone is thrown vertically upward with an initial velocity of\u00a040 m\/s. Taking $g = 10 ms^{2}$ , find the maximum height reached\u00a0by the stone. What is the net displacement and the total\u00a0distance covered by the stone?<\/p>\n Answer<\/strong><\/p>\n Equation of motion of object under gravity we have<\/p>\n $v^{2}-u^{2}=2gh$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (1)<\/p>\n From values given in question we have<\/p>\n initial velocity u = 40 m\/s and acceleration due to gravity is given as\u00a0$g = 10 ms^{2}$.<\/p>\n Let $h$ be the maximum height attained and final velocity $v=o$<\/p>\n Rearranging equation (1) we get<\/p>\n $h= \\frac{v^{2}-u^{2}}{2g}$<\/p>\n putting respective values we have<\/p>\n $h=\\frac{40 \\times 40}{20}=80m$<\/p>\n Total distance traveled would be = $2 \\times h = 160m$<\/p>\n Since\u00a0the stone returns back to the original position displacement of the stone during its total upward and downward journey \u00a0= $80 +(-80)=0$<\/p>\n Question 16<\/strong><\/p>\n Calculate the force of gravitation between the earth and the\u00a0Sun, given that the mass of the earth $= 6 \\times\u00a010^{24} kg$ and of the\u00a0Sun $= 2 \\times\u00a010^{30} kg$. The average distance between the two is $1.5 \\times\u00a010^{11} m$.<\/p>\n Answer<\/strong><\/p>\n As given in the question<\/p>\n Mass of earth \u00a0$M_{e}= 6 \\times\u00a010^{24} kg$<\/p>\n Mass of sun\u00a0$M_{s}= 2 \\times\u00a010^{30} kg$<\/p>\n distance between the two is\u00a0$R=1.5 \\times\u00a010^{11} m$<\/p>\n From Universal law of gravitation<\/p>\n $F = G\\frac{{M \\times m}}{{{R^2}}}$<\/p>\n Therefore, putting all the values given in question in above equation we get<\/p>\n $F = 6.67\\times10^{-11}\\frac{{(6 \\times 10^{24}) \\times (2 \\times 10^{30})}}{{{(1.5 \\times 10^{11})^2}}}=3.56\\times10^{22}N$<\/p>\n Question 17<\/strong><\/p>\n A stone is allowed to fall from the top of a tower 100 m high\u00a0and at the same time another stone is projected vertically\u00a0upwards from the ground with a velocity of 25 m\/s. Calculate\u00a0when and where the two stones will meet.<\/p>\n Answer<\/strong><\/p>\n Let $t$ be the point at which two stones meet and let $h$ be their height from the ground. It is given in the question that height of the tower is $H=100m$<\/p>\n Now first consider the stone which falls from the top of the tower. So distance covered by this stone at time $t$ can be calculated using equation of motion<\/p>\n $x-x_{0}=u_{0}t+\\frac{1}{2}gt^{2}$<\/p>\n Since initial velocity $u=0$ so we get<\/p>\n $100-x=\\frac{1}{2}gt^{2}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (1)<\/p>\n The distance covered by the same stone that is thrown in upward direction from ground is<\/p>\n $x=u_{0}t-\\frac{1}{2}gt^{2}$<\/p>\n In this case initial velocity is 25 m\/s . So<\/p>\n $x=25t-\\frac{1}{2}gt^{2}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(2)<\/p>\n Combining equations 1 and 2 we get<\/p>\n $100=25t$<\/p>\n this gives $t=4s$<\/p>\n This gives<\/p>\n $x=25 \\times 4-\\frac{1}{2} \\times 9.8 \\times (4)^{2}=100-78.4=21.6m$<\/p>\n Question 18<\/strong><\/p>\n A ball thrown up vertically returns to the thrower after 6 s. Answer<\/strong><\/p>\n (a) Here we have to find the velocity with which the ball is thrown in upwards direction.<\/p>\n Now, Acceleration due to gravity ,$ g=-9.8 ms^{-2}$<\/p>\n It is given in the question that total time taken by the ball in going upward and in comming back is 6 s.<\/p>\n Therefore time taken in upward journey = t=6\/2=3s<\/p>\n Final velocity v= 0 m\/s<\/p>\n we have to find the initial velocity u<\/p>\n We know that from first equation of motion<\/p>\n $v=u+gt$<\/p>\n so,<\/p>\n putting all the values we gat<\/p>\n $0=u+(9.8) \\times 3$<\/p>\n calculating we get<\/p>\n $u=29.4 m\/s$<\/p>\n So the velocity with which ball is thrown in the upward direction is 29.4 m\/s<\/p>\n (b) Now we have to find the maximum height say $h_{max}$ reached by the ball<\/p>\n From second equation of motion we have<\/p>\n $s=ut+\\frac{1}{2}gt^{2}$<\/p>\n here s (distance) is the maximum height $h_{max}$\u00a0reached<\/p>\n putting all the values in above equation we have<\/p>\n $h_{max}=29.4 \\times\u00a03\u00a0+\\frac{1}{2}(-9.8)(3)^{2}=44.1 m $<\/p>\n (c) Ball reaches its maximum height after 3 s . After reaching this height ball started to move in downward direction. So in this case<\/p>\n initial velocity u=0<\/p>\n Position of the ball after 4 s of the throw is given by the distance at which ball reaches while going downward direction in $4-3=1$ $s$<\/p>\n Again from second equation of motion we have<\/p>\n $s=ut+\\frac{1}{2}gt^{2}$<\/p>\n putting all the values in above equation and calculating we find $s=4.9m$<\/p>\n here\u00a0$h_{max}=44.1 m$<\/p>\n So, ball is 39.2 m (44.1 m- 4.9 m) above the ground after 4 s<\/p>\n Question 19<\/strong><\/p>\n In what direction does the buoyant force on an object immersed in an liquid acts?<\/p>\n Answer<\/strong><\/p>\n Buoyant force in this case will acts in vertically upwards direction.<\/p>\n Question 20<\/strong><\/p>\n Why does a block of plastic released under water come up to\u00a0the surface of water?<\/p>\n Answer<\/strong><\/p>\n For an object immersed in water two force acts on it<\/p>\n 1. gravitational force which tends to pull object in downward direction<\/p>\n 2. buoyant force that pushes the object in upward direction<\/p>\n here in this case buoyant force is greater than the gravitational pull on the plastic block. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water. Note:- this only happens in case where density of block is less than that of water.<\/p>\n Question 21<\/strong><\/p>\n The volume of 50 g of a substance is $20 cm^{3 }$. If the density of\u00a0water is $1 g$ $cm^{\u20133}$ , will the substance float or sink?<\/p>\n Answer<\/strong><\/p>\n If the density of this object is greater then the density of water then it will sink and if it is less than water then t will float in the water. So we calculate the density of this substance<\/p>\n $Density=\\frac{mass}{volume}=\\frac{50}{20}=25 g$ $cm^{-3}$<\/p>\n Since density of this substance is greater then the density of water so it will sink in water.<\/p>\n Question 22<\/strong><\/p>\n The volume of a 500 g sealed packet is $350 cm^{3} $ . Will the packet\u00a0float or sink in water if the density of water is $1 g$ $cm^{\u20133}$ \u00a0What\u00a0will be the mass of the water displaced by this packet?<\/p>\n Answer<\/strong><\/p>\n Density of packet = (mass of packet)\/(volume of packet)<\/p>\n From the question we have<\/p>\n mass of packet = 500 g<\/p>\n volume of packet =350 $cm^{3}$<\/p>\n so density of packet =$\\frac{500}{350}=1.428g$ $ cm^{-3}$<\/p>\n The density of this substance is greater then the density of water. Hence it will sink in water. The mass of water displaced by the packet is equal to the volume of packet that is 350 gm<\/p>\n If you are interested in other links containing class 9 notes in science and maths along with free download links consider following the links given below <\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" This page contains NCERT book Solutions class -9 Gravitation starting from page 143 at the end of the chapter. I have also been writing the notes for this chapter to learn more about notes you can follow this link and to get the in-text questions solution please follow this link . If you want to get the solutions of first 11 questions follow this link. Question 12
\nA ball is thrown vertically upwards with a velocity of 49 m\/s.
\nCalculate
\n(i) the maximum height to which it rises,
\n(ii) the total time it takes to return to the surface of the earth.<\/p>\n
\nFind
\n(a) the velocity with which it was thrown up,
\n(b) the maximum height it reaches, and
\n(c) its position after 4 s.<\/p>\n
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\nGravitational force on the surface of the moon is only 1\/6 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[24,279],"tags":[],"class_list":["post-2459","post","type-post","status-publish","format-standard","hentry","category-general","category-ncert-solutions"],"yoast_head":"\n