","comment":{"@type":"Comment","text":" The general equation for the acceleration due to gravity is to be used here. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$1500\\text{ }km$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":"$$500\\text{ }km$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"$$1600\\text{ }km$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$800\\text{ }km$$","position":3,"answerExplanation":{"@type":"Comment","text":" The acceleration due to gravity at a small height can be written as, $$ \\\\ {{g}_{h}}=g\\left( 1-\\dfrac{2h}{R} \\right) \\\\ $$ Substituting the values in it, $$ \\\\ 7.35=9.8\\left( 1-\\dfrac{2h}{6400} \\right) \\\\ \\Rightarrow 0.75=1-\\dfrac{2h}{6400} \\\\ \\Rightarrow \\dfrac{2h}{6400}=0.25 \\\\ \\therefore h=800\\text{ }km$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Circular motion and gravitation Quiz 1","text":" The centripetal force of a car having a mass of $$1000\\,kg$$ moving around a roundabout is found to be $$360\\,N$$. The radius of the path covered is found to be $$10\\,km$$. What will be the time taken by the car to complete one complete rotation around this roundabout?","comment":{"@type":"Comment","text":" The distance travelled in the circular motion in a rotation can be taken as the circumference of the path taken. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2\\text{ }\\min $$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":"$$1.5\\text{ }\\min $$ ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":"$$5\\text{ }\\min $$ ","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1.75\\text{ }\\min $$","position":1,"answerExplanation":{"@type":"Comment","text":" The centripetal force can be found as, $$ \\\\ F=\\dfrac{m{{v}^{2}}}{r} \\\\ $$ The velocity of the journey can be found by rearranging the equation, $$ \\\\ v=\\sqrt{\\dfrac{Fr}{m}} \\\\ $$ Substituting the values in it, $$ \\\\ v=\\sqrt{\\dfrac{360\\times 10\\times {{10}^{3}}}{1000}} \\\\ \\therefore v=60\\text{ }m{{s}^{-1}} \\\\ $$ The velocity can be written as, $$ \\\\ \\text{v=}\\dfrac{\\text{distance}}{\\text{time}}=\\dfrac{\\text{circumference}}{\\text{time}} \\\\ \\Rightarrow v=\\dfrac{2\\pi r}{t} \\\\ $$ Substituting the values in it, $$ \\\\ 60=\\dfrac{2\\pi \\times 10\\times {{10}^{3}}}{t} \\\\ \\Rightarrow t=\\dfrac{2\\pi \\times {{10}^{4}}}{60}=0.105\\times {{10}^{3}}\\text{ }s \\\\ \\therefore t=105\\text{ }s \\\\ $$ That is, $$t=\\dfrac{105}{60}=1.75\\text{ }\\min $$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Circular motion and gravitation Quiz 1","text":" It has been found that the escape velocity of a satellite to escape from the surface of earth is $$11.19\\text{ }km{{s}^{-1}}$$. Calculate the mean density of the earth. The radius of the earth is mentioned as$$6400\\text{ }km$$. The value of universal gas constant is, $$G=6.674\\times {{10}^{-11}}\\text{ }{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$$.","comment":{"@type":"Comment","text":" The mass is to be substituted in the escape velocity equation as the product of volume and the density."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$6515\\text{ }kg{{m}^{-3}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"$$7515\\text{ }kg{{m}^{-3}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":"$$9515\\text{ }kg{{m}^{-3}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$5515\\text{ }kg{{m}^{-3}}$$","position":0,"answerExplanation":{"@type":"Comment","text":" The equation of escape velocity has been written as, $$ \\\\ {{V}_{e}}=\\sqrt{\\dfrac{2GM}{R}}………………. (1) \\\\ $$ Let us assume that the mean density of the earth is $$\\rho $$. The mass of the earth will be, $$ \\\\ M=\\dfrac{4}{3}\\pi {{R}^{3}}\\times \\rho \\\\ $$ Substituting this in the equation $$(1)$$ will give, $$ \\\\ {{V}_{e}}=\\sqrt{\\dfrac{2\\times G\\times \\dfrac{4}{3}\\pi {{R}^{3}}\\times \\rho }{R}} \\\\ \\Rightarrow {{V}_{e}}=\\sqrt{\\dfrac{8G\\pi \\rho {{R}^{2}}}{3}} \\\\ \\Rightarrow {{V}_{e}}=2R\\sqrt{\\dfrac{2G\\pi \\rho }{3}} \\\\ $$ Now let us substitute the values in this equation will give, $$ \\\\ {{V}_{e}}=2\\times 6400\\times {{10}^{3}}\\sqrt{\\dfrac{2\\times \\pi \\times 6.674\\times {{10}^{-11}}\\times \\rho }{3}} \\\\ \\Rightarrow 11.19\\times {{10}^{3}}=12800\\times {{10}^{3}}\\sqrt{1.3978\\times {{10}^{-10}}\\rho } \\\\\\Rightarrow 1.3978\\times {{10}^{-10}}\\rho ={{\\left( \\dfrac{11.19}{12800} \\right)}^{2}} \\\\ \\therefore \\rho =5467.5\\text{ }kg{{m}^{-3}}\\approx 5515\\text{ }kg{{m}^{-3}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}