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This is one of our many quizzes on Circular motion and gravitation. Click on this to access more such competitive quizzes on Kinematics.

Circular motion and gravitation Quiz 2

physics

 Suppose the radius of the earth has been reduced to ${{\dfrac{1}{6}}^{th}}$ of its initial value and the mass of the earth has been increased twice the initial value. Calculate the ratio of weight experienced for a man with mass $m$ to his initial weight.

 The weight of a man can be found by taking the product of mass of the man and the acceleration due to gravity experienced. 

The acceleration due to gravity can be found by the equation, $g=\dfrac{G{{M}_{E}}}{{{R}_{E}}^{2}}$ As per the question, we can write that, ${{M}_{E}}^{\prime }=2{{M}_{E}} \\ \Rightarrow{{R}_{E}}^{\prime }=\dfrac{1}{6}{{R}_{E}}$ Substituting the values in it, ${g’}=\dfrac{G\times 2{{M}_{E}}}{{{\left( \dfrac{{{R}_{E}}}{6} \right)}^{2}}} \\ \Rightarrow{g’}=72\dfrac{G{{M}_{E}}}{{{R}_{E}}^{2}}=72g $ The weight of a man is found by, $F=mg$ That is, $ \dfrac{{{F}'}}{F}=\dfrac{m{g}'}{mg} \\ \Rightarrow\dfrac{{{F}'}}{F}=\dfrac{{{g}'}}{g} \\ \therefore\dfrac{{{F}'}}{F}=\dfrac{72g}{g}=72 $

 The time period of the satellite $a$ has been given as $T$ and the time period of the other satellite $b$ has been found to be $1.5T$. Find the ratio of the radii of $a$ to that of $b$?

 The Square of the time period will be proportional to the cube of the radius. 

According to Kepler’s third law, we can write that, ${{T}^{2}}\propto {{R}^{3}}$ That is, ${{\left( \dfrac{{{T}_{a}}}{{{T}_{b}}} \right)}^{2}}={{\left( \dfrac{{{R}_{a}}}{{{R}_{b}}} \right)}^{3}}$ Substituting the values, $ {{\left( \dfrac{T}{1.5T} \right)}^{2}}={{\left( \dfrac{{{R}_{a}}}{{{R}_{b}}} \right)}^{3}} \\ \therefore {{\left( \dfrac{2}{3} \right)}^{\dfrac{2}{3}}}=\dfrac{{{R}_{a}}}{{{R}_{b}}}$

What will be the change in potential energy when a body of mass $m$, is taken into a depth $K{{R}_{E}}$ from the surface of earth? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/89b760fe-8164-410a-98ed-90bf8d8181f8929301065609219178.png"/>

 The potential energy is a scalar quantity. So the change in potential energy will be simply the difference between the values.

The potential energy can be found by the equation, $U=-\dfrac{GMm}{r}$ The potential energy at the earth's surface will be, ${{U}_{1}}=-\dfrac{GMm}{{{R}_{E}}}$ The potential energy at a depth $K{{R}_{E}}$, $r={{R}_{E}}-K{{R}_{E}}={{R}_{E}}(1-K)$ That is, ${{U}_{2}}=-\dfrac{GMm}{{{R}_{E}}(1-K)}$ The change in potential energy will be, $ \Delta U={{U}_{2}}-{{U}_{1}} \\ \Rightarrow\Delta U=\dfrac{-GMm}{{{R}_{E}}(1-K)}-\dfrac{-GMm}{{{R}_{E}}}=\dfrac{GMm}{{{R}_{E}}}\left( 1-\dfrac{1}{1-K} \right) \\ \therefore \Delta U=\dfrac{GMm}{{{R}_{E}}}\left( \dfrac{-K}{1-K} \right)$

 Suppose that an electron is in circular motion when placed in a region of electric field $E$. The radius of the circular path is $1.2\mu m$ and the velocity of travel has been given as $6\text{ }m{{s}^{-1}}$. What will be the magnitude of the electric field? 

 The centripetal force will be equivalent to the electrostatic force. 

The centripetal force due to the circular motion can be written as, $F=\dfrac{{{m}_{e}}{{v}^{2}}}{r}$ The electrostatic force of the electron, $F=eE$ This centripetal force will be obtained by the electrostatic force due to the charge in the electric field. That is, $ \dfrac{{{m}_{e}}{{v}^{2}}}{r}=eE \\ \Rightarrow E=\dfrac{{{m}_{e}}{{v}^{2}}}{er} $ Substituting the values, $ \Rightarrow E=\dfrac{9.1\times {{10}^{-31}}{{\left( 6 \right)}^{2}}}{1.6\times {{10}^{-19}}\times 1.2\times {{10}^{-6}}} \\ \Rightarrow E=\dfrac{327.6\times {{10}^{-31}}}{1.92\times {{10}^{-25}}}=170.625\times {{10}^{-6}} \\ \therefore E=1.7\times {{10}^{-4}}\text{ }V{{m}^{-1}}$

 A car has been travelling through a circular track which has been banked at an angle $15{}^\circ $. The car is allowed to travel at an optimum speed in order to avoid the wear and tear. The weight of the car is $980\text{ }N$. The car took $10\text{ }\min $ to complete one rotation. Find out the approximate kinetic energy of the car travelling.

 The kinetic energy can be found by taking the half of the product of the mass and the square of the velocity.

The optimum speed can be found by the equation, $v=\sqrt{rg\tan \theta }$ Where $\theta $ be the angle of banking given as, $\theta =15{}^\circ $ That is, $v=\sqrt{rg\tan 15}=\sqrt{0.268rg}$…………… (1) The velocity can be otherwise found by equation, $v=\dfrac{\text{circumference}}{\text{time}}=\dfrac{2\pi r}{t}$ The time taken will be, $t=10\text{ }\min =600\text{ }s$ Substituting the values in it, $r=\dfrac{600v}{2\pi }$…………………… (2) The weight of the car be, $ mg=980\text{ }N \\ \Rightarrow m=\dfrac{980}{9.8}=100\text{ }kg$ The equation (2) can be substituted in equation (1), $ v=\sqrt{\dfrac{600v}{2\pi }\times g\times 0.268} \\ \Rightarrow{{v}^{2}}=\dfrac{600v}{2\pi }\times 9.8\times 0.268 \\ \Rightarrow v=250.8\text{ }m{{s}^{-1}}$ Therefore the kinetic energy will be, $ KE=\dfrac{1}{2}m{{v}^{2}} \\ \Rightarrow KE=\dfrac{1}{2}\times 100\times {{\left( 250.8 \right)}^{2}}=3.14\times {{10}^{6}}J \\ \Rightarrow KE\approx \pi \times {{10}^{6}}J $