To view explanation, please take trial in the course below. NEET 2023 - Target Batch - Aryan Raj Singh To view explanation, please take trial in the course below. NEET 2023 - Target Batch - Aryan Raj Singh Please attempt this question first. Add Note 
With Turito Academy.Get 7 days Free Trail NowWith Turito Foundation.Get an Expert Advice From Turito.With Turito Academy.Get 7 days Free Trail NowWith Turito Foundation.Answer Hint- In this question we have to simply relate the weight with density and radius of earth by some equation. Then only we will be able to see what happens with changing the density and radius of earth.Formula used: \[g = \dfrac{{GM}}{{{R^2}}}\], M = $\rho \times V$ Complete step-by-step solution - Since, we know that our weight is a pull force acting on us by the earth is given by W = mg,Where m- is one's massAnd g- is the acceleration due to gravity.Since mass is an amount of our body which is a constant term. Therefore only we have to check for g on changing the density and radius of earth which is as shown below:Here, we have used the terms,G- Gravitational constant ,R- Radius of the earth,$\rho $ - is the density ,V- is the volume of the earth,${g_1}$ - acceleration due to gravity before variation,${g_2}$ - acceleration due to gravity after changing the densities and radius.The formula for acceleration due to gravity is given by \[g = \dfrac{{GM}}{{{R^2}}}\]The relation between mass and density is given by Mass = density x volumeM = $\rho \times V$Therefore \[ g = \dfrac{{G\rho V}}{{{R^2}}} = \dfrac{{G\rho }}{{{R^2}}} \times \dfrac{{4\pi {R^3}}}{3} \\ g = \dfrac{4}{3}\pi R\rho G \\ \]From the above equation it can be seen that $g \propto \rho R$\[ \dfrac{{{g_2}}}{{{g_1}}} = \dfrac{{{\rho _2}{R_2}}}{{{\rho _1}{R_1}}} \\ {g_2} = {g_1}\left[ {\dfrac{{{\rho _2}{R_2}}}{{{\rho _1}{R_1}}}} \right] \\ \]Substitute the value of ${\rho _2} = 4{\rho _1}\& {R_2} = \dfrac{{{R_1}}}{2}$ in the above equation$ {g_2} = {g_1}\left[ {4 \times \dfrac{1}{2}} \right] \\ {g_2} = 2{g_1} \\ $Multiply both sides by m, we get$ m{g_2} = 2m{g_1} \\ {w_2} = 2{w_1} \\ $Therefore the weight is doubled.Hence, the correct option is B.Note- In order to solve these types of problems, the first step is to extract the data given in the questions and then see which formula is to use to find the quantity and solve for the relation we need. Remember all the formula and relation between capital G and small g. where G is the gravitational constant and g is the acceleration due to gravity.Four times it present value Open in App 1 |
If the density of the earth becomes 4 times and radius becomes half of what it is, our weight will
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