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Directions : The following question contains 

statemen-1 and statement -2  of the four choices   given, choose the one that best describes the two statements.

statemen-1  For a mass  M kept at the centre of a cube of side  a  The flux of gravitational field  passing through its sides                     is  4\pi GM     

statement -2     If the direction of a field due to a point source is radial and its dependence on the distance r  from the source is given as 1/r^{2}, its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface   

 

 

 

  • Option 1)

    statemen-1 is true ,and statement -2 false .

  • Option 2)

    statemen-1 is false ,and statement -2 true .

  • Option 3)

    statement-1 is true ,and statement -2 true  ;

            statement -2 is a correct explanation for statement -1

  • Option 4)

    statemen-1 is true ,and statement -2 true  ;

            statement -2 is not a correct explanation for statement -1

 

Answers (1)

best_answer

As we learnt in 

Let A \; be \; the \; Gaussian\; surface\; enclosing \; a spherical \; charge\; Q

\vec{E}.4\pi r^{2}=\frac{Q}{\varepsilon _{0}}

\vec{E}=\frac{Q}{4\pi \varepsilon _{0}.r^{2}}

Flux\; \phi =\vec{E}.4\pi r^{2}=\frac{Q}{\varepsilon _{0}}

Every line passing through A has to pass through B, whether B is a cube or any surface. It is only for Gaussian surface, the lines of field should be normal. Assuming the mass is a point mass. \vec{g} , gravitational \; field =-\frac{GM}{r^{2}}

Flux\; \phi _{g}=\left | \vec{g}.4\pi r^{2} \right |=\frac{4\pi r^{2}.GM}{r^{2}}=4\pi GM.

Here B s a cube. As explained earlier, whatever be the shape, all the lines passing through A are passing through B, although all the lines are not normal.

Statement­ 2 is correct because when the shape of the earth is spherical, area of the Gaussian surface is 4\pi r^{2}. This ensures inverse square law.

Let A be the Guassion surface enclosing a sperical charge Q.

\vec{E}4\pi r^{2}=\frac{q}{\epsilon _{0}}

\vec{E}=\frac{Q}{4\pi\epsilon _{0}r^{2}}

\phi =\vec{E}\left ( 4\pi r^{2} \right )=\frac{Q}{\epsilon _{0}}

Everyline passing through A has pass through B, whether B is a cube or any surface, it is only for guassion surface, the lines of fixed should be normal, Assuming the mass is a point mass.

g=\frac{-Gm}{r^{2}}

thus \phi _{g}= \left | \vec{g.4\pi r^{2}} \right |

\Rightarrow 4\pi r^{2}\frac{Gm}{r^{2}}=4 \pi Gm

Here B is a cube. As explained earlier, whatever be the shape, all the lines passing through A are passing through B. Although all the lines are not normal.

Statement 2 is covered because when the shape of the earth is spherical, area of the gaussian surface is 4\pi r^{2}


Option 1)

statemen-1 is true ,and statement -2 false .

Incorrect option

Option 2)

statemen-1 is false ,and statement -2 true .

Incorrect option

Option 3)

statement-1 is true ,and statement -2 true  ;

        statement -2 is a correct explanation for statement -1

Correct option

Option 4)

statemen-1 is true ,and statement -2 true  ;

        statement -2 is not a correct explanation for statement -1

Incorrect option

Posted by

prateek

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