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  • A body weighed with a spring balance in a train at reset, shows a weight W0. When the train begins to move with a velocity v around the equator from west to east and if the angular velocity of eart

A body weighed with a spring balance in a train at reset, shows a weight W0. When the train begins to move with a velocity v around the equator from west to east and if the angular velocity of earth is then the weight recorded by the spring balance is

Option: 1

\mathrm{W}_0

 


Option: 2

\mathrm{W}_0\left(1+\frac{\mathrm{v}^2}{\mathrm{R}}\right)
 


Option: 3

\mathrm{W}_0\left(1+\frac{2 \mathrm{v} \omega}{\mathrm{g}}\right)
 


Option: 4

\mathrm{W}_0\left(1-\frac{2 \mathrm{v} \omega}{\mathrm{g}}\right)


Answers (1)

If \mathrm{W} represents weight of the body when earth were not rotating, then weight of body, \mathrm{W}_0 in train at rest is given by

\mathrm{ \mathrm{W}_0=\mathrm{W}-\frac{\mathrm{mv}_{\mathrm{e}}^2}{\mathrm{R}} }           ............(1)

When the train moves from west to east with velocity \mathrm{v}, the velocity of the body relative to earth gets added to the velocity of earth.

Therefore, the weight of the body,

\mathrm{\mathrm{W}_1=\mathrm{W}-\mathrm{m} \frac{\left(\mathrm{v}_{\mathrm{e}}+\mathrm{v}\right)^2}{\mathrm{R}}}     ...........(2)

Subtracting (2) from (1)

\mathrm{ \mathrm{W}_0-\mathrm{W}_1=\frac{\mathrm{m}}{\mathrm{R}}\left[\left(\mathrm{v}_{\mathrm{e}}+\mathrm{v}\right)^2-\mathrm{v}_{\mathrm{e}}^2\right] }

\mathrm{ =\frac{\mathrm{m}}{\mathrm{R}}\left[\mathrm{v}_{\mathrm{e}}^2+\mathrm{v}^2+2 \mathrm{v}_{\mathrm{e}} \mathrm{v}-\mathrm{v}_{\mathrm{e}}^2\right] }

\mathrm{ \mathrm{W}_0-\mathrm{W}_1=\frac{\mathrm{m}}{\mathrm{R}}\left(\mathrm{v}^2+2 \mathrm{v}_{\mathrm{e}} \mathrm{v}\right)}

\mathrm{ \text { Since } \mathrm{v}^2<2 \mathrm{v}_{\mathrm{e}} \mathrm{v} \text { and } \frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{R}}=\omega }

\mathrm{ \mathrm{W}_0-\mathrm{W}_1=\frac{\mathrm{mg}}{\mathrm{Rg}}\left(2 \mathrm{v}_{\mathrm{e}} \mathrm{v}\right)=\frac{\mathrm{mg}}{\mathrm{R}}(2 \mathrm{vR} \omega) }

\mathrm{ \mathrm{W}_0-\mathrm{W}_1=\mathrm{W}_0 2 \mathrm{v} \frac{\omega}{\mathrm{g}} }

\mathrm{ \mathrm{W}_1=\mathrm{W}_0\left(1-\frac{2 \mathrm{v} \omega}{\mathrm{g}}\right) }

Hence option 4 is correct.










 

Posted by

Kshitij

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