Answer
Verified
469.8k+ views
Hint: The weight of the body in the absence of rotational motion is mg. This downward force mg is subjected to changes if centripetal force acts too. Velocities of bodies differ if their distances are different from the axis.
Complete step by step answer:
If the earth were not spinning, the weight of the body would have been mg. The earth rotating about its polar axis will make the body undergo a circular motion too. Suppose the object is present at an angle $\phi$ with the polar axis, then, the distance (or radius) from the axis will be $R\sin \phi$ . The centripetal force is given by $mv^2/r$ or $m \omega^2 r$, if the object is at a distance r from the axis. Therefore, we may write:
$mg' = mg - m \omega^2 R\sin \phi$
Canceling the m,
$g' = g - \omega^2 R\sin \phi$.
In the question it has been asked to find what happens to the g when our object is at the equator. So, we are given $\phi = 90^{\circ}$.
This clearly helps us see that the object at the equator will have a g' that is smaller than the actual g. Therefore the correct answer is option (C). The weight of the body decreases at the equator.
So, the correct answer is “Option C”.
Note:
To clear up the formula, consider drawing two perpendicular axes at the point where the object is present. Now, see what angle the centripetal force makes with these axes. mg will be always directed towards the centre of the earth, no matter where it is present and centripetal force points out from the circle that the object makes.
Complete step by step answer:
If the earth were not spinning, the weight of the body would have been mg. The earth rotating about its polar axis will make the body undergo a circular motion too. Suppose the object is present at an angle $\phi$ with the polar axis, then, the distance (or radius) from the axis will be $R\sin \phi$ . The centripetal force is given by $mv^2/r$ or $m \omega^2 r$, if the object is at a distance r from the axis. Therefore, we may write:
$mg' = mg - m \omega^2 R\sin \phi$
Canceling the m,
$g' = g - \omega^2 R\sin \phi$.
In the question it has been asked to find what happens to the g when our object is at the equator. So, we are given $\phi = 90^{\circ}$.
This clearly helps us see that the object at the equator will have a g' that is smaller than the actual g. Therefore the correct answer is option (C). The weight of the body decreases at the equator.
So, the correct answer is “Option C”.
Note:
To clear up the formula, consider drawing two perpendicular axes at the point where the object is present. Now, see what angle the centripetal force makes with these axes. mg will be always directed towards the centre of the earth, no matter where it is present and centripetal force points out from the circle that the object makes.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE