JEE Main 12 Jan 2019 Evening Question 27
Question: The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is I(x). Which one of the graphs represents the variation of I(x) with x correctly? [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A)
B)
C)
D)
Show Answer
Answer:
Correct Answer: B
Solution:
The moment of inertia I(x) at distance x is $ I(x)=\frac{2}{5}MR^{2}+Mx^{2} $ $ x^{2}=\frac{1}{M}(I(x)-\frac{2}{5}MR^{2}) $
This equation resembles the standard equation of parabola i.e., $ {{(x-h)}^{2}}=4p(y-k) $
Hence the curve will be parabolic.
