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CBSE Class 12-science Answered

A metallic solid sphere is rotating about its diameter as axis of rotation . If the temperature is increased by 200∘C200∘C, the percentage increase in its moment of inertia is : (Coefficient of linear expansion of the metal =10−5/∘C=10−5/∘C)
Asked by Atulcaald | 31 May, 2018, 09:30: PM
answered-by-expert Expert Answer
Change in volume δV due to thermal expansion is given by
 
begin mathsize 12px style delta V space equals space V subscript 0 gamma capital delta T space................. left parenthesis 1 right parenthesis end style
where V0 is initial volume, γ is volume expansion coefficient and ΔT is increase of temperature.
 
change in volume as a function of change in radius is given by
 
begin mathsize 12px style delta V space equals space 4 over 3 pi open parentheses R plus delta R close parentheses cubed space minus 4 over 3 pi R cubed space equals space 4 over 3 pi R cubed open square brackets open parentheses 1 plus fraction numerator delta R over denominator R end fraction close parentheses cubed minus space 1 close square brackets space equals V subscript 0 open square brackets open parentheses 1 plus fraction numerator 3 delta R over denominator R end fraction close parentheses minus space 1 close square brackets space because delta R space less than less than space R h e n c e space space delta V space equals space V subscript 0 fraction numerator 3 delta R over denominator R end fraction............................ left parenthesis 2 right parenthesis end style
 
from (1) and (2), we get begin mathsize 12px style fraction numerator delta R over denominator R end fraction space equals space gamma over 3 capital delta T space equals space alpha capital delta T space space............................. left parenthesis 3 right parenthesis end style
where α is linear expansion coefficient ( γ = 3α )
 
Moment of inertia I is given by, I = K×R2 ....................(4)
where all constants are clubbed together as K.
 
change in moment of inertia δI due to change in radius δR is given by,
δI = 2×R×K×δR ...................(5)
 
from (4) and (5), we get begin mathsize 12px style fraction numerator delta I over denominator I end fraction space equals space fraction numerator 2 delta R over denominator R end fraction end style...................(6)
using eqn(3), we rewrite eqn(6) to get percentage change as,
begin mathsize 12px style fraction numerator delta I over denominator I end fraction cross times 100 space equals space 2 fraction numerator delta R over denominator R end fraction cross times 100 space equals space 2 cross times alpha cross times capital delta T cross times 100 space equals space 2 cross times 10 to the power of negative 5 end exponent cross times 200 cross times 100 space equals space 0.4 percent sign end style
Answered by Thiyagarajan K | 09 Jun, 2018, 11:09: AM

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