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Question Number 50808 by Tinkutara last updated on 20/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Dec/18

at temparature t M.I is  I=(1/2)(πR^2 hρ)R^2    +Md^2  at temparature t  d=distance between axis of cylinder and axis  of rotation...  I=((πρ)/2)(R^4 h)  (dI/dt)=((πρ)/2)(R^4 (dh/dt)+4hR^3 (dR/dt))+((dM/dt))d^2   dM=0   no change of mass with temparature  rise  α=(dl/(ldt))=((change of length)/(original length×change of temparsture))  so   (dI/dt)=((πρ)/2)(R^4 h×(dh/(hdt))+4hR^4 ×(dR/(Rdt)))  dI=((πρR^4 h)/2)αdt+4×((πρhR^4 )/2)×(−α)dt[dR=−ve]  dI=Iαdt−4×Iαdt=−3Iαdt  since value of h increases so  value of R decreases  hence M.I get decreases...  pls check is it right...

$${at}\:{temparature}\:{t}\:{M}.{I}\:{is} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi{R}^{\mathrm{2}} {h}\rho\right){R}^{\mathrm{2}} \:\:\:+{Md}^{\mathrm{2}} \:{at}\:{temparature}\:{t} \\ $$$${d}={distance}\:{between}\:{axis}\:{of}\:{cylinder}\:{and}\:{axis} \\ $$$${of}\:{rotation}... \\ $$$${I}=\frac{\pi\rho}{\mathrm{2}}\left({R}^{\mathrm{4}} {h}\right) \\ $$$$\frac{{dI}}{{dt}}=\frac{\pi\rho}{\mathrm{2}}\left({R}^{\mathrm{4}} \frac{{dh}}{{dt}}+\mathrm{4}{hR}^{\mathrm{3}} \frac{{dR}}{{dt}}\right)+\left(\frac{{dM}}{{dt}}\right){d}^{\mathrm{2}} \\ $$$${dM}=\mathrm{0}\:\:\:{no}\:{change}\:{of}\:{mass}\:{with}\:{temparature} \\ $$$${rise} \\ $$$$\alpha=\frac{{dl}}{{ldt}}=\frac{{change}\:{of}\:{length}}{{original}\:{length}×{change}\:{of}\:{temparsture}} \\ $$$${so}\: \\ $$$$\frac{{dI}}{{dt}}=\frac{\pi\rho}{\mathrm{2}}\left({R}^{\mathrm{4}} {h}×\frac{{dh}}{{hdt}}+\mathrm{4}{hR}^{\mathrm{4}} ×\frac{{dR}}{{Rdt}}\right) \\ $$$${dI}=\frac{\pi\rho{R}^{\mathrm{4}} {h}}{\mathrm{2}}\alpha{dt}+\mathrm{4}×\frac{\pi\rho{hR}^{\mathrm{4}} }{\mathrm{2}}×\left(−\alpha\right){dt}\left[{dR}=−{ve}\right] \\ $$$${dI}={I}\alpha{dt}−\mathrm{4}×{I}\alpha{dt}=−\mathrm{3}{I}\alpha{dt} \\ $$$${since}\:{value}\:{of}\:{h}\:{increases}\:{so}\:\:{value}\:{of}\:{R}\:{decreases} \\ $$$${hence}\:{M}.{I}\:{get}\:{decreases}... \\ $$$${pls}\:{check}\:{is}\:{it}\:{right}... \\ $$

Commented by Tinkutara last updated on 22/Dec/18

No sir your answer is wrong.

Answered by ajfour last updated on 22/Dec/18

I = ((mr^2 )/2)      △I = (m/2)(2αr^2 )△t  ⇒   △I =  2αI△t     [option (c)].

$${I}\:=\:\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\: \\ $$$$\bigtriangleup{I}\:=\:\frac{{m}}{\mathrm{2}}\left(\mathrm{2}\alpha{r}^{\mathrm{2}} \right)\bigtriangleup{t} \\ $$$$\Rightarrow\:\:\:\bigtriangleup{I}\:=\:\:\mathrm{2}\alpha{I}\bigtriangleup{t}\:\:\:\:\:\left[{option}\:\left({c}\right)\right]. \\ $$

Commented by Tinkutara last updated on 23/Dec/18

Thanks Sir!

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