Question Number 50808 by Tinkutara last updated on 20/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 21/Dec/18 $${at}\:{temparature}\:{t}\:{M}.{I}\:{is} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi{R}^{\mathrm{2}} {h}\rho\right){R}^{\mathrm{2}} \:\:\:+{Md}^{\mathrm{2}} \:{at}\:{temparature}\:{t} \\ $$$${d}={distance}\:{between}\:{axis}\:{of}\:{cylinder}\:{and}\:{axis} \\…
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Question Number 47377 by rahul 19 last updated on 09/Nov/18 Commented by rahul 19 last updated on 09/Nov/18 $${Ans}:\:\mathrm{10}^{\mathrm{5}} {Pa}. \\ $$ Answered by ajfour…
Question Number 47314 by rahul 19 last updated on 08/Nov/18 $${During}\:{random}\:{motion},\:{gas}\:{molecules} \\ $$$${do}\:{not}\:{interact}\:{with}\:{each}\:{other}\:. \\ $$$${Hence}\:\:{Potential}\:{energy}\:=\mathrm{0}\:…… \\ $$$${Pls}\:{explain}\:{why}\:{P}.{E}\:=\:\mathrm{0}? \\ $$$${P}.{E}\:{is}\:{stored}\:{form}\:{of}\:{energy},{right}? \\ $$ Commented by rahul 19…
Question Number 46790 by rahul 19 last updated on 31/Oct/18 Commented by rahul 19 last updated on 31/Oct/18 $${Why}\:{volume}\:{is}\:{not}\:{converted}\:{in}\:{Litres} \\ $$$${when}\:{everyrthing}\:{is}\:{in}\:{S}.{I}\:{units}\:? \\ $$ Commented by…
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Question Number 45127 by Necxx last updated on 09/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18 $$\left.\mathrm{23}\right)\bigtriangleup{L}_{{Al}} ={L}\propto_{{Al}} \bigtriangleup{T} \\ $$$$\bigtriangleup{L}_{{steel}} ={L}\propto_{{steel}} \bigtriangleup{T} \\ $$$${reauired}\:{space}=\frac{\mathrm{1}}{\mathrm{2}}{L}\bigtriangleup{T}\left(\propto_{{Al}}…
Question Number 45126 by Necxx last updated on 09/Oct/18 Commented by Necxx last updated on 09/Oct/18 $${pls}\:{help}\:{with}\:{this} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 45125 by Necxx last updated on 09/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18 $${at}\:{temparature}\:{T}\:\:{cube}\:{edge}={L} \\ $$$${at}\:{temp}\:{T}+{dT}\:\:{cube}\:{edge}={L}+{dL} \\ $$$${surface}\:{area}\:{at}\:{tepm}\:{T}\:\:{is}\:{L}^{\mathrm{2}} \\ $$$${at}\:{temp}\:{T}+{dT}\:{surface}\:{area}=\left({L}+{dL}\right)^{\mathrm{2}} \\ $$$${given}\:\alpha=\frac{{dL}}{{L}×{dT}}\:\:\:\:\:\:\:{dL}=\alpha{LdT}…