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Question Number 59347 by Tinkutara last updated on 08/May/19
Commented by Tinkutara last updated on 08/May/19
B part only
Commented by Tinkutara last updated on 09/May/19
Yes I am too sad with mine also. Got 4.5k. Yours? But I don't think it's a mistake by NTA, it's an era of tough competition!
Commented by Tinkutara last updated on 09/May/19
Mine was 230 One of my friend got 300 marks but 2.6k AIR What was your marks?
Commented by rahul 19 last updated on 09/May/19
201. U r very lucky! 4530 at 230 marks !!! as far as I know for a rank under 5,000 ,you need 260+ ...
Commented by Tinkutara last updated on 09/May/19
But marks have no meaning now. I got 290 in April but low percentile so January marks were considered. Tmhari kabki %ile maani hai?
Commented by prakash jain last updated on 10/May/19
$$\mathrm{What}\:\mathrm{shift}\:\mathrm{you}\:\mathrm{were}\:\mathrm{in}\:\mathrm{April}? \\ $$$$\mathrm{9th}\:\mathrm{April}\:\mathrm{morning}? \\ $$$$\mathrm{All}\:\mathrm{other}\:\mathrm{shift}\:\mathrm{would}\:\mathrm{have}\:\mathrm{fetched} \\ $$$$\mathrm{you}\:\mathrm{rank}\:\mathrm{under}\:\:\mathrm{2k}\:\mathrm{with}\:\mathrm{290}\:\mathrm{marks} \\ $$
Commented by Tinkutara last updated on 10/May/19
Yes 9th morning in April
Answered by ajfour last updated on 09/May/19
Commented by ajfour last updated on 11/May/19
![m(dv/dt)=−(√(2gh)) (dm/dt) (as m(dv/dt)=v_(rel) (dm/dt)+F_(ext) ) (ρAh)(dv/dt)=−(√(2gh))[(d/dt)(ρAh)] (dv/dt)= −(√(2g))×(1/(√h))(dh/dt) ....(i) further −(A/(100))(√(2g(H/4))) =A(dh/dt) or (dh/dt)=−(1/(100))(√((gH)/2)) (dv/dt)= (√(2g))×(2/(√H))×(1/(100))(√((gH)/2)) a= (g/(50)) And from (i) dv=−(√(2g))×(dh/(√h)) v= 2(√(2g))((√H)−(√h)) when h=(H/4) v=(√(2gH)) = (√((2m_0 g)/(ρA))) .](Q59461.png)
$$\:\:\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=−\sqrt{\mathrm{2gh}}\:\frac{\mathrm{dm}}{\mathrm{dt}}\:\:\:\:\left(\mathrm{as}\:\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{rel}} \frac{\mathrm{dm}}{\mathrm{dt}}+\mathrm{F}_{\mathrm{ext}} \right) \\ $$$$\:\:\:\left(\rho\mathrm{Ah}\right)\frac{\mathrm{dv}}{\mathrm{dt}}=−\sqrt{\mathrm{2gh}}\left[\frac{\mathrm{d}}{\mathrm{dt}}\left(\rho\mathrm{Ah}\right)\right] \\ $$$$\:\:\:\:\:\:\frac{\mathrm{dv}}{\mathrm{dt}}=\:−\sqrt{\mathrm{2g}}×\frac{\mathrm{1}}{\sqrt{\mathrm{h}}}\frac{\mathrm{dh}}{\mathrm{dt}}\:\:\:\:\:\:\:\:\:\:\:....\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{further} \\ $$$$\:\:\:\:\:\:−\frac{\mathrm{A}}{\mathrm{100}}\sqrt{\mathrm{2g}\frac{\mathrm{H}}{\mathrm{4}}}\:=\mathrm{A}\frac{\mathrm{dh}}{\mathrm{dt}} \\ $$$$\mathrm{or}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{dh}}{\mathrm{dt}}=−\frac{\mathrm{1}}{\mathrm{100}}\sqrt{\frac{\mathrm{gH}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\frac{\mathrm{dv}}{\mathrm{dt}}=\:\sqrt{\mathrm{2g}}×\frac{\mathrm{2}}{\sqrt{\mathrm{H}}}×\frac{\mathrm{1}}{\mathrm{100}}\sqrt{\frac{\mathrm{gH}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\mathrm{a}=\:\frac{\mathrm{g}}{\mathrm{50}} \\ $$$$\:\mathrm{And}\:\mathrm{from}\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\mathrm{dv}=−\sqrt{\mathrm{2g}}×\frac{\mathrm{dh}}{\sqrt{\mathrm{h}}} \\ $$$$\:\:\:\mathrm{v}=\:\mathrm{2}\sqrt{\mathrm{2g}}\left(\sqrt{\mathrm{H}}−\sqrt{\mathrm{h}}\right) \\ $$$$\mathrm{when}\:\:\mathrm{h}=\frac{\mathrm{H}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\mathrm{v}=\sqrt{\mathrm{2gH}}\:=\:\sqrt{\frac{\mathrm{2m}_{\mathrm{0}} \mathrm{g}}{\rho\mathrm{A}}}\:. \\ $$
Commented by Tinkutara last updated on 12/May/19
$$\:\mathrm{as}\:\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{rel}} \frac{\mathrm{dm}}{\mathrm{dt}}+\mathrm{F}_{\mathrm{ext}} \\ $$$${Please}\:{explain}\:{this}\:{Sir}. \\ $$
Commented by ajfour last updated on 12/May/19
$$\mathrm{consider}\:\mathrm{a}\:\mathrm{rocket}\:\mathrm{entering}\:\mathrm{an} \\ $$$$\mathrm{atmosphere}\:\mathrm{and}\:\mathrm{accumulating} \\ $$$$\mathrm{mass}\:\mathrm{as}\:\mathrm{it}\:\mathrm{proceeds}; \\ $$$$\:\:\:\:\mathrm{let}\:\mathrm{rocket}\:\mathrm{speed}\:\mathrm{be}\:\mathrm{v}, \\ $$$$\:\:\:\:\mathrm{External}\:\mathrm{force}\:\mathrm{acting}\:\mathrm{F}_{\mathrm{ext}} \\ $$$$\:\:\mathrm{mass}\:\mathrm{dm}\:\mathrm{of}\:\mathrm{atmosphere},\:\mathrm{moving} \\ $$$$\:\:\mathrm{at}\:\mathrm{speed}\:\mathrm{u},\:\mathrm{and} \\ $$$$\mathrm{rocket}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\:\mathrm{be}\:\mathrm{our}\:\mathrm{system}, \\ $$$$\:\:\:\:\:\:\:\mathrm{p}_{\mathrm{i}} =\:\mathrm{udm}+\mathrm{mv} \\ $$$$\:\:\:\:\:\:\:\mathrm{p}_{\mathrm{f}} =\:\left(\mathrm{m}+\mathrm{dm}\right)\left(\mathrm{v}+\mathrm{dv}\right) \\ $$$$\:\:\:\mathrm{dp}=\:\left(\mathrm{v}−\mathrm{u}\right)\mathrm{dm}+\mathrm{mdv} \\ $$$$\:\:\:\mathrm{F}_{\mathrm{ext}} \mathrm{dt}\:+\left(\mathrm{u}−\mathrm{v}\right)\mathrm{dm}\:=\:\mathrm{mdv} \\ $$$$\Rightarrow\:\:\:\frac{{mdv}}{{dt}}=\:{v}_{{rel}} \frac{{dm}}{{dt}}+{F}_{{ext}} \:. \\ $$
Commented by Tinkutara last updated on 13/May/19
Thank you Sir!