Question Number 39755 by rahul 19 last updated on 10/Jul/18
Answered by ajfour last updated on 11/Jul/18
$${a}={g}\mathrm{cot}\:\theta \\ $$$${as}\:\:\:{a}=\frac{{M}\:'{g}}{{M}+{M}\:'}\:=\:{g}\mathrm{cot}\:\theta \\ $$$$\Rightarrow\:\:\:\mathrm{1}+\frac{{M}}{{M}\:'}\:=\mathrm{tan}\:\theta \\ $$$$\Rightarrow\:\:\:\:\:{M}\:'\:=\:\frac{{M}}{\mathrm{tan}\:\theta−\mathrm{1}}\:=\:\frac{{M}\mathrm{cot}\:\theta}{\mathrm{1}−\mathrm{cot}\:\theta} \\ $$$$\:\:{option}\:\left({a}\right)\:{correct} \\ $$$${T}\:=\:{Ma}\:=\:{Mg}\mathrm{cot}\:\theta\:=\:\frac{{Mg}}{\mathrm{tan}\:\theta} \\ $$$${option}\:\left({c}\right)\:{is}\:{correct}\:{too}. \\ $$$$−−−−−−−−−−−−−−− \\ $$
Commented by rahul 19 last updated on 13/Jul/18
Thank You sir ����