question_answer

An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of 12\[=\frac{\sigma }{{{\varepsilon }_{0}}}\left( \frac{{{a}^{2}}}{c}-b+c \right)\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,(\because c=a+b)\]and 2 kg second part moving with a velocity of 8 \[{{V}_{C}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sigma 4\pi {{a}^{2}}}{c}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{b}^{2}}}{c}\]. If the third part flies off with a velocity of 4 \[+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sigma 4\pi {{c}^{2}}}{c}\], its mass would be

A) 5kg                                                        

B) 7kg                                        

C) 17kg                                      

D) 3kg

Correct Answer: A

Solution :

Key Idea Apply law of conservation of linear momentum. Momentum of first part \[Ga<In<Al<tl\] Momentum of the second part \[\text{ethanol}\xrightarrow[{}]{\text{PB}{{\text{r}}_{\text{3}}}}\text{X}\xrightarrow[{}]{\text{alc}\text{.KOH}}\text{Y}\] \[\xrightarrow[\text{(ii)}{{\text{H}}_{\text{2}}}\text{O,heat}]{\text{(i)}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{room}\,\text{temperature}}Z;\]Resultant momentum \[C{{H}_{2}}=C{{H}_{2}}\] The third part should also have the same momentum. Let the mass of the third part be M, then \[C{{H}_{3}}C{{H}_{2}}-O-C{{H}_{2}}-C{{H}_{3}}\] \[C{{H}_{3}}-C{{H}_{2}}-O-S{{O}_{3}}H\] Alternative: \[C{{H}_{3}}C{{H}_{2}}OH\] \[Co{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})Cl\] \[({{k}_{f}}=-{{1.86}^{o}}C/m)\] \[C{{H}_{3}}OH\]