A) \[90{}^\circ \]
B) \[\gamma \text{=}\frac{dV}{VdT}=\frac{3}{T}\]
C) \[Q=\frac{V}{t'}=\frac{\pi p{{r}^{4}}}{8\eta l}\]
D) \[30{}^\circ \]
Correct Answer: B
Solution :
Let \[{{n}_{1}}\]and \[{{n}_{2}}\] be the vectors normal to the faces OAB and ABC. Then, \[{{n}_{1}}=OA\times OB=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \\ \end{matrix} \right|\] \[=(6-1)\hat{i}-(3-2)\hat{j}+(1-4)\hat{k}=5\hat{i}-\hat{j}-3\hat{k}\]and \[{{n}_{2}}=AB\times AC=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \\ \end{matrix} \right|\] \[=(-1+2)\hat{i}=(1+4)\hat{j}+(-1-2)\hat{k}\] \[=\hat{i}-5\hat{j}-3\hat{k}\] If \[\theta \] Is the angle between the faces OAB and ABC, then \[\cos \theta =\frac{{{n}_{1}}.{{n}_{2}}}{|{{n}_{1}}||{{n}_{2}}|}\] \[\Rightarrow \] \[\cos \theta =\frac{(5\hat{i}-\hat{j}-3\hat{k}).(\hat{i}-5\hat{j}-3\hat{k})}{\sqrt{{{5}^{2}}+{{(-1)}^{2}}+{{(-3)}^{2}}}\sqrt{{{1}^{2}}{{(-5)}^{2}}+{{(-3)}^{2}}}}\] \[=\frac{5+5+9}{\sqrt{25+1+9}\sqrt{1+25+9}}=\frac{19}{35}\] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}\left( \frac{19}{35} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
