Question
If $\Delta=\left|\begin{array}{ccc}x+s & x+t & x+u \\ x+s+1 & x+t+1 & x+u+1 \\ x+p & x+q & x+r\end{array}\right|$, then $\Delta$ does not depend on(a) $\mathrm{p}$(b) $\mathrm{q}$(c) $\mathrm{r}$(d) $x$
Step 1
We can simplify this determinant by performing row operations. Show more…
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$$ \begin{aligned} &\Delta S=-\frac{m q_{1}}{T_{2}}-m c \ln \frac{T_{2}}{T_{1}}+\frac{M q_{i c e}}{T_{1}} \\ &\text { where } \quad M q_{i c e}=m\left(q_{2}+c\left(T_{2}-T_{1}\right)\right) \\ &=m q_{2}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)+m c\left(\frac{T_{2}}{T_{1}}-1-\frac{T_{2}}{T_{1}}\right) \\ &=0.2245+0.2564 \sim 0.48 \mathrm{~J} / \mathrm{K} \end{aligned} $$
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If $\Delta=\left|\begin{array}{lll}b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2}\end{array}\right|$ then $\Delta$ equals (a) $(b-c)(c-a)(a-b)$ (b) abe $(b-c)(c-a)(a-b)$ (c) $(a+b+c)(b-c)(c-a)(a-b)$ (d) 0
Which of the following is correct equation? (a) $\Delta \mathrm{U}=\Delta \mathrm{Q}-\mathrm{W}$ (b) $\Delta \mathrm{W}=\Delta \mathrm{U}+\Delta \mathrm{Q}$ (c) $\Delta \mathrm{U}=\Delta \mathrm{W}+\Delta \mathrm{Q}$ (d) None of these
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