MCQ
Twelve conducting rods form the riders of a uniform cube of side $'l'.$ If in steady state, $B$ and $H$ ends of the rod are at $100^o C$ and $0^o C$. Find the temperature of the junction $'A'$ ....... $^oC$


- A$80$
- ✓$60$
- C$40$
- D$70$

(Where $R=$ thermal resistance of each side)
$H=\frac{100-0}{5 R / 6}$
For side $A$ $\frac{H}{3}=\frac{100-\theta_{A}}{R} \Rightarrow \theta_{A}=60^{\circ} C$
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$F=F_{0}\left(1-\left(\frac{t-T}{T}\right)^{2}\right)$
Where $F_{0}$ and $T$ are constants. The force acts only for the time internal $2 T$. The velocity $v$ of the particle after time $2 {T}$ is -

(image)
$(A)$ Process $I$ is an isochoric process $(B)$ In process $II$, gas absorbs heat
$(C)$ In process $IV$, gas releases heat $(D)$ Processes $I$ and $III$ are $not$ isobaric