
- A$\frac{{600}}{{13}}\ ^oC$
- ✓$\frac{{500}}{{7}}\ ^oC$
- C$\frac{{400}}{{13}}\ ^oC$
- D$\frac{{700}}{{6}}\ ^oC$

$\Delta \mathrm{T}=100^\circ \mathrm{C}$
Expanding it on a plan
$\text { By symmetry } \quad \mathrm{T}_{\mathrm{H}}=\mathrm{T}_{\mathrm{B}}$
$\mathrm{T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{E}}$
$\Rightarrow$ Points $\mathrm{B}$ and $\mathrm{H} \& \mathrm{E}$ and $\mathrm{C}$ coule
ai ned toget her.
olving it like normal current circuit
Temp. difference across $\mathrm{BC}$
$\frac{{{{100}^ \circ } - {0^ \circ }C}}{{\frac{{7R}}{2}}} \times \frac{{5R}}{2} = \frac{{100}}{7} \times 5 = \frac{{500}}{7}{\,^ \circ }C$
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$1.$ If the total energy of the particle is $E$, it will perform periodic motion only if
$(A)$ $E$ $<0$ $(B)$ $E$ $>0$ $(C)$ $\mathrm{V}_0 > \mathrm{E}>0$ $(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $\mathrm{A}$, the time period $\mathrm{T}$ of this particle is proportional to
$(A)$ $\mathrm{A} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(B)$ $\frac{1}{\mathrm{~A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(C)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$ $(D)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$
$3.$ The acceleration of this particle for $|\mathrm{x}|>\mathrm{X}_0$ is
$(A)$ proportional to $\mathrm{V}_0$
$(B)$ proportional to $\frac{\mathrm{V}_0}{\mathrm{mX}_0}$
$(C)$ proportional to $\sqrt{\frac{\mathrm{V}_0}{\mathrm{mX}_0}}$
$(D)$ zero
Give the answer qustion $1,2$ and $3.$
| List$-I$ | List$-II$ |
| $(a)$ Torque | $(i)$ ${MLT}^{-1}$ |
| $(b)$ Impulse | $(ii)$ ${MT}^{-2}$ |
| $(c)$ Tension | $(iii)$ ${ML}^{2} {T}^{-2}$ |
| $(d)$ Surface Tension | $(iv)$ ${MI} {T}^{-2}$ |
Choose the most appropriate answer from the option given below :
