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Standard XII

Chemistry

Abnormal Colligative Properties

26 V W 3 Ω ii...

Question

There are three setups, shown in figure. Voltage of ends are given, heat developed in 1, 2 and 3 are $H_{1}, H_{2}$ and $H_{3}$ , then

H_{1} > H_{2} > H_{3}

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H_{1} = H_{2} > H_{3}

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H_{1} > H_{2} = H_{3}

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H_{1} = H_{2} = H_{3}

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Solution

The correct option is D $H_{1} = H_{2} = H_{3}$
$Δ V$ and $R$ are same in all cases.
$H = \frac{(Δ V)^{2}}{R}$
$H_{1} = H_{2} = H_{3}$

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Then

\frac{h_{1} + r}{h_{1} - r} + \frac{h_{2} + r}{h_{2} - r} + \frac{h_{3} + r}{h_{3} - r}

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There are three setups, shown in figure. Voltage of ends are given, heat developed in 1, 2 and 3 are H1,H2 and H3, then

The correct option is D H1=H2=H3ΔV and R are same in all cases. H=(ΔV)2R H1=H2=H3

There are three setups, shown in figure. Voltage of ends are given, heat developed in 1, 2 and 3 are $H_{1}, H_{2}$ and $H_{3}$ , then

The correct option is D $H_{1} = H_{2} = H_{3}$
$Δ V$ and $R$ are same in all cases.
$H = \frac{(Δ V)^{2}}{R}$
$H_{1} = H_{2} = H_{3}$