PhysicsQ1–2 of 2 questions
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1
PhysicsMediumMCQIIT JEE · 1975
To double the volume of a given mass of an ideal gas at $27^{\circ}C$ keeping the pressure constant,one must raise the temperature in degrees Celsius to ....... $^{\circ}C$.
A
$54$
B
$270$
C
$327$
D
$600$
Solution
(C) According to Charles's Law,for a given mass of an ideal gas at constant pressure,the volume is directly proportional to the absolute temperature: $V \propto T$.
This implies: $\frac{V_1}{V_2} = \frac{T_1}{T_2}$.
Given: $V_1 = V$,$V_2 = 2V$,and $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Substituting the values: $\frac{V}{2V} = \frac{300}{T_2}$.
Solving for $T_2$: $T_2 = 300 \times 2 = 600 \ K$.
Converting back to degrees Celsius: $T_2(^{\circ}C) = 600 - 273 = 327^{\circ}C$.
This implies: $\frac{V_1}{V_2} = \frac{T_1}{T_2}$.
Given: $V_1 = V$,$V_2 = 2V$,and $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Substituting the values: $\frac{V}{2V} = \frac{300}{T_2}$.
Solving for $T_2$: $T_2 = 300 \times 2 = 600 \ K$.
Converting back to degrees Celsius: $T_2(^{\circ}C) = 600 - 273 = 327^{\circ}C$.
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2
PhysicsDifficultMCQIIT JEE · 1975
Silver and copper voltameters are connected in parallel with a battery of $e.m.f.$ $12\, V$. In $30$ minutes,$1\, g$ of silver and $1.8\, g$ of copper are liberated. The power supplied by the battery is ...................... $J/s$ ($Z_{Cu} = 6.6 \times 10^{-4}\,g/C$ and $Z_{Ag} = 11.2 \times 10^{-4}\,g/C$).
A
$24.13$
B
$2.413$
C
$0.2413$
D
$2413$
Solution
(A) The current $I_1$ through the silver voltameter is given by Faraday's law of electrolysis: $I_1 = \frac{m_1}{Z_{Ag} t} = \frac{1}{11.2 \times 10^{-4} \times 30 \times 60} \approx 0.496\, A$.
The current $I_2$ through the copper voltameter is: $I_2 = \frac{m_2}{Z_{Cu} t} = \frac{1.8}{6.6 \times 10^{-4} \times 30 \times 60} \approx 1.515\, A$.
The total current $I$ supplied by the battery is $I = I_1 + I_2 = 0.496 + 1.515 = 2.011\, A$.
The power supplied by the battery is $P = I \times V = 2.011 \times 12 = 24.132\, J/s$.
The current $I_2$ through the copper voltameter is: $I_2 = \frac{m_2}{Z_{Cu} t} = \frac{1.8}{6.6 \times 10^{-4} \times 30 \times 60} \approx 1.515\, A$.
The total current $I$ supplied by the battery is $I = I_1 + I_2 = 0.496 + 1.515 = 2.011\, A$.
The power supplied by the battery is $P = I \times V = 2.011 \times 12 = 24.132\, J/s$.
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