268319
An electric kettle has two coils. W hen one coil isswitched on it takes 15 minutes and theother takes 30 minutes to boil certain mass of water. The ratio of times taken by them, when connected in series and in parallel to boil the same mass of water is
268320
A resistance coil of \(60 \Omega\) is immersed in \(42 \mathrm{~kg}\) of water. A current of 7A is passed through it. The rise in temperature of water per minute is
1 \(4^{\circ} \mathrm{C}\)
2 \(8^{\circ} \mathrm{C}\)
3 \(13 C\)
4 \(12^{\circ} \mathrm{C}\)
Explanation:
\(\mathrm{JQ}=\mathrm{i}^{2} R t, m S \Delta t=i^{2} R t\)
Current Electricity
268321
What is the required resistance of the heater coil of an immersion heater that will increase the temperature of \(1.50 \mathrm{~kg}\) of water from \(10^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in 10 minutes while operating at \(240 \mathrm{~V}\) ?
1 \(25 \Omega\)
2 \(12.5 \Omega\)
3 \(250 \Omega\)
4 \(137.2 \Omega\)
Explanation:
use Joule's law \(Q=m \Delta \Delta T \Rightarrow\) but \(Q=i^{2} R T \Rightarrow \Delta T \alpha i^{2}\)
Current Electricity
268322
A \(5^{\circ} \mathrm{C}\) rise in the temperature is observed in a conductor by passing some current. When the current is doubled, then rise in tem perature will be equal to
268319
An electric kettle has two coils. W hen one coil isswitched on it takes 15 minutes and theother takes 30 minutes to boil certain mass of water. The ratio of times taken by them, when connected in series and in parallel to boil the same mass of water is
268320
A resistance coil of \(60 \Omega\) is immersed in \(42 \mathrm{~kg}\) of water. A current of 7A is passed through it. The rise in temperature of water per minute is
1 \(4^{\circ} \mathrm{C}\)
2 \(8^{\circ} \mathrm{C}\)
3 \(13 C\)
4 \(12^{\circ} \mathrm{C}\)
Explanation:
\(\mathrm{JQ}=\mathrm{i}^{2} R t, m S \Delta t=i^{2} R t\)
Current Electricity
268321
What is the required resistance of the heater coil of an immersion heater that will increase the temperature of \(1.50 \mathrm{~kg}\) of water from \(10^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in 10 minutes while operating at \(240 \mathrm{~V}\) ?
1 \(25 \Omega\)
2 \(12.5 \Omega\)
3 \(250 \Omega\)
4 \(137.2 \Omega\)
Explanation:
use Joule's law \(Q=m \Delta \Delta T \Rightarrow\) but \(Q=i^{2} R T \Rightarrow \Delta T \alpha i^{2}\)
Current Electricity
268322
A \(5^{\circ} \mathrm{C}\) rise in the temperature is observed in a conductor by passing some current. When the current is doubled, then rise in tem perature will be equal to
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Current Electricity
268319
An electric kettle has two coils. W hen one coil isswitched on it takes 15 minutes and theother takes 30 minutes to boil certain mass of water. The ratio of times taken by them, when connected in series and in parallel to boil the same mass of water is
268320
A resistance coil of \(60 \Omega\) is immersed in \(42 \mathrm{~kg}\) of water. A current of 7A is passed through it. The rise in temperature of water per minute is
1 \(4^{\circ} \mathrm{C}\)
2 \(8^{\circ} \mathrm{C}\)
3 \(13 C\)
4 \(12^{\circ} \mathrm{C}\)
Explanation:
\(\mathrm{JQ}=\mathrm{i}^{2} R t, m S \Delta t=i^{2} R t\)
Current Electricity
268321
What is the required resistance of the heater coil of an immersion heater that will increase the temperature of \(1.50 \mathrm{~kg}\) of water from \(10^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in 10 minutes while operating at \(240 \mathrm{~V}\) ?
1 \(25 \Omega\)
2 \(12.5 \Omega\)
3 \(250 \Omega\)
4 \(137.2 \Omega\)
Explanation:
use Joule's law \(Q=m \Delta \Delta T \Rightarrow\) but \(Q=i^{2} R T \Rightarrow \Delta T \alpha i^{2}\)
Current Electricity
268322
A \(5^{\circ} \mathrm{C}\) rise in the temperature is observed in a conductor by passing some current. When the current is doubled, then rise in tem perature will be equal to
268319
An electric kettle has two coils. W hen one coil isswitched on it takes 15 minutes and theother takes 30 minutes to boil certain mass of water. The ratio of times taken by them, when connected in series and in parallel to boil the same mass of water is
268320
A resistance coil of \(60 \Omega\) is immersed in \(42 \mathrm{~kg}\) of water. A current of 7A is passed through it. The rise in temperature of water per minute is
1 \(4^{\circ} \mathrm{C}\)
2 \(8^{\circ} \mathrm{C}\)
3 \(13 C\)
4 \(12^{\circ} \mathrm{C}\)
Explanation:
\(\mathrm{JQ}=\mathrm{i}^{2} R t, m S \Delta t=i^{2} R t\)
Current Electricity
268321
What is the required resistance of the heater coil of an immersion heater that will increase the temperature of \(1.50 \mathrm{~kg}\) of water from \(10^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in 10 minutes while operating at \(240 \mathrm{~V}\) ?
1 \(25 \Omega\)
2 \(12.5 \Omega\)
3 \(250 \Omega\)
4 \(137.2 \Omega\)
Explanation:
use Joule's law \(Q=m \Delta \Delta T \Rightarrow\) but \(Q=i^{2} R T \Rightarrow \Delta T \alpha i^{2}\)
Current Electricity
268322
A \(5^{\circ} \mathrm{C}\) rise in the temperature is observed in a conductor by passing some current. When the current is doubled, then rise in tem perature will be equal to