NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
268315
A \(200 \mathrm{~W}-200 \mathrm{~V}\) lamp is connected to \(250 \mathrm{~V}\) mains. It power consumption is
1 \(300 \mathrm{~W}\)
2 \(312.5 \mathrm{~W}\)
3 \(292 \mathrm{~W}\)
4 \(250 \mathrm{~W}\)
Explanation:
\(P=\frac{V^{2}}{R}\)
Current Electricity
268316
If the current in a heater increases by \(10 \%\), the eercentage change in the power consumption
1 \(19 \%\)
2 \(21 \%\)
3 \(25 \%\)
4 \(17 \%\)
Explanation:
\(P \propto i^{2}\)
Current Electricity
268317
The power of a heating coil is \(P\). It is cut into two equal parts. The power of one of them across same mains is
1 \(2 \mathrm{P}\)
2 \(3 P\)
3 \(P / 2\)
4 \(4 P\)
Explanation:
\(P=\frac{V^{2}}{R} ; R=\frac{\rho l}{A}\)
Current Electricity
268318
In a house there are four bulbs each of \(50 \mathrm{~W}\) and 5 fans each of \(60 \mathrm{~W}\). If they are used at the rate of 6 hours a day, the electrical energy consumed in a month of 30 days is
1 \(64 \mathrm{KWH}\)
2 \(90.8 \mathrm{KWH}\)
3 \(72 \mathrm{KWH}\)
4 \(42 \mathrm{KWH}\)
Explanation:
1 unit \(=\frac{\text { no. watts } \times \text { no. of hours }}{1000}\)
268315
A \(200 \mathrm{~W}-200 \mathrm{~V}\) lamp is connected to \(250 \mathrm{~V}\) mains. It power consumption is
1 \(300 \mathrm{~W}\)
2 \(312.5 \mathrm{~W}\)
3 \(292 \mathrm{~W}\)
4 \(250 \mathrm{~W}\)
Explanation:
\(P=\frac{V^{2}}{R}\)
Current Electricity
268316
If the current in a heater increases by \(10 \%\), the eercentage change in the power consumption
1 \(19 \%\)
2 \(21 \%\)
3 \(25 \%\)
4 \(17 \%\)
Explanation:
\(P \propto i^{2}\)
Current Electricity
268317
The power of a heating coil is \(P\). It is cut into two equal parts. The power of one of them across same mains is
1 \(2 \mathrm{P}\)
2 \(3 P\)
3 \(P / 2\)
4 \(4 P\)
Explanation:
\(P=\frac{V^{2}}{R} ; R=\frac{\rho l}{A}\)
Current Electricity
268318
In a house there are four bulbs each of \(50 \mathrm{~W}\) and 5 fans each of \(60 \mathrm{~W}\). If they are used at the rate of 6 hours a day, the electrical energy consumed in a month of 30 days is
1 \(64 \mathrm{KWH}\)
2 \(90.8 \mathrm{KWH}\)
3 \(72 \mathrm{KWH}\)
4 \(42 \mathrm{KWH}\)
Explanation:
1 unit \(=\frac{\text { no. watts } \times \text { no. of hours }}{1000}\)
268315
A \(200 \mathrm{~W}-200 \mathrm{~V}\) lamp is connected to \(250 \mathrm{~V}\) mains. It power consumption is
1 \(300 \mathrm{~W}\)
2 \(312.5 \mathrm{~W}\)
3 \(292 \mathrm{~W}\)
4 \(250 \mathrm{~W}\)
Explanation:
\(P=\frac{V^{2}}{R}\)
Current Electricity
268316
If the current in a heater increases by \(10 \%\), the eercentage change in the power consumption
1 \(19 \%\)
2 \(21 \%\)
3 \(25 \%\)
4 \(17 \%\)
Explanation:
\(P \propto i^{2}\)
Current Electricity
268317
The power of a heating coil is \(P\). It is cut into two equal parts. The power of one of them across same mains is
1 \(2 \mathrm{P}\)
2 \(3 P\)
3 \(P / 2\)
4 \(4 P\)
Explanation:
\(P=\frac{V^{2}}{R} ; R=\frac{\rho l}{A}\)
Current Electricity
268318
In a house there are four bulbs each of \(50 \mathrm{~W}\) and 5 fans each of \(60 \mathrm{~W}\). If they are used at the rate of 6 hours a day, the electrical energy consumed in a month of 30 days is
1 \(64 \mathrm{KWH}\)
2 \(90.8 \mathrm{KWH}\)
3 \(72 \mathrm{KWH}\)
4 \(42 \mathrm{KWH}\)
Explanation:
1 unit \(=\frac{\text { no. watts } \times \text { no. of hours }}{1000}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
268315
A \(200 \mathrm{~W}-200 \mathrm{~V}\) lamp is connected to \(250 \mathrm{~V}\) mains. It power consumption is
1 \(300 \mathrm{~W}\)
2 \(312.5 \mathrm{~W}\)
3 \(292 \mathrm{~W}\)
4 \(250 \mathrm{~W}\)
Explanation:
\(P=\frac{V^{2}}{R}\)
Current Electricity
268316
If the current in a heater increases by \(10 \%\), the eercentage change in the power consumption
1 \(19 \%\)
2 \(21 \%\)
3 \(25 \%\)
4 \(17 \%\)
Explanation:
\(P \propto i^{2}\)
Current Electricity
268317
The power of a heating coil is \(P\). It is cut into two equal parts. The power of one of them across same mains is
1 \(2 \mathrm{P}\)
2 \(3 P\)
3 \(P / 2\)
4 \(4 P\)
Explanation:
\(P=\frac{V^{2}}{R} ; R=\frac{\rho l}{A}\)
Current Electricity
268318
In a house there are four bulbs each of \(50 \mathrm{~W}\) and 5 fans each of \(60 \mathrm{~W}\). If they are used at the rate of 6 hours a day, the electrical energy consumed in a month of 30 days is
1 \(64 \mathrm{KWH}\)
2 \(90.8 \mathrm{KWH}\)
3 \(72 \mathrm{KWH}\)
4 \(42 \mathrm{KWH}\)
Explanation:
1 unit \(=\frac{\text { no. watts } \times \text { no. of hours }}{1000}\)