NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
357159
Assertion : Heater wire must have high resistance. Reason : Heater wires should have high melting point.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Heater wires are designed with high resistance because, in a series circuit, the current remains the same, and according to Joule's law, heat produced is directly proportional to the resistance \(H=I^{2} R t\). Therefore, a high resistance results in higher heat production. A high melting point is necessary ensure that the wire does not melt or become damaged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357160
In the circuit diagram, all the bulbs are identical. Which bulb will be the brightest?
1 \(A\)
2 \(B\)
3 \(C\)
4 \(D\)
Explanation:
The resistor with the most current will be the brightest. As bulb \(C\) has the same current as that through the battery, the current through bulb \(C\) is the greatest.
PHXII03:CURRENT ELECTRICITY
357161
The number of electrons flowing per second in the filament of a \(110\,W\) bulb operating at \(220\,V\) is (Given \(e = 1.6 \times {10^{ - 19}}C\,)\)
1 \(1.25 \times 10^{19}\)
2 \(6.25 \times 10^{18}\)
3 \(6.25 \times 10^{17}\)
4 \(31.25 \times 10^{17}\)
Explanation:
\(P = 110\;W,\;V = 220\;V\) \(P = VI \Rightarrow I = \frac{P}{V} = \frac{{110}}{{220}} = 0.5\;A\) \(I=\dfrac{n e}{t} \Rightarrow \dfrac{n}{t}=\dfrac{\text { No. of electrons }}{\text { second }}=\dfrac{I}{e}\) \(=\dfrac{0.5}{1.6 \times 10^{-19}}=31.25 \times 10^{17}\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357162
A piece of fuse wire melts when a current of 15 ampere flows through it. With this current, if it dissipates 22.5 \(W\) , the resistance of fuse wire will be
1 zero
2 \(10\,\Omega \)
3 \(1\,\Omega \)
4 \(0.1\,\Omega \)
Explanation:
Current flow \({I=15 {~A}}\) Power dissipated \({=22.5 {~W}}\) \({P=l^{2} R}\) \({R=\dfrac{P}{I^{2}}=\dfrac{22.5}{225}=0.1 \Omega}\). So correct option is (4)
357159
Assertion : Heater wire must have high resistance. Reason : Heater wires should have high melting point.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Heater wires are designed with high resistance because, in a series circuit, the current remains the same, and according to Joule's law, heat produced is directly proportional to the resistance \(H=I^{2} R t\). Therefore, a high resistance results in higher heat production. A high melting point is necessary ensure that the wire does not melt or become damaged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357160
In the circuit diagram, all the bulbs are identical. Which bulb will be the brightest?
1 \(A\)
2 \(B\)
3 \(C\)
4 \(D\)
Explanation:
The resistor with the most current will be the brightest. As bulb \(C\) has the same current as that through the battery, the current through bulb \(C\) is the greatest.
PHXII03:CURRENT ELECTRICITY
357161
The number of electrons flowing per second in the filament of a \(110\,W\) bulb operating at \(220\,V\) is (Given \(e = 1.6 \times {10^{ - 19}}C\,)\)
1 \(1.25 \times 10^{19}\)
2 \(6.25 \times 10^{18}\)
3 \(6.25 \times 10^{17}\)
4 \(31.25 \times 10^{17}\)
Explanation:
\(P = 110\;W,\;V = 220\;V\) \(P = VI \Rightarrow I = \frac{P}{V} = \frac{{110}}{{220}} = 0.5\;A\) \(I=\dfrac{n e}{t} \Rightarrow \dfrac{n}{t}=\dfrac{\text { No. of electrons }}{\text { second }}=\dfrac{I}{e}\) \(=\dfrac{0.5}{1.6 \times 10^{-19}}=31.25 \times 10^{17}\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357162
A piece of fuse wire melts when a current of 15 ampere flows through it. With this current, if it dissipates 22.5 \(W\) , the resistance of fuse wire will be
1 zero
2 \(10\,\Omega \)
3 \(1\,\Omega \)
4 \(0.1\,\Omega \)
Explanation:
Current flow \({I=15 {~A}}\) Power dissipated \({=22.5 {~W}}\) \({P=l^{2} R}\) \({R=\dfrac{P}{I^{2}}=\dfrac{22.5}{225}=0.1 \Omega}\). So correct option is (4)
357159
Assertion : Heater wire must have high resistance. Reason : Heater wires should have high melting point.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Heater wires are designed with high resistance because, in a series circuit, the current remains the same, and according to Joule's law, heat produced is directly proportional to the resistance \(H=I^{2} R t\). Therefore, a high resistance results in higher heat production. A high melting point is necessary ensure that the wire does not melt or become damaged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357160
In the circuit diagram, all the bulbs are identical. Which bulb will be the brightest?
1 \(A\)
2 \(B\)
3 \(C\)
4 \(D\)
Explanation:
The resistor with the most current will be the brightest. As bulb \(C\) has the same current as that through the battery, the current through bulb \(C\) is the greatest.
PHXII03:CURRENT ELECTRICITY
357161
The number of electrons flowing per second in the filament of a \(110\,W\) bulb operating at \(220\,V\) is (Given \(e = 1.6 \times {10^{ - 19}}C\,)\)
1 \(1.25 \times 10^{19}\)
2 \(6.25 \times 10^{18}\)
3 \(6.25 \times 10^{17}\)
4 \(31.25 \times 10^{17}\)
Explanation:
\(P = 110\;W,\;V = 220\;V\) \(P = VI \Rightarrow I = \frac{P}{V} = \frac{{110}}{{220}} = 0.5\;A\) \(I=\dfrac{n e}{t} \Rightarrow \dfrac{n}{t}=\dfrac{\text { No. of electrons }}{\text { second }}=\dfrac{I}{e}\) \(=\dfrac{0.5}{1.6 \times 10^{-19}}=31.25 \times 10^{17}\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357162
A piece of fuse wire melts when a current of 15 ampere flows through it. With this current, if it dissipates 22.5 \(W\) , the resistance of fuse wire will be
1 zero
2 \(10\,\Omega \)
3 \(1\,\Omega \)
4 \(0.1\,\Omega \)
Explanation:
Current flow \({I=15 {~A}}\) Power dissipated \({=22.5 {~W}}\) \({P=l^{2} R}\) \({R=\dfrac{P}{I^{2}}=\dfrac{22.5}{225}=0.1 \Omega}\). So correct option is (4)
357159
Assertion : Heater wire must have high resistance. Reason : Heater wires should have high melting point.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Heater wires are designed with high resistance because, in a series circuit, the current remains the same, and according to Joule's law, heat produced is directly proportional to the resistance \(H=I^{2} R t\). Therefore, a high resistance results in higher heat production. A high melting point is necessary ensure that the wire does not melt or become damaged. So correct option is (2).
PHXII03:CURRENT ELECTRICITY
357160
In the circuit diagram, all the bulbs are identical. Which bulb will be the brightest?
1 \(A\)
2 \(B\)
3 \(C\)
4 \(D\)
Explanation:
The resistor with the most current will be the brightest. As bulb \(C\) has the same current as that through the battery, the current through bulb \(C\) is the greatest.
PHXII03:CURRENT ELECTRICITY
357161
The number of electrons flowing per second in the filament of a \(110\,W\) bulb operating at \(220\,V\) is (Given \(e = 1.6 \times {10^{ - 19}}C\,)\)
1 \(1.25 \times 10^{19}\)
2 \(6.25 \times 10^{18}\)
3 \(6.25 \times 10^{17}\)
4 \(31.25 \times 10^{17}\)
Explanation:
\(P = 110\;W,\;V = 220\;V\) \(P = VI \Rightarrow I = \frac{P}{V} = \frac{{110}}{{220}} = 0.5\;A\) \(I=\dfrac{n e}{t} \Rightarrow \dfrac{n}{t}=\dfrac{\text { No. of electrons }}{\text { second }}=\dfrac{I}{e}\) \(=\dfrac{0.5}{1.6 \times 10^{-19}}=31.25 \times 10^{17}\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357162
A piece of fuse wire melts when a current of 15 ampere flows through it. With this current, if it dissipates 22.5 \(W\) , the resistance of fuse wire will be
1 zero
2 \(10\,\Omega \)
3 \(1\,\Omega \)
4 \(0.1\,\Omega \)
Explanation:
Current flow \({I=15 {~A}}\) Power dissipated \({=22.5 {~W}}\) \({P=l^{2} R}\) \({R=\dfrac{P}{I^{2}}=\dfrac{22.5}{225}=0.1 \Omega}\). So correct option is (4)
