357111
Two bulbs \(X\) and \(Y\) having same voltage rating and of power 40 watt and 60 watt respectively are connected in series across a potential difference of 300 volts, then
1 \(X\) will glow brighter
2 Resistance of \(Y\) is greater than \(X\)
3 Heat produced in \(Y\) will be greater than \(X\)
4 Voltage drop in \(Y\) will be greater than \(X\)
Explanation:
For a given constant voltage Resistance \( \propto \frac{1}{{\operatorname{Power} }}.\) Thus, 40 \(W\) bulb has a high resistance because of which there will be more potential drop across 40 \(W\) bulb. Thus 40 \(W\) bulb will glow brighter.
PHXII03:CURRENT ELECTRICITY
357112
Each of three resistors in figure has a resistance of \({2.4\,\, \Omega}\) and can dissipate a maximum of 36 \(W\) without becoming excessively heated. What is the maximum power the circuit can dissipate?
357113
A heater coil is cut into two parts of equal length and only one of them is used in the heater. The ratio of the heat produced by this half-coil to that by the original coil is
357114
If percentage change in current through a resistor is \(1 \%\), then the change in power through it would be
1 \(1 \%\)
2 \(2 \%\)
3 \(1.7 \%\)
4 \(0.5 \%\)
Explanation:
Maximum percentage error arises due to limit of accuracy of the measured value. By Joule's law of heating, the power change due to current \((i)\), through resistor \((R)\) is given by \(P=i^{2} R\) Taking partial differentiation, we have \(\frac{{\Delta P}}{P} \times 100 = 2\frac{{\Delta i}}{i} \times 100 + \frac{{\Delta R}}{R} \times 100\) \(\therefore \quad \frac{{\Delta P}}{P} \times 100 = 2 \times 1\,\% = 2\,\% \) \(\quad \left( {{\text{ given, }}\frac{{\Delta i}}{i} = 1{\text{ and }}\frac{{\Delta R}}{R} = 0} \right)\)
PHXII03:CURRENT ELECTRICITY
357115
Assertion : When current through a bulb is increased by \(20 \%\) power increases by \(36 \%\). Reason : Illumination of lamp depends on square of current.
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:
Illumination \(\propto\) Power from lamp \(P=I^{2} R\) \(\dfrac{P_{1}}{P_{2}}=\left(\dfrac{I_{1}}{I_{2}}\right)^{2}\) \(I_{1}=I\)\(I_{2}=1.2 I\) \(\Rightarrow P_{2}=(1.2)^{2} P_{1}\) Percentage increase in Power \(=\dfrac{P_{2}-P_{1}}{P_{1}} \times 100\) \(=\left(\dfrac{1.44-1}{1}\right) \times 100=44 \%\) So, correct option is (4).
357111
Two bulbs \(X\) and \(Y\) having same voltage rating and of power 40 watt and 60 watt respectively are connected in series across a potential difference of 300 volts, then
1 \(X\) will glow brighter
2 Resistance of \(Y\) is greater than \(X\)
3 Heat produced in \(Y\) will be greater than \(X\)
4 Voltage drop in \(Y\) will be greater than \(X\)
Explanation:
For a given constant voltage Resistance \( \propto \frac{1}{{\operatorname{Power} }}.\) Thus, 40 \(W\) bulb has a high resistance because of which there will be more potential drop across 40 \(W\) bulb. Thus 40 \(W\) bulb will glow brighter.
PHXII03:CURRENT ELECTRICITY
357112
Each of three resistors in figure has a resistance of \({2.4\,\, \Omega}\) and can dissipate a maximum of 36 \(W\) without becoming excessively heated. What is the maximum power the circuit can dissipate?
357113
A heater coil is cut into two parts of equal length and only one of them is used in the heater. The ratio of the heat produced by this half-coil to that by the original coil is
357114
If percentage change in current through a resistor is \(1 \%\), then the change in power through it would be
1 \(1 \%\)
2 \(2 \%\)
3 \(1.7 \%\)
4 \(0.5 \%\)
Explanation:
Maximum percentage error arises due to limit of accuracy of the measured value. By Joule's law of heating, the power change due to current \((i)\), through resistor \((R)\) is given by \(P=i^{2} R\) Taking partial differentiation, we have \(\frac{{\Delta P}}{P} \times 100 = 2\frac{{\Delta i}}{i} \times 100 + \frac{{\Delta R}}{R} \times 100\) \(\therefore \quad \frac{{\Delta P}}{P} \times 100 = 2 \times 1\,\% = 2\,\% \) \(\quad \left( {{\text{ given, }}\frac{{\Delta i}}{i} = 1{\text{ and }}\frac{{\Delta R}}{R} = 0} \right)\)
PHXII03:CURRENT ELECTRICITY
357115
Assertion : When current through a bulb is increased by \(20 \%\) power increases by \(36 \%\). Reason : Illumination of lamp depends on square of current.
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:
Illumination \(\propto\) Power from lamp \(P=I^{2} R\) \(\dfrac{P_{1}}{P_{2}}=\left(\dfrac{I_{1}}{I_{2}}\right)^{2}\) \(I_{1}=I\)\(I_{2}=1.2 I\) \(\Rightarrow P_{2}=(1.2)^{2} P_{1}\) Percentage increase in Power \(=\dfrac{P_{2}-P_{1}}{P_{1}} \times 100\) \(=\left(\dfrac{1.44-1}{1}\right) \times 100=44 \%\) So, correct option is (4).
357111
Two bulbs \(X\) and \(Y\) having same voltage rating and of power 40 watt and 60 watt respectively are connected in series across a potential difference of 300 volts, then
1 \(X\) will glow brighter
2 Resistance of \(Y\) is greater than \(X\)
3 Heat produced in \(Y\) will be greater than \(X\)
4 Voltage drop in \(Y\) will be greater than \(X\)
Explanation:
For a given constant voltage Resistance \( \propto \frac{1}{{\operatorname{Power} }}.\) Thus, 40 \(W\) bulb has a high resistance because of which there will be more potential drop across 40 \(W\) bulb. Thus 40 \(W\) bulb will glow brighter.
PHXII03:CURRENT ELECTRICITY
357112
Each of three resistors in figure has a resistance of \({2.4\,\, \Omega}\) and can dissipate a maximum of 36 \(W\) without becoming excessively heated. What is the maximum power the circuit can dissipate?
357113
A heater coil is cut into two parts of equal length and only one of them is used in the heater. The ratio of the heat produced by this half-coil to that by the original coil is
357114
If percentage change in current through a resistor is \(1 \%\), then the change in power through it would be
1 \(1 \%\)
2 \(2 \%\)
3 \(1.7 \%\)
4 \(0.5 \%\)
Explanation:
Maximum percentage error arises due to limit of accuracy of the measured value. By Joule's law of heating, the power change due to current \((i)\), through resistor \((R)\) is given by \(P=i^{2} R\) Taking partial differentiation, we have \(\frac{{\Delta P}}{P} \times 100 = 2\frac{{\Delta i}}{i} \times 100 + \frac{{\Delta R}}{R} \times 100\) \(\therefore \quad \frac{{\Delta P}}{P} \times 100 = 2 \times 1\,\% = 2\,\% \) \(\quad \left( {{\text{ given, }}\frac{{\Delta i}}{i} = 1{\text{ and }}\frac{{\Delta R}}{R} = 0} \right)\)
PHXII03:CURRENT ELECTRICITY
357115
Assertion : When current through a bulb is increased by \(20 \%\) power increases by \(36 \%\). Reason : Illumination of lamp depends on square of current.
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:
Illumination \(\propto\) Power from lamp \(P=I^{2} R\) \(\dfrac{P_{1}}{P_{2}}=\left(\dfrac{I_{1}}{I_{2}}\right)^{2}\) \(I_{1}=I\)\(I_{2}=1.2 I\) \(\Rightarrow P_{2}=(1.2)^{2} P_{1}\) Percentage increase in Power \(=\dfrac{P_{2}-P_{1}}{P_{1}} \times 100\) \(=\left(\dfrac{1.44-1}{1}\right) \times 100=44 \%\) So, correct option is (4).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
357111
Two bulbs \(X\) and \(Y\) having same voltage rating and of power 40 watt and 60 watt respectively are connected in series across a potential difference of 300 volts, then
1 \(X\) will glow brighter
2 Resistance of \(Y\) is greater than \(X\)
3 Heat produced in \(Y\) will be greater than \(X\)
4 Voltage drop in \(Y\) will be greater than \(X\)
Explanation:
For a given constant voltage Resistance \( \propto \frac{1}{{\operatorname{Power} }}.\) Thus, 40 \(W\) bulb has a high resistance because of which there will be more potential drop across 40 \(W\) bulb. Thus 40 \(W\) bulb will glow brighter.
PHXII03:CURRENT ELECTRICITY
357112
Each of three resistors in figure has a resistance of \({2.4\,\, \Omega}\) and can dissipate a maximum of 36 \(W\) without becoming excessively heated. What is the maximum power the circuit can dissipate?
357113
A heater coil is cut into two parts of equal length and only one of them is used in the heater. The ratio of the heat produced by this half-coil to that by the original coil is
357114
If percentage change in current through a resistor is \(1 \%\), then the change in power through it would be
1 \(1 \%\)
2 \(2 \%\)
3 \(1.7 \%\)
4 \(0.5 \%\)
Explanation:
Maximum percentage error arises due to limit of accuracy of the measured value. By Joule's law of heating, the power change due to current \((i)\), through resistor \((R)\) is given by \(P=i^{2} R\) Taking partial differentiation, we have \(\frac{{\Delta P}}{P} \times 100 = 2\frac{{\Delta i}}{i} \times 100 + \frac{{\Delta R}}{R} \times 100\) \(\therefore \quad \frac{{\Delta P}}{P} \times 100 = 2 \times 1\,\% = 2\,\% \) \(\quad \left( {{\text{ given, }}\frac{{\Delta i}}{i} = 1{\text{ and }}\frac{{\Delta R}}{R} = 0} \right)\)
PHXII03:CURRENT ELECTRICITY
357115
Assertion : When current through a bulb is increased by \(20 \%\) power increases by \(36 \%\). Reason : Illumination of lamp depends on square of current.
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:
Illumination \(\propto\) Power from lamp \(P=I^{2} R\) \(\dfrac{P_{1}}{P_{2}}=\left(\dfrac{I_{1}}{I_{2}}\right)^{2}\) \(I_{1}=I\)\(I_{2}=1.2 I\) \(\Rightarrow P_{2}=(1.2)^{2} P_{1}\) Percentage increase in Power \(=\dfrac{P_{2}-P_{1}}{P_{1}} \times 100\) \(=\left(\dfrac{1.44-1}{1}\right) \times 100=44 \%\) So, correct option is (4).
357111
Two bulbs \(X\) and \(Y\) having same voltage rating and of power 40 watt and 60 watt respectively are connected in series across a potential difference of 300 volts, then
1 \(X\) will glow brighter
2 Resistance of \(Y\) is greater than \(X\)
3 Heat produced in \(Y\) will be greater than \(X\)
4 Voltage drop in \(Y\) will be greater than \(X\)
Explanation:
For a given constant voltage Resistance \( \propto \frac{1}{{\operatorname{Power} }}.\) Thus, 40 \(W\) bulb has a high resistance because of which there will be more potential drop across 40 \(W\) bulb. Thus 40 \(W\) bulb will glow brighter.
PHXII03:CURRENT ELECTRICITY
357112
Each of three resistors in figure has a resistance of \({2.4\,\, \Omega}\) and can dissipate a maximum of 36 \(W\) without becoming excessively heated. What is the maximum power the circuit can dissipate?
357113
A heater coil is cut into two parts of equal length and only one of them is used in the heater. The ratio of the heat produced by this half-coil to that by the original coil is
357114
If percentage change in current through a resistor is \(1 \%\), then the change in power through it would be
1 \(1 \%\)
2 \(2 \%\)
3 \(1.7 \%\)
4 \(0.5 \%\)
Explanation:
Maximum percentage error arises due to limit of accuracy of the measured value. By Joule's law of heating, the power change due to current \((i)\), through resistor \((R)\) is given by \(P=i^{2} R\) Taking partial differentiation, we have \(\frac{{\Delta P}}{P} \times 100 = 2\frac{{\Delta i}}{i} \times 100 + \frac{{\Delta R}}{R} \times 100\) \(\therefore \quad \frac{{\Delta P}}{P} \times 100 = 2 \times 1\,\% = 2\,\% \) \(\quad \left( {{\text{ given, }}\frac{{\Delta i}}{i} = 1{\text{ and }}\frac{{\Delta R}}{R} = 0} \right)\)
PHXII03:CURRENT ELECTRICITY
357115
Assertion : When current through a bulb is increased by \(20 \%\) power increases by \(36 \%\). Reason : Illumination of lamp depends on square of current.
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:
Illumination \(\propto\) Power from lamp \(P=I^{2} R\) \(\dfrac{P_{1}}{P_{2}}=\left(\dfrac{I_{1}}{I_{2}}\right)^{2}\) \(I_{1}=I\)\(I_{2}=1.2 I\) \(\Rightarrow P_{2}=(1.2)^{2} P_{1}\) Percentage increase in Power \(=\dfrac{P_{2}-P_{1}}{P_{1}} \times 100\) \(=\left(\dfrac{1.44-1}{1}\right) \times 100=44 \%\) So, correct option is (4).
