152747
Assertion: Long distance power transmission is done at high voltage. Reason: At high voltage supply power losses are less.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A As to transmit power over long distance the transmission is done high voltages supply power losses are less. $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R}=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{R}$ Power loss is inversely proportional to the square of the voltage.
AIIMS-2014
Current Electricity
152748
Assertion: When current through a bulb decreases by $0.5 \%$, the glow of bulb decreases by $1 \%$. Reason: Glow (Power) which is directly proportional to square of current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{dP}}{\mathrm{P}} =2 \frac{\mathrm{dI}}{\mathrm{I}}$ $=2 \times 0.5 \%=1 \%$ Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 \%$ when current through the bulb decreases by $0.5 \%$.
AIIMS-2015
Current Electricity
152750
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if
1 both the length and radius of wire are halved
2 both length and radius of wire are doubled
3 the radius of wire is doubled
4 the length of the wire is doubled.
Explanation:
B The heat produced is given by, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ We know that, $\mathrm{R}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ Where, Length of wire $=l$ $\text { Radius of wire }=r$ $\rho=\text { resistivity }$ On putting the value $\mathrm{R}$ in equation (i), We get - $\text { Then, } \mathrm{H} =\frac{\mathrm{V}^{2}}{\frac{\rho l}{\pi \mathrm{r}^{2}}} \mathrm{t}$ $\mathrm{H} =\mathrm{V}^{2}\left(\frac{\pi \mathrm{r}^{2}}{\rho l}\right) \mathrm{t}$ When, $l$ and $\mathrm{r}$ are doubled. Then, $\quad \mathrm{H}^{\prime}=\frac{\mathrm{V}^{2} \pi}{\rho}\left(\frac{(2 \mathrm{r})^{2}}{2 l}\right) \mathrm{t}$ $\mathrm{H}^{\prime}=\frac{\mathrm{V}^{2}}{\rho}\left(\frac{\pi \mathrm{r}^{2}}{l}\right) \mathrm{t} \times 2$ $\mathrm{H}^{\prime}=2 \mathrm{H}$ So, heat produced will be doubled when both the length and radius of the wire are doubled.
AIIMS-2012
Current Electricity
152749
Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on. Reason: Dimness decreases after sometime.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B After some time, when heater coil becomes hot its resistance increases. As a result, current through the heater coil decreases and the current through the bulb filament increases and thus dimness of the bulb decreases. Therefore, if both Assertion and Reason are correct but reason in not a correct explanation of the assertion.
AIIMS-2008
Current Electricity
152751
In the circuit shown in figure, the $5 \Omega$ resistance develops $20.00 \mathrm{cal} / \mathrm{s}$ due to the current flowing through it. The heat developed in $2 \Omega$ resistance (in $\mathrm{cal} / \mathrm{s}$ ) is
1 23.8
2 14.2
3 11.9
4 7.1
Explanation:
B Given, $\text { Heat, } \mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $\mathrm{R}=5 \Omega$, and $\frac{\mathrm{H}}{\mathrm{t}}=20.00 \mathrm{cal} / \mathrm{s}$ $\therefore \quad 20=\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1}=2 \mathrm{~A}$ Voltage across $5 \Omega=$ Voltage across $(6+9) \Omega$ Now, $V=I_{1} \times 5=I_{2} \times(6+9)$ $5 \mathrm{I}_{1}=15 \mathrm{I}_{2}$ $\therefore \quad \mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{15}$ $I_{2}=\frac{5 \times 2}{15}$ $I_{2}=\frac{2}{3} A$ Hence, heat in $2 \Omega$ resistor per second, $\frac{\mathrm{H}}{\mathrm{t}} =\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(2+\frac{2}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(\frac{8}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =14.2 \mathrm{cal} / \mathrm{s}$
152747
Assertion: Long distance power transmission is done at high voltage. Reason: At high voltage supply power losses are less.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A As to transmit power over long distance the transmission is done high voltages supply power losses are less. $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R}=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{R}$ Power loss is inversely proportional to the square of the voltage.
AIIMS-2014
Current Electricity
152748
Assertion: When current through a bulb decreases by $0.5 \%$, the glow of bulb decreases by $1 \%$. Reason: Glow (Power) which is directly proportional to square of current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{dP}}{\mathrm{P}} =2 \frac{\mathrm{dI}}{\mathrm{I}}$ $=2 \times 0.5 \%=1 \%$ Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 \%$ when current through the bulb decreases by $0.5 \%$.
AIIMS-2015
Current Electricity
152750
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if
1 both the length and radius of wire are halved
2 both length and radius of wire are doubled
3 the radius of wire is doubled
4 the length of the wire is doubled.
Explanation:
B The heat produced is given by, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ We know that, $\mathrm{R}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ Where, Length of wire $=l$ $\text { Radius of wire }=r$ $\rho=\text { resistivity }$ On putting the value $\mathrm{R}$ in equation (i), We get - $\text { Then, } \mathrm{H} =\frac{\mathrm{V}^{2}}{\frac{\rho l}{\pi \mathrm{r}^{2}}} \mathrm{t}$ $\mathrm{H} =\mathrm{V}^{2}\left(\frac{\pi \mathrm{r}^{2}}{\rho l}\right) \mathrm{t}$ When, $l$ and $\mathrm{r}$ are doubled. Then, $\quad \mathrm{H}^{\prime}=\frac{\mathrm{V}^{2} \pi}{\rho}\left(\frac{(2 \mathrm{r})^{2}}{2 l}\right) \mathrm{t}$ $\mathrm{H}^{\prime}=\frac{\mathrm{V}^{2}}{\rho}\left(\frac{\pi \mathrm{r}^{2}}{l}\right) \mathrm{t} \times 2$ $\mathrm{H}^{\prime}=2 \mathrm{H}$ So, heat produced will be doubled when both the length and radius of the wire are doubled.
AIIMS-2012
Current Electricity
152749
Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on. Reason: Dimness decreases after sometime.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B After some time, when heater coil becomes hot its resistance increases. As a result, current through the heater coil decreases and the current through the bulb filament increases and thus dimness of the bulb decreases. Therefore, if both Assertion and Reason are correct but reason in not a correct explanation of the assertion.
AIIMS-2008
Current Electricity
152751
In the circuit shown in figure, the $5 \Omega$ resistance develops $20.00 \mathrm{cal} / \mathrm{s}$ due to the current flowing through it. The heat developed in $2 \Omega$ resistance (in $\mathrm{cal} / \mathrm{s}$ ) is
1 23.8
2 14.2
3 11.9
4 7.1
Explanation:
B Given, $\text { Heat, } \mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $\mathrm{R}=5 \Omega$, and $\frac{\mathrm{H}}{\mathrm{t}}=20.00 \mathrm{cal} / \mathrm{s}$ $\therefore \quad 20=\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1}=2 \mathrm{~A}$ Voltage across $5 \Omega=$ Voltage across $(6+9) \Omega$ Now, $V=I_{1} \times 5=I_{2} \times(6+9)$ $5 \mathrm{I}_{1}=15 \mathrm{I}_{2}$ $\therefore \quad \mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{15}$ $I_{2}=\frac{5 \times 2}{15}$ $I_{2}=\frac{2}{3} A$ Hence, heat in $2 \Omega$ resistor per second, $\frac{\mathrm{H}}{\mathrm{t}} =\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(2+\frac{2}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(\frac{8}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =14.2 \mathrm{cal} / \mathrm{s}$
152747
Assertion: Long distance power transmission is done at high voltage. Reason: At high voltage supply power losses are less.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A As to transmit power over long distance the transmission is done high voltages supply power losses are less. $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R}=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{R}$ Power loss is inversely proportional to the square of the voltage.
AIIMS-2014
Current Electricity
152748
Assertion: When current through a bulb decreases by $0.5 \%$, the glow of bulb decreases by $1 \%$. Reason: Glow (Power) which is directly proportional to square of current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{dP}}{\mathrm{P}} =2 \frac{\mathrm{dI}}{\mathrm{I}}$ $=2 \times 0.5 \%=1 \%$ Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 \%$ when current through the bulb decreases by $0.5 \%$.
AIIMS-2015
Current Electricity
152750
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if
1 both the length and radius of wire are halved
2 both length and radius of wire are doubled
3 the radius of wire is doubled
4 the length of the wire is doubled.
Explanation:
B The heat produced is given by, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ We know that, $\mathrm{R}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ Where, Length of wire $=l$ $\text { Radius of wire }=r$ $\rho=\text { resistivity }$ On putting the value $\mathrm{R}$ in equation (i), We get - $\text { Then, } \mathrm{H} =\frac{\mathrm{V}^{2}}{\frac{\rho l}{\pi \mathrm{r}^{2}}} \mathrm{t}$ $\mathrm{H} =\mathrm{V}^{2}\left(\frac{\pi \mathrm{r}^{2}}{\rho l}\right) \mathrm{t}$ When, $l$ and $\mathrm{r}$ are doubled. Then, $\quad \mathrm{H}^{\prime}=\frac{\mathrm{V}^{2} \pi}{\rho}\left(\frac{(2 \mathrm{r})^{2}}{2 l}\right) \mathrm{t}$ $\mathrm{H}^{\prime}=\frac{\mathrm{V}^{2}}{\rho}\left(\frac{\pi \mathrm{r}^{2}}{l}\right) \mathrm{t} \times 2$ $\mathrm{H}^{\prime}=2 \mathrm{H}$ So, heat produced will be doubled when both the length and radius of the wire are doubled.
AIIMS-2012
Current Electricity
152749
Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on. Reason: Dimness decreases after sometime.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B After some time, when heater coil becomes hot its resistance increases. As a result, current through the heater coil decreases and the current through the bulb filament increases and thus dimness of the bulb decreases. Therefore, if both Assertion and Reason are correct but reason in not a correct explanation of the assertion.
AIIMS-2008
Current Electricity
152751
In the circuit shown in figure, the $5 \Omega$ resistance develops $20.00 \mathrm{cal} / \mathrm{s}$ due to the current flowing through it. The heat developed in $2 \Omega$ resistance (in $\mathrm{cal} / \mathrm{s}$ ) is
1 23.8
2 14.2
3 11.9
4 7.1
Explanation:
B Given, $\text { Heat, } \mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $\mathrm{R}=5 \Omega$, and $\frac{\mathrm{H}}{\mathrm{t}}=20.00 \mathrm{cal} / \mathrm{s}$ $\therefore \quad 20=\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1}=2 \mathrm{~A}$ Voltage across $5 \Omega=$ Voltage across $(6+9) \Omega$ Now, $V=I_{1} \times 5=I_{2} \times(6+9)$ $5 \mathrm{I}_{1}=15 \mathrm{I}_{2}$ $\therefore \quad \mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{15}$ $I_{2}=\frac{5 \times 2}{15}$ $I_{2}=\frac{2}{3} A$ Hence, heat in $2 \Omega$ resistor per second, $\frac{\mathrm{H}}{\mathrm{t}} =\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(2+\frac{2}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(\frac{8}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =14.2 \mathrm{cal} / \mathrm{s}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152747
Assertion: Long distance power transmission is done at high voltage. Reason: At high voltage supply power losses are less.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A As to transmit power over long distance the transmission is done high voltages supply power losses are less. $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R}=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{R}$ Power loss is inversely proportional to the square of the voltage.
AIIMS-2014
Current Electricity
152748
Assertion: When current through a bulb decreases by $0.5 \%$, the glow of bulb decreases by $1 \%$. Reason: Glow (Power) which is directly proportional to square of current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{dP}}{\mathrm{P}} =2 \frac{\mathrm{dI}}{\mathrm{I}}$ $=2 \times 0.5 \%=1 \%$ Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 \%$ when current through the bulb decreases by $0.5 \%$.
AIIMS-2015
Current Electricity
152750
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if
1 both the length and radius of wire are halved
2 both length and radius of wire are doubled
3 the radius of wire is doubled
4 the length of the wire is doubled.
Explanation:
B The heat produced is given by, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ We know that, $\mathrm{R}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ Where, Length of wire $=l$ $\text { Radius of wire }=r$ $\rho=\text { resistivity }$ On putting the value $\mathrm{R}$ in equation (i), We get - $\text { Then, } \mathrm{H} =\frac{\mathrm{V}^{2}}{\frac{\rho l}{\pi \mathrm{r}^{2}}} \mathrm{t}$ $\mathrm{H} =\mathrm{V}^{2}\left(\frac{\pi \mathrm{r}^{2}}{\rho l}\right) \mathrm{t}$ When, $l$ and $\mathrm{r}$ are doubled. Then, $\quad \mathrm{H}^{\prime}=\frac{\mathrm{V}^{2} \pi}{\rho}\left(\frac{(2 \mathrm{r})^{2}}{2 l}\right) \mathrm{t}$ $\mathrm{H}^{\prime}=\frac{\mathrm{V}^{2}}{\rho}\left(\frac{\pi \mathrm{r}^{2}}{l}\right) \mathrm{t} \times 2$ $\mathrm{H}^{\prime}=2 \mathrm{H}$ So, heat produced will be doubled when both the length and radius of the wire are doubled.
AIIMS-2012
Current Electricity
152749
Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on. Reason: Dimness decreases after sometime.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B After some time, when heater coil becomes hot its resistance increases. As a result, current through the heater coil decreases and the current through the bulb filament increases and thus dimness of the bulb decreases. Therefore, if both Assertion and Reason are correct but reason in not a correct explanation of the assertion.
AIIMS-2008
Current Electricity
152751
In the circuit shown in figure, the $5 \Omega$ resistance develops $20.00 \mathrm{cal} / \mathrm{s}$ due to the current flowing through it. The heat developed in $2 \Omega$ resistance (in $\mathrm{cal} / \mathrm{s}$ ) is
1 23.8
2 14.2
3 11.9
4 7.1
Explanation:
B Given, $\text { Heat, } \mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $\mathrm{R}=5 \Omega$, and $\frac{\mathrm{H}}{\mathrm{t}}=20.00 \mathrm{cal} / \mathrm{s}$ $\therefore \quad 20=\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1}=2 \mathrm{~A}$ Voltage across $5 \Omega=$ Voltage across $(6+9) \Omega$ Now, $V=I_{1} \times 5=I_{2} \times(6+9)$ $5 \mathrm{I}_{1}=15 \mathrm{I}_{2}$ $\therefore \quad \mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{15}$ $I_{2}=\frac{5 \times 2}{15}$ $I_{2}=\frac{2}{3} A$ Hence, heat in $2 \Omega$ resistor per second, $\frac{\mathrm{H}}{\mathrm{t}} =\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(2+\frac{2}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(\frac{8}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =14.2 \mathrm{cal} / \mathrm{s}$
152747
Assertion: Long distance power transmission is done at high voltage. Reason: At high voltage supply power losses are less.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A As to transmit power over long distance the transmission is done high voltages supply power losses are less. $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R}=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{R}$ Power loss is inversely proportional to the square of the voltage.
AIIMS-2014
Current Electricity
152748
Assertion: When current through a bulb decreases by $0.5 \%$, the glow of bulb decreases by $1 \%$. Reason: Glow (Power) which is directly proportional to square of current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
A We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{dP}}{\mathrm{P}} =2 \frac{\mathrm{dI}}{\mathrm{I}}$ $=2 \times 0.5 \%=1 \%$ Glow (Power) is directly proportional to square of current. That is why glow of bulb decreases by $1 \%$ when current through the bulb decreases by $0.5 \%$.
AIIMS-2015
Current Electricity
152750
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is double if
1 both the length and radius of wire are halved
2 both length and radius of wire are doubled
3 the radius of wire is doubled
4 the length of the wire is doubled.
Explanation:
B The heat produced is given by, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}$ We know that, $\mathrm{R}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ Where, Length of wire $=l$ $\text { Radius of wire }=r$ $\rho=\text { resistivity }$ On putting the value $\mathrm{R}$ in equation (i), We get - $\text { Then, } \mathrm{H} =\frac{\mathrm{V}^{2}}{\frac{\rho l}{\pi \mathrm{r}^{2}}} \mathrm{t}$ $\mathrm{H} =\mathrm{V}^{2}\left(\frac{\pi \mathrm{r}^{2}}{\rho l}\right) \mathrm{t}$ When, $l$ and $\mathrm{r}$ are doubled. Then, $\quad \mathrm{H}^{\prime}=\frac{\mathrm{V}^{2} \pi}{\rho}\left(\frac{(2 \mathrm{r})^{2}}{2 l}\right) \mathrm{t}$ $\mathrm{H}^{\prime}=\frac{\mathrm{V}^{2}}{\rho}\left(\frac{\pi \mathrm{r}^{2}}{l}\right) \mathrm{t} \times 2$ $\mathrm{H}^{\prime}=2 \mathrm{H}$ So, heat produced will be doubled when both the length and radius of the wire are doubled.
AIIMS-2012
Current Electricity
152749
Assertion: An electric bulb becomes dim, when the electric heater in parallel circuit is switched on. Reason: Dimness decreases after sometime.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B After some time, when heater coil becomes hot its resistance increases. As a result, current through the heater coil decreases and the current through the bulb filament increases and thus dimness of the bulb decreases. Therefore, if both Assertion and Reason are correct but reason in not a correct explanation of the assertion.
AIIMS-2008
Current Electricity
152751
In the circuit shown in figure, the $5 \Omega$ resistance develops $20.00 \mathrm{cal} / \mathrm{s}$ due to the current flowing through it. The heat developed in $2 \Omega$ resistance (in $\mathrm{cal} / \mathrm{s}$ ) is
1 23.8
2 14.2
3 11.9
4 7.1
Explanation:
B Given, $\text { Heat, } \mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $\mathrm{R}=5 \Omega$, and $\frac{\mathrm{H}}{\mathrm{t}}=20.00 \mathrm{cal} / \mathrm{s}$ $\therefore \quad 20=\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1}=2 \mathrm{~A}$ Voltage across $5 \Omega=$ Voltage across $(6+9) \Omega$ Now, $V=I_{1} \times 5=I_{2} \times(6+9)$ $5 \mathrm{I}_{1}=15 \mathrm{I}_{2}$ $\therefore \quad \mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{15}$ $I_{2}=\frac{5 \times 2}{15}$ $I_{2}=\frac{2}{3} A$ Hence, heat in $2 \Omega$ resistor per second, $\frac{\mathrm{H}}{\mathrm{t}} =\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(2+\frac{2}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =\left(\frac{8}{3}\right)^{2} \times 2$ $\frac{\mathrm{H}}{\mathrm{t}} =14.2 \mathrm{cal} / \mathrm{s}$
