NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152658
A $500 \mathrm{~W}$ heater is designed to operate at $200 \mathrm{~V}$ potential difference. If it is connected across 160 $\mathrm{V}$ line, the heat it will produce in $\mathbf{2 0}$ minutes is
1 $384 \mathrm{~kJ}$
2 $483 \mathrm{~kJ}$
3 $843 \mathrm{~kJ}$
4 $348 \mathrm{~kJ}$
Explanation:
A Given, Power $\mathrm{P}_{1}=500 \mathrm{~W}$ Voltage $\mathrm{V}_{1}=200 \mathrm{~V}$ Voltage $\mathrm{V}_{2}=160 \mathrm{~V}$ Time $(t)=20 \mathrm{~min}=20 \times 60=1200 \mathrm{sec}$ We know that, $\text { Power }(\mathrm{P})=\mathrm{VI}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, Case I, $\mathrm{P}_{1} =\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{1}}$ $\therefore \quad \mathrm{R}_{1} =\frac{200^{2}}{500}$ $\mathrm{R}_{1} =80 \Omega$ Case II, $\mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}=\frac{160^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{160 \times 160}{80}$ $\mathrm{P}_{2}=320 \mathrm{~W}$ $\therefore$ Now, heat produced in 20 minutes $\mathrm{dQ}=\mathrm{P}_{2} \times \text { time }=320 \times 20 \times 60$ $\mathrm{Q}=384 \mathrm{~kJ}$
AP EAMCET (22.04.2019) Shift-II
Current Electricity
152659
It takes 12 minutes to boil 1 litre of water in an electric kettle. Due to some defect it becomes necessary to remove $20 \%$ turns of heating coil of the kettle. After repair, how much time will it take to boil 1 litre of water?
1 9.6 minute
2 14.4 minute
3 16.8 minute
4 18.2 minute
Explanation:
A Given, If takes 12 minutes to boil 1 litre of water in an electric kettle. Let, Resistance $=\mathrm{R}$ $\mathrm{n} \text { turns } =\mathrm{R}$ $\therefore \quad 1 \text { turns } =\left(\frac{\mathrm{R}}{\mathrm{n}}\right)$ $\text { So, } 20 \% \text { turns } =\frac{20}{100} \times \mathrm{n} \times \frac{\mathrm{R}}{\mathrm{n}}$ $=\frac{20}{100} \mathrm{R}$ Net remaining resistance $=R-\frac{20}{100} R$ $=\frac{80}{100} \mathrm{R}=0.80 \mathrm{R}$ Heat is supplied, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}=\frac{(\mathrm{IR})^{2}}{\mathrm{R}} \mathrm{t}=\mathrm{I}^{2} \mathrm{Rt}$ Similarly, $\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \times \mathrm{t}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \times \mathrm{t}_{2}$ $\frac{\mathrm{t}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{t}_{2}}{\mathrm{R}_{2}}$ Then, $\frac{12}{\mathrm{R}}=\frac{\mathrm{t}_{2}}{0.8 \mathrm{R}}$ $\mathrm{t}_{2}=12 \times 0.8$ $\mathrm{t}_{2}=9.6 \text { minutes }$
AIIMS-26.05.2019(M) Shift-1
Current Electricity
152660
The heat produced per unit time, on passing electric current through a conductor at a given temperature, is directly proportional to the
1 Electric current
2 Reciprocal of electric current
3 Square of electric current
4 Reciprocal of square of electric current
Explanation:
C According to Joule's law of heating effect, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ Where, $\mathrm{I}=$ current $\mathrm{R}=$ resistance of conductor $\mathrm{t}=$ time $\mathrm{H}=$ heat produced $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{H}}{\mathrm{t}} \propto \mathrm{I}^{2}$ Hence, heat produced per unit time is directly proportional to the square of electric current.
GUJCET 2019
Current Electricity
152661
Two electric bulbs whose resistances are in the ratio $1: 2$ are connected in parallel to a constant voltage source. The power dissipated in them have the ratio
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
Explanation:
C Given, $\mathrm{R}_{1}: \mathrm{R}_{2}=1: 2$ Let the resistance of bulb are $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ respectively. In parallel connection, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{2}{1}$ The power dissipated in them have the ratio $=2: 1$
152658
A $500 \mathrm{~W}$ heater is designed to operate at $200 \mathrm{~V}$ potential difference. If it is connected across 160 $\mathrm{V}$ line, the heat it will produce in $\mathbf{2 0}$ minutes is
1 $384 \mathrm{~kJ}$
2 $483 \mathrm{~kJ}$
3 $843 \mathrm{~kJ}$
4 $348 \mathrm{~kJ}$
Explanation:
A Given, Power $\mathrm{P}_{1}=500 \mathrm{~W}$ Voltage $\mathrm{V}_{1}=200 \mathrm{~V}$ Voltage $\mathrm{V}_{2}=160 \mathrm{~V}$ Time $(t)=20 \mathrm{~min}=20 \times 60=1200 \mathrm{sec}$ We know that, $\text { Power }(\mathrm{P})=\mathrm{VI}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, Case I, $\mathrm{P}_{1} =\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{1}}$ $\therefore \quad \mathrm{R}_{1} =\frac{200^{2}}{500}$ $\mathrm{R}_{1} =80 \Omega$ Case II, $\mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}=\frac{160^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{160 \times 160}{80}$ $\mathrm{P}_{2}=320 \mathrm{~W}$ $\therefore$ Now, heat produced in 20 minutes $\mathrm{dQ}=\mathrm{P}_{2} \times \text { time }=320 \times 20 \times 60$ $\mathrm{Q}=384 \mathrm{~kJ}$
AP EAMCET (22.04.2019) Shift-II
Current Electricity
152659
It takes 12 minutes to boil 1 litre of water in an electric kettle. Due to some defect it becomes necessary to remove $20 \%$ turns of heating coil of the kettle. After repair, how much time will it take to boil 1 litre of water?
1 9.6 minute
2 14.4 minute
3 16.8 minute
4 18.2 minute
Explanation:
A Given, If takes 12 minutes to boil 1 litre of water in an electric kettle. Let, Resistance $=\mathrm{R}$ $\mathrm{n} \text { turns } =\mathrm{R}$ $\therefore \quad 1 \text { turns } =\left(\frac{\mathrm{R}}{\mathrm{n}}\right)$ $\text { So, } 20 \% \text { turns } =\frac{20}{100} \times \mathrm{n} \times \frac{\mathrm{R}}{\mathrm{n}}$ $=\frac{20}{100} \mathrm{R}$ Net remaining resistance $=R-\frac{20}{100} R$ $=\frac{80}{100} \mathrm{R}=0.80 \mathrm{R}$ Heat is supplied, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}=\frac{(\mathrm{IR})^{2}}{\mathrm{R}} \mathrm{t}=\mathrm{I}^{2} \mathrm{Rt}$ Similarly, $\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \times \mathrm{t}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \times \mathrm{t}_{2}$ $\frac{\mathrm{t}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{t}_{2}}{\mathrm{R}_{2}}$ Then, $\frac{12}{\mathrm{R}}=\frac{\mathrm{t}_{2}}{0.8 \mathrm{R}}$ $\mathrm{t}_{2}=12 \times 0.8$ $\mathrm{t}_{2}=9.6 \text { minutes }$
AIIMS-26.05.2019(M) Shift-1
Current Electricity
152660
The heat produced per unit time, on passing electric current through a conductor at a given temperature, is directly proportional to the
1 Electric current
2 Reciprocal of electric current
3 Square of electric current
4 Reciprocal of square of electric current
Explanation:
C According to Joule's law of heating effect, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ Where, $\mathrm{I}=$ current $\mathrm{R}=$ resistance of conductor $\mathrm{t}=$ time $\mathrm{H}=$ heat produced $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{H}}{\mathrm{t}} \propto \mathrm{I}^{2}$ Hence, heat produced per unit time is directly proportional to the square of electric current.
GUJCET 2019
Current Electricity
152661
Two electric bulbs whose resistances are in the ratio $1: 2$ are connected in parallel to a constant voltage source. The power dissipated in them have the ratio
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
Explanation:
C Given, $\mathrm{R}_{1}: \mathrm{R}_{2}=1: 2$ Let the resistance of bulb are $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ respectively. In parallel connection, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{2}{1}$ The power dissipated in them have the ratio $=2: 1$
152658
A $500 \mathrm{~W}$ heater is designed to operate at $200 \mathrm{~V}$ potential difference. If it is connected across 160 $\mathrm{V}$ line, the heat it will produce in $\mathbf{2 0}$ minutes is
1 $384 \mathrm{~kJ}$
2 $483 \mathrm{~kJ}$
3 $843 \mathrm{~kJ}$
4 $348 \mathrm{~kJ}$
Explanation:
A Given, Power $\mathrm{P}_{1}=500 \mathrm{~W}$ Voltage $\mathrm{V}_{1}=200 \mathrm{~V}$ Voltage $\mathrm{V}_{2}=160 \mathrm{~V}$ Time $(t)=20 \mathrm{~min}=20 \times 60=1200 \mathrm{sec}$ We know that, $\text { Power }(\mathrm{P})=\mathrm{VI}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, Case I, $\mathrm{P}_{1} =\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{1}}$ $\therefore \quad \mathrm{R}_{1} =\frac{200^{2}}{500}$ $\mathrm{R}_{1} =80 \Omega$ Case II, $\mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}=\frac{160^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{160 \times 160}{80}$ $\mathrm{P}_{2}=320 \mathrm{~W}$ $\therefore$ Now, heat produced in 20 minutes $\mathrm{dQ}=\mathrm{P}_{2} \times \text { time }=320 \times 20 \times 60$ $\mathrm{Q}=384 \mathrm{~kJ}$
AP EAMCET (22.04.2019) Shift-II
Current Electricity
152659
It takes 12 minutes to boil 1 litre of water in an electric kettle. Due to some defect it becomes necessary to remove $20 \%$ turns of heating coil of the kettle. After repair, how much time will it take to boil 1 litre of water?
1 9.6 minute
2 14.4 minute
3 16.8 minute
4 18.2 minute
Explanation:
A Given, If takes 12 minutes to boil 1 litre of water in an electric kettle. Let, Resistance $=\mathrm{R}$ $\mathrm{n} \text { turns } =\mathrm{R}$ $\therefore \quad 1 \text { turns } =\left(\frac{\mathrm{R}}{\mathrm{n}}\right)$ $\text { So, } 20 \% \text { turns } =\frac{20}{100} \times \mathrm{n} \times \frac{\mathrm{R}}{\mathrm{n}}$ $=\frac{20}{100} \mathrm{R}$ Net remaining resistance $=R-\frac{20}{100} R$ $=\frac{80}{100} \mathrm{R}=0.80 \mathrm{R}$ Heat is supplied, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}=\frac{(\mathrm{IR})^{2}}{\mathrm{R}} \mathrm{t}=\mathrm{I}^{2} \mathrm{Rt}$ Similarly, $\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \times \mathrm{t}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \times \mathrm{t}_{2}$ $\frac{\mathrm{t}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{t}_{2}}{\mathrm{R}_{2}}$ Then, $\frac{12}{\mathrm{R}}=\frac{\mathrm{t}_{2}}{0.8 \mathrm{R}}$ $\mathrm{t}_{2}=12 \times 0.8$ $\mathrm{t}_{2}=9.6 \text { minutes }$
AIIMS-26.05.2019(M) Shift-1
Current Electricity
152660
The heat produced per unit time, on passing electric current through a conductor at a given temperature, is directly proportional to the
1 Electric current
2 Reciprocal of electric current
3 Square of electric current
4 Reciprocal of square of electric current
Explanation:
C According to Joule's law of heating effect, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ Where, $\mathrm{I}=$ current $\mathrm{R}=$ resistance of conductor $\mathrm{t}=$ time $\mathrm{H}=$ heat produced $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{H}}{\mathrm{t}} \propto \mathrm{I}^{2}$ Hence, heat produced per unit time is directly proportional to the square of electric current.
GUJCET 2019
Current Electricity
152661
Two electric bulbs whose resistances are in the ratio $1: 2$ are connected in parallel to a constant voltage source. The power dissipated in them have the ratio
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
Explanation:
C Given, $\mathrm{R}_{1}: \mathrm{R}_{2}=1: 2$ Let the resistance of bulb are $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ respectively. In parallel connection, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{2}{1}$ The power dissipated in them have the ratio $=2: 1$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152658
A $500 \mathrm{~W}$ heater is designed to operate at $200 \mathrm{~V}$ potential difference. If it is connected across 160 $\mathrm{V}$ line, the heat it will produce in $\mathbf{2 0}$ minutes is
1 $384 \mathrm{~kJ}$
2 $483 \mathrm{~kJ}$
3 $843 \mathrm{~kJ}$
4 $348 \mathrm{~kJ}$
Explanation:
A Given, Power $\mathrm{P}_{1}=500 \mathrm{~W}$ Voltage $\mathrm{V}_{1}=200 \mathrm{~V}$ Voltage $\mathrm{V}_{2}=160 \mathrm{~V}$ Time $(t)=20 \mathrm{~min}=20 \times 60=1200 \mathrm{sec}$ We know that, $\text { Power }(\mathrm{P})=\mathrm{VI}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, Case I, $\mathrm{P}_{1} =\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{1}}$ $\therefore \quad \mathrm{R}_{1} =\frac{200^{2}}{500}$ $\mathrm{R}_{1} =80 \Omega$ Case II, $\mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}=\frac{160^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{160 \times 160}{80}$ $\mathrm{P}_{2}=320 \mathrm{~W}$ $\therefore$ Now, heat produced in 20 minutes $\mathrm{dQ}=\mathrm{P}_{2} \times \text { time }=320 \times 20 \times 60$ $\mathrm{Q}=384 \mathrm{~kJ}$
AP EAMCET (22.04.2019) Shift-II
Current Electricity
152659
It takes 12 minutes to boil 1 litre of water in an electric kettle. Due to some defect it becomes necessary to remove $20 \%$ turns of heating coil of the kettle. After repair, how much time will it take to boil 1 litre of water?
1 9.6 minute
2 14.4 minute
3 16.8 minute
4 18.2 minute
Explanation:
A Given, If takes 12 minutes to boil 1 litre of water in an electric kettle. Let, Resistance $=\mathrm{R}$ $\mathrm{n} \text { turns } =\mathrm{R}$ $\therefore \quad 1 \text { turns } =\left(\frac{\mathrm{R}}{\mathrm{n}}\right)$ $\text { So, } 20 \% \text { turns } =\frac{20}{100} \times \mathrm{n} \times \frac{\mathrm{R}}{\mathrm{n}}$ $=\frac{20}{100} \mathrm{R}$ Net remaining resistance $=R-\frac{20}{100} R$ $=\frac{80}{100} \mathrm{R}=0.80 \mathrm{R}$ Heat is supplied, $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t}=\frac{(\mathrm{IR})^{2}}{\mathrm{R}} \mathrm{t}=\mathrm{I}^{2} \mathrm{Rt}$ Similarly, $\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \times \mathrm{t}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \times \mathrm{t}_{2}$ $\frac{\mathrm{t}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{t}_{2}}{\mathrm{R}_{2}}$ Then, $\frac{12}{\mathrm{R}}=\frac{\mathrm{t}_{2}}{0.8 \mathrm{R}}$ $\mathrm{t}_{2}=12 \times 0.8$ $\mathrm{t}_{2}=9.6 \text { minutes }$
AIIMS-26.05.2019(M) Shift-1
Current Electricity
152660
The heat produced per unit time, on passing electric current through a conductor at a given temperature, is directly proportional to the
1 Electric current
2 Reciprocal of electric current
3 Square of electric current
4 Reciprocal of square of electric current
Explanation:
C According to Joule's law of heating effect, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ Where, $\mathrm{I}=$ current $\mathrm{R}=$ resistance of conductor $\mathrm{t}=$ time $\mathrm{H}=$ heat produced $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ $\frac{\mathrm{H}}{\mathrm{t}} \propto \mathrm{I}^{2}$ Hence, heat produced per unit time is directly proportional to the square of electric current.
GUJCET 2019
Current Electricity
152661
Two electric bulbs whose resistances are in the ratio $1: 2$ are connected in parallel to a constant voltage source. The power dissipated in them have the ratio
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
Explanation:
C Given, $\mathrm{R}_{1}: \mathrm{R}_{2}=1: 2$ Let the resistance of bulb are $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ respectively. In parallel connection, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{2}{1}$ The power dissipated in them have the ratio $=2: 1$