NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152721
Ratio of resistance of two bulbs $40 \mathrm{~W}$ and 60 $\mathrm{W}$ connected across $220 \mathrm{~V}$ source is
1 $3: 2$
2 $3: 8$
3 $4: 3$
4 $9: 4$
Explanation:
A Given, $\mathrm{P}_{1} =40 \mathrm{~W}$ $\mathrm{P}_{2} =60 \mathrm{~W}$ $\mathrm{~V} =220 \mathrm{~V}$ Power of a bulb, $P=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{60}{40}=\frac{3}{2}$ The ratio of resistance of two bulbs $=3: 2$
CG PET- 2007
Current Electricity
152722
An electric heater of resistance $6 \Omega$ is run for 10 min on $120 \mathrm{~V}$ line. The energy liberated in this period of time is
152723
The resistor having equal resistance are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature change in a given time interval
1 equal amounts of thermal energy must be produced in the resistors
2 unequal amounts of thermal energy may be produced
3 the temperature must rise equally in the resistors
4 the temperature must rise unequally in the resistors
Explanation:
A Heat produce in a resistance is given by, $\mathrm{H}=\mathrm{I}^{2} \mathrm{R} \mathrm{t}$ Where, $\mathrm{R}=$ Resistance, $\mathrm{I}=\text { Current }$ $\mathrm{t}=\text { time }$ As the resistors are in series, the current through them will be same. Thus, the amount of thermal energy produced in the resistors is same. The rise in the temperature of the resistor will depend on the shape and size of the resistor. Thus, the rise in the temperature of the two resistances may be equal.
CG PET- 2005
Current Electricity
152725
Two heater wires, similar in all respects, are first connected in series and then in parallel. If the rate of heat produced in the two cases is respectively $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$, then the value of $\mathrm{H}_{1} / \mathrm{H}_{2}$ will be
1 2
2 $1 / 2$
3 4
4 $1 / 4$
Explanation:
D Rate of heat produced in a wire $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ The amount of heat generated on series combination, $\mathrm{H}_{1}=\frac{\mathrm{V}^{2}}{(\mathrm{R}+\mathrm{R})}$ $\mathrm{H}_{1}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}}$ The amount of heat generated on the parallel combination, $\mathrm{H}_{2}=\frac{\mathrm{V}^{2}}{(\mathrm{R} / 2)}$ $\mathrm{H}_{2}=\frac{2 \mathrm{~V}^{2}}{\mathrm{R}}$ The ratio of the amount of heat generated $b / w$ parallel and series combination. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{V}^{2}}{2 \mathrm{R}}}{\frac{2 \mathrm{~V}^{2}}{\mathrm{R}}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}} \times \frac{\mathrm{R}}{2 \mathrm{~V}^{2}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{1}{4}$
152721
Ratio of resistance of two bulbs $40 \mathrm{~W}$ and 60 $\mathrm{W}$ connected across $220 \mathrm{~V}$ source is
1 $3: 2$
2 $3: 8$
3 $4: 3$
4 $9: 4$
Explanation:
A Given, $\mathrm{P}_{1} =40 \mathrm{~W}$ $\mathrm{P}_{2} =60 \mathrm{~W}$ $\mathrm{~V} =220 \mathrm{~V}$ Power of a bulb, $P=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{60}{40}=\frac{3}{2}$ The ratio of resistance of two bulbs $=3: 2$
CG PET- 2007
Current Electricity
152722
An electric heater of resistance $6 \Omega$ is run for 10 min on $120 \mathrm{~V}$ line. The energy liberated in this period of time is
152723
The resistor having equal resistance are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature change in a given time interval
1 equal amounts of thermal energy must be produced in the resistors
2 unequal amounts of thermal energy may be produced
3 the temperature must rise equally in the resistors
4 the temperature must rise unequally in the resistors
Explanation:
A Heat produce in a resistance is given by, $\mathrm{H}=\mathrm{I}^{2} \mathrm{R} \mathrm{t}$ Where, $\mathrm{R}=$ Resistance, $\mathrm{I}=\text { Current }$ $\mathrm{t}=\text { time }$ As the resistors are in series, the current through them will be same. Thus, the amount of thermal energy produced in the resistors is same. The rise in the temperature of the resistor will depend on the shape and size of the resistor. Thus, the rise in the temperature of the two resistances may be equal.
CG PET- 2005
Current Electricity
152725
Two heater wires, similar in all respects, are first connected in series and then in parallel. If the rate of heat produced in the two cases is respectively $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$, then the value of $\mathrm{H}_{1} / \mathrm{H}_{2}$ will be
1 2
2 $1 / 2$
3 4
4 $1 / 4$
Explanation:
D Rate of heat produced in a wire $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ The amount of heat generated on series combination, $\mathrm{H}_{1}=\frac{\mathrm{V}^{2}}{(\mathrm{R}+\mathrm{R})}$ $\mathrm{H}_{1}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}}$ The amount of heat generated on the parallel combination, $\mathrm{H}_{2}=\frac{\mathrm{V}^{2}}{(\mathrm{R} / 2)}$ $\mathrm{H}_{2}=\frac{2 \mathrm{~V}^{2}}{\mathrm{R}}$ The ratio of the amount of heat generated $b / w$ parallel and series combination. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{V}^{2}}{2 \mathrm{R}}}{\frac{2 \mathrm{~V}^{2}}{\mathrm{R}}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}} \times \frac{\mathrm{R}}{2 \mathrm{~V}^{2}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{1}{4}$
152721
Ratio of resistance of two bulbs $40 \mathrm{~W}$ and 60 $\mathrm{W}$ connected across $220 \mathrm{~V}$ source is
1 $3: 2$
2 $3: 8$
3 $4: 3$
4 $9: 4$
Explanation:
A Given, $\mathrm{P}_{1} =40 \mathrm{~W}$ $\mathrm{P}_{2} =60 \mathrm{~W}$ $\mathrm{~V} =220 \mathrm{~V}$ Power of a bulb, $P=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{60}{40}=\frac{3}{2}$ The ratio of resistance of two bulbs $=3: 2$
CG PET- 2007
Current Electricity
152722
An electric heater of resistance $6 \Omega$ is run for 10 min on $120 \mathrm{~V}$ line. The energy liberated in this period of time is
152723
The resistor having equal resistance are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature change in a given time interval
1 equal amounts of thermal energy must be produced in the resistors
2 unequal amounts of thermal energy may be produced
3 the temperature must rise equally in the resistors
4 the temperature must rise unequally in the resistors
Explanation:
A Heat produce in a resistance is given by, $\mathrm{H}=\mathrm{I}^{2} \mathrm{R} \mathrm{t}$ Where, $\mathrm{R}=$ Resistance, $\mathrm{I}=\text { Current }$ $\mathrm{t}=\text { time }$ As the resistors are in series, the current through them will be same. Thus, the amount of thermal energy produced in the resistors is same. The rise in the temperature of the resistor will depend on the shape and size of the resistor. Thus, the rise in the temperature of the two resistances may be equal.
CG PET- 2005
Current Electricity
152725
Two heater wires, similar in all respects, are first connected in series and then in parallel. If the rate of heat produced in the two cases is respectively $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$, then the value of $\mathrm{H}_{1} / \mathrm{H}_{2}$ will be
1 2
2 $1 / 2$
3 4
4 $1 / 4$
Explanation:
D Rate of heat produced in a wire $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ The amount of heat generated on series combination, $\mathrm{H}_{1}=\frac{\mathrm{V}^{2}}{(\mathrm{R}+\mathrm{R})}$ $\mathrm{H}_{1}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}}$ The amount of heat generated on the parallel combination, $\mathrm{H}_{2}=\frac{\mathrm{V}^{2}}{(\mathrm{R} / 2)}$ $\mathrm{H}_{2}=\frac{2 \mathrm{~V}^{2}}{\mathrm{R}}$ The ratio of the amount of heat generated $b / w$ parallel and series combination. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{V}^{2}}{2 \mathrm{R}}}{\frac{2 \mathrm{~V}^{2}}{\mathrm{R}}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}} \times \frac{\mathrm{R}}{2 \mathrm{~V}^{2}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{1}{4}$
152721
Ratio of resistance of two bulbs $40 \mathrm{~W}$ and 60 $\mathrm{W}$ connected across $220 \mathrm{~V}$ source is
1 $3: 2$
2 $3: 8$
3 $4: 3$
4 $9: 4$
Explanation:
A Given, $\mathrm{P}_{1} =40 \mathrm{~W}$ $\mathrm{P}_{2} =60 \mathrm{~W}$ $\mathrm{~V} =220 \mathrm{~V}$ Power of a bulb, $P=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{60}{40}=\frac{3}{2}$ The ratio of resistance of two bulbs $=3: 2$
CG PET- 2007
Current Electricity
152722
An electric heater of resistance $6 \Omega$ is run for 10 min on $120 \mathrm{~V}$ line. The energy liberated in this period of time is
152723
The resistor having equal resistance are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature change in a given time interval
1 equal amounts of thermal energy must be produced in the resistors
2 unequal amounts of thermal energy may be produced
3 the temperature must rise equally in the resistors
4 the temperature must rise unequally in the resistors
Explanation:
A Heat produce in a resistance is given by, $\mathrm{H}=\mathrm{I}^{2} \mathrm{R} \mathrm{t}$ Where, $\mathrm{R}=$ Resistance, $\mathrm{I}=\text { Current }$ $\mathrm{t}=\text { time }$ As the resistors are in series, the current through them will be same. Thus, the amount of thermal energy produced in the resistors is same. The rise in the temperature of the resistor will depend on the shape and size of the resistor. Thus, the rise in the temperature of the two resistances may be equal.
CG PET- 2005
Current Electricity
152725
Two heater wires, similar in all respects, are first connected in series and then in parallel. If the rate of heat produced in the two cases is respectively $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$, then the value of $\mathrm{H}_{1} / \mathrm{H}_{2}$ will be
1 2
2 $1 / 2$
3 4
4 $1 / 4$
Explanation:
D Rate of heat produced in a wire $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ The amount of heat generated on series combination, $\mathrm{H}_{1}=\frac{\mathrm{V}^{2}}{(\mathrm{R}+\mathrm{R})}$ $\mathrm{H}_{1}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}}$ The amount of heat generated on the parallel combination, $\mathrm{H}_{2}=\frac{\mathrm{V}^{2}}{(\mathrm{R} / 2)}$ $\mathrm{H}_{2}=\frac{2 \mathrm{~V}^{2}}{\mathrm{R}}$ The ratio of the amount of heat generated $b / w$ parallel and series combination. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{V}^{2}}{2 \mathrm{R}}}{\frac{2 \mathrm{~V}^{2}}{\mathrm{R}}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}} \times \frac{\mathrm{R}}{2 \mathrm{~V}^{2}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{1}{4}$
