152757
An electric bulb is made of tungsten filament of resistance $R \Omega$. It is marked $100 \mathrm{~W}$ and $230 \mathrm{~V}$. Then, the value of $R$ is-
1 $300 \Omega$
2 $529 \Omega$
3 $739 \Omega$
4 $100 \Omega$
Explanation:
B Given, Resistance $=\mathrm{R} \Omega$ Power $(\mathrm{P})=100 \mathrm{~W}$ Voltage $(\mathrm{V})=230$ volt As we know that the power in terms of voltage and resistance is, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{(230)^{2}}{100}=529 \Omega$
BCECE-2014
Current Electricity
152758
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from room temperature $20^{\circ} \mathrm{C}$ ? (the temperature of boiling water is $100^{\circ} \mathrm{C}$ )
1 $12.6 \mathrm{~min}$
2 $12.8 \mathrm{~min}$
3 $6.3 \mathrm{~min}$
4 $6.4 \mathrm{~min}$
Explanation:
C Given, Current $(\mathrm{I})=4 \mathrm{~A}$ Voltage $(\mathrm{V})=220$ volt Mass of water $(\mathrm{M})=1 \mathrm{~kg}=1000 \mathrm{~g}$ Temperature $\left(\mathrm{T}_{1}\right)=20^{\circ} \mathrm{C}$ Temperature $\left(\mathrm{T}_{2}\right)=100^{\circ} \mathrm{C}$ $\therefore \quad \Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=100-20=80^{\circ} \mathrm{C}$ Specific heat capacity of water $(\mathrm{S})=1 \mathrm{cal} / \mathrm{g}-\mathrm{C}^{\mathrm{o}}$ $\therefore$ Heat required $(\mathrm{Q})=$ M.S. $\Delta \mathrm{t}$ $\mathrm{Q}=1000 \times 1 \times 80$ $\mathrm{Q}=1000 \times 80 \mathrm{cal}$ Now applying joule's law, heat generated by the conductor, $\mathrm{H}=\mathrm{V} \times \mathrm{I} \times \mathrm{t}$ $\mathrm{H}=220 \times 4 \times \mathrm{t}$ $\mathrm{H}=880 \mathrm{t} \text { Joule }$ $\mathrm{H}=\frac{880 \mathrm{t}}{4.18} \mathrm{cal}$ Heat is evolved due to Joule's effect is used up in boiling water such as $\mathrm{Q}=\mathrm{H}$ $1000 \times 80 =\frac{880 \mathrm{t}}{4.18} \mathrm{cal}$ $\mathrm{t} =\frac{1000 \times 80 \times 4.18}{880}$ $\mathrm{t} =380 \mathrm{sec}$ $\mathrm{t} =6.3 \mathrm{~min}$
BCECE-2013
Current Electricity
152759
The current through a bulb is increased by $1 \%$. Assuming that the resistance of the filament remains unchanged the power of the bulb will -
1 increase by $1 \%$
2 decrease by $1 \%$
3 increase by $2 \%$
4 decrease by $2 \%$
Explanation:
C Let, Current $=\mathrm{I}$ $\text { Resistance }=\mathrm{R}$ We know that, $\text { Power }(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $\therefore \mathrm{P} \propto \mathrm{I}^{2}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\left(\mathrm{I}_{1}\right)^{2}}{\left(\mathrm{I}_{2}\right)^{2}}$ $\because$ The current through a bulb is increased by $1 \%$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\left(\mathrm{I}_{1}\right)^{2}}{\left(\frac{101}{100}\right)^{2} \mathrm{I}^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\left(\frac{100}{101}\right)^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{101}{100}\right)^{2}$ Percentage increase in power $\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}-1\right) \%=\left[\left(\frac{101}{100}\right)^{2}-1\right] \%=2.01 \%$ Therefore, the power of the bulb will increase by $2 \%$
BCECE-2009
Current Electricity
152760
The power dissipated across resistance $R$ which is connected across a battery of potential $V$ is $P$. If resistance is doubled, then the power becomes-
1 $1 / 2$
2 2
3 $1 / 4$
4 4
Explanation:
A Power is inversely proportional to resistance provided potential difference remains constant. We know that, $\text { Power } (\mathrm{P}) =\mathrm{Vi}$ $\mathrm{P} =\mathrm{V}\left(\frac{\mathrm{V}}{\mathrm{R}}\right) \quad\left[\because \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}\right]$ $\mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}$ If resistance is doubled (2R). Then, $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}}$ $2 \mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ On comparing equation (i) and equation (ii) $\mathrm{P}=2 \mathrm{P}^{\prime}$ $\mathrm{P}^{\prime}=\frac{\mathrm{P}}{2}$ So, Power becomes $\frac{1}{2}$ of initial value.
152757
An electric bulb is made of tungsten filament of resistance $R \Omega$. It is marked $100 \mathrm{~W}$ and $230 \mathrm{~V}$. Then, the value of $R$ is-
1 $300 \Omega$
2 $529 \Omega$
3 $739 \Omega$
4 $100 \Omega$
Explanation:
B Given, Resistance $=\mathrm{R} \Omega$ Power $(\mathrm{P})=100 \mathrm{~W}$ Voltage $(\mathrm{V})=230$ volt As we know that the power in terms of voltage and resistance is, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{(230)^{2}}{100}=529 \Omega$
BCECE-2014
Current Electricity
152758
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from room temperature $20^{\circ} \mathrm{C}$ ? (the temperature of boiling water is $100^{\circ} \mathrm{C}$ )
1 $12.6 \mathrm{~min}$
2 $12.8 \mathrm{~min}$
3 $6.3 \mathrm{~min}$
4 $6.4 \mathrm{~min}$
Explanation:
C Given, Current $(\mathrm{I})=4 \mathrm{~A}$ Voltage $(\mathrm{V})=220$ volt Mass of water $(\mathrm{M})=1 \mathrm{~kg}=1000 \mathrm{~g}$ Temperature $\left(\mathrm{T}_{1}\right)=20^{\circ} \mathrm{C}$ Temperature $\left(\mathrm{T}_{2}\right)=100^{\circ} \mathrm{C}$ $\therefore \quad \Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=100-20=80^{\circ} \mathrm{C}$ Specific heat capacity of water $(\mathrm{S})=1 \mathrm{cal} / \mathrm{g}-\mathrm{C}^{\mathrm{o}}$ $\therefore$ Heat required $(\mathrm{Q})=$ M.S. $\Delta \mathrm{t}$ $\mathrm{Q}=1000 \times 1 \times 80$ $\mathrm{Q}=1000 \times 80 \mathrm{cal}$ Now applying joule's law, heat generated by the conductor, $\mathrm{H}=\mathrm{V} \times \mathrm{I} \times \mathrm{t}$ $\mathrm{H}=220 \times 4 \times \mathrm{t}$ $\mathrm{H}=880 \mathrm{t} \text { Joule }$ $\mathrm{H}=\frac{880 \mathrm{t}}{4.18} \mathrm{cal}$ Heat is evolved due to Joule's effect is used up in boiling water such as $\mathrm{Q}=\mathrm{H}$ $1000 \times 80 =\frac{880 \mathrm{t}}{4.18} \mathrm{cal}$ $\mathrm{t} =\frac{1000 \times 80 \times 4.18}{880}$ $\mathrm{t} =380 \mathrm{sec}$ $\mathrm{t} =6.3 \mathrm{~min}$
BCECE-2013
Current Electricity
152759
The current through a bulb is increased by $1 \%$. Assuming that the resistance of the filament remains unchanged the power of the bulb will -
1 increase by $1 \%$
2 decrease by $1 \%$
3 increase by $2 \%$
4 decrease by $2 \%$
Explanation:
C Let, Current $=\mathrm{I}$ $\text { Resistance }=\mathrm{R}$ We know that, $\text { Power }(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $\therefore \mathrm{P} \propto \mathrm{I}^{2}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\left(\mathrm{I}_{1}\right)^{2}}{\left(\mathrm{I}_{2}\right)^{2}}$ $\because$ The current through a bulb is increased by $1 \%$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\left(\mathrm{I}_{1}\right)^{2}}{\left(\frac{101}{100}\right)^{2} \mathrm{I}^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\left(\frac{100}{101}\right)^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{101}{100}\right)^{2}$ Percentage increase in power $\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}-1\right) \%=\left[\left(\frac{101}{100}\right)^{2}-1\right] \%=2.01 \%$ Therefore, the power of the bulb will increase by $2 \%$
BCECE-2009
Current Electricity
152760
The power dissipated across resistance $R$ which is connected across a battery of potential $V$ is $P$. If resistance is doubled, then the power becomes-
1 $1 / 2$
2 2
3 $1 / 4$
4 4
Explanation:
A Power is inversely proportional to resistance provided potential difference remains constant. We know that, $\text { Power } (\mathrm{P}) =\mathrm{Vi}$ $\mathrm{P} =\mathrm{V}\left(\frac{\mathrm{V}}{\mathrm{R}}\right) \quad\left[\because \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}\right]$ $\mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}$ If resistance is doubled (2R). Then, $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}}$ $2 \mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ On comparing equation (i) and equation (ii) $\mathrm{P}=2 \mathrm{P}^{\prime}$ $\mathrm{P}^{\prime}=\frac{\mathrm{P}}{2}$ So, Power becomes $\frac{1}{2}$ of initial value.
152757
An electric bulb is made of tungsten filament of resistance $R \Omega$. It is marked $100 \mathrm{~W}$ and $230 \mathrm{~V}$. Then, the value of $R$ is-
1 $300 \Omega$
2 $529 \Omega$
3 $739 \Omega$
4 $100 \Omega$
Explanation:
B Given, Resistance $=\mathrm{R} \Omega$ Power $(\mathrm{P})=100 \mathrm{~W}$ Voltage $(\mathrm{V})=230$ volt As we know that the power in terms of voltage and resistance is, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{(230)^{2}}{100}=529 \Omega$
BCECE-2014
Current Electricity
152758
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from room temperature $20^{\circ} \mathrm{C}$ ? (the temperature of boiling water is $100^{\circ} \mathrm{C}$ )
1 $12.6 \mathrm{~min}$
2 $12.8 \mathrm{~min}$
3 $6.3 \mathrm{~min}$
4 $6.4 \mathrm{~min}$
Explanation:
C Given, Current $(\mathrm{I})=4 \mathrm{~A}$ Voltage $(\mathrm{V})=220$ volt Mass of water $(\mathrm{M})=1 \mathrm{~kg}=1000 \mathrm{~g}$ Temperature $\left(\mathrm{T}_{1}\right)=20^{\circ} \mathrm{C}$ Temperature $\left(\mathrm{T}_{2}\right)=100^{\circ} \mathrm{C}$ $\therefore \quad \Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=100-20=80^{\circ} \mathrm{C}$ Specific heat capacity of water $(\mathrm{S})=1 \mathrm{cal} / \mathrm{g}-\mathrm{C}^{\mathrm{o}}$ $\therefore$ Heat required $(\mathrm{Q})=$ M.S. $\Delta \mathrm{t}$ $\mathrm{Q}=1000 \times 1 \times 80$ $\mathrm{Q}=1000 \times 80 \mathrm{cal}$ Now applying joule's law, heat generated by the conductor, $\mathrm{H}=\mathrm{V} \times \mathrm{I} \times \mathrm{t}$ $\mathrm{H}=220 \times 4 \times \mathrm{t}$ $\mathrm{H}=880 \mathrm{t} \text { Joule }$ $\mathrm{H}=\frac{880 \mathrm{t}}{4.18} \mathrm{cal}$ Heat is evolved due to Joule's effect is used up in boiling water such as $\mathrm{Q}=\mathrm{H}$ $1000 \times 80 =\frac{880 \mathrm{t}}{4.18} \mathrm{cal}$ $\mathrm{t} =\frac{1000 \times 80 \times 4.18}{880}$ $\mathrm{t} =380 \mathrm{sec}$ $\mathrm{t} =6.3 \mathrm{~min}$
BCECE-2013
Current Electricity
152759
The current through a bulb is increased by $1 \%$. Assuming that the resistance of the filament remains unchanged the power of the bulb will -
1 increase by $1 \%$
2 decrease by $1 \%$
3 increase by $2 \%$
4 decrease by $2 \%$
Explanation:
C Let, Current $=\mathrm{I}$ $\text { Resistance }=\mathrm{R}$ We know that, $\text { Power }(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $\therefore \mathrm{P} \propto \mathrm{I}^{2}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\left(\mathrm{I}_{1}\right)^{2}}{\left(\mathrm{I}_{2}\right)^{2}}$ $\because$ The current through a bulb is increased by $1 \%$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\left(\mathrm{I}_{1}\right)^{2}}{\left(\frac{101}{100}\right)^{2} \mathrm{I}^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\left(\frac{100}{101}\right)^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{101}{100}\right)^{2}$ Percentage increase in power $\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}-1\right) \%=\left[\left(\frac{101}{100}\right)^{2}-1\right] \%=2.01 \%$ Therefore, the power of the bulb will increase by $2 \%$
BCECE-2009
Current Electricity
152760
The power dissipated across resistance $R$ which is connected across a battery of potential $V$ is $P$. If resistance is doubled, then the power becomes-
1 $1 / 2$
2 2
3 $1 / 4$
4 4
Explanation:
A Power is inversely proportional to resistance provided potential difference remains constant. We know that, $\text { Power } (\mathrm{P}) =\mathrm{Vi}$ $\mathrm{P} =\mathrm{V}\left(\frac{\mathrm{V}}{\mathrm{R}}\right) \quad\left[\because \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}\right]$ $\mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}$ If resistance is doubled (2R). Then, $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}}$ $2 \mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ On comparing equation (i) and equation (ii) $\mathrm{P}=2 \mathrm{P}^{\prime}$ $\mathrm{P}^{\prime}=\frac{\mathrm{P}}{2}$ So, Power becomes $\frac{1}{2}$ of initial value.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152757
An electric bulb is made of tungsten filament of resistance $R \Omega$. It is marked $100 \mathrm{~W}$ and $230 \mathrm{~V}$. Then, the value of $R$ is-
1 $300 \Omega$
2 $529 \Omega$
3 $739 \Omega$
4 $100 \Omega$
Explanation:
B Given, Resistance $=\mathrm{R} \Omega$ Power $(\mathrm{P})=100 \mathrm{~W}$ Voltage $(\mathrm{V})=230$ volt As we know that the power in terms of voltage and resistance is, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{(230)^{2}}{100}=529 \Omega$
BCECE-2014
Current Electricity
152758
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from room temperature $20^{\circ} \mathrm{C}$ ? (the temperature of boiling water is $100^{\circ} \mathrm{C}$ )
1 $12.6 \mathrm{~min}$
2 $12.8 \mathrm{~min}$
3 $6.3 \mathrm{~min}$
4 $6.4 \mathrm{~min}$
Explanation:
C Given, Current $(\mathrm{I})=4 \mathrm{~A}$ Voltage $(\mathrm{V})=220$ volt Mass of water $(\mathrm{M})=1 \mathrm{~kg}=1000 \mathrm{~g}$ Temperature $\left(\mathrm{T}_{1}\right)=20^{\circ} \mathrm{C}$ Temperature $\left(\mathrm{T}_{2}\right)=100^{\circ} \mathrm{C}$ $\therefore \quad \Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}=100-20=80^{\circ} \mathrm{C}$ Specific heat capacity of water $(\mathrm{S})=1 \mathrm{cal} / \mathrm{g}-\mathrm{C}^{\mathrm{o}}$ $\therefore$ Heat required $(\mathrm{Q})=$ M.S. $\Delta \mathrm{t}$ $\mathrm{Q}=1000 \times 1 \times 80$ $\mathrm{Q}=1000 \times 80 \mathrm{cal}$ Now applying joule's law, heat generated by the conductor, $\mathrm{H}=\mathrm{V} \times \mathrm{I} \times \mathrm{t}$ $\mathrm{H}=220 \times 4 \times \mathrm{t}$ $\mathrm{H}=880 \mathrm{t} \text { Joule }$ $\mathrm{H}=\frac{880 \mathrm{t}}{4.18} \mathrm{cal}$ Heat is evolved due to Joule's effect is used up in boiling water such as $\mathrm{Q}=\mathrm{H}$ $1000 \times 80 =\frac{880 \mathrm{t}}{4.18} \mathrm{cal}$ $\mathrm{t} =\frac{1000 \times 80 \times 4.18}{880}$ $\mathrm{t} =380 \mathrm{sec}$ $\mathrm{t} =6.3 \mathrm{~min}$
BCECE-2013
Current Electricity
152759
The current through a bulb is increased by $1 \%$. Assuming that the resistance of the filament remains unchanged the power of the bulb will -
1 increase by $1 \%$
2 decrease by $1 \%$
3 increase by $2 \%$
4 decrease by $2 \%$
Explanation:
C Let, Current $=\mathrm{I}$ $\text { Resistance }=\mathrm{R}$ We know that, $\text { Power }(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $\therefore \mathrm{P} \propto \mathrm{I}^{2}$ Then, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\left(\mathrm{I}_{1}\right)^{2}}{\left(\mathrm{I}_{2}\right)^{2}}$ $\because$ The current through a bulb is increased by $1 \%$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\left(\mathrm{I}_{1}\right)^{2}}{\left(\frac{101}{100}\right)^{2} \mathrm{I}^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\left(\frac{100}{101}\right)^{2}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{101}{100}\right)^{2}$ Percentage increase in power $\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}-1\right) \%=\left[\left(\frac{101}{100}\right)^{2}-1\right] \%=2.01 \%$ Therefore, the power of the bulb will increase by $2 \%$
BCECE-2009
Current Electricity
152760
The power dissipated across resistance $R$ which is connected across a battery of potential $V$ is $P$. If resistance is doubled, then the power becomes-
1 $1 / 2$
2 2
3 $1 / 4$
4 4
Explanation:
A Power is inversely proportional to resistance provided potential difference remains constant. We know that, $\text { Power } (\mathrm{P}) =\mathrm{Vi}$ $\mathrm{P} =\mathrm{V}\left(\frac{\mathrm{V}}{\mathrm{R}}\right) \quad\left[\because \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}\right]$ $\mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}$ If resistance is doubled (2R). Then, $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{2 \mathrm{R}}$ $2 \mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ On comparing equation (i) and equation (ii) $\mathrm{P}=2 \mathrm{P}^{\prime}$ $\mathrm{P}^{\prime}=\frac{\mathrm{P}}{2}$ So, Power becomes $\frac{1}{2}$ of initial value.
