152007
An immersion heater of $1 \mathrm{~kW}$ will take the time for increasing the temperature of 1 litre of water through $10^{\circ} \mathrm{C}$ equal to
1 $10 \mathrm{~s}$
2 $24 \mathrm{~s}$
3 $42 \mathrm{~s}$
4 $100 \mathrm{~s}$
Explanation:
C Given, Mass, $\mathrm{m}=$ mass in gram $=10$ litre $=10^{3} \text { gram }$ Specific heat for water $=\mathrm{s}=4.186 \approx 4.2 \mathrm{~J} / \mathrm{gram} /{ }^{\circ} \mathrm{C}$ Temperature difference, $\Delta \mathrm{T}=10^{\circ} \mathrm{C}$ Power, $\mathrm{P}=1 \mathrm{~kW}=1000$ watt $=1000 \mathrm{~J} / \mathrm{sec}$ $\because$ Time, $\mathrm{t}=\frac{\mathrm{ms} \Delta \mathrm{T}}{\mathrm{P}}=\frac{10^{3} \times 4.186 \times 10}{1000}$ $=\frac{4.2 \times 10 \times 10^{3}}{10^{3}}$ $\mathrm{t}=42 \mathrm{sec}$.
J and K CET- 1997
Current Electricity
152008
A silver wire has a resistance of $1.6 \Omega$ at 25.5 ${ }^{\circ} \mathrm{C}$ and a resistance of $2.5 \Omega$ at $100^{\circ} \mathrm{C}$, then temperature coefficient of resistivity of silver is
1 $5.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
2 $7.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
3 $11.75 \times 10^{-2}{ }^{\circ} \mathrm{C}$
4 $15.5 \times 10^{-3}{ }^{\circ} \mathrm{C}$
Explanation:
B $\mathrm{R}_{1}=1.6 \Omega$ $\mathrm{R}_{2}=2.5 \Omega$ $\mathrm{T}_{1}=25.5^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ Temperature coefficient of silver $(\alpha)=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}$ $=\frac{2.5-1.6}{1.6(100-25.5)}=\frac{0.9}{1.6 \times 74.5}=\frac{0.9}{119.2}$ $=7.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
J and K-CET-2014
Current Electricity
152009
Two bulbs operating on standard voltage 110 volt have resistances in the ratio $9: 16$. The ratio of brightness of light from them is
1 $9: 16$
2 $16: 9$
3 $3: 4$
4 $4: 3$
Explanation:
B V $=110 \mathrm{~V}$ Ratio of resistance $=9: 16$ Brightness depend on current $I=\frac{P}{V}$ $I=\frac{V^{2} / R}{V}$ $I=\frac{V}{R}$ $\therefore \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{16}{9} \quad(\because$ voltage same $)$
J and K-CET-2014
Current Electricity
152011
In a circuit, 3 resistors of resistances $1.2 \Omega, 2 \Omega$, $3 \Omega$ are connected in parallel. The value of equivalent resistance is
1 Less than $1.2 \Omega$
2 Greater than $1.2 \Omega$
3 Between $1.2 \Omega$ and $2 \Omega$
4 Between $2 \Omega$ and $3 \Omega$
Explanation:
A The three resistors are $1.2 \Omega, 2 \Omega, 3 \Omega$ Apply the formula of parallel- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{1.2}+\frac{1}{2}+\frac{1}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{10+6+4}{12}=\frac{20}{12}$ $\mathrm{R}_{\mathrm{eq}}=12 / 20=0.6 \Omega$ Hence, value of equivalent resistance is less than $1.2 \Omega$.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152007
An immersion heater of $1 \mathrm{~kW}$ will take the time for increasing the temperature of 1 litre of water through $10^{\circ} \mathrm{C}$ equal to
1 $10 \mathrm{~s}$
2 $24 \mathrm{~s}$
3 $42 \mathrm{~s}$
4 $100 \mathrm{~s}$
Explanation:
C Given, Mass, $\mathrm{m}=$ mass in gram $=10$ litre $=10^{3} \text { gram }$ Specific heat for water $=\mathrm{s}=4.186 \approx 4.2 \mathrm{~J} / \mathrm{gram} /{ }^{\circ} \mathrm{C}$ Temperature difference, $\Delta \mathrm{T}=10^{\circ} \mathrm{C}$ Power, $\mathrm{P}=1 \mathrm{~kW}=1000$ watt $=1000 \mathrm{~J} / \mathrm{sec}$ $\because$ Time, $\mathrm{t}=\frac{\mathrm{ms} \Delta \mathrm{T}}{\mathrm{P}}=\frac{10^{3} \times 4.186 \times 10}{1000}$ $=\frac{4.2 \times 10 \times 10^{3}}{10^{3}}$ $\mathrm{t}=42 \mathrm{sec}$.
J and K CET- 1997
Current Electricity
152008
A silver wire has a resistance of $1.6 \Omega$ at 25.5 ${ }^{\circ} \mathrm{C}$ and a resistance of $2.5 \Omega$ at $100^{\circ} \mathrm{C}$, then temperature coefficient of resistivity of silver is
1 $5.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
2 $7.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
3 $11.75 \times 10^{-2}{ }^{\circ} \mathrm{C}$
4 $15.5 \times 10^{-3}{ }^{\circ} \mathrm{C}$
Explanation:
B $\mathrm{R}_{1}=1.6 \Omega$ $\mathrm{R}_{2}=2.5 \Omega$ $\mathrm{T}_{1}=25.5^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ Temperature coefficient of silver $(\alpha)=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}$ $=\frac{2.5-1.6}{1.6(100-25.5)}=\frac{0.9}{1.6 \times 74.5}=\frac{0.9}{119.2}$ $=7.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
J and K-CET-2014
Current Electricity
152009
Two bulbs operating on standard voltage 110 volt have resistances in the ratio $9: 16$. The ratio of brightness of light from them is
1 $9: 16$
2 $16: 9$
3 $3: 4$
4 $4: 3$
Explanation:
B V $=110 \mathrm{~V}$ Ratio of resistance $=9: 16$ Brightness depend on current $I=\frac{P}{V}$ $I=\frac{V^{2} / R}{V}$ $I=\frac{V}{R}$ $\therefore \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{16}{9} \quad(\because$ voltage same $)$
J and K-CET-2014
Current Electricity
152011
In a circuit, 3 resistors of resistances $1.2 \Omega, 2 \Omega$, $3 \Omega$ are connected in parallel. The value of equivalent resistance is
1 Less than $1.2 \Omega$
2 Greater than $1.2 \Omega$
3 Between $1.2 \Omega$ and $2 \Omega$
4 Between $2 \Omega$ and $3 \Omega$
Explanation:
A The three resistors are $1.2 \Omega, 2 \Omega, 3 \Omega$ Apply the formula of parallel- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{1.2}+\frac{1}{2}+\frac{1}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{10+6+4}{12}=\frac{20}{12}$ $\mathrm{R}_{\mathrm{eq}}=12 / 20=0.6 \Omega$ Hence, value of equivalent resistance is less than $1.2 \Omega$.
152007
An immersion heater of $1 \mathrm{~kW}$ will take the time for increasing the temperature of 1 litre of water through $10^{\circ} \mathrm{C}$ equal to
1 $10 \mathrm{~s}$
2 $24 \mathrm{~s}$
3 $42 \mathrm{~s}$
4 $100 \mathrm{~s}$
Explanation:
C Given, Mass, $\mathrm{m}=$ mass in gram $=10$ litre $=10^{3} \text { gram }$ Specific heat for water $=\mathrm{s}=4.186 \approx 4.2 \mathrm{~J} / \mathrm{gram} /{ }^{\circ} \mathrm{C}$ Temperature difference, $\Delta \mathrm{T}=10^{\circ} \mathrm{C}$ Power, $\mathrm{P}=1 \mathrm{~kW}=1000$ watt $=1000 \mathrm{~J} / \mathrm{sec}$ $\because$ Time, $\mathrm{t}=\frac{\mathrm{ms} \Delta \mathrm{T}}{\mathrm{P}}=\frac{10^{3} \times 4.186 \times 10}{1000}$ $=\frac{4.2 \times 10 \times 10^{3}}{10^{3}}$ $\mathrm{t}=42 \mathrm{sec}$.
J and K CET- 1997
Current Electricity
152008
A silver wire has a resistance of $1.6 \Omega$ at 25.5 ${ }^{\circ} \mathrm{C}$ and a resistance of $2.5 \Omega$ at $100^{\circ} \mathrm{C}$, then temperature coefficient of resistivity of silver is
1 $5.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
2 $7.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
3 $11.75 \times 10^{-2}{ }^{\circ} \mathrm{C}$
4 $15.5 \times 10^{-3}{ }^{\circ} \mathrm{C}$
Explanation:
B $\mathrm{R}_{1}=1.6 \Omega$ $\mathrm{R}_{2}=2.5 \Omega$ $\mathrm{T}_{1}=25.5^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ Temperature coefficient of silver $(\alpha)=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}$ $=\frac{2.5-1.6}{1.6(100-25.5)}=\frac{0.9}{1.6 \times 74.5}=\frac{0.9}{119.2}$ $=7.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
J and K-CET-2014
Current Electricity
152009
Two bulbs operating on standard voltage 110 volt have resistances in the ratio $9: 16$. The ratio of brightness of light from them is
1 $9: 16$
2 $16: 9$
3 $3: 4$
4 $4: 3$
Explanation:
B V $=110 \mathrm{~V}$ Ratio of resistance $=9: 16$ Brightness depend on current $I=\frac{P}{V}$ $I=\frac{V^{2} / R}{V}$ $I=\frac{V}{R}$ $\therefore \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{16}{9} \quad(\because$ voltage same $)$
J and K-CET-2014
Current Electricity
152011
In a circuit, 3 resistors of resistances $1.2 \Omega, 2 \Omega$, $3 \Omega$ are connected in parallel. The value of equivalent resistance is
1 Less than $1.2 \Omega$
2 Greater than $1.2 \Omega$
3 Between $1.2 \Omega$ and $2 \Omega$
4 Between $2 \Omega$ and $3 \Omega$
Explanation:
A The three resistors are $1.2 \Omega, 2 \Omega, 3 \Omega$ Apply the formula of parallel- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{1.2}+\frac{1}{2}+\frac{1}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{10+6+4}{12}=\frac{20}{12}$ $\mathrm{R}_{\mathrm{eq}}=12 / 20=0.6 \Omega$ Hence, value of equivalent resistance is less than $1.2 \Omega$.
152007
An immersion heater of $1 \mathrm{~kW}$ will take the time for increasing the temperature of 1 litre of water through $10^{\circ} \mathrm{C}$ equal to
1 $10 \mathrm{~s}$
2 $24 \mathrm{~s}$
3 $42 \mathrm{~s}$
4 $100 \mathrm{~s}$
Explanation:
C Given, Mass, $\mathrm{m}=$ mass in gram $=10$ litre $=10^{3} \text { gram }$ Specific heat for water $=\mathrm{s}=4.186 \approx 4.2 \mathrm{~J} / \mathrm{gram} /{ }^{\circ} \mathrm{C}$ Temperature difference, $\Delta \mathrm{T}=10^{\circ} \mathrm{C}$ Power, $\mathrm{P}=1 \mathrm{~kW}=1000$ watt $=1000 \mathrm{~J} / \mathrm{sec}$ $\because$ Time, $\mathrm{t}=\frac{\mathrm{ms} \Delta \mathrm{T}}{\mathrm{P}}=\frac{10^{3} \times 4.186 \times 10}{1000}$ $=\frac{4.2 \times 10 \times 10^{3}}{10^{3}}$ $\mathrm{t}=42 \mathrm{sec}$.
J and K CET- 1997
Current Electricity
152008
A silver wire has a resistance of $1.6 \Omega$ at 25.5 ${ }^{\circ} \mathrm{C}$ and a resistance of $2.5 \Omega$ at $100^{\circ} \mathrm{C}$, then temperature coefficient of resistivity of silver is
1 $5.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
2 $7.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
3 $11.75 \times 10^{-2}{ }^{\circ} \mathrm{C}$
4 $15.5 \times 10^{-3}{ }^{\circ} \mathrm{C}$
Explanation:
B $\mathrm{R}_{1}=1.6 \Omega$ $\mathrm{R}_{2}=2.5 \Omega$ $\mathrm{T}_{1}=25.5^{\circ} \mathrm{C}$ $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ Temperature coefficient of silver $(\alpha)=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}$ $=\frac{2.5-1.6}{1.6(100-25.5)}=\frac{0.9}{1.6 \times 74.5}=\frac{0.9}{119.2}$ $=7.55 \times 10^{-3}{ }^{\circ} \mathrm{C}$
J and K-CET-2014
Current Electricity
152009
Two bulbs operating on standard voltage 110 volt have resistances in the ratio $9: 16$. The ratio of brightness of light from them is
1 $9: 16$
2 $16: 9$
3 $3: 4$
4 $4: 3$
Explanation:
B V $=110 \mathrm{~V}$ Ratio of resistance $=9: 16$ Brightness depend on current $I=\frac{P}{V}$ $I=\frac{V^{2} / R}{V}$ $I=\frac{V}{R}$ $\therefore \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{16}{9} \quad(\because$ voltage same $)$
J and K-CET-2014
Current Electricity
152011
In a circuit, 3 resistors of resistances $1.2 \Omega, 2 \Omega$, $3 \Omega$ are connected in parallel. The value of equivalent resistance is
1 Less than $1.2 \Omega$
2 Greater than $1.2 \Omega$
3 Between $1.2 \Omega$ and $2 \Omega$
4 Between $2 \Omega$ and $3 \Omega$
Explanation:
A The three resistors are $1.2 \Omega, 2 \Omega, 3 \Omega$ Apply the formula of parallel- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{1.2}+\frac{1}{2}+\frac{1}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{10+6+4}{12}=\frac{20}{12}$ $\mathrm{R}_{\mathrm{eq}}=12 / 20=0.6 \Omega$ Hence, value of equivalent resistance is less than $1.2 \Omega$.