151749
The current passing through the AB branch of the circuit shown in the figure is.
1 $\frac{10}{17} \mathrm{~A}$
2 $\frac{4}{17} \mathrm{~A}$
3 $\frac{16}{17} \mathrm{~A}$
4 $\frac{6}{17} \mathrm{~A}$
Explanation:
B Applying Kirchhoff's voltage law in loop CDBA - $12-4 \mathrm{I}_{1}-12+6 \mathrm{I}_{2}=0$ $-4 \mathrm{I}_{1}+6 \mathrm{I}_{2}=0$ Again applying Kirchhoff's voltage law in loop ABEF - $12-6 \mathrm{I}_{2}-18\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)=0$ $-18 \mathrm{I}_{1}-24 \mathrm{I}_{2}=-12$ $3 \mathrm{I}_{1}+4 \mathrm{I}_{2}=2$ Solving equation (i) \& (ii), $\mathrm{I}_{2}=\frac{4}{17} \mathrm{~A}$ So, current in branch $\mathrm{AB}$ is $=\frac{4}{17} \mathrm{~A}$
TS EAMCET (Medical)-02.05.2018
Current Electricity
151745
Assertion: For a conductor resistivity increases with increase in temperature. Reason: Since $\rho=\frac{m}{n^{2} \tau}$, when temperature increases the random motion of free electrons increases and vibration of ions increases which decreases $\tau$.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When temperature increases, free electrons collide frequently which results conductivity decreases and resistivity increases hence, both statements are true and reason is correct explanation of assertion.
AIIMS-26.05.2019(E) Shift-2
Current Electricity
151751
A copper wire with a cross-section area of $2 \times 10^{-6} \mathrm{~m}^{2}$ has a free electron density equal to $5 \times 10^{22} / \mathrm{cm}^{3}$. If this wire carries a current of $16 \mathrm{~A}$, the drift velocity of the electron is
1 $1 \mathrm{~m} / \mathrm{s}$
2 $0.1 \mathrm{~m} / \mathrm{s}$
3 $0.01 \mathrm{~m} / \mathrm{s}$
4 $0.001 \mathrm{~m} / \mathrm{s}$
5 $0.0001 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Copper wire cross-sectional area, $A=2 \times 10^{-6} \mathrm{~m}^{2}$ Free electron density, $\mathrm{n}=5 \times 10^{22} / \mathrm{cm}^{3}=5 \times 10^{28} / \mathrm{m}^{3}$ Current (I) $=16$ A The drift velocity, $v_{d}=\frac{I}{n \times e \times A}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{1.6 \times 10^{4}}$ $\mathrm{v}_{\mathrm{d}}=\frac{1}{10^{3}}$ $\mathrm{v}_{\mathrm{d}}=0.001 \mathrm{~m} / \mathrm{s}$
Kerala CEE - 2017
Current Electricity
151752
In a neon discharge tube $2.9 \times 10^{18} \mathrm{Ne}^{+}$ions move to right each second while $1.2 \times 10^{18}$ electrons move to the left per second, electrons charge is $1.6 \times 10^{-19} \mathrm{C}$. The current in the discharge tube is
1 $1 \mathrm{~A}$, towards right
2 $0.66 \mathrm{~A}$, towards right
3 $0.66 \mathrm{~A}$, towards left
4 zero
Explanation:
B Given, Charge on neon $=2.9 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.464 \mathrm{C}$ Charge on electrons $=1.2 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.192 \mathrm{C}$ We know that, $\text { Current }=\frac{\text { Total charge }}{\text { Time }}=\frac{0.464+0.192}{1}$ $=0.656=0.66 \mathrm{~A}$ So, the current in the discharge tube is $0.66 \mathrm{~A}$ along the direction in which positive charges flow. Hence, The current in the discharge tube flows towards right.
Manipal UGET-2011
Current Electricity
151755
If in the below circuit, key $(k)$ is pressed, what will be the effect on the hanging portions $A B$ and $C D$ of the wire?
1 Both will attract each other
2 Both will repel each other
3 They will break
4 None of the above
Explanation:
B The given circuit is shown as - As the current in the portions $\mathrm{AB}$ and $\mathrm{CD}$ are in opposite direction, so both will repel each other.
151749
The current passing through the AB branch of the circuit shown in the figure is.
1 $\frac{10}{17} \mathrm{~A}$
2 $\frac{4}{17} \mathrm{~A}$
3 $\frac{16}{17} \mathrm{~A}$
4 $\frac{6}{17} \mathrm{~A}$
Explanation:
B Applying Kirchhoff's voltage law in loop CDBA - $12-4 \mathrm{I}_{1}-12+6 \mathrm{I}_{2}=0$ $-4 \mathrm{I}_{1}+6 \mathrm{I}_{2}=0$ Again applying Kirchhoff's voltage law in loop ABEF - $12-6 \mathrm{I}_{2}-18\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)=0$ $-18 \mathrm{I}_{1}-24 \mathrm{I}_{2}=-12$ $3 \mathrm{I}_{1}+4 \mathrm{I}_{2}=2$ Solving equation (i) \& (ii), $\mathrm{I}_{2}=\frac{4}{17} \mathrm{~A}$ So, current in branch $\mathrm{AB}$ is $=\frac{4}{17} \mathrm{~A}$
TS EAMCET (Medical)-02.05.2018
Current Electricity
151745
Assertion: For a conductor resistivity increases with increase in temperature. Reason: Since $\rho=\frac{m}{n^{2} \tau}$, when temperature increases the random motion of free electrons increases and vibration of ions increases which decreases $\tau$.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When temperature increases, free electrons collide frequently which results conductivity decreases and resistivity increases hence, both statements are true and reason is correct explanation of assertion.
AIIMS-26.05.2019(E) Shift-2
Current Electricity
151751
A copper wire with a cross-section area of $2 \times 10^{-6} \mathrm{~m}^{2}$ has a free electron density equal to $5 \times 10^{22} / \mathrm{cm}^{3}$. If this wire carries a current of $16 \mathrm{~A}$, the drift velocity of the electron is
1 $1 \mathrm{~m} / \mathrm{s}$
2 $0.1 \mathrm{~m} / \mathrm{s}$
3 $0.01 \mathrm{~m} / \mathrm{s}$
4 $0.001 \mathrm{~m} / \mathrm{s}$
5 $0.0001 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Copper wire cross-sectional area, $A=2 \times 10^{-6} \mathrm{~m}^{2}$ Free electron density, $\mathrm{n}=5 \times 10^{22} / \mathrm{cm}^{3}=5 \times 10^{28} / \mathrm{m}^{3}$ Current (I) $=16$ A The drift velocity, $v_{d}=\frac{I}{n \times e \times A}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{1.6 \times 10^{4}}$ $\mathrm{v}_{\mathrm{d}}=\frac{1}{10^{3}}$ $\mathrm{v}_{\mathrm{d}}=0.001 \mathrm{~m} / \mathrm{s}$
Kerala CEE - 2017
Current Electricity
151752
In a neon discharge tube $2.9 \times 10^{18} \mathrm{Ne}^{+}$ions move to right each second while $1.2 \times 10^{18}$ electrons move to the left per second, electrons charge is $1.6 \times 10^{-19} \mathrm{C}$. The current in the discharge tube is
1 $1 \mathrm{~A}$, towards right
2 $0.66 \mathrm{~A}$, towards right
3 $0.66 \mathrm{~A}$, towards left
4 zero
Explanation:
B Given, Charge on neon $=2.9 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.464 \mathrm{C}$ Charge on electrons $=1.2 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.192 \mathrm{C}$ We know that, $\text { Current }=\frac{\text { Total charge }}{\text { Time }}=\frac{0.464+0.192}{1}$ $=0.656=0.66 \mathrm{~A}$ So, the current in the discharge tube is $0.66 \mathrm{~A}$ along the direction in which positive charges flow. Hence, The current in the discharge tube flows towards right.
Manipal UGET-2011
Current Electricity
151755
If in the below circuit, key $(k)$ is pressed, what will be the effect on the hanging portions $A B$ and $C D$ of the wire?
1 Both will attract each other
2 Both will repel each other
3 They will break
4 None of the above
Explanation:
B The given circuit is shown as - As the current in the portions $\mathrm{AB}$ and $\mathrm{CD}$ are in opposite direction, so both will repel each other.
151749
The current passing through the AB branch of the circuit shown in the figure is.
1 $\frac{10}{17} \mathrm{~A}$
2 $\frac{4}{17} \mathrm{~A}$
3 $\frac{16}{17} \mathrm{~A}$
4 $\frac{6}{17} \mathrm{~A}$
Explanation:
B Applying Kirchhoff's voltage law in loop CDBA - $12-4 \mathrm{I}_{1}-12+6 \mathrm{I}_{2}=0$ $-4 \mathrm{I}_{1}+6 \mathrm{I}_{2}=0$ Again applying Kirchhoff's voltage law in loop ABEF - $12-6 \mathrm{I}_{2}-18\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)=0$ $-18 \mathrm{I}_{1}-24 \mathrm{I}_{2}=-12$ $3 \mathrm{I}_{1}+4 \mathrm{I}_{2}=2$ Solving equation (i) \& (ii), $\mathrm{I}_{2}=\frac{4}{17} \mathrm{~A}$ So, current in branch $\mathrm{AB}$ is $=\frac{4}{17} \mathrm{~A}$
TS EAMCET (Medical)-02.05.2018
Current Electricity
151745
Assertion: For a conductor resistivity increases with increase in temperature. Reason: Since $\rho=\frac{m}{n^{2} \tau}$, when temperature increases the random motion of free electrons increases and vibration of ions increases which decreases $\tau$.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When temperature increases, free electrons collide frequently which results conductivity decreases and resistivity increases hence, both statements are true and reason is correct explanation of assertion.
AIIMS-26.05.2019(E) Shift-2
Current Electricity
151751
A copper wire with a cross-section area of $2 \times 10^{-6} \mathrm{~m}^{2}$ has a free electron density equal to $5 \times 10^{22} / \mathrm{cm}^{3}$. If this wire carries a current of $16 \mathrm{~A}$, the drift velocity of the electron is
1 $1 \mathrm{~m} / \mathrm{s}$
2 $0.1 \mathrm{~m} / \mathrm{s}$
3 $0.01 \mathrm{~m} / \mathrm{s}$
4 $0.001 \mathrm{~m} / \mathrm{s}$
5 $0.0001 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Copper wire cross-sectional area, $A=2 \times 10^{-6} \mathrm{~m}^{2}$ Free electron density, $\mathrm{n}=5 \times 10^{22} / \mathrm{cm}^{3}=5 \times 10^{28} / \mathrm{m}^{3}$ Current (I) $=16$ A The drift velocity, $v_{d}=\frac{I}{n \times e \times A}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{1.6 \times 10^{4}}$ $\mathrm{v}_{\mathrm{d}}=\frac{1}{10^{3}}$ $\mathrm{v}_{\mathrm{d}}=0.001 \mathrm{~m} / \mathrm{s}$
Kerala CEE - 2017
Current Electricity
151752
In a neon discharge tube $2.9 \times 10^{18} \mathrm{Ne}^{+}$ions move to right each second while $1.2 \times 10^{18}$ electrons move to the left per second, electrons charge is $1.6 \times 10^{-19} \mathrm{C}$. The current in the discharge tube is
1 $1 \mathrm{~A}$, towards right
2 $0.66 \mathrm{~A}$, towards right
3 $0.66 \mathrm{~A}$, towards left
4 zero
Explanation:
B Given, Charge on neon $=2.9 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.464 \mathrm{C}$ Charge on electrons $=1.2 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.192 \mathrm{C}$ We know that, $\text { Current }=\frac{\text { Total charge }}{\text { Time }}=\frac{0.464+0.192}{1}$ $=0.656=0.66 \mathrm{~A}$ So, the current in the discharge tube is $0.66 \mathrm{~A}$ along the direction in which positive charges flow. Hence, The current in the discharge tube flows towards right.
Manipal UGET-2011
Current Electricity
151755
If in the below circuit, key $(k)$ is pressed, what will be the effect on the hanging portions $A B$ and $C D$ of the wire?
1 Both will attract each other
2 Both will repel each other
3 They will break
4 None of the above
Explanation:
B The given circuit is shown as - As the current in the portions $\mathrm{AB}$ and $\mathrm{CD}$ are in opposite direction, so both will repel each other.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
151749
The current passing through the AB branch of the circuit shown in the figure is.
1 $\frac{10}{17} \mathrm{~A}$
2 $\frac{4}{17} \mathrm{~A}$
3 $\frac{16}{17} \mathrm{~A}$
4 $\frac{6}{17} \mathrm{~A}$
Explanation:
B Applying Kirchhoff's voltage law in loop CDBA - $12-4 \mathrm{I}_{1}-12+6 \mathrm{I}_{2}=0$ $-4 \mathrm{I}_{1}+6 \mathrm{I}_{2}=0$ Again applying Kirchhoff's voltage law in loop ABEF - $12-6 \mathrm{I}_{2}-18\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)=0$ $-18 \mathrm{I}_{1}-24 \mathrm{I}_{2}=-12$ $3 \mathrm{I}_{1}+4 \mathrm{I}_{2}=2$ Solving equation (i) \& (ii), $\mathrm{I}_{2}=\frac{4}{17} \mathrm{~A}$ So, current in branch $\mathrm{AB}$ is $=\frac{4}{17} \mathrm{~A}$
TS EAMCET (Medical)-02.05.2018
Current Electricity
151745
Assertion: For a conductor resistivity increases with increase in temperature. Reason: Since $\rho=\frac{m}{n^{2} \tau}$, when temperature increases the random motion of free electrons increases and vibration of ions increases which decreases $\tau$.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When temperature increases, free electrons collide frequently which results conductivity decreases and resistivity increases hence, both statements are true and reason is correct explanation of assertion.
AIIMS-26.05.2019(E) Shift-2
Current Electricity
151751
A copper wire with a cross-section area of $2 \times 10^{-6} \mathrm{~m}^{2}$ has a free electron density equal to $5 \times 10^{22} / \mathrm{cm}^{3}$. If this wire carries a current of $16 \mathrm{~A}$, the drift velocity of the electron is
1 $1 \mathrm{~m} / \mathrm{s}$
2 $0.1 \mathrm{~m} / \mathrm{s}$
3 $0.01 \mathrm{~m} / \mathrm{s}$
4 $0.001 \mathrm{~m} / \mathrm{s}$
5 $0.0001 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Copper wire cross-sectional area, $A=2 \times 10^{-6} \mathrm{~m}^{2}$ Free electron density, $\mathrm{n}=5 \times 10^{22} / \mathrm{cm}^{3}=5 \times 10^{28} / \mathrm{m}^{3}$ Current (I) $=16$ A The drift velocity, $v_{d}=\frac{I}{n \times e \times A}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{1.6 \times 10^{4}}$ $\mathrm{v}_{\mathrm{d}}=\frac{1}{10^{3}}$ $\mathrm{v}_{\mathrm{d}}=0.001 \mathrm{~m} / \mathrm{s}$
Kerala CEE - 2017
Current Electricity
151752
In a neon discharge tube $2.9 \times 10^{18} \mathrm{Ne}^{+}$ions move to right each second while $1.2 \times 10^{18}$ electrons move to the left per second, electrons charge is $1.6 \times 10^{-19} \mathrm{C}$. The current in the discharge tube is
1 $1 \mathrm{~A}$, towards right
2 $0.66 \mathrm{~A}$, towards right
3 $0.66 \mathrm{~A}$, towards left
4 zero
Explanation:
B Given, Charge on neon $=2.9 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.464 \mathrm{C}$ Charge on electrons $=1.2 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.192 \mathrm{C}$ We know that, $\text { Current }=\frac{\text { Total charge }}{\text { Time }}=\frac{0.464+0.192}{1}$ $=0.656=0.66 \mathrm{~A}$ So, the current in the discharge tube is $0.66 \mathrm{~A}$ along the direction in which positive charges flow. Hence, The current in the discharge tube flows towards right.
Manipal UGET-2011
Current Electricity
151755
If in the below circuit, key $(k)$ is pressed, what will be the effect on the hanging portions $A B$ and $C D$ of the wire?
1 Both will attract each other
2 Both will repel each other
3 They will break
4 None of the above
Explanation:
B The given circuit is shown as - As the current in the portions $\mathrm{AB}$ and $\mathrm{CD}$ are in opposite direction, so both will repel each other.
151749
The current passing through the AB branch of the circuit shown in the figure is.
1 $\frac{10}{17} \mathrm{~A}$
2 $\frac{4}{17} \mathrm{~A}$
3 $\frac{16}{17} \mathrm{~A}$
4 $\frac{6}{17} \mathrm{~A}$
Explanation:
B Applying Kirchhoff's voltage law in loop CDBA - $12-4 \mathrm{I}_{1}-12+6 \mathrm{I}_{2}=0$ $-4 \mathrm{I}_{1}+6 \mathrm{I}_{2}=0$ Again applying Kirchhoff's voltage law in loop ABEF - $12-6 \mathrm{I}_{2}-18\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)=0$ $-18 \mathrm{I}_{1}-24 \mathrm{I}_{2}=-12$ $3 \mathrm{I}_{1}+4 \mathrm{I}_{2}=2$ Solving equation (i) \& (ii), $\mathrm{I}_{2}=\frac{4}{17} \mathrm{~A}$ So, current in branch $\mathrm{AB}$ is $=\frac{4}{17} \mathrm{~A}$
TS EAMCET (Medical)-02.05.2018
Current Electricity
151745
Assertion: For a conductor resistivity increases with increase in temperature. Reason: Since $\rho=\frac{m}{n^{2} \tau}$, when temperature increases the random motion of free electrons increases and vibration of ions increases which decreases $\tau$.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A When temperature increases, free electrons collide frequently which results conductivity decreases and resistivity increases hence, both statements are true and reason is correct explanation of assertion.
AIIMS-26.05.2019(E) Shift-2
Current Electricity
151751
A copper wire with a cross-section area of $2 \times 10^{-6} \mathrm{~m}^{2}$ has a free electron density equal to $5 \times 10^{22} / \mathrm{cm}^{3}$. If this wire carries a current of $16 \mathrm{~A}$, the drift velocity of the electron is
1 $1 \mathrm{~m} / \mathrm{s}$
2 $0.1 \mathrm{~m} / \mathrm{s}$
3 $0.01 \mathrm{~m} / \mathrm{s}$
4 $0.001 \mathrm{~m} / \mathrm{s}$
5 $0.0001 \mathrm{~m} / \mathrm{s}$
Explanation:
D Given, Copper wire cross-sectional area, $A=2 \times 10^{-6} \mathrm{~m}^{2}$ Free electron density, $\mathrm{n}=5 \times 10^{22} / \mathrm{cm}^{3}=5 \times 10^{28} / \mathrm{m}^{3}$ Current (I) $=16$ A The drift velocity, $v_{d}=\frac{I}{n \times e \times A}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}$ $\mathrm{v}_{\mathrm{d}}=\frac{16}{1.6 \times 10^{4}}$ $\mathrm{v}_{\mathrm{d}}=\frac{1}{10^{3}}$ $\mathrm{v}_{\mathrm{d}}=0.001 \mathrm{~m} / \mathrm{s}$
Kerala CEE - 2017
Current Electricity
151752
In a neon discharge tube $2.9 \times 10^{18} \mathrm{Ne}^{+}$ions move to right each second while $1.2 \times 10^{18}$ electrons move to the left per second, electrons charge is $1.6 \times 10^{-19} \mathrm{C}$. The current in the discharge tube is
1 $1 \mathrm{~A}$, towards right
2 $0.66 \mathrm{~A}$, towards right
3 $0.66 \mathrm{~A}$, towards left
4 zero
Explanation:
B Given, Charge on neon $=2.9 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.464 \mathrm{C}$ Charge on electrons $=1.2 \times 10^{18} \times 1.6 \times 10^{-19} \mathrm{C}$ $=0.192 \mathrm{C}$ We know that, $\text { Current }=\frac{\text { Total charge }}{\text { Time }}=\frac{0.464+0.192}{1}$ $=0.656=0.66 \mathrm{~A}$ So, the current in the discharge tube is $0.66 \mathrm{~A}$ along the direction in which positive charges flow. Hence, The current in the discharge tube flows towards right.
Manipal UGET-2011
Current Electricity
151755
If in the below circuit, key $(k)$ is pressed, what will be the effect on the hanging portions $A B$ and $C D$ of the wire?
1 Both will attract each other
2 Both will repel each other
3 They will break
4 None of the above
Explanation:
B The given circuit is shown as - As the current in the portions $\mathrm{AB}$ and $\mathrm{CD}$ are in opposite direction, so both will repel each other.
