151787
In the circuit shown below, the switch is kept in position $a$ for a long time and is then thrown to position $b$. The amplitude of this resulting oscillating current is given by
1 $\mathrm{E} \sqrt{\mathrm{L} / \mathrm{C}}$
2 $E / R$
3 infinity
4 $\mathrm{E} \sqrt{\mathrm{C} / \mathrm{L}}$
Explanation:
D When switch is connected in position a, then capacitor will be charged. And charge on capacitor is - $\mathrm{q}=\mathrm{C} . \mathrm{E} \quad(\mathrm{E}=\mathrm{V}) \quad \text { Potential in battery }$ We know that, Energy store in capacitor $(\mathrm{U})=\frac{\mathrm{q}^{2}}{2 \mathrm{C}}$ Again, When switch is position $b$ then circuit behaves as L.C combination. and amplitude of current in LC circuit is $\mathrm{I}_{\mathrm{o}}$ Hence, apply energy conservation, Maximum electrical energy $=$ maximum magnetic energy $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \mathrm{LI}_{\mathrm{o}}^{2}$ $\frac{\mathrm{C}^{2} \mathrm{E}^{2}}{\mathrm{C}}=\mathrm{LI}_{\mathrm{o}}^{2}$ $\mathrm{I}_{\mathrm{o}}=\mathrm{E} \sqrt{\mathrm{C} / \mathrm{L}}$
WB JEE 2015
Current Electricity
151788
An electric cell of emf $E$ is connected across a copper wire of diameter $\mathrm{d}$ and length $l$. The drift velocity of electrons in the wire is $v_{d}$. If the length of the wire is changed to $2 l$, the new drift velocity of electrons in the copper wire will be
1 $v_{d}$
2 $2 v_{d}$
3 $v_{d} / 2$
4 $\mathrm{v}_{\mathrm{d}} / 4$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ Where, $n=$ no. of density $E =I . R$ $I =\frac{E}{R}$ $R =\frac{\rho L}{A}$ From equation (i), $I=\frac{E}{\frac{\rho L}{A}}=n e A v_{d}$ $\frac{E A}{\rho . L}=n e A v_{d}$ $v_{d}=\frac{E}{n e \rho L}$ $v_{d} \propto \frac{1}{L}$ When, length of wire charged to $2 l$, the new drift velocity. $\frac{\mathrm{v}_{\mathrm{d}}}{\mathrm{v}_{\mathrm{d}}^{\prime}}=\frac{2 l}{l}$ $\mathrm{v}_{\mathrm{d}}^{\prime}=\frac{\mathrm{v}_{\mathrm{d}}}{2}$
WB JEE 2013
Current Electricity
151793
The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the two rods of these materials of same thickness is same. then the ratio of the length of these rods will be:
1 $5: 3$
2 $3: 5$
3 $9: 25$
4 $25: 9$
Explanation:
A The thermal resistance of rod of length $l$, area of cross - section A and thermal conductivity $\mathrm{K}$, $\mathrm{R}=\frac{l}{\mathrm{KA}}$ Given, Thermal resistance of two rods are same, $\therefore \quad\left(\mathrm{R}_{1}=\mathrm{R}_{2}\right)$ $\frac{l_{1}}{\mathrm{~K}_{1} \mathrm{~A}_{1}}=\frac{l_{2}}{\mathrm{~K}_{2} \mathrm{~A}_{2}}$ $\frac{l_{1}}{\mathrm{~K}_{1}}=\frac{l_{2}}{\mathrm{~K}_{2}}$ $\left(\because \mathrm{A}_{1}=\mathrm{A}_{2}\right)$ $\frac{l_{1}}{l_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{5}{3}$ $\therefore \quad l_{1}: l_{2}=5: 3$
MP PET -2013
Current Electricity
151794
At room temperature, copper has free electron density of $8.4 \times 10^{28} \mathrm{~m}^{-3}$. The electron drift velocity in a copper conductor of crosssectional area of $10^{-6} \mathrm{~m}^{2}$ and carrying a current of $5.4 \mathrm{~A}$, will be
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
151787
In the circuit shown below, the switch is kept in position $a$ for a long time and is then thrown to position $b$. The amplitude of this resulting oscillating current is given by
1 $\mathrm{E} \sqrt{\mathrm{L} / \mathrm{C}}$
2 $E / R$
3 infinity
4 $\mathrm{E} \sqrt{\mathrm{C} / \mathrm{L}}$
Explanation:
D When switch is connected in position a, then capacitor will be charged. And charge on capacitor is - $\mathrm{q}=\mathrm{C} . \mathrm{E} \quad(\mathrm{E}=\mathrm{V}) \quad \text { Potential in battery }$ We know that, Energy store in capacitor $(\mathrm{U})=\frac{\mathrm{q}^{2}}{2 \mathrm{C}}$ Again, When switch is position $b$ then circuit behaves as L.C combination. and amplitude of current in LC circuit is $\mathrm{I}_{\mathrm{o}}$ Hence, apply energy conservation, Maximum electrical energy $=$ maximum magnetic energy $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \mathrm{LI}_{\mathrm{o}}^{2}$ $\frac{\mathrm{C}^{2} \mathrm{E}^{2}}{\mathrm{C}}=\mathrm{LI}_{\mathrm{o}}^{2}$ $\mathrm{I}_{\mathrm{o}}=\mathrm{E} \sqrt{\mathrm{C} / \mathrm{L}}$
WB JEE 2015
Current Electricity
151788
An electric cell of emf $E$ is connected across a copper wire of diameter $\mathrm{d}$ and length $l$. The drift velocity of electrons in the wire is $v_{d}$. If the length of the wire is changed to $2 l$, the new drift velocity of electrons in the copper wire will be
1 $v_{d}$
2 $2 v_{d}$
3 $v_{d} / 2$
4 $\mathrm{v}_{\mathrm{d}} / 4$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ Where, $n=$ no. of density $E =I . R$ $I =\frac{E}{R}$ $R =\frac{\rho L}{A}$ From equation (i), $I=\frac{E}{\frac{\rho L}{A}}=n e A v_{d}$ $\frac{E A}{\rho . L}=n e A v_{d}$ $v_{d}=\frac{E}{n e \rho L}$ $v_{d} \propto \frac{1}{L}$ When, length of wire charged to $2 l$, the new drift velocity. $\frac{\mathrm{v}_{\mathrm{d}}}{\mathrm{v}_{\mathrm{d}}^{\prime}}=\frac{2 l}{l}$ $\mathrm{v}_{\mathrm{d}}^{\prime}=\frac{\mathrm{v}_{\mathrm{d}}}{2}$
WB JEE 2013
Current Electricity
151793
The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the two rods of these materials of same thickness is same. then the ratio of the length of these rods will be:
1 $5: 3$
2 $3: 5$
3 $9: 25$
4 $25: 9$
Explanation:
A The thermal resistance of rod of length $l$, area of cross - section A and thermal conductivity $\mathrm{K}$, $\mathrm{R}=\frac{l}{\mathrm{KA}}$ Given, Thermal resistance of two rods are same, $\therefore \quad\left(\mathrm{R}_{1}=\mathrm{R}_{2}\right)$ $\frac{l_{1}}{\mathrm{~K}_{1} \mathrm{~A}_{1}}=\frac{l_{2}}{\mathrm{~K}_{2} \mathrm{~A}_{2}}$ $\frac{l_{1}}{\mathrm{~K}_{1}}=\frac{l_{2}}{\mathrm{~K}_{2}}$ $\left(\because \mathrm{A}_{1}=\mathrm{A}_{2}\right)$ $\frac{l_{1}}{l_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{5}{3}$ $\therefore \quad l_{1}: l_{2}=5: 3$
MP PET -2013
Current Electricity
151794
At room temperature, copper has free electron density of $8.4 \times 10^{28} \mathrm{~m}^{-3}$. The electron drift velocity in a copper conductor of crosssectional area of $10^{-6} \mathrm{~m}^{2}$ and carrying a current of $5.4 \mathrm{~A}$, will be
151787
In the circuit shown below, the switch is kept in position $a$ for a long time and is then thrown to position $b$. The amplitude of this resulting oscillating current is given by
1 $\mathrm{E} \sqrt{\mathrm{L} / \mathrm{C}}$
2 $E / R$
3 infinity
4 $\mathrm{E} \sqrt{\mathrm{C} / \mathrm{L}}$
Explanation:
D When switch is connected in position a, then capacitor will be charged. And charge on capacitor is - $\mathrm{q}=\mathrm{C} . \mathrm{E} \quad(\mathrm{E}=\mathrm{V}) \quad \text { Potential in battery }$ We know that, Energy store in capacitor $(\mathrm{U})=\frac{\mathrm{q}^{2}}{2 \mathrm{C}}$ Again, When switch is position $b$ then circuit behaves as L.C combination. and amplitude of current in LC circuit is $\mathrm{I}_{\mathrm{o}}$ Hence, apply energy conservation, Maximum electrical energy $=$ maximum magnetic energy $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \mathrm{LI}_{\mathrm{o}}^{2}$ $\frac{\mathrm{C}^{2} \mathrm{E}^{2}}{\mathrm{C}}=\mathrm{LI}_{\mathrm{o}}^{2}$ $\mathrm{I}_{\mathrm{o}}=\mathrm{E} \sqrt{\mathrm{C} / \mathrm{L}}$
WB JEE 2015
Current Electricity
151788
An electric cell of emf $E$ is connected across a copper wire of diameter $\mathrm{d}$ and length $l$. The drift velocity of electrons in the wire is $v_{d}$. If the length of the wire is changed to $2 l$, the new drift velocity of electrons in the copper wire will be
1 $v_{d}$
2 $2 v_{d}$
3 $v_{d} / 2$
4 $\mathrm{v}_{\mathrm{d}} / 4$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ Where, $n=$ no. of density $E =I . R$ $I =\frac{E}{R}$ $R =\frac{\rho L}{A}$ From equation (i), $I=\frac{E}{\frac{\rho L}{A}}=n e A v_{d}$ $\frac{E A}{\rho . L}=n e A v_{d}$ $v_{d}=\frac{E}{n e \rho L}$ $v_{d} \propto \frac{1}{L}$ When, length of wire charged to $2 l$, the new drift velocity. $\frac{\mathrm{v}_{\mathrm{d}}}{\mathrm{v}_{\mathrm{d}}^{\prime}}=\frac{2 l}{l}$ $\mathrm{v}_{\mathrm{d}}^{\prime}=\frac{\mathrm{v}_{\mathrm{d}}}{2}$
WB JEE 2013
Current Electricity
151793
The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the two rods of these materials of same thickness is same. then the ratio of the length of these rods will be:
1 $5: 3$
2 $3: 5$
3 $9: 25$
4 $25: 9$
Explanation:
A The thermal resistance of rod of length $l$, area of cross - section A and thermal conductivity $\mathrm{K}$, $\mathrm{R}=\frac{l}{\mathrm{KA}}$ Given, Thermal resistance of two rods are same, $\therefore \quad\left(\mathrm{R}_{1}=\mathrm{R}_{2}\right)$ $\frac{l_{1}}{\mathrm{~K}_{1} \mathrm{~A}_{1}}=\frac{l_{2}}{\mathrm{~K}_{2} \mathrm{~A}_{2}}$ $\frac{l_{1}}{\mathrm{~K}_{1}}=\frac{l_{2}}{\mathrm{~K}_{2}}$ $\left(\because \mathrm{A}_{1}=\mathrm{A}_{2}\right)$ $\frac{l_{1}}{l_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{5}{3}$ $\therefore \quad l_{1}: l_{2}=5: 3$
MP PET -2013
Current Electricity
151794
At room temperature, copper has free electron density of $8.4 \times 10^{28} \mathrm{~m}^{-3}$. The electron drift velocity in a copper conductor of crosssectional area of $10^{-6} \mathrm{~m}^{2}$ and carrying a current of $5.4 \mathrm{~A}$, will be
151787
In the circuit shown below, the switch is kept in position $a$ for a long time and is then thrown to position $b$. The amplitude of this resulting oscillating current is given by
1 $\mathrm{E} \sqrt{\mathrm{L} / \mathrm{C}}$
2 $E / R$
3 infinity
4 $\mathrm{E} \sqrt{\mathrm{C} / \mathrm{L}}$
Explanation:
D When switch is connected in position a, then capacitor will be charged. And charge on capacitor is - $\mathrm{q}=\mathrm{C} . \mathrm{E} \quad(\mathrm{E}=\mathrm{V}) \quad \text { Potential in battery }$ We know that, Energy store in capacitor $(\mathrm{U})=\frac{\mathrm{q}^{2}}{2 \mathrm{C}}$ Again, When switch is position $b$ then circuit behaves as L.C combination. and amplitude of current in LC circuit is $\mathrm{I}_{\mathrm{o}}$ Hence, apply energy conservation, Maximum electrical energy $=$ maximum magnetic energy $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \mathrm{LI}_{\mathrm{o}}^{2}$ $\frac{\mathrm{C}^{2} \mathrm{E}^{2}}{\mathrm{C}}=\mathrm{LI}_{\mathrm{o}}^{2}$ $\mathrm{I}_{\mathrm{o}}=\mathrm{E} \sqrt{\mathrm{C} / \mathrm{L}}$
WB JEE 2015
Current Electricity
151788
An electric cell of emf $E$ is connected across a copper wire of diameter $\mathrm{d}$ and length $l$. The drift velocity of electrons in the wire is $v_{d}$. If the length of the wire is changed to $2 l$, the new drift velocity of electrons in the copper wire will be
1 $v_{d}$
2 $2 v_{d}$
3 $v_{d} / 2$
4 $\mathrm{v}_{\mathrm{d}} / 4$
Explanation:
C We know that, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$ Where, $n=$ no. of density $E =I . R$ $I =\frac{E}{R}$ $R =\frac{\rho L}{A}$ From equation (i), $I=\frac{E}{\frac{\rho L}{A}}=n e A v_{d}$ $\frac{E A}{\rho . L}=n e A v_{d}$ $v_{d}=\frac{E}{n e \rho L}$ $v_{d} \propto \frac{1}{L}$ When, length of wire charged to $2 l$, the new drift velocity. $\frac{\mathrm{v}_{\mathrm{d}}}{\mathrm{v}_{\mathrm{d}}^{\prime}}=\frac{2 l}{l}$ $\mathrm{v}_{\mathrm{d}}^{\prime}=\frac{\mathrm{v}_{\mathrm{d}}}{2}$
WB JEE 2013
Current Electricity
151793
The ratio of the coefficient of thermal conductivity of two different materials is $5: 3$. If the thermal resistance of the two rods of these materials of same thickness is same. then the ratio of the length of these rods will be:
1 $5: 3$
2 $3: 5$
3 $9: 25$
4 $25: 9$
Explanation:
A The thermal resistance of rod of length $l$, area of cross - section A and thermal conductivity $\mathrm{K}$, $\mathrm{R}=\frac{l}{\mathrm{KA}}$ Given, Thermal resistance of two rods are same, $\therefore \quad\left(\mathrm{R}_{1}=\mathrm{R}_{2}\right)$ $\frac{l_{1}}{\mathrm{~K}_{1} \mathrm{~A}_{1}}=\frac{l_{2}}{\mathrm{~K}_{2} \mathrm{~A}_{2}}$ $\frac{l_{1}}{\mathrm{~K}_{1}}=\frac{l_{2}}{\mathrm{~K}_{2}}$ $\left(\because \mathrm{A}_{1}=\mathrm{A}_{2}\right)$ $\frac{l_{1}}{l_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{5}{3}$ $\therefore \quad l_{1}: l_{2}=5: 3$
MP PET -2013
Current Electricity
151794
At room temperature, copper has free electron density of $8.4 \times 10^{28} \mathrm{~m}^{-3}$. The electron drift velocity in a copper conductor of crosssectional area of $10^{-6} \mathrm{~m}^{2}$ and carrying a current of $5.4 \mathrm{~A}$, will be
