NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
151860
The current through $2 \Omega$ resistance for the given circuit is
1 $1.4 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 zero
4 $4.2 \mathrm{~A}$
Explanation:
C Applying KCL at junction A Current $\mathrm{I}_{1}$ is passing through the battery and through the $10 \Omega$ resistor and current $I_{2}$ is passing through $2 \Omega$ resistor. By applying $\mathrm{KCL}-$ $I_{1}=I_{2}+I_{1}$ $I_{2}=0$ Applying KCL at junction B, Current $I_{1}$ is passing through battery and through the $15 \Omega$ resistor and current $I_{2}$ is passing through $2 \Omega$ resistor. By applying $\mathrm{KCL}-$ $\mathrm{I}_{1}=\mathrm{I}_{2}+\mathrm{I}_{1}$ $\mathrm{I}_{2}=0$ So, current through $2 \Omega$ resistor is zero.
MHT-CET 2019
Current Electricity
151861
A wire has a length of $2 \mathrm{~m}$ and resistance of $10 \Omega$. It is connected in series with a resistance of $990 \Omega$ and a cell of e.m.f. $2 \mathrm{~V}$. The potential gradient along the wire will be
151862
The current -voltage graph for a given metallic wire at two different temperatures $T_{1}$ and $T_{2}$ is shown in the figure. The temperatures $T_{1}$ and $T_{2}$ are related as
1 $\mathrm{T}_{1}>\mathrm{T}_{2}$
2 $\mathrm{T}_{1}\lt\mathrm{T}_{2}$
3 $\mathrm{T}_{1}=\mathrm{T}_{2}$
4 $\mathrm{T}_{1}>2 \mathrm{~T}_{2}$
Explanation:
B According to Ohm's Law, V = IR $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ The slope of the given graph is the inverse of resistance. In the figure, $T_{2}$ has smaller slope and hence corresponding to higher resistance. Therefore, $\quad \mathrm{T}_{2}>\mathrm{T}_{1}$ or $\mathrm{T}_{1}\lt\mathrm{T}_{2}$
Assam CEE- 2019
Current Electricity
151863
The resistance of a platinum wire is $100 \Omega$ at $0^{\circ} \mathrm{C}$. If its temperature coefficient of resistance is $0.0045 /{ }^{\circ} \mathrm{C}$ then its resistance at $60^{\circ} \mathrm{C}$ temperature will be
1 $127 \Omega$
2 $73 \Omega$
3 $370 \Omega$
4 $2800 \Omega$
Explanation:
A Given, $\text { Resistance at } 0^{\circ} \mathrm{C} \text { - }$ $\mathrm{R}_{0}=100 \Omega$ Coefficient of resistance $(\alpha)=0.0045 /{ }^{\circ} \mathrm{C}$ Resistance at $60{ }^{\circ} \mathrm{C}$ -$ \mathrm{R}_{60^{\circ}}=\text { ? } We know that, $\mathrm{R}_{\mathrm{T}}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})$ $\mathrm{R}_{60^{\circ}}=100[(1+0.045 \times(60-0)]$ $\mathrm{R}_{60^{\circ}}=100(1+0.270)$ $\mathrm{R}_{60^{\circ}}=100 \times 1.27$ $\mathrm{R}_{60^{\circ}}=127 \Omega$
151860
The current through $2 \Omega$ resistance for the given circuit is
1 $1.4 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 zero
4 $4.2 \mathrm{~A}$
Explanation:
C Applying KCL at junction A Current $\mathrm{I}_{1}$ is passing through the battery and through the $10 \Omega$ resistor and current $I_{2}$ is passing through $2 \Omega$ resistor. By applying $\mathrm{KCL}-$ $I_{1}=I_{2}+I_{1}$ $I_{2}=0$ Applying KCL at junction B, Current $I_{1}$ is passing through battery and through the $15 \Omega$ resistor and current $I_{2}$ is passing through $2 \Omega$ resistor. By applying $\mathrm{KCL}-$ $\mathrm{I}_{1}=\mathrm{I}_{2}+\mathrm{I}_{1}$ $\mathrm{I}_{2}=0$ So, current through $2 \Omega$ resistor is zero.
MHT-CET 2019
Current Electricity
151861
A wire has a length of $2 \mathrm{~m}$ and resistance of $10 \Omega$. It is connected in series with a resistance of $990 \Omega$ and a cell of e.m.f. $2 \mathrm{~V}$. The potential gradient along the wire will be
151862
The current -voltage graph for a given metallic wire at two different temperatures $T_{1}$ and $T_{2}$ is shown in the figure. The temperatures $T_{1}$ and $T_{2}$ are related as
1 $\mathrm{T}_{1}>\mathrm{T}_{2}$
2 $\mathrm{T}_{1}\lt\mathrm{T}_{2}$
3 $\mathrm{T}_{1}=\mathrm{T}_{2}$
4 $\mathrm{T}_{1}>2 \mathrm{~T}_{2}$
Explanation:
B According to Ohm's Law, V = IR $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ The slope of the given graph is the inverse of resistance. In the figure, $T_{2}$ has smaller slope and hence corresponding to higher resistance. Therefore, $\quad \mathrm{T}_{2}>\mathrm{T}_{1}$ or $\mathrm{T}_{1}\lt\mathrm{T}_{2}$
Assam CEE- 2019
Current Electricity
151863
The resistance of a platinum wire is $100 \Omega$ at $0^{\circ} \mathrm{C}$. If its temperature coefficient of resistance is $0.0045 /{ }^{\circ} \mathrm{C}$ then its resistance at $60^{\circ} \mathrm{C}$ temperature will be
1 $127 \Omega$
2 $73 \Omega$
3 $370 \Omega$
4 $2800 \Omega$
Explanation:
A Given, $\text { Resistance at } 0^{\circ} \mathrm{C} \text { - }$ $\mathrm{R}_{0}=100 \Omega$ Coefficient of resistance $(\alpha)=0.0045 /{ }^{\circ} \mathrm{C}$ Resistance at $60{ }^{\circ} \mathrm{C}$ -$ \mathrm{R}_{60^{\circ}}=\text { ? } We know that, $\mathrm{R}_{\mathrm{T}}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})$ $\mathrm{R}_{60^{\circ}}=100[(1+0.045 \times(60-0)]$ $\mathrm{R}_{60^{\circ}}=100(1+0.270)$ $\mathrm{R}_{60^{\circ}}=100 \times 1.27$ $\mathrm{R}_{60^{\circ}}=127 \Omega$
151860
The current through $2 \Omega$ resistance for the given circuit is
1 $1.4 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 zero
4 $4.2 \mathrm{~A}$
Explanation:
C Applying KCL at junction A Current $\mathrm{I}_{1}$ is passing through the battery and through the $10 \Omega$ resistor and current $I_{2}$ is passing through $2 \Omega$ resistor. By applying $\mathrm{KCL}-$ $I_{1}=I_{2}+I_{1}$ $I_{2}=0$ Applying KCL at junction B, Current $I_{1}$ is passing through battery and through the $15 \Omega$ resistor and current $I_{2}$ is passing through $2 \Omega$ resistor. By applying $\mathrm{KCL}-$ $\mathrm{I}_{1}=\mathrm{I}_{2}+\mathrm{I}_{1}$ $\mathrm{I}_{2}=0$ So, current through $2 \Omega$ resistor is zero.
MHT-CET 2019
Current Electricity
151861
A wire has a length of $2 \mathrm{~m}$ and resistance of $10 \Omega$. It is connected in series with a resistance of $990 \Omega$ and a cell of e.m.f. $2 \mathrm{~V}$. The potential gradient along the wire will be
151862
The current -voltage graph for a given metallic wire at two different temperatures $T_{1}$ and $T_{2}$ is shown in the figure. The temperatures $T_{1}$ and $T_{2}$ are related as
1 $\mathrm{T}_{1}>\mathrm{T}_{2}$
2 $\mathrm{T}_{1}\lt\mathrm{T}_{2}$
3 $\mathrm{T}_{1}=\mathrm{T}_{2}$
4 $\mathrm{T}_{1}>2 \mathrm{~T}_{2}$
Explanation:
B According to Ohm's Law, V = IR $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ The slope of the given graph is the inverse of resistance. In the figure, $T_{2}$ has smaller slope and hence corresponding to higher resistance. Therefore, $\quad \mathrm{T}_{2}>\mathrm{T}_{1}$ or $\mathrm{T}_{1}\lt\mathrm{T}_{2}$
Assam CEE- 2019
Current Electricity
151863
The resistance of a platinum wire is $100 \Omega$ at $0^{\circ} \mathrm{C}$. If its temperature coefficient of resistance is $0.0045 /{ }^{\circ} \mathrm{C}$ then its resistance at $60^{\circ} \mathrm{C}$ temperature will be
1 $127 \Omega$
2 $73 \Omega$
3 $370 \Omega$
4 $2800 \Omega$
Explanation:
A Given, $\text { Resistance at } 0^{\circ} \mathrm{C} \text { - }$ $\mathrm{R}_{0}=100 \Omega$ Coefficient of resistance $(\alpha)=0.0045 /{ }^{\circ} \mathrm{C}$ Resistance at $60{ }^{\circ} \mathrm{C}$ -$ \mathrm{R}_{60^{\circ}}=\text { ? } We know that, $\mathrm{R}_{\mathrm{T}}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})$ $\mathrm{R}_{60^{\circ}}=100[(1+0.045 \times(60-0)]$ $\mathrm{R}_{60^{\circ}}=100(1+0.270)$ $\mathrm{R}_{60^{\circ}}=100 \times 1.27$ $\mathrm{R}_{60^{\circ}}=127 \Omega$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
151860
The current through $2 \Omega$ resistance for the given circuit is
1 $1.4 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 zero
4 $4.2 \mathrm{~A}$
Explanation:
C Applying KCL at junction A Current $\mathrm{I}_{1}$ is passing through the battery and through the $10 \Omega$ resistor and current $I_{2}$ is passing through $2 \Omega$ resistor. By applying $\mathrm{KCL}-$ $I_{1}=I_{2}+I_{1}$ $I_{2}=0$ Applying KCL at junction B, Current $I_{1}$ is passing through battery and through the $15 \Omega$ resistor and current $I_{2}$ is passing through $2 \Omega$ resistor. By applying $\mathrm{KCL}-$ $\mathrm{I}_{1}=\mathrm{I}_{2}+\mathrm{I}_{1}$ $\mathrm{I}_{2}=0$ So, current through $2 \Omega$ resistor is zero.
MHT-CET 2019
Current Electricity
151861
A wire has a length of $2 \mathrm{~m}$ and resistance of $10 \Omega$. It is connected in series with a resistance of $990 \Omega$ and a cell of e.m.f. $2 \mathrm{~V}$. The potential gradient along the wire will be
151862
The current -voltage graph for a given metallic wire at two different temperatures $T_{1}$ and $T_{2}$ is shown in the figure. The temperatures $T_{1}$ and $T_{2}$ are related as
1 $\mathrm{T}_{1}>\mathrm{T}_{2}$
2 $\mathrm{T}_{1}\lt\mathrm{T}_{2}$
3 $\mathrm{T}_{1}=\mathrm{T}_{2}$
4 $\mathrm{T}_{1}>2 \mathrm{~T}_{2}$
Explanation:
B According to Ohm's Law, V = IR $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ The slope of the given graph is the inverse of resistance. In the figure, $T_{2}$ has smaller slope and hence corresponding to higher resistance. Therefore, $\quad \mathrm{T}_{2}>\mathrm{T}_{1}$ or $\mathrm{T}_{1}\lt\mathrm{T}_{2}$
Assam CEE- 2019
Current Electricity
151863
The resistance of a platinum wire is $100 \Omega$ at $0^{\circ} \mathrm{C}$. If its temperature coefficient of resistance is $0.0045 /{ }^{\circ} \mathrm{C}$ then its resistance at $60^{\circ} \mathrm{C}$ temperature will be
1 $127 \Omega$
2 $73 \Omega$
3 $370 \Omega$
4 $2800 \Omega$
Explanation:
A Given, $\text { Resistance at } 0^{\circ} \mathrm{C} \text { - }$ $\mathrm{R}_{0}=100 \Omega$ Coefficient of resistance $(\alpha)=0.0045 /{ }^{\circ} \mathrm{C}$ Resistance at $60{ }^{\circ} \mathrm{C}$ -$ \mathrm{R}_{60^{\circ}}=\text { ? } We know that, $\mathrm{R}_{\mathrm{T}}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})$ $\mathrm{R}_{60^{\circ}}=100[(1+0.045 \times(60-0)]$ $\mathrm{R}_{60^{\circ}}=100(1+0.270)$ $\mathrm{R}_{60^{\circ}}=100 \times 1.27$ $\mathrm{R}_{60^{\circ}}=127 \Omega$
