151994
Variation of resistance of the conductor with temperature is as shown : The temperature coefficient $(\alpha)$ of the conductor is :
1 $\frac{\mathrm{R}_{0}}{\mathrm{~m}}$
2 $\mathrm{mR}_{0}$
3 $\mathrm{m}^{2} \mathrm{R}_{0}$
4 $\frac{\mathrm{m}}{\mathrm{R}_{0}}$
Explanation:
D From resistance with temperature, $\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})=\mathrm{R}_{0}+\mathrm{R}_{0} \alpha \Delta \mathrm{T}$ Where, $\mathrm{R}_{0}=$ Resistance at reference temperature $\alpha=$ Coefficient of temperature of conductor $\Delta \mathrm{T}=$ Chang in temperature It is a straight line having slope $=\alpha R_{0}=\mathrm{m}$ So, $\quad \alpha=\frac{\mathrm{m}}{\mathrm{R}_{0}}$
Karnataka CET-2016
Current Electricity
151995
Three resistances $2 \Omega, 3 \Omega$ and $4 \Omega$ are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be :
1 $5: 4: 3$
2 $6: 3: 2$
3 $4: 3: 2$
4 $6: 4: 3$
Explanation:
D There resistance, $\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=3 \Omega, \mathrm{R}_{3}=4 \Omega$ $\because \quad \mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\mathrm{I} \propto \frac{1}{\mathrm{R}}$ Ratio of current, $\mathrm{I}_{1}: \mathrm{I}_{2}: \mathrm{I}_{3}=\frac{1}{\mathrm{R}_{1}}: \frac{1}{\mathrm{R}_{2}}: \frac{1}{\mathrm{R}_{3}}=\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ $\mathrm{I}_{1}: \mathrm{I}_{2}: \mathrm{I}_{3}=6: 4: 3$
Karnataka CET-2015
Current Electricity
151996
A carbon film resistor has colour code Green Black Violet Gold. The value of the resistor is:
1 $50 \mathrm{M} \Omega$
2 $500 \mathrm{M} \Omega$
3 $500 \pm 5 \% \mathrm{M} \Omega$
4 $500 \pm 10 \% \mathrm{M} \Omega$
Explanation:
C First digit colour $=$ Green $=5$ Second digit colour $=$ Black $=0$ The multiplier colour $=$ violet $= \pm 10^{7}$ Tolerance colour $=$ Golden $= \pm 5 \%$ So, $\quad \mathrm{R}=50 \times 10^{7} \pm 5$ $\mathrm{R}=500 \times 10^{6} \pm 5 \Omega$ $\mathrm{R}=500 \pm 5 \% \mathrm{M} \Omega$ Current Electricity
Karnataka CET-2014
Current Electricity
151997
A battery consists of a variable number (n) of identical cells, each having an internal resistance $r$ connected in series. The terminals of the battery are short-circuited. A graph of current (I) in the circuit verses the number of cells will be as shown in figure.
1
2
3
4
Explanation:
C Relationship between current (I), emf (E) and internal resistance (r), Total voltage of the circuit $\left(\mathrm{V}_{\text {Total }}\right)=\mathrm{nE}$ Total resistance of the circuit $\left(\mathrm{R}_{\text {Total }}\right)=\mathrm{nr}$ $\text { Total current (I) } =\frac{\mathrm{V}_{\text {Total }}}{\mathrm{R}_{\text {Total }}}$ $=\frac{\mathrm{nE}}{\mathrm{nr}}$ - It is clear that total current in the circuit is independent of variable number (n).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
151994
Variation of resistance of the conductor with temperature is as shown : The temperature coefficient $(\alpha)$ of the conductor is :
1 $\frac{\mathrm{R}_{0}}{\mathrm{~m}}$
2 $\mathrm{mR}_{0}$
3 $\mathrm{m}^{2} \mathrm{R}_{0}$
4 $\frac{\mathrm{m}}{\mathrm{R}_{0}}$
Explanation:
D From resistance with temperature, $\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})=\mathrm{R}_{0}+\mathrm{R}_{0} \alpha \Delta \mathrm{T}$ Where, $\mathrm{R}_{0}=$ Resistance at reference temperature $\alpha=$ Coefficient of temperature of conductor $\Delta \mathrm{T}=$ Chang in temperature It is a straight line having slope $=\alpha R_{0}=\mathrm{m}$ So, $\quad \alpha=\frac{\mathrm{m}}{\mathrm{R}_{0}}$
Karnataka CET-2016
Current Electricity
151995
Three resistances $2 \Omega, 3 \Omega$ and $4 \Omega$ are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be :
1 $5: 4: 3$
2 $6: 3: 2$
3 $4: 3: 2$
4 $6: 4: 3$
Explanation:
D There resistance, $\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=3 \Omega, \mathrm{R}_{3}=4 \Omega$ $\because \quad \mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\mathrm{I} \propto \frac{1}{\mathrm{R}}$ Ratio of current, $\mathrm{I}_{1}: \mathrm{I}_{2}: \mathrm{I}_{3}=\frac{1}{\mathrm{R}_{1}}: \frac{1}{\mathrm{R}_{2}}: \frac{1}{\mathrm{R}_{3}}=\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ $\mathrm{I}_{1}: \mathrm{I}_{2}: \mathrm{I}_{3}=6: 4: 3$
Karnataka CET-2015
Current Electricity
151996
A carbon film resistor has colour code Green Black Violet Gold. The value of the resistor is:
1 $50 \mathrm{M} \Omega$
2 $500 \mathrm{M} \Omega$
3 $500 \pm 5 \% \mathrm{M} \Omega$
4 $500 \pm 10 \% \mathrm{M} \Omega$
Explanation:
C First digit colour $=$ Green $=5$ Second digit colour $=$ Black $=0$ The multiplier colour $=$ violet $= \pm 10^{7}$ Tolerance colour $=$ Golden $= \pm 5 \%$ So, $\quad \mathrm{R}=50 \times 10^{7} \pm 5$ $\mathrm{R}=500 \times 10^{6} \pm 5 \Omega$ $\mathrm{R}=500 \pm 5 \% \mathrm{M} \Omega$ Current Electricity
Karnataka CET-2014
Current Electricity
151997
A battery consists of a variable number (n) of identical cells, each having an internal resistance $r$ connected in series. The terminals of the battery are short-circuited. A graph of current (I) in the circuit verses the number of cells will be as shown in figure.
1
2
3
4
Explanation:
C Relationship between current (I), emf (E) and internal resistance (r), Total voltage of the circuit $\left(\mathrm{V}_{\text {Total }}\right)=\mathrm{nE}$ Total resistance of the circuit $\left(\mathrm{R}_{\text {Total }}\right)=\mathrm{nr}$ $\text { Total current (I) } =\frac{\mathrm{V}_{\text {Total }}}{\mathrm{R}_{\text {Total }}}$ $=\frac{\mathrm{nE}}{\mathrm{nr}}$ - It is clear that total current in the circuit is independent of variable number (n).
151994
Variation of resistance of the conductor with temperature is as shown : The temperature coefficient $(\alpha)$ of the conductor is :
1 $\frac{\mathrm{R}_{0}}{\mathrm{~m}}$
2 $\mathrm{mR}_{0}$
3 $\mathrm{m}^{2} \mathrm{R}_{0}$
4 $\frac{\mathrm{m}}{\mathrm{R}_{0}}$
Explanation:
D From resistance with temperature, $\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})=\mathrm{R}_{0}+\mathrm{R}_{0} \alpha \Delta \mathrm{T}$ Where, $\mathrm{R}_{0}=$ Resistance at reference temperature $\alpha=$ Coefficient of temperature of conductor $\Delta \mathrm{T}=$ Chang in temperature It is a straight line having slope $=\alpha R_{0}=\mathrm{m}$ So, $\quad \alpha=\frac{\mathrm{m}}{\mathrm{R}_{0}}$
Karnataka CET-2016
Current Electricity
151995
Three resistances $2 \Omega, 3 \Omega$ and $4 \Omega$ are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be :
1 $5: 4: 3$
2 $6: 3: 2$
3 $4: 3: 2$
4 $6: 4: 3$
Explanation:
D There resistance, $\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=3 \Omega, \mathrm{R}_{3}=4 \Omega$ $\because \quad \mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\mathrm{I} \propto \frac{1}{\mathrm{R}}$ Ratio of current, $\mathrm{I}_{1}: \mathrm{I}_{2}: \mathrm{I}_{3}=\frac{1}{\mathrm{R}_{1}}: \frac{1}{\mathrm{R}_{2}}: \frac{1}{\mathrm{R}_{3}}=\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ $\mathrm{I}_{1}: \mathrm{I}_{2}: \mathrm{I}_{3}=6: 4: 3$
Karnataka CET-2015
Current Electricity
151996
A carbon film resistor has colour code Green Black Violet Gold. The value of the resistor is:
1 $50 \mathrm{M} \Omega$
2 $500 \mathrm{M} \Omega$
3 $500 \pm 5 \% \mathrm{M} \Omega$
4 $500 \pm 10 \% \mathrm{M} \Omega$
Explanation:
C First digit colour $=$ Green $=5$ Second digit colour $=$ Black $=0$ The multiplier colour $=$ violet $= \pm 10^{7}$ Tolerance colour $=$ Golden $= \pm 5 \%$ So, $\quad \mathrm{R}=50 \times 10^{7} \pm 5$ $\mathrm{R}=500 \times 10^{6} \pm 5 \Omega$ $\mathrm{R}=500 \pm 5 \% \mathrm{M} \Omega$ Current Electricity
Karnataka CET-2014
Current Electricity
151997
A battery consists of a variable number (n) of identical cells, each having an internal resistance $r$ connected in series. The terminals of the battery are short-circuited. A graph of current (I) in the circuit verses the number of cells will be as shown in figure.
1
2
3
4
Explanation:
C Relationship between current (I), emf (E) and internal resistance (r), Total voltage of the circuit $\left(\mathrm{V}_{\text {Total }}\right)=\mathrm{nE}$ Total resistance of the circuit $\left(\mathrm{R}_{\text {Total }}\right)=\mathrm{nr}$ $\text { Total current (I) } =\frac{\mathrm{V}_{\text {Total }}}{\mathrm{R}_{\text {Total }}}$ $=\frac{\mathrm{nE}}{\mathrm{nr}}$ - It is clear that total current in the circuit is independent of variable number (n).
151994
Variation of resistance of the conductor with temperature is as shown : The temperature coefficient $(\alpha)$ of the conductor is :
1 $\frac{\mathrm{R}_{0}}{\mathrm{~m}}$
2 $\mathrm{mR}_{0}$
3 $\mathrm{m}^{2} \mathrm{R}_{0}$
4 $\frac{\mathrm{m}}{\mathrm{R}_{0}}$
Explanation:
D From resistance with temperature, $\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{T})=\mathrm{R}_{0}+\mathrm{R}_{0} \alpha \Delta \mathrm{T}$ Where, $\mathrm{R}_{0}=$ Resistance at reference temperature $\alpha=$ Coefficient of temperature of conductor $\Delta \mathrm{T}=$ Chang in temperature It is a straight line having slope $=\alpha R_{0}=\mathrm{m}$ So, $\quad \alpha=\frac{\mathrm{m}}{\mathrm{R}_{0}}$
Karnataka CET-2016
Current Electricity
151995
Three resistances $2 \Omega, 3 \Omega$ and $4 \Omega$ are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be :
1 $5: 4: 3$
2 $6: 3: 2$
3 $4: 3: 2$
4 $6: 4: 3$
Explanation:
D There resistance, $\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=3 \Omega, \mathrm{R}_{3}=4 \Omega$ $\because \quad \mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$ $\mathrm{I} \propto \frac{1}{\mathrm{R}}$ Ratio of current, $\mathrm{I}_{1}: \mathrm{I}_{2}: \mathrm{I}_{3}=\frac{1}{\mathrm{R}_{1}}: \frac{1}{\mathrm{R}_{2}}: \frac{1}{\mathrm{R}_{3}}=\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ $\mathrm{I}_{1}: \mathrm{I}_{2}: \mathrm{I}_{3}=6: 4: 3$
Karnataka CET-2015
Current Electricity
151996
A carbon film resistor has colour code Green Black Violet Gold. The value of the resistor is:
1 $50 \mathrm{M} \Omega$
2 $500 \mathrm{M} \Omega$
3 $500 \pm 5 \% \mathrm{M} \Omega$
4 $500 \pm 10 \% \mathrm{M} \Omega$
Explanation:
C First digit colour $=$ Green $=5$ Second digit colour $=$ Black $=0$ The multiplier colour $=$ violet $= \pm 10^{7}$ Tolerance colour $=$ Golden $= \pm 5 \%$ So, $\quad \mathrm{R}=50 \times 10^{7} \pm 5$ $\mathrm{R}=500 \times 10^{6} \pm 5 \Omega$ $\mathrm{R}=500 \pm 5 \% \mathrm{M} \Omega$ Current Electricity
Karnataka CET-2014
Current Electricity
151997
A battery consists of a variable number (n) of identical cells, each having an internal resistance $r$ connected in series. The terminals of the battery are short-circuited. A graph of current (I) in the circuit verses the number of cells will be as shown in figure.
1
2
3
4
Explanation:
C Relationship between current (I), emf (E) and internal resistance (r), Total voltage of the circuit $\left(\mathrm{V}_{\text {Total }}\right)=\mathrm{nE}$ Total resistance of the circuit $\left(\mathrm{R}_{\text {Total }}\right)=\mathrm{nr}$ $\text { Total current (I) } =\frac{\mathrm{V}_{\text {Total }}}{\mathrm{R}_{\text {Total }}}$ $=\frac{\mathrm{nE}}{\mathrm{nr}}$ - It is clear that total current in the circuit is independent of variable number (n).