152596
A battery of $6 \mathrm{~V}$ and internal resistance $2 \Omega$ is connected to a silver voltmeter. If the current of 1.5 A flows through the circuit, the resistance of the voltmeter is
1 $4 \Omega$
2 $2 \Omega$
3 $6 \Omega$
4 $1 \Omega$
5 $5 \Omega$
Explanation:
B Given, $\mathrm{E}=6 \mathrm{~V}, \mathrm{r}=2 \Omega$ and $\mathrm{I}=1.5 \mathrm{~A}$ We know, current across a circuit $I=\frac{E}{r+R}$ $1.5=\frac{6}{2+R}$ $R=2 \Omega$
Kerala CEE - 2015
Current Electricity
152598
When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found to be $540 \mathrm{~cm}$. If the balancing length becomes $550 \mathrm{~cm}$ when the cell is short circuited with $1 \Omega$, the internal of the cell is
1 $0.08 \Omega$
2 $0.04 \Omega$
3 $1.0 \Omega$
4 $1.08 \Omega$
5 $1.45 \Omega$
Explanation:
A Given $l_{1}=540 \mathrm{~cm}, l_{2}=500 \mathrm{~cm}$ $\mathrm{R}=1 \Omega$ Internal Resistance $\mathrm{r}=\mathrm{R}\left(\frac{l_{1}-l_{2}}{l_{2}}\right)$ $=1\left(\frac{540-500}{500}\right)$ $=\frac{40}{500}=0.08 \Omega$
Kerala CEE - 2008
Current Electricity
152609
The cold junction of a thermocouple is maintained at $10^{\circ} \mathrm{C}$. No thermo e.m.f. is developed when the hot junction is maintained at $530^{\circ} \mathrm{C}$. The natural temperature is
1 $260^{\circ} \mathrm{C}$
2 $265^{\circ} \mathrm{C}$
3 $270^{\circ} \mathrm{C}$
4 $520^{\circ} \mathrm{C}$
Explanation:
C When thermo emf is zero, the corresponding temperature is the inversion temperature, $\theta_{\mathrm{i}}$ Here, $\theta_{\mathrm{i}}=530^{\circ} \mathrm{C}$ and cold junction temperature $\theta_{\mathrm{c}}$ is $0^{\circ} \mathrm{C}$ $\therefore \quad$ Neutral temperature, $\theta_{\mathrm{n}}=\frac{\theta_{1}+\theta_{\mathrm{c}}}{2}=\frac{530+10}{2}=270^{\circ} \mathrm{C}$
AIIMS-2014
Current Electricity
152619
Which of the following graphs represent variation of thermo emf (E) of a thermocouple with temperature $t$ of the junction, the cold junction beings kept at $0^{\circ} \mathrm{C}$ ?
1
2
3
4
Explanation:
C Relation between thermo emf and $t_{\mathrm{emp}}$ diff. $\mathrm{E}=\mathrm{at}+\frac{1}{2} \mathrm{bt}^{2}$ Where $\mathrm{a}, \mathrm{b}$ are thermo - electric constants. Equation (i) represents a parabola.
152596
A battery of $6 \mathrm{~V}$ and internal resistance $2 \Omega$ is connected to a silver voltmeter. If the current of 1.5 A flows through the circuit, the resistance of the voltmeter is
1 $4 \Omega$
2 $2 \Omega$
3 $6 \Omega$
4 $1 \Omega$
5 $5 \Omega$
Explanation:
B Given, $\mathrm{E}=6 \mathrm{~V}, \mathrm{r}=2 \Omega$ and $\mathrm{I}=1.5 \mathrm{~A}$ We know, current across a circuit $I=\frac{E}{r+R}$ $1.5=\frac{6}{2+R}$ $R=2 \Omega$
Kerala CEE - 2015
Current Electricity
152598
When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found to be $540 \mathrm{~cm}$. If the balancing length becomes $550 \mathrm{~cm}$ when the cell is short circuited with $1 \Omega$, the internal of the cell is
1 $0.08 \Omega$
2 $0.04 \Omega$
3 $1.0 \Omega$
4 $1.08 \Omega$
5 $1.45 \Omega$
Explanation:
A Given $l_{1}=540 \mathrm{~cm}, l_{2}=500 \mathrm{~cm}$ $\mathrm{R}=1 \Omega$ Internal Resistance $\mathrm{r}=\mathrm{R}\left(\frac{l_{1}-l_{2}}{l_{2}}\right)$ $=1\left(\frac{540-500}{500}\right)$ $=\frac{40}{500}=0.08 \Omega$
Kerala CEE - 2008
Current Electricity
152609
The cold junction of a thermocouple is maintained at $10^{\circ} \mathrm{C}$. No thermo e.m.f. is developed when the hot junction is maintained at $530^{\circ} \mathrm{C}$. The natural temperature is
1 $260^{\circ} \mathrm{C}$
2 $265^{\circ} \mathrm{C}$
3 $270^{\circ} \mathrm{C}$
4 $520^{\circ} \mathrm{C}$
Explanation:
C When thermo emf is zero, the corresponding temperature is the inversion temperature, $\theta_{\mathrm{i}}$ Here, $\theta_{\mathrm{i}}=530^{\circ} \mathrm{C}$ and cold junction temperature $\theta_{\mathrm{c}}$ is $0^{\circ} \mathrm{C}$ $\therefore \quad$ Neutral temperature, $\theta_{\mathrm{n}}=\frac{\theta_{1}+\theta_{\mathrm{c}}}{2}=\frac{530+10}{2}=270^{\circ} \mathrm{C}$
AIIMS-2014
Current Electricity
152619
Which of the following graphs represent variation of thermo emf (E) of a thermocouple with temperature $t$ of the junction, the cold junction beings kept at $0^{\circ} \mathrm{C}$ ?
1
2
3
4
Explanation:
C Relation between thermo emf and $t_{\mathrm{emp}}$ diff. $\mathrm{E}=\mathrm{at}+\frac{1}{2} \mathrm{bt}^{2}$ Where $\mathrm{a}, \mathrm{b}$ are thermo - electric constants. Equation (i) represents a parabola.
152596
A battery of $6 \mathrm{~V}$ and internal resistance $2 \Omega$ is connected to a silver voltmeter. If the current of 1.5 A flows through the circuit, the resistance of the voltmeter is
1 $4 \Omega$
2 $2 \Omega$
3 $6 \Omega$
4 $1 \Omega$
5 $5 \Omega$
Explanation:
B Given, $\mathrm{E}=6 \mathrm{~V}, \mathrm{r}=2 \Omega$ and $\mathrm{I}=1.5 \mathrm{~A}$ We know, current across a circuit $I=\frac{E}{r+R}$ $1.5=\frac{6}{2+R}$ $R=2 \Omega$
Kerala CEE - 2015
Current Electricity
152598
When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found to be $540 \mathrm{~cm}$. If the balancing length becomes $550 \mathrm{~cm}$ when the cell is short circuited with $1 \Omega$, the internal of the cell is
1 $0.08 \Omega$
2 $0.04 \Omega$
3 $1.0 \Omega$
4 $1.08 \Omega$
5 $1.45 \Omega$
Explanation:
A Given $l_{1}=540 \mathrm{~cm}, l_{2}=500 \mathrm{~cm}$ $\mathrm{R}=1 \Omega$ Internal Resistance $\mathrm{r}=\mathrm{R}\left(\frac{l_{1}-l_{2}}{l_{2}}\right)$ $=1\left(\frac{540-500}{500}\right)$ $=\frac{40}{500}=0.08 \Omega$
Kerala CEE - 2008
Current Electricity
152609
The cold junction of a thermocouple is maintained at $10^{\circ} \mathrm{C}$. No thermo e.m.f. is developed when the hot junction is maintained at $530^{\circ} \mathrm{C}$. The natural temperature is
1 $260^{\circ} \mathrm{C}$
2 $265^{\circ} \mathrm{C}$
3 $270^{\circ} \mathrm{C}$
4 $520^{\circ} \mathrm{C}$
Explanation:
C When thermo emf is zero, the corresponding temperature is the inversion temperature, $\theta_{\mathrm{i}}$ Here, $\theta_{\mathrm{i}}=530^{\circ} \mathrm{C}$ and cold junction temperature $\theta_{\mathrm{c}}$ is $0^{\circ} \mathrm{C}$ $\therefore \quad$ Neutral temperature, $\theta_{\mathrm{n}}=\frac{\theta_{1}+\theta_{\mathrm{c}}}{2}=\frac{530+10}{2}=270^{\circ} \mathrm{C}$
AIIMS-2014
Current Electricity
152619
Which of the following graphs represent variation of thermo emf (E) of a thermocouple with temperature $t$ of the junction, the cold junction beings kept at $0^{\circ} \mathrm{C}$ ?
1
2
3
4
Explanation:
C Relation between thermo emf and $t_{\mathrm{emp}}$ diff. $\mathrm{E}=\mathrm{at}+\frac{1}{2} \mathrm{bt}^{2}$ Where $\mathrm{a}, \mathrm{b}$ are thermo - electric constants. Equation (i) represents a parabola.
152596
A battery of $6 \mathrm{~V}$ and internal resistance $2 \Omega$ is connected to a silver voltmeter. If the current of 1.5 A flows through the circuit, the resistance of the voltmeter is
1 $4 \Omega$
2 $2 \Omega$
3 $6 \Omega$
4 $1 \Omega$
5 $5 \Omega$
Explanation:
B Given, $\mathrm{E}=6 \mathrm{~V}, \mathrm{r}=2 \Omega$ and $\mathrm{I}=1.5 \mathrm{~A}$ We know, current across a circuit $I=\frac{E}{r+R}$ $1.5=\frac{6}{2+R}$ $R=2 \Omega$
Kerala CEE - 2015
Current Electricity
152598
When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found to be $540 \mathrm{~cm}$. If the balancing length becomes $550 \mathrm{~cm}$ when the cell is short circuited with $1 \Omega$, the internal of the cell is
1 $0.08 \Omega$
2 $0.04 \Omega$
3 $1.0 \Omega$
4 $1.08 \Omega$
5 $1.45 \Omega$
Explanation:
A Given $l_{1}=540 \mathrm{~cm}, l_{2}=500 \mathrm{~cm}$ $\mathrm{R}=1 \Omega$ Internal Resistance $\mathrm{r}=\mathrm{R}\left(\frac{l_{1}-l_{2}}{l_{2}}\right)$ $=1\left(\frac{540-500}{500}\right)$ $=\frac{40}{500}=0.08 \Omega$
Kerala CEE - 2008
Current Electricity
152609
The cold junction of a thermocouple is maintained at $10^{\circ} \mathrm{C}$. No thermo e.m.f. is developed when the hot junction is maintained at $530^{\circ} \mathrm{C}$. The natural temperature is
1 $260^{\circ} \mathrm{C}$
2 $265^{\circ} \mathrm{C}$
3 $270^{\circ} \mathrm{C}$
4 $520^{\circ} \mathrm{C}$
Explanation:
C When thermo emf is zero, the corresponding temperature is the inversion temperature, $\theta_{\mathrm{i}}$ Here, $\theta_{\mathrm{i}}=530^{\circ} \mathrm{C}$ and cold junction temperature $\theta_{\mathrm{c}}$ is $0^{\circ} \mathrm{C}$ $\therefore \quad$ Neutral temperature, $\theta_{\mathrm{n}}=\frac{\theta_{1}+\theta_{\mathrm{c}}}{2}=\frac{530+10}{2}=270^{\circ} \mathrm{C}$
AIIMS-2014
Current Electricity
152619
Which of the following graphs represent variation of thermo emf (E) of a thermocouple with temperature $t$ of the junction, the cold junction beings kept at $0^{\circ} \mathrm{C}$ ?
1
2
3
4
Explanation:
C Relation between thermo emf and $t_{\mathrm{emp}}$ diff. $\mathrm{E}=\mathrm{at}+\frac{1}{2} \mathrm{bt}^{2}$ Where $\mathrm{a}, \mathrm{b}$ are thermo - electric constants. Equation (i) represents a parabola.
