152885
The meter bridge shown in the balance position with $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{l_{1}}{l_{2}}$.If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?
D The meter bridge is balanced against length $l_{1}$ and $l_{2}$ are- When galvanometer and cells are interchanged, the balance condition remains unchanged. For balancing condition, $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{l_{1}}{l_{2}}$ Yes, the bridge will work for balanced condition.
NEET Odisha-2019
Current Electricity
152887
You are given an ammeter, a galvanometer and a voltmeter. From these, the device having maximum resistance is:
1 ammeter
2 galvanometer
3 voltmeter
4 all will have the same resistance
Explanation:
C You are given an ammeter, a galvanometer and voltmeter. From these the device having maximum resistance is voltmeter. (i) Voltmeter Equivalent resistance of voltmeter $\left(\mathrm{R}_{\mathrm{eq}}\right)=\mathrm{R}_{\mathrm{G}}+\mathrm{R}_{\mathrm{V}}$ (ii) Galvanometer Equivalent resistance of galvanometer, $R_{e g}(G)=R_{G}$ (iii) Ammeter Equivalent resistance of ammeter, $R_{e q}(A)=\frac{R_{S} R_{G}}{R_{G}+R_{S}}$ Hence, the order of resistance $\mathrm{R}_{\mathrm{eq}(\mathrm{V})}>\mathrm{R}_{\mathrm{eq}(\mathrm{G})}>\mathrm{R}_{\mathrm{eq}(\mathrm{A})}$ So, that maximum resistance device is voltmeter.
UPSEE 2019
Current Electricity
152888
In a potentiometer experiment, the balancing point with a cell is at a length $240 \mathrm{~cm}$. On shunting the cell with a resistance of $2 \Omega$. the balancing length becomes $120 \mathrm{~cm}$. The internal resistance of the cell is:
1 $4 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
B Given that, $l_{1}=240$ $l_{2}=120$ $\mathrm{R}=2 \Omega$ $\mathrm{r}=?$ $\mathrm{r}=\mathrm{R}\left[\frac{l_{1}}{l_{2}}-1\right]=2\left[\frac{240}{120}-1\right]$ $=2[2-1]=2 \times 1$ $=2 \Omega$
Karnataka CET-2019
Current Electricity
152889
The number of turns in a coil of galvanometer is tripled, then :
1 voltage sensitivity increases 3 times and current sensitivity remains constant
2 voltage sensitivity remains constant and current sensitivity increases 3 times
3 both voltage and current sensitivity remains constant
4 both voltage and current sensitivity decreases by $33 \%$
Explanation:
B Current sensitivity $\left(I_{s}\right)=\frac{N B A}{K}$ Where, N- number of turns B- magnetic field A- area of coils $\mathrm{K}$ - figure of miret Now, $\quad I_{\mathrm{s}} \propto \mathrm{N}$ Therefore, current sensitivity will increase 3 times And, voltage sensitivity $\left(V_{\mathrm{s}}\right)=\frac{I_{s}}{R}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{NBA}}{\mathrm{K}} \times \frac{1}{\mathrm{R}}$ From above expression we can conclude $\mathrm{N} \propto \mathrm{R} \text { and } \mathrm{I}_{\mathrm{s}} \propto \mathrm{N}$ Thus, $\mathrm{V}_{\mathrm{s}}$ will remain constant. Voltage sensitivity remains constant and current sensitivity increases 3 times.
152885
The meter bridge shown in the balance position with $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{l_{1}}{l_{2}}$.If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?
D The meter bridge is balanced against length $l_{1}$ and $l_{2}$ are- When galvanometer and cells are interchanged, the balance condition remains unchanged. For balancing condition, $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{l_{1}}{l_{2}}$ Yes, the bridge will work for balanced condition.
NEET Odisha-2019
Current Electricity
152887
You are given an ammeter, a galvanometer and a voltmeter. From these, the device having maximum resistance is:
1 ammeter
2 galvanometer
3 voltmeter
4 all will have the same resistance
Explanation:
C You are given an ammeter, a galvanometer and voltmeter. From these the device having maximum resistance is voltmeter. (i) Voltmeter Equivalent resistance of voltmeter $\left(\mathrm{R}_{\mathrm{eq}}\right)=\mathrm{R}_{\mathrm{G}}+\mathrm{R}_{\mathrm{V}}$ (ii) Galvanometer Equivalent resistance of galvanometer, $R_{e g}(G)=R_{G}$ (iii) Ammeter Equivalent resistance of ammeter, $R_{e q}(A)=\frac{R_{S} R_{G}}{R_{G}+R_{S}}$ Hence, the order of resistance $\mathrm{R}_{\mathrm{eq}(\mathrm{V})}>\mathrm{R}_{\mathrm{eq}(\mathrm{G})}>\mathrm{R}_{\mathrm{eq}(\mathrm{A})}$ So, that maximum resistance device is voltmeter.
UPSEE 2019
Current Electricity
152888
In a potentiometer experiment, the balancing point with a cell is at a length $240 \mathrm{~cm}$. On shunting the cell with a resistance of $2 \Omega$. the balancing length becomes $120 \mathrm{~cm}$. The internal resistance of the cell is:
1 $4 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
B Given that, $l_{1}=240$ $l_{2}=120$ $\mathrm{R}=2 \Omega$ $\mathrm{r}=?$ $\mathrm{r}=\mathrm{R}\left[\frac{l_{1}}{l_{2}}-1\right]=2\left[\frac{240}{120}-1\right]$ $=2[2-1]=2 \times 1$ $=2 \Omega$
Karnataka CET-2019
Current Electricity
152889
The number of turns in a coil of galvanometer is tripled, then :
1 voltage sensitivity increases 3 times and current sensitivity remains constant
2 voltage sensitivity remains constant and current sensitivity increases 3 times
3 both voltage and current sensitivity remains constant
4 both voltage and current sensitivity decreases by $33 \%$
Explanation:
B Current sensitivity $\left(I_{s}\right)=\frac{N B A}{K}$ Where, N- number of turns B- magnetic field A- area of coils $\mathrm{K}$ - figure of miret Now, $\quad I_{\mathrm{s}} \propto \mathrm{N}$ Therefore, current sensitivity will increase 3 times And, voltage sensitivity $\left(V_{\mathrm{s}}\right)=\frac{I_{s}}{R}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{NBA}}{\mathrm{K}} \times \frac{1}{\mathrm{R}}$ From above expression we can conclude $\mathrm{N} \propto \mathrm{R} \text { and } \mathrm{I}_{\mathrm{s}} \propto \mathrm{N}$ Thus, $\mathrm{V}_{\mathrm{s}}$ will remain constant. Voltage sensitivity remains constant and current sensitivity increases 3 times.
152885
The meter bridge shown in the balance position with $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{l_{1}}{l_{2}}$.If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?
D The meter bridge is balanced against length $l_{1}$ and $l_{2}$ are- When galvanometer and cells are interchanged, the balance condition remains unchanged. For balancing condition, $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{l_{1}}{l_{2}}$ Yes, the bridge will work for balanced condition.
NEET Odisha-2019
Current Electricity
152887
You are given an ammeter, a galvanometer and a voltmeter. From these, the device having maximum resistance is:
1 ammeter
2 galvanometer
3 voltmeter
4 all will have the same resistance
Explanation:
C You are given an ammeter, a galvanometer and voltmeter. From these the device having maximum resistance is voltmeter. (i) Voltmeter Equivalent resistance of voltmeter $\left(\mathrm{R}_{\mathrm{eq}}\right)=\mathrm{R}_{\mathrm{G}}+\mathrm{R}_{\mathrm{V}}$ (ii) Galvanometer Equivalent resistance of galvanometer, $R_{e g}(G)=R_{G}$ (iii) Ammeter Equivalent resistance of ammeter, $R_{e q}(A)=\frac{R_{S} R_{G}}{R_{G}+R_{S}}$ Hence, the order of resistance $\mathrm{R}_{\mathrm{eq}(\mathrm{V})}>\mathrm{R}_{\mathrm{eq}(\mathrm{G})}>\mathrm{R}_{\mathrm{eq}(\mathrm{A})}$ So, that maximum resistance device is voltmeter.
UPSEE 2019
Current Electricity
152888
In a potentiometer experiment, the balancing point with a cell is at a length $240 \mathrm{~cm}$. On shunting the cell with a resistance of $2 \Omega$. the balancing length becomes $120 \mathrm{~cm}$. The internal resistance of the cell is:
1 $4 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
B Given that, $l_{1}=240$ $l_{2}=120$ $\mathrm{R}=2 \Omega$ $\mathrm{r}=?$ $\mathrm{r}=\mathrm{R}\left[\frac{l_{1}}{l_{2}}-1\right]=2\left[\frac{240}{120}-1\right]$ $=2[2-1]=2 \times 1$ $=2 \Omega$
Karnataka CET-2019
Current Electricity
152889
The number of turns in a coil of galvanometer is tripled, then :
1 voltage sensitivity increases 3 times and current sensitivity remains constant
2 voltage sensitivity remains constant and current sensitivity increases 3 times
3 both voltage and current sensitivity remains constant
4 both voltage and current sensitivity decreases by $33 \%$
Explanation:
B Current sensitivity $\left(I_{s}\right)=\frac{N B A}{K}$ Where, N- number of turns B- magnetic field A- area of coils $\mathrm{K}$ - figure of miret Now, $\quad I_{\mathrm{s}} \propto \mathrm{N}$ Therefore, current sensitivity will increase 3 times And, voltage sensitivity $\left(V_{\mathrm{s}}\right)=\frac{I_{s}}{R}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{NBA}}{\mathrm{K}} \times \frac{1}{\mathrm{R}}$ From above expression we can conclude $\mathrm{N} \propto \mathrm{R} \text { and } \mathrm{I}_{\mathrm{s}} \propto \mathrm{N}$ Thus, $\mathrm{V}_{\mathrm{s}}$ will remain constant. Voltage sensitivity remains constant and current sensitivity increases 3 times.
152885
The meter bridge shown in the balance position with $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{l_{1}}{l_{2}}$.If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?
D The meter bridge is balanced against length $l_{1}$ and $l_{2}$ are- When galvanometer and cells are interchanged, the balance condition remains unchanged. For balancing condition, $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{l_{1}}{l_{2}}$ Yes, the bridge will work for balanced condition.
NEET Odisha-2019
Current Electricity
152887
You are given an ammeter, a galvanometer and a voltmeter. From these, the device having maximum resistance is:
1 ammeter
2 galvanometer
3 voltmeter
4 all will have the same resistance
Explanation:
C You are given an ammeter, a galvanometer and voltmeter. From these the device having maximum resistance is voltmeter. (i) Voltmeter Equivalent resistance of voltmeter $\left(\mathrm{R}_{\mathrm{eq}}\right)=\mathrm{R}_{\mathrm{G}}+\mathrm{R}_{\mathrm{V}}$ (ii) Galvanometer Equivalent resistance of galvanometer, $R_{e g}(G)=R_{G}$ (iii) Ammeter Equivalent resistance of ammeter, $R_{e q}(A)=\frac{R_{S} R_{G}}{R_{G}+R_{S}}$ Hence, the order of resistance $\mathrm{R}_{\mathrm{eq}(\mathrm{V})}>\mathrm{R}_{\mathrm{eq}(\mathrm{G})}>\mathrm{R}_{\mathrm{eq}(\mathrm{A})}$ So, that maximum resistance device is voltmeter.
UPSEE 2019
Current Electricity
152888
In a potentiometer experiment, the balancing point with a cell is at a length $240 \mathrm{~cm}$. On shunting the cell with a resistance of $2 \Omega$. the balancing length becomes $120 \mathrm{~cm}$. The internal resistance of the cell is:
1 $4 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
B Given that, $l_{1}=240$ $l_{2}=120$ $\mathrm{R}=2 \Omega$ $\mathrm{r}=?$ $\mathrm{r}=\mathrm{R}\left[\frac{l_{1}}{l_{2}}-1\right]=2\left[\frac{240}{120}-1\right]$ $=2[2-1]=2 \times 1$ $=2 \Omega$
Karnataka CET-2019
Current Electricity
152889
The number of turns in a coil of galvanometer is tripled, then :
1 voltage sensitivity increases 3 times and current sensitivity remains constant
2 voltage sensitivity remains constant and current sensitivity increases 3 times
3 both voltage and current sensitivity remains constant
4 both voltage and current sensitivity decreases by $33 \%$
Explanation:
B Current sensitivity $\left(I_{s}\right)=\frac{N B A}{K}$ Where, N- number of turns B- magnetic field A- area of coils $\mathrm{K}$ - figure of miret Now, $\quad I_{\mathrm{s}} \propto \mathrm{N}$ Therefore, current sensitivity will increase 3 times And, voltage sensitivity $\left(V_{\mathrm{s}}\right)=\frac{I_{s}}{R}$ $\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{NBA}}{\mathrm{K}} \times \frac{1}{\mathrm{R}}$ From above expression we can conclude $\mathrm{N} \propto \mathrm{R} \text { and } \mathrm{I}_{\mathrm{s}} \propto \mathrm{N}$ Thus, $\mathrm{V}_{\mathrm{s}}$ will remain constant. Voltage sensitivity remains constant and current sensitivity increases 3 times.
