151837
A cylindrical metallic wire is stretched to increase its length. If the resistance of the wire is increased by $4 \%$ then the percentage increase in its length is
1 $4 \%$
2 $8 \%$
3 $1 \%$
4 $2 \%$
Explanation:
D The change in resistance is due to both increase in length and decrease in cross-sectional area. We know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}} \text { or } \frac{\rho l}{\mathrm{~V}} {\left[\because \mathrm{A}=\frac{\mathrm{V}}{l}\right]}$ $\therefore \mathrm{R} \propto l^{2}$ $\text { Or } \frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\left(\frac{l^{\prime}}{l}\right)^{2} \ldots . .(\mathrm{i})$ According to the question, Resistance of the wire is increased by $4 \%$ $\therefore \quad \mathrm{R}^{\prime} =\mathrm{R}+\frac{4}{100} \mathrm{R}$ $\mathrm{R}^{\prime} =1.04 \mathrm{R}$ From equation (i) $\frac{1.04 \mathrm{R}}{\mathrm{R}}=\left(\frac{l^{\prime}}{l}\right)^{2}$ $\frac{l^{\prime}}{l}=\sqrt{1.04}$ $\frac{l^{\prime}}{l}=1.0198$ $l^{\prime}=1.02 l$ Now, percentage increase in its length, $\Delta l=\frac{l^{\prime}-l}{l} \times 100=\frac{1.02 l-l}{l} \times 100$ $\Delta l=2 \%$
TS EAMCET 05.08.2021
Current Electricity
151838 A carbon resistor with color code is shown in the figure. There is no fourth band in the resistor. The value of the resistance is
1 $24 \mathrm{M} \Omega \pm 20 \%$
2 $14 \mathrm{k} \Omega \pm 5 \%$
3 $24 \mathrm{k} \Omega \pm 20 \%$
4 $34 \mathrm{k} \Omega \pm 10 \%$
Explanation:
C $\operatorname{Red}=2$, yellow $=4$, Orange $=10^{3}$, No colour $=20 \%$ Therefore, resistance of carbon $=24 \times 10^{3} \pm 20 \%$ $=24 \mathrm{k} \Omega \pm 20 \%$
WB JEE 2021
Current Electricity
151839
The length of a potentiometer wire is $1200 \mathrm{~cm}$ and it carries a current of $60 \mathrm{~mA}$. For a cell of emf $5 \mathrm{~V}$ and internal resistance of $20 \Omega$, the null point on it is found to be at $1000 \mathrm{~cm}$. the resistance of whole wire is
1 $80 \Omega$
2 $100 \Omega$
3 $60 \Omega$
4 $120 \Omega$
Explanation:
B Given, $l_{\mathrm{AB}}=1200 \mathrm{~cm}, \mathrm{~V}_{\mathrm{AC}}=5 \mathrm{~V}$ $l_{\mathrm{AC}}=1000 \mathrm{~cm}$ $\mathrm{AB}$ is the potentiometer wire and $\mathrm{C}$ is the null point. $\mathrm{I}=$ current passing through the potentiometer formula. $\frac{\mathrm{V}_{\mathrm{AB}}}{l_{\mathrm{AB}}} =\frac{\mathrm{V}_{\mathrm{AC}}}{l_{\mathrm{AC}}}$ $\mathrm{V}_{\mathrm{AB}} =\frac{\mathrm{V}_{\mathrm{AC}}}{l_{\mathrm{AC}}} \times l_{\mathrm{AB}}$ $\mathrm{V}_{\mathrm{AB}} =\frac{5}{1000} \times 1200$ $\mathrm{~V}_{\mathrm{AB}} =6 \mathrm{~V}$ Resistance, $\mathrm{R}_{\mathrm{AB}}=\frac{6 \mathrm{~V}}{60 \times 10^{-3} \mathrm{~A}}=100 \Omega$
JEE Mains- 2020
Current Electricity
151841
The color code of a resistance is given below The values of resistance and tolerance respectively, are
1 $47 \mathrm{~K} \Omega, 10 \%$
2 $4.7 \mathrm{~K} \Omega, 5 \%$
3 $470 \Omega, 5 \%$
4 $47 \mathrm{~K} \Omega, 5 \%$
Explanation:
C The value of resistance and tolerance respectively, $\text { Yellow } \text { Violet } \text { Brown } \text { Gold }$ $4 7 1 5$ The colour coding information for electric resistances, we have $\mathrm{R}=47 \times 10^{1} \pm 5 \%$ $\mathrm{R}=470 \Omega \pm 5 \%$ Hence, the value of resistance and tolerance is $470 \Omega$ and $5 \%$.
151837
A cylindrical metallic wire is stretched to increase its length. If the resistance of the wire is increased by $4 \%$ then the percentage increase in its length is
1 $4 \%$
2 $8 \%$
3 $1 \%$
4 $2 \%$
Explanation:
D The change in resistance is due to both increase in length and decrease in cross-sectional area. We know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}} \text { or } \frac{\rho l}{\mathrm{~V}} {\left[\because \mathrm{A}=\frac{\mathrm{V}}{l}\right]}$ $\therefore \mathrm{R} \propto l^{2}$ $\text { Or } \frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\left(\frac{l^{\prime}}{l}\right)^{2} \ldots . .(\mathrm{i})$ According to the question, Resistance of the wire is increased by $4 \%$ $\therefore \quad \mathrm{R}^{\prime} =\mathrm{R}+\frac{4}{100} \mathrm{R}$ $\mathrm{R}^{\prime} =1.04 \mathrm{R}$ From equation (i) $\frac{1.04 \mathrm{R}}{\mathrm{R}}=\left(\frac{l^{\prime}}{l}\right)^{2}$ $\frac{l^{\prime}}{l}=\sqrt{1.04}$ $\frac{l^{\prime}}{l}=1.0198$ $l^{\prime}=1.02 l$ Now, percentage increase in its length, $\Delta l=\frac{l^{\prime}-l}{l} \times 100=\frac{1.02 l-l}{l} \times 100$ $\Delta l=2 \%$
TS EAMCET 05.08.2021
Current Electricity
151838 A carbon resistor with color code is shown in the figure. There is no fourth band in the resistor. The value of the resistance is
1 $24 \mathrm{M} \Omega \pm 20 \%$
2 $14 \mathrm{k} \Omega \pm 5 \%$
3 $24 \mathrm{k} \Omega \pm 20 \%$
4 $34 \mathrm{k} \Omega \pm 10 \%$
Explanation:
C $\operatorname{Red}=2$, yellow $=4$, Orange $=10^{3}$, No colour $=20 \%$ Therefore, resistance of carbon $=24 \times 10^{3} \pm 20 \%$ $=24 \mathrm{k} \Omega \pm 20 \%$
WB JEE 2021
Current Electricity
151839
The length of a potentiometer wire is $1200 \mathrm{~cm}$ and it carries a current of $60 \mathrm{~mA}$. For a cell of emf $5 \mathrm{~V}$ and internal resistance of $20 \Omega$, the null point on it is found to be at $1000 \mathrm{~cm}$. the resistance of whole wire is
1 $80 \Omega$
2 $100 \Omega$
3 $60 \Omega$
4 $120 \Omega$
Explanation:
B Given, $l_{\mathrm{AB}}=1200 \mathrm{~cm}, \mathrm{~V}_{\mathrm{AC}}=5 \mathrm{~V}$ $l_{\mathrm{AC}}=1000 \mathrm{~cm}$ $\mathrm{AB}$ is the potentiometer wire and $\mathrm{C}$ is the null point. $\mathrm{I}=$ current passing through the potentiometer formula. $\frac{\mathrm{V}_{\mathrm{AB}}}{l_{\mathrm{AB}}} =\frac{\mathrm{V}_{\mathrm{AC}}}{l_{\mathrm{AC}}}$ $\mathrm{V}_{\mathrm{AB}} =\frac{\mathrm{V}_{\mathrm{AC}}}{l_{\mathrm{AC}}} \times l_{\mathrm{AB}}$ $\mathrm{V}_{\mathrm{AB}} =\frac{5}{1000} \times 1200$ $\mathrm{~V}_{\mathrm{AB}} =6 \mathrm{~V}$ Resistance, $\mathrm{R}_{\mathrm{AB}}=\frac{6 \mathrm{~V}}{60 \times 10^{-3} \mathrm{~A}}=100 \Omega$
JEE Mains- 2020
Current Electricity
151841
The color code of a resistance is given below The values of resistance and tolerance respectively, are
1 $47 \mathrm{~K} \Omega, 10 \%$
2 $4.7 \mathrm{~K} \Omega, 5 \%$
3 $470 \Omega, 5 \%$
4 $47 \mathrm{~K} \Omega, 5 \%$
Explanation:
C The value of resistance and tolerance respectively, $\text { Yellow } \text { Violet } \text { Brown } \text { Gold }$ $4 7 1 5$ The colour coding information for electric resistances, we have $\mathrm{R}=47 \times 10^{1} \pm 5 \%$ $\mathrm{R}=470 \Omega \pm 5 \%$ Hence, the value of resistance and tolerance is $470 \Omega$ and $5 \%$.
151837
A cylindrical metallic wire is stretched to increase its length. If the resistance of the wire is increased by $4 \%$ then the percentage increase in its length is
1 $4 \%$
2 $8 \%$
3 $1 \%$
4 $2 \%$
Explanation:
D The change in resistance is due to both increase in length and decrease in cross-sectional area. We know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}} \text { or } \frac{\rho l}{\mathrm{~V}} {\left[\because \mathrm{A}=\frac{\mathrm{V}}{l}\right]}$ $\therefore \mathrm{R} \propto l^{2}$ $\text { Or } \frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\left(\frac{l^{\prime}}{l}\right)^{2} \ldots . .(\mathrm{i})$ According to the question, Resistance of the wire is increased by $4 \%$ $\therefore \quad \mathrm{R}^{\prime} =\mathrm{R}+\frac{4}{100} \mathrm{R}$ $\mathrm{R}^{\prime} =1.04 \mathrm{R}$ From equation (i) $\frac{1.04 \mathrm{R}}{\mathrm{R}}=\left(\frac{l^{\prime}}{l}\right)^{2}$ $\frac{l^{\prime}}{l}=\sqrt{1.04}$ $\frac{l^{\prime}}{l}=1.0198$ $l^{\prime}=1.02 l$ Now, percentage increase in its length, $\Delta l=\frac{l^{\prime}-l}{l} \times 100=\frac{1.02 l-l}{l} \times 100$ $\Delta l=2 \%$
TS EAMCET 05.08.2021
Current Electricity
151838 A carbon resistor with color code is shown in the figure. There is no fourth band in the resistor. The value of the resistance is
1 $24 \mathrm{M} \Omega \pm 20 \%$
2 $14 \mathrm{k} \Omega \pm 5 \%$
3 $24 \mathrm{k} \Omega \pm 20 \%$
4 $34 \mathrm{k} \Omega \pm 10 \%$
Explanation:
C $\operatorname{Red}=2$, yellow $=4$, Orange $=10^{3}$, No colour $=20 \%$ Therefore, resistance of carbon $=24 \times 10^{3} \pm 20 \%$ $=24 \mathrm{k} \Omega \pm 20 \%$
WB JEE 2021
Current Electricity
151839
The length of a potentiometer wire is $1200 \mathrm{~cm}$ and it carries a current of $60 \mathrm{~mA}$. For a cell of emf $5 \mathrm{~V}$ and internal resistance of $20 \Omega$, the null point on it is found to be at $1000 \mathrm{~cm}$. the resistance of whole wire is
1 $80 \Omega$
2 $100 \Omega$
3 $60 \Omega$
4 $120 \Omega$
Explanation:
B Given, $l_{\mathrm{AB}}=1200 \mathrm{~cm}, \mathrm{~V}_{\mathrm{AC}}=5 \mathrm{~V}$ $l_{\mathrm{AC}}=1000 \mathrm{~cm}$ $\mathrm{AB}$ is the potentiometer wire and $\mathrm{C}$ is the null point. $\mathrm{I}=$ current passing through the potentiometer formula. $\frac{\mathrm{V}_{\mathrm{AB}}}{l_{\mathrm{AB}}} =\frac{\mathrm{V}_{\mathrm{AC}}}{l_{\mathrm{AC}}}$ $\mathrm{V}_{\mathrm{AB}} =\frac{\mathrm{V}_{\mathrm{AC}}}{l_{\mathrm{AC}}} \times l_{\mathrm{AB}}$ $\mathrm{V}_{\mathrm{AB}} =\frac{5}{1000} \times 1200$ $\mathrm{~V}_{\mathrm{AB}} =6 \mathrm{~V}$ Resistance, $\mathrm{R}_{\mathrm{AB}}=\frac{6 \mathrm{~V}}{60 \times 10^{-3} \mathrm{~A}}=100 \Omega$
JEE Mains- 2020
Current Electricity
151841
The color code of a resistance is given below The values of resistance and tolerance respectively, are
1 $47 \mathrm{~K} \Omega, 10 \%$
2 $4.7 \mathrm{~K} \Omega, 5 \%$
3 $470 \Omega, 5 \%$
4 $47 \mathrm{~K} \Omega, 5 \%$
Explanation:
C The value of resistance and tolerance respectively, $\text { Yellow } \text { Violet } \text { Brown } \text { Gold }$ $4 7 1 5$ The colour coding information for electric resistances, we have $\mathrm{R}=47 \times 10^{1} \pm 5 \%$ $\mathrm{R}=470 \Omega \pm 5 \%$ Hence, the value of resistance and tolerance is $470 \Omega$ and $5 \%$.
151837
A cylindrical metallic wire is stretched to increase its length. If the resistance of the wire is increased by $4 \%$ then the percentage increase in its length is
1 $4 \%$
2 $8 \%$
3 $1 \%$
4 $2 \%$
Explanation:
D The change in resistance is due to both increase in length and decrease in cross-sectional area. We know that, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}} \text { or } \frac{\rho l}{\mathrm{~V}} {\left[\because \mathrm{A}=\frac{\mathrm{V}}{l}\right]}$ $\therefore \mathrm{R} \propto l^{2}$ $\text { Or } \frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\left(\frac{l^{\prime}}{l}\right)^{2} \ldots . .(\mathrm{i})$ According to the question, Resistance of the wire is increased by $4 \%$ $\therefore \quad \mathrm{R}^{\prime} =\mathrm{R}+\frac{4}{100} \mathrm{R}$ $\mathrm{R}^{\prime} =1.04 \mathrm{R}$ From equation (i) $\frac{1.04 \mathrm{R}}{\mathrm{R}}=\left(\frac{l^{\prime}}{l}\right)^{2}$ $\frac{l^{\prime}}{l}=\sqrt{1.04}$ $\frac{l^{\prime}}{l}=1.0198$ $l^{\prime}=1.02 l$ Now, percentage increase in its length, $\Delta l=\frac{l^{\prime}-l}{l} \times 100=\frac{1.02 l-l}{l} \times 100$ $\Delta l=2 \%$
TS EAMCET 05.08.2021
Current Electricity
151838 A carbon resistor with color code is shown in the figure. There is no fourth band in the resistor. The value of the resistance is
1 $24 \mathrm{M} \Omega \pm 20 \%$
2 $14 \mathrm{k} \Omega \pm 5 \%$
3 $24 \mathrm{k} \Omega \pm 20 \%$
4 $34 \mathrm{k} \Omega \pm 10 \%$
Explanation:
C $\operatorname{Red}=2$, yellow $=4$, Orange $=10^{3}$, No colour $=20 \%$ Therefore, resistance of carbon $=24 \times 10^{3} \pm 20 \%$ $=24 \mathrm{k} \Omega \pm 20 \%$
WB JEE 2021
Current Electricity
151839
The length of a potentiometer wire is $1200 \mathrm{~cm}$ and it carries a current of $60 \mathrm{~mA}$. For a cell of emf $5 \mathrm{~V}$ and internal resistance of $20 \Omega$, the null point on it is found to be at $1000 \mathrm{~cm}$. the resistance of whole wire is
1 $80 \Omega$
2 $100 \Omega$
3 $60 \Omega$
4 $120 \Omega$
Explanation:
B Given, $l_{\mathrm{AB}}=1200 \mathrm{~cm}, \mathrm{~V}_{\mathrm{AC}}=5 \mathrm{~V}$ $l_{\mathrm{AC}}=1000 \mathrm{~cm}$ $\mathrm{AB}$ is the potentiometer wire and $\mathrm{C}$ is the null point. $\mathrm{I}=$ current passing through the potentiometer formula. $\frac{\mathrm{V}_{\mathrm{AB}}}{l_{\mathrm{AB}}} =\frac{\mathrm{V}_{\mathrm{AC}}}{l_{\mathrm{AC}}}$ $\mathrm{V}_{\mathrm{AB}} =\frac{\mathrm{V}_{\mathrm{AC}}}{l_{\mathrm{AC}}} \times l_{\mathrm{AB}}$ $\mathrm{V}_{\mathrm{AB}} =\frac{5}{1000} \times 1200$ $\mathrm{~V}_{\mathrm{AB}} =6 \mathrm{~V}$ Resistance, $\mathrm{R}_{\mathrm{AB}}=\frac{6 \mathrm{~V}}{60 \times 10^{-3} \mathrm{~A}}=100 \Omega$
JEE Mains- 2020
Current Electricity
151841
The color code of a resistance is given below The values of resistance and tolerance respectively, are
1 $47 \mathrm{~K} \Omega, 10 \%$
2 $4.7 \mathrm{~K} \Omega, 5 \%$
3 $470 \Omega, 5 \%$
4 $47 \mathrm{~K} \Omega, 5 \%$
Explanation:
C The value of resistance and tolerance respectively, $\text { Yellow } \text { Violet } \text { Brown } \text { Gold }$ $4 7 1 5$ The colour coding information for electric resistances, we have $\mathrm{R}=47 \times 10^{1} \pm 5 \%$ $\mathrm{R}=470 \Omega \pm 5 \%$ Hence, the value of resistance and tolerance is $470 \Omega$ and $5 \%$.
