151915
The resistance $R_{t}$ of a conductor varies with temperature $t$ as shown in figure. If the variation is represented by $R_{t}=R_{0}\left(1+\alpha t+\beta t^{2}\right)$. Then,
1 $\alpha$ and $\beta$ both negative
2 $\alpha$ is positive and $\beta$ is negative
3 $\alpha$ and $\beta$ both are positive
4 $\alpha$ is negative and $\beta$ is negative
Explanation:
C Given, $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{0}\left(1+\alpha \mathrm{t}+\beta \mathrm{t}^{2}\right)$ For $\mathrm{R}$ to increase continuously $\alpha$ and $\beta$ both should be positive. Graph is upward, parabola to satisfy this condition $\alpha$ and $\beta$ both should be positive.
BCECE-2017
Current Electricity
151916
A carbon resistor is marked with the rings coloured brown, black, green and gold. The resistance in ohm is
1 $3.2 \times 10^{5} \pm 5 \%$
2 $1 \times 10^{6} \pm 10 \%$
3 $1 \times 10^{7} \pm 5 \%$
4 $1 \times 10^{6} \pm 5 \%$
Explanation:
D | Colour | Digit | | :--- | :--- | | Black | 0 | | Brown | 1 | | Red | 2 | | Orange | 3 | | Yellow | 4 | | Green | 5 | | Blue | 6 | | Violet | 7 | | Grey | 8 | $\text { Brown }=1 \%$ $\text { Red }=2 \%$ $\text { Gold }=5 \%$ $\text { Silver }=10 \%$ $\text { The resistance is }\left(1 \times 10^6 \pm 5 \%\right)$
COMEDK 2017
Current Electricity
151917
The external diameter of a $314 \mathrm{~m}$ long copper tube is $1.2 \mathrm{~cm}$ and the internal diameter is 1 $\mathrm{cm}$. Calculate its resistance if the specific resistance of copper is \(2.2 \times 10^{-8} \mathrm{ohm}\)-meter.
1 $5.0 \times 10^{-2} \mathrm{ohm}$
2 $4.4 \times 10^{-2} \mathrm{ohm}$
3 $3.14 \times 10^{-2} \mathrm{ohm}$
4 $2 \times 10^{-1} \mathrm{ohm}$
Explanation:
D Given, $l=314 \mathrm{~m}$ $\rho=2.2 \times 10^{-8} \mathrm{ohm}-$ meter $\mathrm{D}_{1}=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}$ $\mathrm{D}_{2}=1 \times 10^{-2} \mathrm{~m}$ We know that Resistance of tube $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R}=\frac{\rho l}{\pi\left[\mathrm{D}_{1}^{2}-\mathrm{D}_{2}^{2}\right]}$ $\mathrm{R}=\frac{2.2 \times 10^{-8} \times 314}{\frac{\pi}{4}\left(1.2^{2}-1^{2}\right) \times 10^{-4}}$ $\mathrm{R}=\frac{40 \times 2.2 \times 10^{-8}}{0.44 \times 10^{-4}}$ $\mathrm{R}=2 \times 10^{-1} \mathrm{ohm}$
COMEDK 2017
Current Electricity
151920
Two cylindrical rods $A$ and $B$ have same resistivity and same lengths. Diameter of $\operatorname{rod} A$ is twice the diameter of the rod $B$. Ratio of voltage drop across $\operatorname{rod} A$ to $\operatorname{rod} B$ is
1 $\frac{1}{2}$
2 2
3 4
4 $\frac{1}{4}$
Explanation:
D Given, Two cylinder A and B same length and same resistivity Diameter of $\operatorname{rod}(\mathrm{A})=2 \mathrm{R}$ We know that, Diameter of $\operatorname{rod}(\mathrm{B})=\mathrm{R}$ Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi r^{2}} \quad\left(\because \mathrm{A}=\pi \mathrm{r}^{2}\right)$ For $\operatorname{rod} \mathrm{A}$, $\mathrm{R}_{\mathrm{A}}=\frac{\rho l}{\pi(2 \mathrm{R})^{2}}$ For $\operatorname{rod} \mathrm{B}$, $\mathrm{R}_{\mathrm{B}}=\frac{\rho l}{\pi \mathrm{R}^{2}}$ From equation (i) and (ii), $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho l}{\pi(2 \mathrm{R})^{2}} \times \frac{\pi(\mathrm{R})^{2}}{\rho l}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho l}{\pi 4 \mathrm{R}^{2}} \times \frac{\pi \mathrm{R}^{2}}{\rho l}, \quad \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{4}$ $\therefore$ The ratio of voltage drop across $\operatorname{rod} \mathrm{A}$ to $\operatorname{rod} \mathrm{B}$ is, $\frac{(\mathrm{I} \times \mathrm{R})_{\mathrm{A}}}{(\mathrm{I} \times \mathrm{R})_{\mathrm{B}}}=\frac{1}{4}$ $\Rightarrow \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{1}{4}$
151915
The resistance $R_{t}$ of a conductor varies with temperature $t$ as shown in figure. If the variation is represented by $R_{t}=R_{0}\left(1+\alpha t+\beta t^{2}\right)$. Then,
1 $\alpha$ and $\beta$ both negative
2 $\alpha$ is positive and $\beta$ is negative
3 $\alpha$ and $\beta$ both are positive
4 $\alpha$ is negative and $\beta$ is negative
Explanation:
C Given, $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{0}\left(1+\alpha \mathrm{t}+\beta \mathrm{t}^{2}\right)$ For $\mathrm{R}$ to increase continuously $\alpha$ and $\beta$ both should be positive. Graph is upward, parabola to satisfy this condition $\alpha$ and $\beta$ both should be positive.
BCECE-2017
Current Electricity
151916
A carbon resistor is marked with the rings coloured brown, black, green and gold. The resistance in ohm is
1 $3.2 \times 10^{5} \pm 5 \%$
2 $1 \times 10^{6} \pm 10 \%$
3 $1 \times 10^{7} \pm 5 \%$
4 $1 \times 10^{6} \pm 5 \%$
Explanation:
D | Colour | Digit | | :--- | :--- | | Black | 0 | | Brown | 1 | | Red | 2 | | Orange | 3 | | Yellow | 4 | | Green | 5 | | Blue | 6 | | Violet | 7 | | Grey | 8 | $\text { Brown }=1 \%$ $\text { Red }=2 \%$ $\text { Gold }=5 \%$ $\text { Silver }=10 \%$ $\text { The resistance is }\left(1 \times 10^6 \pm 5 \%\right)$
COMEDK 2017
Current Electricity
151917
The external diameter of a $314 \mathrm{~m}$ long copper tube is $1.2 \mathrm{~cm}$ and the internal diameter is 1 $\mathrm{cm}$. Calculate its resistance if the specific resistance of copper is \(2.2 \times 10^{-8} \mathrm{ohm}\)-meter.
1 $5.0 \times 10^{-2} \mathrm{ohm}$
2 $4.4 \times 10^{-2} \mathrm{ohm}$
3 $3.14 \times 10^{-2} \mathrm{ohm}$
4 $2 \times 10^{-1} \mathrm{ohm}$
Explanation:
D Given, $l=314 \mathrm{~m}$ $\rho=2.2 \times 10^{-8} \mathrm{ohm}-$ meter $\mathrm{D}_{1}=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}$ $\mathrm{D}_{2}=1 \times 10^{-2} \mathrm{~m}$ We know that Resistance of tube $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R}=\frac{\rho l}{\pi\left[\mathrm{D}_{1}^{2}-\mathrm{D}_{2}^{2}\right]}$ $\mathrm{R}=\frac{2.2 \times 10^{-8} \times 314}{\frac{\pi}{4}\left(1.2^{2}-1^{2}\right) \times 10^{-4}}$ $\mathrm{R}=\frac{40 \times 2.2 \times 10^{-8}}{0.44 \times 10^{-4}}$ $\mathrm{R}=2 \times 10^{-1} \mathrm{ohm}$
COMEDK 2017
Current Electricity
151920
Two cylindrical rods $A$ and $B$ have same resistivity and same lengths. Diameter of $\operatorname{rod} A$ is twice the diameter of the rod $B$. Ratio of voltage drop across $\operatorname{rod} A$ to $\operatorname{rod} B$ is
1 $\frac{1}{2}$
2 2
3 4
4 $\frac{1}{4}$
Explanation:
D Given, Two cylinder A and B same length and same resistivity Diameter of $\operatorname{rod}(\mathrm{A})=2 \mathrm{R}$ We know that, Diameter of $\operatorname{rod}(\mathrm{B})=\mathrm{R}$ Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi r^{2}} \quad\left(\because \mathrm{A}=\pi \mathrm{r}^{2}\right)$ For $\operatorname{rod} \mathrm{A}$, $\mathrm{R}_{\mathrm{A}}=\frac{\rho l}{\pi(2 \mathrm{R})^{2}}$ For $\operatorname{rod} \mathrm{B}$, $\mathrm{R}_{\mathrm{B}}=\frac{\rho l}{\pi \mathrm{R}^{2}}$ From equation (i) and (ii), $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho l}{\pi(2 \mathrm{R})^{2}} \times \frac{\pi(\mathrm{R})^{2}}{\rho l}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho l}{\pi 4 \mathrm{R}^{2}} \times \frac{\pi \mathrm{R}^{2}}{\rho l}, \quad \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{4}$ $\therefore$ The ratio of voltage drop across $\operatorname{rod} \mathrm{A}$ to $\operatorname{rod} \mathrm{B}$ is, $\frac{(\mathrm{I} \times \mathrm{R})_{\mathrm{A}}}{(\mathrm{I} \times \mathrm{R})_{\mathrm{B}}}=\frac{1}{4}$ $\Rightarrow \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{1}{4}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
151915
The resistance $R_{t}$ of a conductor varies with temperature $t$ as shown in figure. If the variation is represented by $R_{t}=R_{0}\left(1+\alpha t+\beta t^{2}\right)$. Then,
1 $\alpha$ and $\beta$ both negative
2 $\alpha$ is positive and $\beta$ is negative
3 $\alpha$ and $\beta$ both are positive
4 $\alpha$ is negative and $\beta$ is negative
Explanation:
C Given, $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{0}\left(1+\alpha \mathrm{t}+\beta \mathrm{t}^{2}\right)$ For $\mathrm{R}$ to increase continuously $\alpha$ and $\beta$ both should be positive. Graph is upward, parabola to satisfy this condition $\alpha$ and $\beta$ both should be positive.
BCECE-2017
Current Electricity
151916
A carbon resistor is marked with the rings coloured brown, black, green and gold. The resistance in ohm is
1 $3.2 \times 10^{5} \pm 5 \%$
2 $1 \times 10^{6} \pm 10 \%$
3 $1 \times 10^{7} \pm 5 \%$
4 $1 \times 10^{6} \pm 5 \%$
Explanation:
D | Colour | Digit | | :--- | :--- | | Black | 0 | | Brown | 1 | | Red | 2 | | Orange | 3 | | Yellow | 4 | | Green | 5 | | Blue | 6 | | Violet | 7 | | Grey | 8 | $\text { Brown }=1 \%$ $\text { Red }=2 \%$ $\text { Gold }=5 \%$ $\text { Silver }=10 \%$ $\text { The resistance is }\left(1 \times 10^6 \pm 5 \%\right)$
COMEDK 2017
Current Electricity
151917
The external diameter of a $314 \mathrm{~m}$ long copper tube is $1.2 \mathrm{~cm}$ and the internal diameter is 1 $\mathrm{cm}$. Calculate its resistance if the specific resistance of copper is \(2.2 \times 10^{-8} \mathrm{ohm}\)-meter.
1 $5.0 \times 10^{-2} \mathrm{ohm}$
2 $4.4 \times 10^{-2} \mathrm{ohm}$
3 $3.14 \times 10^{-2} \mathrm{ohm}$
4 $2 \times 10^{-1} \mathrm{ohm}$
Explanation:
D Given, $l=314 \mathrm{~m}$ $\rho=2.2 \times 10^{-8} \mathrm{ohm}-$ meter $\mathrm{D}_{1}=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}$ $\mathrm{D}_{2}=1 \times 10^{-2} \mathrm{~m}$ We know that Resistance of tube $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R}=\frac{\rho l}{\pi\left[\mathrm{D}_{1}^{2}-\mathrm{D}_{2}^{2}\right]}$ $\mathrm{R}=\frac{2.2 \times 10^{-8} \times 314}{\frac{\pi}{4}\left(1.2^{2}-1^{2}\right) \times 10^{-4}}$ $\mathrm{R}=\frac{40 \times 2.2 \times 10^{-8}}{0.44 \times 10^{-4}}$ $\mathrm{R}=2 \times 10^{-1} \mathrm{ohm}$
COMEDK 2017
Current Electricity
151920
Two cylindrical rods $A$ and $B$ have same resistivity and same lengths. Diameter of $\operatorname{rod} A$ is twice the diameter of the rod $B$. Ratio of voltage drop across $\operatorname{rod} A$ to $\operatorname{rod} B$ is
1 $\frac{1}{2}$
2 2
3 4
4 $\frac{1}{4}$
Explanation:
D Given, Two cylinder A and B same length and same resistivity Diameter of $\operatorname{rod}(\mathrm{A})=2 \mathrm{R}$ We know that, Diameter of $\operatorname{rod}(\mathrm{B})=\mathrm{R}$ Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi r^{2}} \quad\left(\because \mathrm{A}=\pi \mathrm{r}^{2}\right)$ For $\operatorname{rod} \mathrm{A}$, $\mathrm{R}_{\mathrm{A}}=\frac{\rho l}{\pi(2 \mathrm{R})^{2}}$ For $\operatorname{rod} \mathrm{B}$, $\mathrm{R}_{\mathrm{B}}=\frac{\rho l}{\pi \mathrm{R}^{2}}$ From equation (i) and (ii), $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho l}{\pi(2 \mathrm{R})^{2}} \times \frac{\pi(\mathrm{R})^{2}}{\rho l}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho l}{\pi 4 \mathrm{R}^{2}} \times \frac{\pi \mathrm{R}^{2}}{\rho l}, \quad \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{4}$ $\therefore$ The ratio of voltage drop across $\operatorname{rod} \mathrm{A}$ to $\operatorname{rod} \mathrm{B}$ is, $\frac{(\mathrm{I} \times \mathrm{R})_{\mathrm{A}}}{(\mathrm{I} \times \mathrm{R})_{\mathrm{B}}}=\frac{1}{4}$ $\Rightarrow \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{1}{4}$
151915
The resistance $R_{t}$ of a conductor varies with temperature $t$ as shown in figure. If the variation is represented by $R_{t}=R_{0}\left(1+\alpha t+\beta t^{2}\right)$. Then,
1 $\alpha$ and $\beta$ both negative
2 $\alpha$ is positive and $\beta$ is negative
3 $\alpha$ and $\beta$ both are positive
4 $\alpha$ is negative and $\beta$ is negative
Explanation:
C Given, $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{0}\left(1+\alpha \mathrm{t}+\beta \mathrm{t}^{2}\right)$ For $\mathrm{R}$ to increase continuously $\alpha$ and $\beta$ both should be positive. Graph is upward, parabola to satisfy this condition $\alpha$ and $\beta$ both should be positive.
BCECE-2017
Current Electricity
151916
A carbon resistor is marked with the rings coloured brown, black, green and gold. The resistance in ohm is
1 $3.2 \times 10^{5} \pm 5 \%$
2 $1 \times 10^{6} \pm 10 \%$
3 $1 \times 10^{7} \pm 5 \%$
4 $1 \times 10^{6} \pm 5 \%$
Explanation:
D | Colour | Digit | | :--- | :--- | | Black | 0 | | Brown | 1 | | Red | 2 | | Orange | 3 | | Yellow | 4 | | Green | 5 | | Blue | 6 | | Violet | 7 | | Grey | 8 | $\text { Brown }=1 \%$ $\text { Red }=2 \%$ $\text { Gold }=5 \%$ $\text { Silver }=10 \%$ $\text { The resistance is }\left(1 \times 10^6 \pm 5 \%\right)$
COMEDK 2017
Current Electricity
151917
The external diameter of a $314 \mathrm{~m}$ long copper tube is $1.2 \mathrm{~cm}$ and the internal diameter is 1 $\mathrm{cm}$. Calculate its resistance if the specific resistance of copper is \(2.2 \times 10^{-8} \mathrm{ohm}\)-meter.
1 $5.0 \times 10^{-2} \mathrm{ohm}$
2 $4.4 \times 10^{-2} \mathrm{ohm}$
3 $3.14 \times 10^{-2} \mathrm{ohm}$
4 $2 \times 10^{-1} \mathrm{ohm}$
Explanation:
D Given, $l=314 \mathrm{~m}$ $\rho=2.2 \times 10^{-8} \mathrm{ohm}-$ meter $\mathrm{D}_{1}=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}$ $\mathrm{D}_{2}=1 \times 10^{-2} \mathrm{~m}$ We know that Resistance of tube $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ $\mathrm{R}=\frac{\rho l}{\pi\left[\mathrm{D}_{1}^{2}-\mathrm{D}_{2}^{2}\right]}$ $\mathrm{R}=\frac{2.2 \times 10^{-8} \times 314}{\frac{\pi}{4}\left(1.2^{2}-1^{2}\right) \times 10^{-4}}$ $\mathrm{R}=\frac{40 \times 2.2 \times 10^{-8}}{0.44 \times 10^{-4}}$ $\mathrm{R}=2 \times 10^{-1} \mathrm{ohm}$
COMEDK 2017
Current Electricity
151920
Two cylindrical rods $A$ and $B$ have same resistivity and same lengths. Diameter of $\operatorname{rod} A$ is twice the diameter of the rod $B$. Ratio of voltage drop across $\operatorname{rod} A$ to $\operatorname{rod} B$ is
1 $\frac{1}{2}$
2 2
3 4
4 $\frac{1}{4}$
Explanation:
D Given, Two cylinder A and B same length and same resistivity Diameter of $\operatorname{rod}(\mathrm{A})=2 \mathrm{R}$ We know that, Diameter of $\operatorname{rod}(\mathrm{B})=\mathrm{R}$ Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi r^{2}} \quad\left(\because \mathrm{A}=\pi \mathrm{r}^{2}\right)$ For $\operatorname{rod} \mathrm{A}$, $\mathrm{R}_{\mathrm{A}}=\frac{\rho l}{\pi(2 \mathrm{R})^{2}}$ For $\operatorname{rod} \mathrm{B}$, $\mathrm{R}_{\mathrm{B}}=\frac{\rho l}{\pi \mathrm{R}^{2}}$ From equation (i) and (ii), $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho l}{\pi(2 \mathrm{R})^{2}} \times \frac{\pi(\mathrm{R})^{2}}{\rho l}$ $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho l}{\pi 4 \mathrm{R}^{2}} \times \frac{\pi \mathrm{R}^{2}}{\rho l}, \quad \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{4}$ $\therefore$ The ratio of voltage drop across $\operatorname{rod} \mathrm{A}$ to $\operatorname{rod} \mathrm{B}$ is, $\frac{(\mathrm{I} \times \mathrm{R})_{\mathrm{A}}}{(\mathrm{I} \times \mathrm{R})_{\mathrm{B}}}=\frac{1}{4}$ $\Rightarrow \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{1}{4}$
