151870
Six equal resistances are connected between the point $A, B$ and $C$ as shown in figure below. If $\mathbf{R}_{\mathrm{AB}}, \mathbf{R}_{\mathrm{AC}}$ and $\mathbf{R}_{\mathrm{BC}}$ are resistances between points $A-B, A-C$ and $B-C$ respectively, then correct option is
D Equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$, $\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{\mathrm{R}}+\frac{1}{\left(\frac{\mathrm{R}}{2}+\frac{\mathrm{R}}{3}\right)}=\frac{1}{\mathrm{R}}+\frac{6}{5 \mathrm{R}}=\frac{11}{5 \mathrm{R}}$ $\mathrm{R}_{\mathrm{AB}}=\frac{5 \mathrm{R}}{11} \Omega$ Equivalent resistance between $\mathrm{B}$ and $\mathrm{C}$, $\frac{1}{\mathrm{R}_{B C}}=\frac{1}{\mathrm{R} / 2}+\frac{1}{\left(\mathrm{R}+\frac{\mathrm{R}}{3}\right)}=\frac{2}{\mathrm{R}}+\frac{3}{4 \mathrm{R}}=\frac{11}{4 \mathrm{R}}$ $\mathrm{R}_{\mathrm{BC}}=\frac{4 \mathrm{R}}{11} \Omega$ Similarly, $\mathrm{R}_{\mathrm{CA}}=\frac{3 \mathrm{R}}{11} \Omega$ Comparing equation (i), (ii) and (iii), we get- $\mathrm{R}_{\mathrm{AB}}: \mathrm{R}_{\mathrm{BC}}: \mathrm{R}_{\mathrm{CA}}=5: 4: 3$ Then, $R_{A B}>R_{B C}>R_{C A}$
TS EAMCET 08.05.2019
Current Electricity
151871
The current $I_{3}$ in the following circuit is
1 $\frac{-2}{35} \mathrm{~A}$
2 $\frac{4}{35} \mathrm{~A}$
3 $\frac{2}{35} \mathrm{~A}$
4 $-\frac{4}{35} \mathrm{~A}$
Explanation:
D From figure- $\mathrm{I}_{2}=\mathrm{I}_{1}+\mathrm{I}_{3}$ Applying Kirchhoff's voltage law in loop BCDEB, $25 \mathrm{I}_{3}+5+50 \mathrm{I}_{2}=0$ $25 \mathrm{I}_{3}+5+50\left(\mathrm{I}_{3}+\mathrm{I}_{1}\right)=0$ $75 \mathrm{I}_{3}+5+50 \mathrm{I}_{1}=0$ Again applying Kirchhoff's voltage law in loop ABCDEFA, $-100 \mathrm{I}_{1}+25 \mathrm{I}_{3}=-10$ From equation (i) and (ii), we get- $I_{3}=-\frac{4}{35} A$
TS EAMCET 08.05.2019
Current Electricity
151873
What will be the resistance of a wire at 0 degree Celsius. If its resistance is $5 \mathrm{ohm}$ at 50 degree Celsius and $6 \mathrm{ohm}$ at 100 degree celsius?
1 $3 \mathrm{ohm}$
2 $2 \mathrm{ohm}$
3 $1 \mathrm{ohm}$
4 $4 \mathrm{ohm}$
Explanation:
D Given that, $\mathrm{R}_{1}=5 \mathrm{ohm}$ $\mathrm{T}_{1}=50^{\circ} \mathrm{C}$ $\mathrm{R}_{2}=6 \mathrm{ohm}$ $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ $\mathrm{R}_{0}=$ ? We know that, $\mathrm{R}=\mathrm{R}_{0}[1+\alpha \mathrm{T}]$ $\mathrm{R}_{1}=\mathrm{R}_{0}\left[1+\alpha \mathrm{T}_{1}\right]$ $5=\mathrm{R}_{0}[1+\alpha \times 50]$ $\mathrm{R}_{2}=\mathrm{R}_{0}\left[1+\alpha \mathrm{T}_{2}\right]$ $6=\mathrm{R}_{0}[1+\alpha \times 100]$ Dividing equation (i) by (ii), we get - $\frac{5}{6}=\frac{\mathrm{R}_{0}[1+50 \alpha]}{\mathrm{R}_{0}[1+100 \alpha]}$ $\frac{5}{6}=\frac{1+50 \alpha}{1+100 \alpha}$ $5+500 \alpha=6+300 \alpha$ $200 \alpha=1$ $\alpha=\frac{1}{200}$ Putting the value of ' $\alpha$ ' in equation (i), $5=\mathrm{R}_{0}[1+50 \alpha]$ $5=\mathrm{R}_{0}\left[1+50 \times \frac{1}{200}\right]$ $5=\mathrm{R}_{0}\left[1+\frac{1}{4}\right]=\frac{5 \mathrm{R}_{0}}{4}$ $\mathrm{R}_{0}=4 \mathrm{ohm}$
J and K-CET-2019
Current Electricity
151874
The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then
1 the resistance and the specific resistance, will both remain unchanged
2 the resistance will be doubled and the specific resistance will be halved
3 The resistance will be halved and the specific resistance will remain unchanged
4 The resistance will be halved and the specific resistance will be doubled
Explanation:
C Given, $l_{1}=l, l_{2}=2 l$ $\mathrm{r}_{1}=\mathrm{r}, \mathrm{r}_{2}=2 \mathrm{r}$ Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\pi \mathrm{r}_{2}^{2}}{\pi \mathrm{r}_{1}^{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l}{2 l} \times\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{2} \times 4$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=2$ $\mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{2}$ The resistance will become half but specific resistance will remains same due to same material.
151870
Six equal resistances are connected between the point $A, B$ and $C$ as shown in figure below. If $\mathbf{R}_{\mathrm{AB}}, \mathbf{R}_{\mathrm{AC}}$ and $\mathbf{R}_{\mathrm{BC}}$ are resistances between points $A-B, A-C$ and $B-C$ respectively, then correct option is
D Equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$, $\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{\mathrm{R}}+\frac{1}{\left(\frac{\mathrm{R}}{2}+\frac{\mathrm{R}}{3}\right)}=\frac{1}{\mathrm{R}}+\frac{6}{5 \mathrm{R}}=\frac{11}{5 \mathrm{R}}$ $\mathrm{R}_{\mathrm{AB}}=\frac{5 \mathrm{R}}{11} \Omega$ Equivalent resistance between $\mathrm{B}$ and $\mathrm{C}$, $\frac{1}{\mathrm{R}_{B C}}=\frac{1}{\mathrm{R} / 2}+\frac{1}{\left(\mathrm{R}+\frac{\mathrm{R}}{3}\right)}=\frac{2}{\mathrm{R}}+\frac{3}{4 \mathrm{R}}=\frac{11}{4 \mathrm{R}}$ $\mathrm{R}_{\mathrm{BC}}=\frac{4 \mathrm{R}}{11} \Omega$ Similarly, $\mathrm{R}_{\mathrm{CA}}=\frac{3 \mathrm{R}}{11} \Omega$ Comparing equation (i), (ii) and (iii), we get- $\mathrm{R}_{\mathrm{AB}}: \mathrm{R}_{\mathrm{BC}}: \mathrm{R}_{\mathrm{CA}}=5: 4: 3$ Then, $R_{A B}>R_{B C}>R_{C A}$
TS EAMCET 08.05.2019
Current Electricity
151871
The current $I_{3}$ in the following circuit is
1 $\frac{-2}{35} \mathrm{~A}$
2 $\frac{4}{35} \mathrm{~A}$
3 $\frac{2}{35} \mathrm{~A}$
4 $-\frac{4}{35} \mathrm{~A}$
Explanation:
D From figure- $\mathrm{I}_{2}=\mathrm{I}_{1}+\mathrm{I}_{3}$ Applying Kirchhoff's voltage law in loop BCDEB, $25 \mathrm{I}_{3}+5+50 \mathrm{I}_{2}=0$ $25 \mathrm{I}_{3}+5+50\left(\mathrm{I}_{3}+\mathrm{I}_{1}\right)=0$ $75 \mathrm{I}_{3}+5+50 \mathrm{I}_{1}=0$ Again applying Kirchhoff's voltage law in loop ABCDEFA, $-100 \mathrm{I}_{1}+25 \mathrm{I}_{3}=-10$ From equation (i) and (ii), we get- $I_{3}=-\frac{4}{35} A$
TS EAMCET 08.05.2019
Current Electricity
151873
What will be the resistance of a wire at 0 degree Celsius. If its resistance is $5 \mathrm{ohm}$ at 50 degree Celsius and $6 \mathrm{ohm}$ at 100 degree celsius?
1 $3 \mathrm{ohm}$
2 $2 \mathrm{ohm}$
3 $1 \mathrm{ohm}$
4 $4 \mathrm{ohm}$
Explanation:
D Given that, $\mathrm{R}_{1}=5 \mathrm{ohm}$ $\mathrm{T}_{1}=50^{\circ} \mathrm{C}$ $\mathrm{R}_{2}=6 \mathrm{ohm}$ $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ $\mathrm{R}_{0}=$ ? We know that, $\mathrm{R}=\mathrm{R}_{0}[1+\alpha \mathrm{T}]$ $\mathrm{R}_{1}=\mathrm{R}_{0}\left[1+\alpha \mathrm{T}_{1}\right]$ $5=\mathrm{R}_{0}[1+\alpha \times 50]$ $\mathrm{R}_{2}=\mathrm{R}_{0}\left[1+\alpha \mathrm{T}_{2}\right]$ $6=\mathrm{R}_{0}[1+\alpha \times 100]$ Dividing equation (i) by (ii), we get - $\frac{5}{6}=\frac{\mathrm{R}_{0}[1+50 \alpha]}{\mathrm{R}_{0}[1+100 \alpha]}$ $\frac{5}{6}=\frac{1+50 \alpha}{1+100 \alpha}$ $5+500 \alpha=6+300 \alpha$ $200 \alpha=1$ $\alpha=\frac{1}{200}$ Putting the value of ' $\alpha$ ' in equation (i), $5=\mathrm{R}_{0}[1+50 \alpha]$ $5=\mathrm{R}_{0}\left[1+50 \times \frac{1}{200}\right]$ $5=\mathrm{R}_{0}\left[1+\frac{1}{4}\right]=\frac{5 \mathrm{R}_{0}}{4}$ $\mathrm{R}_{0}=4 \mathrm{ohm}$
J and K-CET-2019
Current Electricity
151874
The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then
1 the resistance and the specific resistance, will both remain unchanged
2 the resistance will be doubled and the specific resistance will be halved
3 The resistance will be halved and the specific resistance will remain unchanged
4 The resistance will be halved and the specific resistance will be doubled
Explanation:
C Given, $l_{1}=l, l_{2}=2 l$ $\mathrm{r}_{1}=\mathrm{r}, \mathrm{r}_{2}=2 \mathrm{r}$ Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\pi \mathrm{r}_{2}^{2}}{\pi \mathrm{r}_{1}^{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l}{2 l} \times\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{2} \times 4$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=2$ $\mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{2}$ The resistance will become half but specific resistance will remains same due to same material.
151870
Six equal resistances are connected between the point $A, B$ and $C$ as shown in figure below. If $\mathbf{R}_{\mathrm{AB}}, \mathbf{R}_{\mathrm{AC}}$ and $\mathbf{R}_{\mathrm{BC}}$ are resistances between points $A-B, A-C$ and $B-C$ respectively, then correct option is
D Equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$, $\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{\mathrm{R}}+\frac{1}{\left(\frac{\mathrm{R}}{2}+\frac{\mathrm{R}}{3}\right)}=\frac{1}{\mathrm{R}}+\frac{6}{5 \mathrm{R}}=\frac{11}{5 \mathrm{R}}$ $\mathrm{R}_{\mathrm{AB}}=\frac{5 \mathrm{R}}{11} \Omega$ Equivalent resistance between $\mathrm{B}$ and $\mathrm{C}$, $\frac{1}{\mathrm{R}_{B C}}=\frac{1}{\mathrm{R} / 2}+\frac{1}{\left(\mathrm{R}+\frac{\mathrm{R}}{3}\right)}=\frac{2}{\mathrm{R}}+\frac{3}{4 \mathrm{R}}=\frac{11}{4 \mathrm{R}}$ $\mathrm{R}_{\mathrm{BC}}=\frac{4 \mathrm{R}}{11} \Omega$ Similarly, $\mathrm{R}_{\mathrm{CA}}=\frac{3 \mathrm{R}}{11} \Omega$ Comparing equation (i), (ii) and (iii), we get- $\mathrm{R}_{\mathrm{AB}}: \mathrm{R}_{\mathrm{BC}}: \mathrm{R}_{\mathrm{CA}}=5: 4: 3$ Then, $R_{A B}>R_{B C}>R_{C A}$
TS EAMCET 08.05.2019
Current Electricity
151871
The current $I_{3}$ in the following circuit is
1 $\frac{-2}{35} \mathrm{~A}$
2 $\frac{4}{35} \mathrm{~A}$
3 $\frac{2}{35} \mathrm{~A}$
4 $-\frac{4}{35} \mathrm{~A}$
Explanation:
D From figure- $\mathrm{I}_{2}=\mathrm{I}_{1}+\mathrm{I}_{3}$ Applying Kirchhoff's voltage law in loop BCDEB, $25 \mathrm{I}_{3}+5+50 \mathrm{I}_{2}=0$ $25 \mathrm{I}_{3}+5+50\left(\mathrm{I}_{3}+\mathrm{I}_{1}\right)=0$ $75 \mathrm{I}_{3}+5+50 \mathrm{I}_{1}=0$ Again applying Kirchhoff's voltage law in loop ABCDEFA, $-100 \mathrm{I}_{1}+25 \mathrm{I}_{3}=-10$ From equation (i) and (ii), we get- $I_{3}=-\frac{4}{35} A$
TS EAMCET 08.05.2019
Current Electricity
151873
What will be the resistance of a wire at 0 degree Celsius. If its resistance is $5 \mathrm{ohm}$ at 50 degree Celsius and $6 \mathrm{ohm}$ at 100 degree celsius?
1 $3 \mathrm{ohm}$
2 $2 \mathrm{ohm}$
3 $1 \mathrm{ohm}$
4 $4 \mathrm{ohm}$
Explanation:
D Given that, $\mathrm{R}_{1}=5 \mathrm{ohm}$ $\mathrm{T}_{1}=50^{\circ} \mathrm{C}$ $\mathrm{R}_{2}=6 \mathrm{ohm}$ $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ $\mathrm{R}_{0}=$ ? We know that, $\mathrm{R}=\mathrm{R}_{0}[1+\alpha \mathrm{T}]$ $\mathrm{R}_{1}=\mathrm{R}_{0}\left[1+\alpha \mathrm{T}_{1}\right]$ $5=\mathrm{R}_{0}[1+\alpha \times 50]$ $\mathrm{R}_{2}=\mathrm{R}_{0}\left[1+\alpha \mathrm{T}_{2}\right]$ $6=\mathrm{R}_{0}[1+\alpha \times 100]$ Dividing equation (i) by (ii), we get - $\frac{5}{6}=\frac{\mathrm{R}_{0}[1+50 \alpha]}{\mathrm{R}_{0}[1+100 \alpha]}$ $\frac{5}{6}=\frac{1+50 \alpha}{1+100 \alpha}$ $5+500 \alpha=6+300 \alpha$ $200 \alpha=1$ $\alpha=\frac{1}{200}$ Putting the value of ' $\alpha$ ' in equation (i), $5=\mathrm{R}_{0}[1+50 \alpha]$ $5=\mathrm{R}_{0}\left[1+50 \times \frac{1}{200}\right]$ $5=\mathrm{R}_{0}\left[1+\frac{1}{4}\right]=\frac{5 \mathrm{R}_{0}}{4}$ $\mathrm{R}_{0}=4 \mathrm{ohm}$
J and K-CET-2019
Current Electricity
151874
The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then
1 the resistance and the specific resistance, will both remain unchanged
2 the resistance will be doubled and the specific resistance will be halved
3 The resistance will be halved and the specific resistance will remain unchanged
4 The resistance will be halved and the specific resistance will be doubled
Explanation:
C Given, $l_{1}=l, l_{2}=2 l$ $\mathrm{r}_{1}=\mathrm{r}, \mathrm{r}_{2}=2 \mathrm{r}$ Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\pi \mathrm{r}_{2}^{2}}{\pi \mathrm{r}_{1}^{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l}{2 l} \times\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{2} \times 4$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=2$ $\mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{2}$ The resistance will become half but specific resistance will remains same due to same material.
151870
Six equal resistances are connected between the point $A, B$ and $C$ as shown in figure below. If $\mathbf{R}_{\mathrm{AB}}, \mathbf{R}_{\mathrm{AC}}$ and $\mathbf{R}_{\mathrm{BC}}$ are resistances between points $A-B, A-C$ and $B-C$ respectively, then correct option is
D Equivalent resistance between $\mathrm{A}$ and $\mathrm{B}$, $\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{\mathrm{R}}+\frac{1}{\left(\frac{\mathrm{R}}{2}+\frac{\mathrm{R}}{3}\right)}=\frac{1}{\mathrm{R}}+\frac{6}{5 \mathrm{R}}=\frac{11}{5 \mathrm{R}}$ $\mathrm{R}_{\mathrm{AB}}=\frac{5 \mathrm{R}}{11} \Omega$ Equivalent resistance between $\mathrm{B}$ and $\mathrm{C}$, $\frac{1}{\mathrm{R}_{B C}}=\frac{1}{\mathrm{R} / 2}+\frac{1}{\left(\mathrm{R}+\frac{\mathrm{R}}{3}\right)}=\frac{2}{\mathrm{R}}+\frac{3}{4 \mathrm{R}}=\frac{11}{4 \mathrm{R}}$ $\mathrm{R}_{\mathrm{BC}}=\frac{4 \mathrm{R}}{11} \Omega$ Similarly, $\mathrm{R}_{\mathrm{CA}}=\frac{3 \mathrm{R}}{11} \Omega$ Comparing equation (i), (ii) and (iii), we get- $\mathrm{R}_{\mathrm{AB}}: \mathrm{R}_{\mathrm{BC}}: \mathrm{R}_{\mathrm{CA}}=5: 4: 3$ Then, $R_{A B}>R_{B C}>R_{C A}$
TS EAMCET 08.05.2019
Current Electricity
151871
The current $I_{3}$ in the following circuit is
1 $\frac{-2}{35} \mathrm{~A}$
2 $\frac{4}{35} \mathrm{~A}$
3 $\frac{2}{35} \mathrm{~A}$
4 $-\frac{4}{35} \mathrm{~A}$
Explanation:
D From figure- $\mathrm{I}_{2}=\mathrm{I}_{1}+\mathrm{I}_{3}$ Applying Kirchhoff's voltage law in loop BCDEB, $25 \mathrm{I}_{3}+5+50 \mathrm{I}_{2}=0$ $25 \mathrm{I}_{3}+5+50\left(\mathrm{I}_{3}+\mathrm{I}_{1}\right)=0$ $75 \mathrm{I}_{3}+5+50 \mathrm{I}_{1}=0$ Again applying Kirchhoff's voltage law in loop ABCDEFA, $-100 \mathrm{I}_{1}+25 \mathrm{I}_{3}=-10$ From equation (i) and (ii), we get- $I_{3}=-\frac{4}{35} A$
TS EAMCET 08.05.2019
Current Electricity
151873
What will be the resistance of a wire at 0 degree Celsius. If its resistance is $5 \mathrm{ohm}$ at 50 degree Celsius and $6 \mathrm{ohm}$ at 100 degree celsius?
1 $3 \mathrm{ohm}$
2 $2 \mathrm{ohm}$
3 $1 \mathrm{ohm}$
4 $4 \mathrm{ohm}$
Explanation:
D Given that, $\mathrm{R}_{1}=5 \mathrm{ohm}$ $\mathrm{T}_{1}=50^{\circ} \mathrm{C}$ $\mathrm{R}_{2}=6 \mathrm{ohm}$ $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$ $\mathrm{R}_{0}=$ ? We know that, $\mathrm{R}=\mathrm{R}_{0}[1+\alpha \mathrm{T}]$ $\mathrm{R}_{1}=\mathrm{R}_{0}\left[1+\alpha \mathrm{T}_{1}\right]$ $5=\mathrm{R}_{0}[1+\alpha \times 50]$ $\mathrm{R}_{2}=\mathrm{R}_{0}\left[1+\alpha \mathrm{T}_{2}\right]$ $6=\mathrm{R}_{0}[1+\alpha \times 100]$ Dividing equation (i) by (ii), we get - $\frac{5}{6}=\frac{\mathrm{R}_{0}[1+50 \alpha]}{\mathrm{R}_{0}[1+100 \alpha]}$ $\frac{5}{6}=\frac{1+50 \alpha}{1+100 \alpha}$ $5+500 \alpha=6+300 \alpha$ $200 \alpha=1$ $\alpha=\frac{1}{200}$ Putting the value of ' $\alpha$ ' in equation (i), $5=\mathrm{R}_{0}[1+50 \alpha]$ $5=\mathrm{R}_{0}\left[1+50 \times \frac{1}{200}\right]$ $5=\mathrm{R}_{0}\left[1+\frac{1}{4}\right]=\frac{5 \mathrm{R}_{0}}{4}$ $\mathrm{R}_{0}=4 \mathrm{ohm}$
J and K-CET-2019
Current Electricity
151874
The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then
1 the resistance and the specific resistance, will both remain unchanged
2 the resistance will be doubled and the specific resistance will be halved
3 The resistance will be halved and the specific resistance will remain unchanged
4 The resistance will be halved and the specific resistance will be doubled
Explanation:
C Given, $l_{1}=l, l_{2}=2 l$ $\mathrm{r}_{1}=\mathrm{r}, \mathrm{r}_{2}=2 \mathrm{r}$ Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\pi \mathrm{r}_{2}^{2}}{\pi \mathrm{r}_{1}^{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l_{1}}{l_{2}} \times \frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{l}{2 l} \times\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{2} \times 4$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=2$ $\mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{2}$ The resistance will become half but specific resistance will remains same due to same material.
