151814
If the potential at $A$ is greater than the potential at $B$, then the equivalent resistance of the circuit across $A B$ is
1 $4.4 \Omega$
2 $5.2 \Omega$
3 $6 \Omega$
4 $9 \Omega$
5 $3.6 \Omega$
Explanation:
E Given circuit – Redraw the figure We can see that it behave like a Wheat Stone's Bridge, $\frac{3}{6}=\frac{2}{4} \quad[\because 4 \Omega \text { circuit is negligible }]$ Resistance across $\mathrm{AB}$, $\mathrm{R}=\frac{9 \times 6}{9+6}=\frac{54}{15}=3.6 \Omega$
Kerala CEE 04.07.2022
Current Electricity
151817
Consider $\mathbf{N}$ resistors each with equal resistance $R$. If the ratio between the highest value of resistance and the lowest value of resistance that can be obtained by combining these resistors is equal to 289 , then the value of $\mathrm{N}$ is
1 289
2 145
3 17
4 None of (a), (b), (c)
Explanation:
C We know that, maximum resistance will get in series combination - $\therefore \quad \mathrm{R}_{\mathrm{S}}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\ldots \ldots$ $\mathrm{R}_{\mathrm{S}}=\mathrm{NR}$ ( $\mathrm{N}$ times) And minimum resistance will get in parallel combination $\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{P}}} =\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\ldots \ldots . .$ $\frac{1}{\mathrm{R}_{\mathrm{P}}} =\frac{\mathrm{N}}{\mathrm{R}} \Rightarrow \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{\mathrm{N}}$ According to question - $\frac{\mathrm{R}_{\mathrm{S}}}{\mathrm{R}_{\mathrm{P}}}=289$ $\frac{\mathrm{NR}}{\left(\frac{\mathrm{R}}{\mathrm{N}}\right)}=289$ $\mathrm{~N}^{2}=289, \quad \mathrm{~N}=17$
Assam CEE-31.07.2022
Current Electricity
151818
Two metallic wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of the these wires respectively, the effective conductivity of the combination is:
B Let, the resistance of wire be $R_{1}$ and $R_{2}$. Which connected in series then- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ Since, wires are identical dimensions so they have same cross-section area (A) and length $(l)$. $\because \quad$ Conductivity $(\sigma)=\frac{1}{\operatorname{Resistivity}(\rho)}$ $\therefore \quad \mathrm{R}=\frac{1}{\sigma} \times \frac{l}{\mathrm{~A}}$ So, $\quad \mathrm{R}_{1}=\frac{1}{\sigma_{1}}\left(\frac{l}{\mathrm{~A}}\right) \& \mathrm{R}_{2}=\frac{1}{\sigma_{2}}\left(\frac{l}{\mathrm{~A}}\right)$ From equation (i)- $\frac{1}{\sigma_{\text {eff }}}\left(\frac{2 l}{\mathrm{~A}}\right)=\frac{1}{\sigma_{1}}\left(\frac{l}{\mathrm{~A}}\right)+\frac{1}{\sigma_{2}}\left(\frac{l}{\mathrm{~A}}\right)$ $\frac{2}{\sigma_{\text {eff }}}=\frac{\sigma_{1}+\sigma_{2}}{\sigma_{1} \sigma_{2}}$ $\sigma_{\text {eff }}=\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$
JEE Main-29.07.2022
Current Electricity
151819
A $1 \mathrm{~m}$ long wire is broken into two unequal parts $X$ and $Y$. The $X$ part of the wire is stretched into another wire $W$. Length of $W$ is twice the length of $X$ and the resistance of $W$ is twice that of $Y$. Find the ratio of length of $X$ and $Y$.
1 $1: 4$
2 $1: 2$
3 $4: 1$
4 $2: 1$
Explanation:
B Let, the length of wire $\mathrm{X}$ be $l_{\mathrm{x}}$ and resistance $\mathrm{R}_{\mathrm{x}}$ Similarly length and resistance of $\mathrm{Y}$ is $l_{\mathrm{y}} \& \mathrm{R}_{\mathrm{y}}$ $\because \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ According to question $\rho_{\mathrm{x}}=\rho_{\mathrm{y}} \& \mathrm{~A}_{\mathrm{x}}=\mathrm{A}_{\mathrm{y}}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}}=\frac{l_{\mathrm{x}}}{l_{\mathrm{y}}}$ Now, when wire $\mathrm{X}$ is stretched double in length Then, $\mathrm{R}_{\mathrm{w}}=\rho_{\mathrm{w}} \times \frac{2 l_{\mathrm{x}}}{\left(\frac{\mathrm{A}_{\mathrm{x}}}{2}\right)} \Rightarrow \mathrm{R}_{\mathrm{w}}=\frac{4 \rho_{\mathrm{x}} l_{\mathrm{x}}}{\mathrm{A}_{\mathrm{x}}} \quad\left\{\because \rho_{\mathrm{w}}=\rho_{\mathrm{x}}\right\}$ $\mathrm{R}_{\mathrm{w}}=4 \mathrm{R}_{\mathrm{x}}$ $\mathrm{R}_{\mathrm{w}}=2 \mathrm{R}_{\mathrm{y}}$ According to question, From equation (i) and (ii)- $2 \mathrm{R}_{\mathrm{y}} =4 \mathrm{R}_{\mathrm{x}}$ $\frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}} =\frac{1}{2}$ $\frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}} =\frac{l_{\mathrm{x}}}{l_{\mathrm{y}}}=\frac{1}{2}$
151814
If the potential at $A$ is greater than the potential at $B$, then the equivalent resistance of the circuit across $A B$ is
1 $4.4 \Omega$
2 $5.2 \Omega$
3 $6 \Omega$
4 $9 \Omega$
5 $3.6 \Omega$
Explanation:
E Given circuit – Redraw the figure We can see that it behave like a Wheat Stone's Bridge, $\frac{3}{6}=\frac{2}{4} \quad[\because 4 \Omega \text { circuit is negligible }]$ Resistance across $\mathrm{AB}$, $\mathrm{R}=\frac{9 \times 6}{9+6}=\frac{54}{15}=3.6 \Omega$
Kerala CEE 04.07.2022
Current Electricity
151817
Consider $\mathbf{N}$ resistors each with equal resistance $R$. If the ratio between the highest value of resistance and the lowest value of resistance that can be obtained by combining these resistors is equal to 289 , then the value of $\mathrm{N}$ is
1 289
2 145
3 17
4 None of (a), (b), (c)
Explanation:
C We know that, maximum resistance will get in series combination - $\therefore \quad \mathrm{R}_{\mathrm{S}}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\ldots \ldots$ $\mathrm{R}_{\mathrm{S}}=\mathrm{NR}$ ( $\mathrm{N}$ times) And minimum resistance will get in parallel combination $\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{P}}} =\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\ldots \ldots . .$ $\frac{1}{\mathrm{R}_{\mathrm{P}}} =\frac{\mathrm{N}}{\mathrm{R}} \Rightarrow \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{\mathrm{N}}$ According to question - $\frac{\mathrm{R}_{\mathrm{S}}}{\mathrm{R}_{\mathrm{P}}}=289$ $\frac{\mathrm{NR}}{\left(\frac{\mathrm{R}}{\mathrm{N}}\right)}=289$ $\mathrm{~N}^{2}=289, \quad \mathrm{~N}=17$
Assam CEE-31.07.2022
Current Electricity
151818
Two metallic wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of the these wires respectively, the effective conductivity of the combination is:
B Let, the resistance of wire be $R_{1}$ and $R_{2}$. Which connected in series then- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ Since, wires are identical dimensions so they have same cross-section area (A) and length $(l)$. $\because \quad$ Conductivity $(\sigma)=\frac{1}{\operatorname{Resistivity}(\rho)}$ $\therefore \quad \mathrm{R}=\frac{1}{\sigma} \times \frac{l}{\mathrm{~A}}$ So, $\quad \mathrm{R}_{1}=\frac{1}{\sigma_{1}}\left(\frac{l}{\mathrm{~A}}\right) \& \mathrm{R}_{2}=\frac{1}{\sigma_{2}}\left(\frac{l}{\mathrm{~A}}\right)$ From equation (i)- $\frac{1}{\sigma_{\text {eff }}}\left(\frac{2 l}{\mathrm{~A}}\right)=\frac{1}{\sigma_{1}}\left(\frac{l}{\mathrm{~A}}\right)+\frac{1}{\sigma_{2}}\left(\frac{l}{\mathrm{~A}}\right)$ $\frac{2}{\sigma_{\text {eff }}}=\frac{\sigma_{1}+\sigma_{2}}{\sigma_{1} \sigma_{2}}$ $\sigma_{\text {eff }}=\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$
JEE Main-29.07.2022
Current Electricity
151819
A $1 \mathrm{~m}$ long wire is broken into two unequal parts $X$ and $Y$. The $X$ part of the wire is stretched into another wire $W$. Length of $W$ is twice the length of $X$ and the resistance of $W$ is twice that of $Y$. Find the ratio of length of $X$ and $Y$.
1 $1: 4$
2 $1: 2$
3 $4: 1$
4 $2: 1$
Explanation:
B Let, the length of wire $\mathrm{X}$ be $l_{\mathrm{x}}$ and resistance $\mathrm{R}_{\mathrm{x}}$ Similarly length and resistance of $\mathrm{Y}$ is $l_{\mathrm{y}} \& \mathrm{R}_{\mathrm{y}}$ $\because \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ According to question $\rho_{\mathrm{x}}=\rho_{\mathrm{y}} \& \mathrm{~A}_{\mathrm{x}}=\mathrm{A}_{\mathrm{y}}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}}=\frac{l_{\mathrm{x}}}{l_{\mathrm{y}}}$ Now, when wire $\mathrm{X}$ is stretched double in length Then, $\mathrm{R}_{\mathrm{w}}=\rho_{\mathrm{w}} \times \frac{2 l_{\mathrm{x}}}{\left(\frac{\mathrm{A}_{\mathrm{x}}}{2}\right)} \Rightarrow \mathrm{R}_{\mathrm{w}}=\frac{4 \rho_{\mathrm{x}} l_{\mathrm{x}}}{\mathrm{A}_{\mathrm{x}}} \quad\left\{\because \rho_{\mathrm{w}}=\rho_{\mathrm{x}}\right\}$ $\mathrm{R}_{\mathrm{w}}=4 \mathrm{R}_{\mathrm{x}}$ $\mathrm{R}_{\mathrm{w}}=2 \mathrm{R}_{\mathrm{y}}$ According to question, From equation (i) and (ii)- $2 \mathrm{R}_{\mathrm{y}} =4 \mathrm{R}_{\mathrm{x}}$ $\frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}} =\frac{1}{2}$ $\frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}} =\frac{l_{\mathrm{x}}}{l_{\mathrm{y}}}=\frac{1}{2}$
151814
If the potential at $A$ is greater than the potential at $B$, then the equivalent resistance of the circuit across $A B$ is
1 $4.4 \Omega$
2 $5.2 \Omega$
3 $6 \Omega$
4 $9 \Omega$
5 $3.6 \Omega$
Explanation:
E Given circuit – Redraw the figure We can see that it behave like a Wheat Stone's Bridge, $\frac{3}{6}=\frac{2}{4} \quad[\because 4 \Omega \text { circuit is negligible }]$ Resistance across $\mathrm{AB}$, $\mathrm{R}=\frac{9 \times 6}{9+6}=\frac{54}{15}=3.6 \Omega$
Kerala CEE 04.07.2022
Current Electricity
151817
Consider $\mathbf{N}$ resistors each with equal resistance $R$. If the ratio between the highest value of resistance and the lowest value of resistance that can be obtained by combining these resistors is equal to 289 , then the value of $\mathrm{N}$ is
1 289
2 145
3 17
4 None of (a), (b), (c)
Explanation:
C We know that, maximum resistance will get in series combination - $\therefore \quad \mathrm{R}_{\mathrm{S}}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\ldots \ldots$ $\mathrm{R}_{\mathrm{S}}=\mathrm{NR}$ ( $\mathrm{N}$ times) And minimum resistance will get in parallel combination $\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{P}}} =\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\ldots \ldots . .$ $\frac{1}{\mathrm{R}_{\mathrm{P}}} =\frac{\mathrm{N}}{\mathrm{R}} \Rightarrow \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{\mathrm{N}}$ According to question - $\frac{\mathrm{R}_{\mathrm{S}}}{\mathrm{R}_{\mathrm{P}}}=289$ $\frac{\mathrm{NR}}{\left(\frac{\mathrm{R}}{\mathrm{N}}\right)}=289$ $\mathrm{~N}^{2}=289, \quad \mathrm{~N}=17$
Assam CEE-31.07.2022
Current Electricity
151818
Two metallic wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of the these wires respectively, the effective conductivity of the combination is:
B Let, the resistance of wire be $R_{1}$ and $R_{2}$. Which connected in series then- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ Since, wires are identical dimensions so they have same cross-section area (A) and length $(l)$. $\because \quad$ Conductivity $(\sigma)=\frac{1}{\operatorname{Resistivity}(\rho)}$ $\therefore \quad \mathrm{R}=\frac{1}{\sigma} \times \frac{l}{\mathrm{~A}}$ So, $\quad \mathrm{R}_{1}=\frac{1}{\sigma_{1}}\left(\frac{l}{\mathrm{~A}}\right) \& \mathrm{R}_{2}=\frac{1}{\sigma_{2}}\left(\frac{l}{\mathrm{~A}}\right)$ From equation (i)- $\frac{1}{\sigma_{\text {eff }}}\left(\frac{2 l}{\mathrm{~A}}\right)=\frac{1}{\sigma_{1}}\left(\frac{l}{\mathrm{~A}}\right)+\frac{1}{\sigma_{2}}\left(\frac{l}{\mathrm{~A}}\right)$ $\frac{2}{\sigma_{\text {eff }}}=\frac{\sigma_{1}+\sigma_{2}}{\sigma_{1} \sigma_{2}}$ $\sigma_{\text {eff }}=\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$
JEE Main-29.07.2022
Current Electricity
151819
A $1 \mathrm{~m}$ long wire is broken into two unequal parts $X$ and $Y$. The $X$ part of the wire is stretched into another wire $W$. Length of $W$ is twice the length of $X$ and the resistance of $W$ is twice that of $Y$. Find the ratio of length of $X$ and $Y$.
1 $1: 4$
2 $1: 2$
3 $4: 1$
4 $2: 1$
Explanation:
B Let, the length of wire $\mathrm{X}$ be $l_{\mathrm{x}}$ and resistance $\mathrm{R}_{\mathrm{x}}$ Similarly length and resistance of $\mathrm{Y}$ is $l_{\mathrm{y}} \& \mathrm{R}_{\mathrm{y}}$ $\because \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ According to question $\rho_{\mathrm{x}}=\rho_{\mathrm{y}} \& \mathrm{~A}_{\mathrm{x}}=\mathrm{A}_{\mathrm{y}}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}}=\frac{l_{\mathrm{x}}}{l_{\mathrm{y}}}$ Now, when wire $\mathrm{X}$ is stretched double in length Then, $\mathrm{R}_{\mathrm{w}}=\rho_{\mathrm{w}} \times \frac{2 l_{\mathrm{x}}}{\left(\frac{\mathrm{A}_{\mathrm{x}}}{2}\right)} \Rightarrow \mathrm{R}_{\mathrm{w}}=\frac{4 \rho_{\mathrm{x}} l_{\mathrm{x}}}{\mathrm{A}_{\mathrm{x}}} \quad\left\{\because \rho_{\mathrm{w}}=\rho_{\mathrm{x}}\right\}$ $\mathrm{R}_{\mathrm{w}}=4 \mathrm{R}_{\mathrm{x}}$ $\mathrm{R}_{\mathrm{w}}=2 \mathrm{R}_{\mathrm{y}}$ According to question, From equation (i) and (ii)- $2 \mathrm{R}_{\mathrm{y}} =4 \mathrm{R}_{\mathrm{x}}$ $\frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}} =\frac{1}{2}$ $\frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}} =\frac{l_{\mathrm{x}}}{l_{\mathrm{y}}}=\frac{1}{2}$
151814
If the potential at $A$ is greater than the potential at $B$, then the equivalent resistance of the circuit across $A B$ is
1 $4.4 \Omega$
2 $5.2 \Omega$
3 $6 \Omega$
4 $9 \Omega$
5 $3.6 \Omega$
Explanation:
E Given circuit – Redraw the figure We can see that it behave like a Wheat Stone's Bridge, $\frac{3}{6}=\frac{2}{4} \quad[\because 4 \Omega \text { circuit is negligible }]$ Resistance across $\mathrm{AB}$, $\mathrm{R}=\frac{9 \times 6}{9+6}=\frac{54}{15}=3.6 \Omega$
Kerala CEE 04.07.2022
Current Electricity
151817
Consider $\mathbf{N}$ resistors each with equal resistance $R$. If the ratio between the highest value of resistance and the lowest value of resistance that can be obtained by combining these resistors is equal to 289 , then the value of $\mathrm{N}$ is
1 289
2 145
3 17
4 None of (a), (b), (c)
Explanation:
C We know that, maximum resistance will get in series combination - $\therefore \quad \mathrm{R}_{\mathrm{S}}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\ldots \ldots$ $\mathrm{R}_{\mathrm{S}}=\mathrm{NR}$ ( $\mathrm{N}$ times) And minimum resistance will get in parallel combination $\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{P}}} =\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\ldots \ldots . .$ $\frac{1}{\mathrm{R}_{\mathrm{P}}} =\frac{\mathrm{N}}{\mathrm{R}} \Rightarrow \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{\mathrm{N}}$ According to question - $\frac{\mathrm{R}_{\mathrm{S}}}{\mathrm{R}_{\mathrm{P}}}=289$ $\frac{\mathrm{NR}}{\left(\frac{\mathrm{R}}{\mathrm{N}}\right)}=289$ $\mathrm{~N}^{2}=289, \quad \mathrm{~N}=17$
Assam CEE-31.07.2022
Current Electricity
151818
Two metallic wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of the these wires respectively, the effective conductivity of the combination is:
B Let, the resistance of wire be $R_{1}$ and $R_{2}$. Which connected in series then- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ Since, wires are identical dimensions so they have same cross-section area (A) and length $(l)$. $\because \quad$ Conductivity $(\sigma)=\frac{1}{\operatorname{Resistivity}(\rho)}$ $\therefore \quad \mathrm{R}=\frac{1}{\sigma} \times \frac{l}{\mathrm{~A}}$ So, $\quad \mathrm{R}_{1}=\frac{1}{\sigma_{1}}\left(\frac{l}{\mathrm{~A}}\right) \& \mathrm{R}_{2}=\frac{1}{\sigma_{2}}\left(\frac{l}{\mathrm{~A}}\right)$ From equation (i)- $\frac{1}{\sigma_{\text {eff }}}\left(\frac{2 l}{\mathrm{~A}}\right)=\frac{1}{\sigma_{1}}\left(\frac{l}{\mathrm{~A}}\right)+\frac{1}{\sigma_{2}}\left(\frac{l}{\mathrm{~A}}\right)$ $\frac{2}{\sigma_{\text {eff }}}=\frac{\sigma_{1}+\sigma_{2}}{\sigma_{1} \sigma_{2}}$ $\sigma_{\text {eff }}=\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$
JEE Main-29.07.2022
Current Electricity
151819
A $1 \mathrm{~m}$ long wire is broken into two unequal parts $X$ and $Y$. The $X$ part of the wire is stretched into another wire $W$. Length of $W$ is twice the length of $X$ and the resistance of $W$ is twice that of $Y$. Find the ratio of length of $X$ and $Y$.
1 $1: 4$
2 $1: 2$
3 $4: 1$
4 $2: 1$
Explanation:
B Let, the length of wire $\mathrm{X}$ be $l_{\mathrm{x}}$ and resistance $\mathrm{R}_{\mathrm{x}}$ Similarly length and resistance of $\mathrm{Y}$ is $l_{\mathrm{y}} \& \mathrm{R}_{\mathrm{y}}$ $\because \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ According to question $\rho_{\mathrm{x}}=\rho_{\mathrm{y}} \& \mathrm{~A}_{\mathrm{x}}=\mathrm{A}_{\mathrm{y}}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}}=\frac{l_{\mathrm{x}}}{l_{\mathrm{y}}}$ Now, when wire $\mathrm{X}$ is stretched double in length Then, $\mathrm{R}_{\mathrm{w}}=\rho_{\mathrm{w}} \times \frac{2 l_{\mathrm{x}}}{\left(\frac{\mathrm{A}_{\mathrm{x}}}{2}\right)} \Rightarrow \mathrm{R}_{\mathrm{w}}=\frac{4 \rho_{\mathrm{x}} l_{\mathrm{x}}}{\mathrm{A}_{\mathrm{x}}} \quad\left\{\because \rho_{\mathrm{w}}=\rho_{\mathrm{x}}\right\}$ $\mathrm{R}_{\mathrm{w}}=4 \mathrm{R}_{\mathrm{x}}$ $\mathrm{R}_{\mathrm{w}}=2 \mathrm{R}_{\mathrm{y}}$ According to question, From equation (i) and (ii)- $2 \mathrm{R}_{\mathrm{y}} =4 \mathrm{R}_{\mathrm{x}}$ $\frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}} =\frac{1}{2}$ $\frac{\mathrm{R}_{\mathrm{x}}}{\mathrm{R}_{\mathrm{y}}} =\frac{l_{\mathrm{x}}}{l_{\mathrm{y}}}=\frac{1}{2}$
