151925
In a Wheatstone's bridge, three resistances $P$, $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be
B The wheatstone bridge diagram, according to question- $\mathrm{S}_{\mathrm{eq}}=\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}$ Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}_{\mathrm{eq}}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_{\mathrm{eq}}}=\frac{\mathrm{R}}{\left(\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}\right)}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
UPSEE - 2010
Current Electricity
151926
Two wires of equal length and equal diameter and having resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity of the combination is
1 $\sqrt{\rho_{1} \rho_{2}}$
2 $\frac{\rho_{1}+\rho_{2}}{2}$
3 $\frac{\rho_{1} \rho_{2}}{\rho_{1}+\rho_{2}}$
4 $\left(\rho_{1}+\rho_{2}\right)$
Explanation:
B We know that, Resistance, $R=\frac{\rho L}{A}$ Then, $\mathrm{R}_{1}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}, \mathrm{R}_{2}=\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ For series connection $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}+\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ $\mathrm{R}_{\mathrm{eq}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\frac{\rho_{\mathrm{eq}} \times 2 \mathrm{~L}}{\mathrm{~A}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\rho_{\mathrm{eq}}=\frac{\rho_{1}+\rho_{2}}{2}$
GUJCET 2016
Current Electricity
151927
A uniform wire of resistance $R$, of the radius $r$ is uniformly drawn until its radius is reduced to $\mathbf{r} / \mathbf{n}$. Its new resistance is
151929
The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as
1 $\mathrm{r}^{3 / 2}$
2 $r$
3 $\mathrm{r}^{2 / 3}$
4 $r^{1 / 2}$
Explanation:
A The maximum current that flow in the fuse wire before it blows out varies with the radius $\mathrm{r}^{3 / 2}$ If $\quad l=$ length of wire $\mathrm{r}=$ radius of wire $\mathrm{I}=$ Current $\mathrm{Q}=\mathrm{h}=$ the rate of heat loss then, Resistance, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ At steady state- $\mathrm{I}_{\text {max }}^{2} \mathrm{R}=\mathrm{hA}$ $\mathrm{I}_{\text {max }}^{2} \times\left(\frac{\rho l}{\pi \mathrm{r}^{2}}\right)=\mathrm{h} \times 2 \pi \mathrm{rl}$ $\mathrm{I}_{\text {max }}^{2}=\frac{2 \pi^{2} \mathrm{~h}}{\rho} \mathrm{r}^{3}$ $\mathrm{I}_{\max } \propto \mathrm{r}^{3 / 2}$
Manipal UGET-2013
Current Electricity
151930
The resistance of a wire at $300 \mathrm{~K}$ is found to be $0.3 \Omega$. If the temperature coefficient of resistance of wire is $1.5 \times 10^{-3} \mathrm{~K}^{-1}$ the temperature at which the resistance becomes $0.6 \Omega$ is
151925
In a Wheatstone's bridge, three resistances $P$, $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be
B The wheatstone bridge diagram, according to question- $\mathrm{S}_{\mathrm{eq}}=\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}$ Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}_{\mathrm{eq}}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_{\mathrm{eq}}}=\frac{\mathrm{R}}{\left(\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}\right)}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
UPSEE - 2010
Current Electricity
151926
Two wires of equal length and equal diameter and having resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity of the combination is
1 $\sqrt{\rho_{1} \rho_{2}}$
2 $\frac{\rho_{1}+\rho_{2}}{2}$
3 $\frac{\rho_{1} \rho_{2}}{\rho_{1}+\rho_{2}}$
4 $\left(\rho_{1}+\rho_{2}\right)$
Explanation:
B We know that, Resistance, $R=\frac{\rho L}{A}$ Then, $\mathrm{R}_{1}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}, \mathrm{R}_{2}=\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ For series connection $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}+\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ $\mathrm{R}_{\mathrm{eq}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\frac{\rho_{\mathrm{eq}} \times 2 \mathrm{~L}}{\mathrm{~A}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\rho_{\mathrm{eq}}=\frac{\rho_{1}+\rho_{2}}{2}$
GUJCET 2016
Current Electricity
151927
A uniform wire of resistance $R$, of the radius $r$ is uniformly drawn until its radius is reduced to $\mathbf{r} / \mathbf{n}$. Its new resistance is
151929
The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as
1 $\mathrm{r}^{3 / 2}$
2 $r$
3 $\mathrm{r}^{2 / 3}$
4 $r^{1 / 2}$
Explanation:
A The maximum current that flow in the fuse wire before it blows out varies with the radius $\mathrm{r}^{3 / 2}$ If $\quad l=$ length of wire $\mathrm{r}=$ radius of wire $\mathrm{I}=$ Current $\mathrm{Q}=\mathrm{h}=$ the rate of heat loss then, Resistance, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ At steady state- $\mathrm{I}_{\text {max }}^{2} \mathrm{R}=\mathrm{hA}$ $\mathrm{I}_{\text {max }}^{2} \times\left(\frac{\rho l}{\pi \mathrm{r}^{2}}\right)=\mathrm{h} \times 2 \pi \mathrm{rl}$ $\mathrm{I}_{\text {max }}^{2}=\frac{2 \pi^{2} \mathrm{~h}}{\rho} \mathrm{r}^{3}$ $\mathrm{I}_{\max } \propto \mathrm{r}^{3 / 2}$
Manipal UGET-2013
Current Electricity
151930
The resistance of a wire at $300 \mathrm{~K}$ is found to be $0.3 \Omega$. If the temperature coefficient of resistance of wire is $1.5 \times 10^{-3} \mathrm{~K}^{-1}$ the temperature at which the resistance becomes $0.6 \Omega$ is
151925
In a Wheatstone's bridge, three resistances $P$, $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be
B The wheatstone bridge diagram, according to question- $\mathrm{S}_{\mathrm{eq}}=\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}$ Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}_{\mathrm{eq}}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_{\mathrm{eq}}}=\frac{\mathrm{R}}{\left(\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}\right)}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
UPSEE - 2010
Current Electricity
151926
Two wires of equal length and equal diameter and having resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity of the combination is
1 $\sqrt{\rho_{1} \rho_{2}}$
2 $\frac{\rho_{1}+\rho_{2}}{2}$
3 $\frac{\rho_{1} \rho_{2}}{\rho_{1}+\rho_{2}}$
4 $\left(\rho_{1}+\rho_{2}\right)$
Explanation:
B We know that, Resistance, $R=\frac{\rho L}{A}$ Then, $\mathrm{R}_{1}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}, \mathrm{R}_{2}=\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ For series connection $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}+\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ $\mathrm{R}_{\mathrm{eq}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\frac{\rho_{\mathrm{eq}} \times 2 \mathrm{~L}}{\mathrm{~A}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\rho_{\mathrm{eq}}=\frac{\rho_{1}+\rho_{2}}{2}$
GUJCET 2016
Current Electricity
151927
A uniform wire of resistance $R$, of the radius $r$ is uniformly drawn until its radius is reduced to $\mathbf{r} / \mathbf{n}$. Its new resistance is
151929
The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as
1 $\mathrm{r}^{3 / 2}$
2 $r$
3 $\mathrm{r}^{2 / 3}$
4 $r^{1 / 2}$
Explanation:
A The maximum current that flow in the fuse wire before it blows out varies with the radius $\mathrm{r}^{3 / 2}$ If $\quad l=$ length of wire $\mathrm{r}=$ radius of wire $\mathrm{I}=$ Current $\mathrm{Q}=\mathrm{h}=$ the rate of heat loss then, Resistance, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ At steady state- $\mathrm{I}_{\text {max }}^{2} \mathrm{R}=\mathrm{hA}$ $\mathrm{I}_{\text {max }}^{2} \times\left(\frac{\rho l}{\pi \mathrm{r}^{2}}\right)=\mathrm{h} \times 2 \pi \mathrm{rl}$ $\mathrm{I}_{\text {max }}^{2}=\frac{2 \pi^{2} \mathrm{~h}}{\rho} \mathrm{r}^{3}$ $\mathrm{I}_{\max } \propto \mathrm{r}^{3 / 2}$
Manipal UGET-2013
Current Electricity
151930
The resistance of a wire at $300 \mathrm{~K}$ is found to be $0.3 \Omega$. If the temperature coefficient of resistance of wire is $1.5 \times 10^{-3} \mathrm{~K}^{-1}$ the temperature at which the resistance becomes $0.6 \Omega$ is
151925
In a Wheatstone's bridge, three resistances $P$, $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be
B The wheatstone bridge diagram, according to question- $\mathrm{S}_{\mathrm{eq}}=\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}$ Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}_{\mathrm{eq}}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_{\mathrm{eq}}}=\frac{\mathrm{R}}{\left(\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}\right)}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
UPSEE - 2010
Current Electricity
151926
Two wires of equal length and equal diameter and having resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity of the combination is
1 $\sqrt{\rho_{1} \rho_{2}}$
2 $\frac{\rho_{1}+\rho_{2}}{2}$
3 $\frac{\rho_{1} \rho_{2}}{\rho_{1}+\rho_{2}}$
4 $\left(\rho_{1}+\rho_{2}\right)$
Explanation:
B We know that, Resistance, $R=\frac{\rho L}{A}$ Then, $\mathrm{R}_{1}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}, \mathrm{R}_{2}=\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ For series connection $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}+\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ $\mathrm{R}_{\mathrm{eq}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\frac{\rho_{\mathrm{eq}} \times 2 \mathrm{~L}}{\mathrm{~A}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\rho_{\mathrm{eq}}=\frac{\rho_{1}+\rho_{2}}{2}$
GUJCET 2016
Current Electricity
151927
A uniform wire of resistance $R$, of the radius $r$ is uniformly drawn until its radius is reduced to $\mathbf{r} / \mathbf{n}$. Its new resistance is
151929
The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as
1 $\mathrm{r}^{3 / 2}$
2 $r$
3 $\mathrm{r}^{2 / 3}$
4 $r^{1 / 2}$
Explanation:
A The maximum current that flow in the fuse wire before it blows out varies with the radius $\mathrm{r}^{3 / 2}$ If $\quad l=$ length of wire $\mathrm{r}=$ radius of wire $\mathrm{I}=$ Current $\mathrm{Q}=\mathrm{h}=$ the rate of heat loss then, Resistance, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ At steady state- $\mathrm{I}_{\text {max }}^{2} \mathrm{R}=\mathrm{hA}$ $\mathrm{I}_{\text {max }}^{2} \times\left(\frac{\rho l}{\pi \mathrm{r}^{2}}\right)=\mathrm{h} \times 2 \pi \mathrm{rl}$ $\mathrm{I}_{\text {max }}^{2}=\frac{2 \pi^{2} \mathrm{~h}}{\rho} \mathrm{r}^{3}$ $\mathrm{I}_{\max } \propto \mathrm{r}^{3 / 2}$
Manipal UGET-2013
Current Electricity
151930
The resistance of a wire at $300 \mathrm{~K}$ is found to be $0.3 \Omega$. If the temperature coefficient of resistance of wire is $1.5 \times 10^{-3} \mathrm{~K}^{-1}$ the temperature at which the resistance becomes $0.6 \Omega$ is
151925
In a Wheatstone's bridge, three resistances $P$, $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be
B The wheatstone bridge diagram, according to question- $\mathrm{S}_{\mathrm{eq}}=\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}$ Where, $S_{1}$ and $S_{2}$ are connected in parallel Now, wheat stone bridge is balanced, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\mathrm{S}_{\mathrm{eq}}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_{\mathrm{eq}}}=\frac{\mathrm{R}}{\left(\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}\right)}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
UPSEE - 2010
Current Electricity
151926
Two wires of equal length and equal diameter and having resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity of the combination is
1 $\sqrt{\rho_{1} \rho_{2}}$
2 $\frac{\rho_{1}+\rho_{2}}{2}$
3 $\frac{\rho_{1} \rho_{2}}{\rho_{1}+\rho_{2}}$
4 $\left(\rho_{1}+\rho_{2}\right)$
Explanation:
B We know that, Resistance, $R=\frac{\rho L}{A}$ Then, $\mathrm{R}_{1}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}, \mathrm{R}_{2}=\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ For series connection $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\rho_{1} \mathrm{~L}}{\mathrm{~A}}+\frac{\rho_{2} \mathrm{~L}}{\mathrm{~A}}$ $\mathrm{R}_{\mathrm{eq}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\frac{\rho_{\mathrm{eq}} \times 2 \mathrm{~L}}{\mathrm{~A}}=\left(\frac{\rho_{1}+\rho_{2}}{\mathrm{~A}}\right) \mathrm{L}$ $\rho_{\mathrm{eq}}=\frac{\rho_{1}+\rho_{2}}{2}$
GUJCET 2016
Current Electricity
151927
A uniform wire of resistance $R$, of the radius $r$ is uniformly drawn until its radius is reduced to $\mathbf{r} / \mathbf{n}$. Its new resistance is
151929
The maximum current that flow in the fuse wire before it blows out, varies with the radius $r$ as
1 $\mathrm{r}^{3 / 2}$
2 $r$
3 $\mathrm{r}^{2 / 3}$
4 $r^{1 / 2}$
Explanation:
A The maximum current that flow in the fuse wire before it blows out varies with the radius $\mathrm{r}^{3 / 2}$ If $\quad l=$ length of wire $\mathrm{r}=$ radius of wire $\mathrm{I}=$ Current $\mathrm{Q}=\mathrm{h}=$ the rate of heat loss then, Resistance, $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi \mathrm{r}^{2}}$ At steady state- $\mathrm{I}_{\text {max }}^{2} \mathrm{R}=\mathrm{hA}$ $\mathrm{I}_{\text {max }}^{2} \times\left(\frac{\rho l}{\pi \mathrm{r}^{2}}\right)=\mathrm{h} \times 2 \pi \mathrm{rl}$ $\mathrm{I}_{\text {max }}^{2}=\frac{2 \pi^{2} \mathrm{~h}}{\rho} \mathrm{r}^{3}$ $\mathrm{I}_{\max } \propto \mathrm{r}^{3 / 2}$
Manipal UGET-2013
Current Electricity
151930
The resistance of a wire at $300 \mathrm{~K}$ is found to be $0.3 \Omega$. If the temperature coefficient of resistance of wire is $1.5 \times 10^{-3} \mathrm{~K}^{-1}$ the temperature at which the resistance becomes $0.6 \Omega$ is
