356962
Six equal resistance are connected between points \(P\),\(Q\) and \(R\) as shown in the figure. Then net resistance will be maximum between:
1 \(P\) and \(R\)
2 \(P\) and \(Q\)
3 \(Q\) and \(R\)
4 Any two points
Explanation:
Resistance between \(P\) and \(Q\) \({r_{PQ}} = \frac{{r \times \frac{5}{6}r}}{{r + \frac{5}{6}r}} = \frac{5}{{11}}r\) Resistance between \(Q\) and \(R\) \({r_{QR}} = \frac{{\frac{r}{2} \times \frac{4}{3}r}}{{\frac{r}{2} + \frac{4}{3}}} = \frac{4}{{11}}r\) Resistance between \(P\) and \(R\) \({r_{PR}} = \frac{{\frac{r}{3} \times \frac{3}{2}r}}{{\frac{r}{3} + \frac{3}{2}}} = \frac{3}{{11}}r\) Hence, it is clear that \({r_{PQ}}\) is maximum.
JEE - 2013
PHXII03:CURRENT ELECTRICITY
356963
In the network shown below, the ring has zero resistance. The equivalent resistance between the point \(A\) and \(B\) is
1 \(2R\)
2 \(4R\)
3 \(7R\)
4 \(10R\)
Explanation:
As the ring has no resistance, the three resistance of 3\(R\) each are in parallel. \( \Rightarrow \frac{1}{{\rm{R}}} = \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} = \frac{1}{{\rm{R}}} \Rightarrow {\rm{R}} = {\rm{R}}\) \(\therefore \) Between point \(A\) and \(B\) equivalent resistance \( = R + R = 2R\)
PHXII03:CURRENT ELECTRICITY
356964
Two resistances are given as \(R_{1}=(10 \pm 0.5) \Omega\) and \(R_{2}=(15 \pm 0.5) \Omega\). The percentage error in the measurement of equivalent resistance when they are connected in parallel is
1 6.33
2 2.33
3 4.33
4 5.33
Explanation:
Here, \(R_{1}=(10 \pm 0.5) \Omega, R_{2}=(15 \pm 0.5) \Omega\) The equivalent resistance in parallel combination is \(R_{p}=\dfrac{R_{1} R_{2}}{R_{1}+R_{2}}=\dfrac{10 \times 15}{10+15}=6 \Omega\) The error in equivalent resistance is given by \(\dfrac{\Delta R_{p}}{R_{p}^{2}}=\dfrac{\Delta R_{1}}{R_{1}^{2}}+\dfrac{\Delta R_{2}}{R_{2}^{2}}\) \(\Rightarrow \Delta R_{p}=\Delta R_{1}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}+\Delta R_{2}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}\) \(=0.5\left(\dfrac{6}{10}\right)^{2}+0.5\left(\dfrac{6}{15}\right)^{2}=0.26\) and \(\dfrac{\Delta R}{R}\) in percentage \(=4.33 \%\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356965
Four conductors of same resistance connected to form a square. If the resistance between diagonally opposite corners is \(8\,ohm\), the resistance between any two adjacent corners is
1 \(32\,ohm\)
2 \(8\,ohm\)
3 \(1/6\,ohm\)
4 \(6\,ohm\)
Explanation:
Let \(r\) be the resistance of each conductor. The resistance between opposite corners (diagonally) is \(\dfrac{1}{R_{A C}}=\dfrac{1}{2 r}+\dfrac{1}{2 r}=\dfrac{2}{2 r}\) \({R_{AC}} = r = 8{\rm{ }}ohms\) \(\left(\right.\) Given \({R_{AC}} = r = 8{\rm{ }}ohms\)) \(\dfrac{1}{R_{A D}}=\dfrac{1}{r}+\dfrac{1}{3 r}=\dfrac{4}{3 r}\) \(\Rightarrow R_{A D}=\dfrac{3 r}{4}=6 o h m s\)
PHXII03:CURRENT ELECTRICITY
356966
The resistance of a wire is ‘\(R\)’ ohm. If it is melted and stretched to ‘\(n\)’ times its original length, its new resistance will be :-
356962
Six equal resistance are connected between points \(P\),\(Q\) and \(R\) as shown in the figure. Then net resistance will be maximum between:
1 \(P\) and \(R\)
2 \(P\) and \(Q\)
3 \(Q\) and \(R\)
4 Any two points
Explanation:
Resistance between \(P\) and \(Q\) \({r_{PQ}} = \frac{{r \times \frac{5}{6}r}}{{r + \frac{5}{6}r}} = \frac{5}{{11}}r\) Resistance between \(Q\) and \(R\) \({r_{QR}} = \frac{{\frac{r}{2} \times \frac{4}{3}r}}{{\frac{r}{2} + \frac{4}{3}}} = \frac{4}{{11}}r\) Resistance between \(P\) and \(R\) \({r_{PR}} = \frac{{\frac{r}{3} \times \frac{3}{2}r}}{{\frac{r}{3} + \frac{3}{2}}} = \frac{3}{{11}}r\) Hence, it is clear that \({r_{PQ}}\) is maximum.
JEE - 2013
PHXII03:CURRENT ELECTRICITY
356963
In the network shown below, the ring has zero resistance. The equivalent resistance between the point \(A\) and \(B\) is
1 \(2R\)
2 \(4R\)
3 \(7R\)
4 \(10R\)
Explanation:
As the ring has no resistance, the three resistance of 3\(R\) each are in parallel. \( \Rightarrow \frac{1}{{\rm{R}}} = \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} = \frac{1}{{\rm{R}}} \Rightarrow {\rm{R}} = {\rm{R}}\) \(\therefore \) Between point \(A\) and \(B\) equivalent resistance \( = R + R = 2R\)
PHXII03:CURRENT ELECTRICITY
356964
Two resistances are given as \(R_{1}=(10 \pm 0.5) \Omega\) and \(R_{2}=(15 \pm 0.5) \Omega\). The percentage error in the measurement of equivalent resistance when they are connected in parallel is
1 6.33
2 2.33
3 4.33
4 5.33
Explanation:
Here, \(R_{1}=(10 \pm 0.5) \Omega, R_{2}=(15 \pm 0.5) \Omega\) The equivalent resistance in parallel combination is \(R_{p}=\dfrac{R_{1} R_{2}}{R_{1}+R_{2}}=\dfrac{10 \times 15}{10+15}=6 \Omega\) The error in equivalent resistance is given by \(\dfrac{\Delta R_{p}}{R_{p}^{2}}=\dfrac{\Delta R_{1}}{R_{1}^{2}}+\dfrac{\Delta R_{2}}{R_{2}^{2}}\) \(\Rightarrow \Delta R_{p}=\Delta R_{1}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}+\Delta R_{2}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}\) \(=0.5\left(\dfrac{6}{10}\right)^{2}+0.5\left(\dfrac{6}{15}\right)^{2}=0.26\) and \(\dfrac{\Delta R}{R}\) in percentage \(=4.33 \%\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356965
Four conductors of same resistance connected to form a square. If the resistance between diagonally opposite corners is \(8\,ohm\), the resistance between any two adjacent corners is
1 \(32\,ohm\)
2 \(8\,ohm\)
3 \(1/6\,ohm\)
4 \(6\,ohm\)
Explanation:
Let \(r\) be the resistance of each conductor. The resistance between opposite corners (diagonally) is \(\dfrac{1}{R_{A C}}=\dfrac{1}{2 r}+\dfrac{1}{2 r}=\dfrac{2}{2 r}\) \({R_{AC}} = r = 8{\rm{ }}ohms\) \(\left(\right.\) Given \({R_{AC}} = r = 8{\rm{ }}ohms\)) \(\dfrac{1}{R_{A D}}=\dfrac{1}{r}+\dfrac{1}{3 r}=\dfrac{4}{3 r}\) \(\Rightarrow R_{A D}=\dfrac{3 r}{4}=6 o h m s\)
PHXII03:CURRENT ELECTRICITY
356966
The resistance of a wire is ‘\(R\)’ ohm. If it is melted and stretched to ‘\(n\)’ times its original length, its new resistance will be :-
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
356962
Six equal resistance are connected between points \(P\),\(Q\) and \(R\) as shown in the figure. Then net resistance will be maximum between:
1 \(P\) and \(R\)
2 \(P\) and \(Q\)
3 \(Q\) and \(R\)
4 Any two points
Explanation:
Resistance between \(P\) and \(Q\) \({r_{PQ}} = \frac{{r \times \frac{5}{6}r}}{{r + \frac{5}{6}r}} = \frac{5}{{11}}r\) Resistance between \(Q\) and \(R\) \({r_{QR}} = \frac{{\frac{r}{2} \times \frac{4}{3}r}}{{\frac{r}{2} + \frac{4}{3}}} = \frac{4}{{11}}r\) Resistance between \(P\) and \(R\) \({r_{PR}} = \frac{{\frac{r}{3} \times \frac{3}{2}r}}{{\frac{r}{3} + \frac{3}{2}}} = \frac{3}{{11}}r\) Hence, it is clear that \({r_{PQ}}\) is maximum.
JEE - 2013
PHXII03:CURRENT ELECTRICITY
356963
In the network shown below, the ring has zero resistance. The equivalent resistance between the point \(A\) and \(B\) is
1 \(2R\)
2 \(4R\)
3 \(7R\)
4 \(10R\)
Explanation:
As the ring has no resistance, the three resistance of 3\(R\) each are in parallel. \( \Rightarrow \frac{1}{{\rm{R}}} = \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} = \frac{1}{{\rm{R}}} \Rightarrow {\rm{R}} = {\rm{R}}\) \(\therefore \) Between point \(A\) and \(B\) equivalent resistance \( = R + R = 2R\)
PHXII03:CURRENT ELECTRICITY
356964
Two resistances are given as \(R_{1}=(10 \pm 0.5) \Omega\) and \(R_{2}=(15 \pm 0.5) \Omega\). The percentage error in the measurement of equivalent resistance when they are connected in parallel is
1 6.33
2 2.33
3 4.33
4 5.33
Explanation:
Here, \(R_{1}=(10 \pm 0.5) \Omega, R_{2}=(15 \pm 0.5) \Omega\) The equivalent resistance in parallel combination is \(R_{p}=\dfrac{R_{1} R_{2}}{R_{1}+R_{2}}=\dfrac{10 \times 15}{10+15}=6 \Omega\) The error in equivalent resistance is given by \(\dfrac{\Delta R_{p}}{R_{p}^{2}}=\dfrac{\Delta R_{1}}{R_{1}^{2}}+\dfrac{\Delta R_{2}}{R_{2}^{2}}\) \(\Rightarrow \Delta R_{p}=\Delta R_{1}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}+\Delta R_{2}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}\) \(=0.5\left(\dfrac{6}{10}\right)^{2}+0.5\left(\dfrac{6}{15}\right)^{2}=0.26\) and \(\dfrac{\Delta R}{R}\) in percentage \(=4.33 \%\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356965
Four conductors of same resistance connected to form a square. If the resistance between diagonally opposite corners is \(8\,ohm\), the resistance between any two adjacent corners is
1 \(32\,ohm\)
2 \(8\,ohm\)
3 \(1/6\,ohm\)
4 \(6\,ohm\)
Explanation:
Let \(r\) be the resistance of each conductor. The resistance between opposite corners (diagonally) is \(\dfrac{1}{R_{A C}}=\dfrac{1}{2 r}+\dfrac{1}{2 r}=\dfrac{2}{2 r}\) \({R_{AC}} = r = 8{\rm{ }}ohms\) \(\left(\right.\) Given \({R_{AC}} = r = 8{\rm{ }}ohms\)) \(\dfrac{1}{R_{A D}}=\dfrac{1}{r}+\dfrac{1}{3 r}=\dfrac{4}{3 r}\) \(\Rightarrow R_{A D}=\dfrac{3 r}{4}=6 o h m s\)
PHXII03:CURRENT ELECTRICITY
356966
The resistance of a wire is ‘\(R\)’ ohm. If it is melted and stretched to ‘\(n\)’ times its original length, its new resistance will be :-
356962
Six equal resistance are connected between points \(P\),\(Q\) and \(R\) as shown in the figure. Then net resistance will be maximum between:
1 \(P\) and \(R\)
2 \(P\) and \(Q\)
3 \(Q\) and \(R\)
4 Any two points
Explanation:
Resistance between \(P\) and \(Q\) \({r_{PQ}} = \frac{{r \times \frac{5}{6}r}}{{r + \frac{5}{6}r}} = \frac{5}{{11}}r\) Resistance between \(Q\) and \(R\) \({r_{QR}} = \frac{{\frac{r}{2} \times \frac{4}{3}r}}{{\frac{r}{2} + \frac{4}{3}}} = \frac{4}{{11}}r\) Resistance between \(P\) and \(R\) \({r_{PR}} = \frac{{\frac{r}{3} \times \frac{3}{2}r}}{{\frac{r}{3} + \frac{3}{2}}} = \frac{3}{{11}}r\) Hence, it is clear that \({r_{PQ}}\) is maximum.
JEE - 2013
PHXII03:CURRENT ELECTRICITY
356963
In the network shown below, the ring has zero resistance. The equivalent resistance between the point \(A\) and \(B\) is
1 \(2R\)
2 \(4R\)
3 \(7R\)
4 \(10R\)
Explanation:
As the ring has no resistance, the three resistance of 3\(R\) each are in parallel. \( \Rightarrow \frac{1}{{\rm{R}}} = \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} = \frac{1}{{\rm{R}}} \Rightarrow {\rm{R}} = {\rm{R}}\) \(\therefore \) Between point \(A\) and \(B\) equivalent resistance \( = R + R = 2R\)
PHXII03:CURRENT ELECTRICITY
356964
Two resistances are given as \(R_{1}=(10 \pm 0.5) \Omega\) and \(R_{2}=(15 \pm 0.5) \Omega\). The percentage error in the measurement of equivalent resistance when they are connected in parallel is
1 6.33
2 2.33
3 4.33
4 5.33
Explanation:
Here, \(R_{1}=(10 \pm 0.5) \Omega, R_{2}=(15 \pm 0.5) \Omega\) The equivalent resistance in parallel combination is \(R_{p}=\dfrac{R_{1} R_{2}}{R_{1}+R_{2}}=\dfrac{10 \times 15}{10+15}=6 \Omega\) The error in equivalent resistance is given by \(\dfrac{\Delta R_{p}}{R_{p}^{2}}=\dfrac{\Delta R_{1}}{R_{1}^{2}}+\dfrac{\Delta R_{2}}{R_{2}^{2}}\) \(\Rightarrow \Delta R_{p}=\Delta R_{1}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}+\Delta R_{2}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}\) \(=0.5\left(\dfrac{6}{10}\right)^{2}+0.5\left(\dfrac{6}{15}\right)^{2}=0.26\) and \(\dfrac{\Delta R}{R}\) in percentage \(=4.33 \%\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356965
Four conductors of same resistance connected to form a square. If the resistance between diagonally opposite corners is \(8\,ohm\), the resistance between any two adjacent corners is
1 \(32\,ohm\)
2 \(8\,ohm\)
3 \(1/6\,ohm\)
4 \(6\,ohm\)
Explanation:
Let \(r\) be the resistance of each conductor. The resistance between opposite corners (diagonally) is \(\dfrac{1}{R_{A C}}=\dfrac{1}{2 r}+\dfrac{1}{2 r}=\dfrac{2}{2 r}\) \({R_{AC}} = r = 8{\rm{ }}ohms\) \(\left(\right.\) Given \({R_{AC}} = r = 8{\rm{ }}ohms\)) \(\dfrac{1}{R_{A D}}=\dfrac{1}{r}+\dfrac{1}{3 r}=\dfrac{4}{3 r}\) \(\Rightarrow R_{A D}=\dfrac{3 r}{4}=6 o h m s\)
PHXII03:CURRENT ELECTRICITY
356966
The resistance of a wire is ‘\(R\)’ ohm. If it is melted and stretched to ‘\(n\)’ times its original length, its new resistance will be :-
356962
Six equal resistance are connected between points \(P\),\(Q\) and \(R\) as shown in the figure. Then net resistance will be maximum between:
1 \(P\) and \(R\)
2 \(P\) and \(Q\)
3 \(Q\) and \(R\)
4 Any two points
Explanation:
Resistance between \(P\) and \(Q\) \({r_{PQ}} = \frac{{r \times \frac{5}{6}r}}{{r + \frac{5}{6}r}} = \frac{5}{{11}}r\) Resistance between \(Q\) and \(R\) \({r_{QR}} = \frac{{\frac{r}{2} \times \frac{4}{3}r}}{{\frac{r}{2} + \frac{4}{3}}} = \frac{4}{{11}}r\) Resistance between \(P\) and \(R\) \({r_{PR}} = \frac{{\frac{r}{3} \times \frac{3}{2}r}}{{\frac{r}{3} + \frac{3}{2}}} = \frac{3}{{11}}r\) Hence, it is clear that \({r_{PQ}}\) is maximum.
JEE - 2013
PHXII03:CURRENT ELECTRICITY
356963
In the network shown below, the ring has zero resistance. The equivalent resistance between the point \(A\) and \(B\) is
1 \(2R\)
2 \(4R\)
3 \(7R\)
4 \(10R\)
Explanation:
As the ring has no resistance, the three resistance of 3\(R\) each are in parallel. \( \Rightarrow \frac{1}{{\rm{R}}} = \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} + \frac{1}{{3{\rm{R}}}} = \frac{1}{{\rm{R}}} \Rightarrow {\rm{R}} = {\rm{R}}\) \(\therefore \) Between point \(A\) and \(B\) equivalent resistance \( = R + R = 2R\)
PHXII03:CURRENT ELECTRICITY
356964
Two resistances are given as \(R_{1}=(10 \pm 0.5) \Omega\) and \(R_{2}=(15 \pm 0.5) \Omega\). The percentage error in the measurement of equivalent resistance when they are connected in parallel is
1 6.33
2 2.33
3 4.33
4 5.33
Explanation:
Here, \(R_{1}=(10 \pm 0.5) \Omega, R_{2}=(15 \pm 0.5) \Omega\) The equivalent resistance in parallel combination is \(R_{p}=\dfrac{R_{1} R_{2}}{R_{1}+R_{2}}=\dfrac{10 \times 15}{10+15}=6 \Omega\) The error in equivalent resistance is given by \(\dfrac{\Delta R_{p}}{R_{p}^{2}}=\dfrac{\Delta R_{1}}{R_{1}^{2}}+\dfrac{\Delta R_{2}}{R_{2}^{2}}\) \(\Rightarrow \Delta R_{p}=\Delta R_{1}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}+\Delta R_{2}\left(\dfrac{R_{p}}{R_{1}}\right)^{2}\) \(=0.5\left(\dfrac{6}{10}\right)^{2}+0.5\left(\dfrac{6}{15}\right)^{2}=0.26\) and \(\dfrac{\Delta R}{R}\) in percentage \(=4.33 \%\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356965
Four conductors of same resistance connected to form a square. If the resistance between diagonally opposite corners is \(8\,ohm\), the resistance between any two adjacent corners is
1 \(32\,ohm\)
2 \(8\,ohm\)
3 \(1/6\,ohm\)
4 \(6\,ohm\)
Explanation:
Let \(r\) be the resistance of each conductor. The resistance between opposite corners (diagonally) is \(\dfrac{1}{R_{A C}}=\dfrac{1}{2 r}+\dfrac{1}{2 r}=\dfrac{2}{2 r}\) \({R_{AC}} = r = 8{\rm{ }}ohms\) \(\left(\right.\) Given \({R_{AC}} = r = 8{\rm{ }}ohms\)) \(\dfrac{1}{R_{A D}}=\dfrac{1}{r}+\dfrac{1}{3 r}=\dfrac{4}{3 r}\) \(\Rightarrow R_{A D}=\dfrac{3 r}{4}=6 o h m s\)
PHXII03:CURRENT ELECTRICITY
356966
The resistance of a wire is ‘\(R\)’ ohm. If it is melted and stretched to ‘\(n\)’ times its original length, its new resistance will be :-
