Q. No.
Choose the correct statement.
| 1. | Reading of the voltmeter \({V}\) is \(9~\text{V}\) |
| 2. | Reading of the ammeter \({A}_1\) is \(18~\text{A}\) |
| 3. | Reading of the ammeter \({A}_1\) is \(2~\text{A}\) |
| 4. | Reading of the voltmeter \({V}\) is \(7~\text{V}\) |
1. \(\dfrac{5R}{2}\)
2. \(3R\)
3. \(\dfrac{5R}{3}\)
4. \(2R\)
In the figure shown, what is the current (in Ampere) drawn from the battery? You are given \(R_1=15~ \Omega, R_2=10~ \Omega, R_3=20~ \Omega,R_4=5~ \Omega, R_5=25~ \Omega, R_6=30~ \Omega, E=15~V\)
1. \(7/18\)
2. \(20/3\)
3. \(9/32\)
4. \(13/24\)
A wire of resistance \(R\) is bent to form a square \(ABCD\) as shown in the figure. The effective resistance between \(E\) and \(C\) is: (\(E\) is mid-point of arm \(CD\))
1. \( \frac{3}{4} R \)
2. \(\frac{1}{16} R \)
3. \( \frac{7}{64} R \)
4. \(R\)
Space between two concentric conducting spheres of radii \(a\) and \(b\) (\(b>a\)) is filled with a medium of resistivity \(\rho\). The resistance between the two spheres will be:
1. \( \frac{p}{2 \pi}\left(\frac{1}{a}+\frac{1}{b}\right) \)
2. \(\frac{p}{4 \pi}\left(\frac{1}{a}+\frac{1}{b}\right) \)
3. \(\frac{p}{2 \pi}\left(\frac{1}{a}-\frac{1}{b}\right) \)
4. \(\frac{p}{4 \pi}\left(\frac{1}{a}-\frac{1}{b}\right)\)
Five equal resistances are connected in a network as shown in the figure. The net resistance between point \(A\) and point \(B\) is:
1. \(2R\)
2. \(\dfrac{ R}{2}\)
3. \(\dfrac{3 R}{2}\)
4. \(R\)
| 1. | Parallel combination of \(A\) and \(C\) connected in series with \(B\). |
| 2. | Parallel combination of \(A\) and \(B\) connected in series with \(C\). |
| 3. | Series combination of \(A\) and \(C\) connected in parallel with \(B\). |
| 4. | Series combination of \(B\) and \(C\) connected in parallel with \(A\). |
1. \(4~\text{A}\)
2. \(1~\text{A}\)
3. \(2~\text{A}\)
4. \(3~\text{A}\)
1. \(65~\Omega\)
2. \(20~\Omega\)
3. \(5~\Omega\)
4. \(2~\Omega\)
1. \(6\)
2. \(9\)
3. \(3\)
4. \(12\)
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