1. | \(220~\text{cm}\) | 2. | \(330~\text{cm}\) |
3. | \(115~\text{cm}\) | 4. | \(332~\text{cm}\) |
In a meter bridge experiment, the null point is at a distance of \(30~\text{cm}\) from \(\mathrm{A}\). If a resistance of \(16~\Omega\) is connected in parallel with resistance \(Y\), the null point occurs at \(50~\text{cm}\) from \(\mathrm{A}\). The value of the resistance \(Y\) is:
1. | \(\dfrac{112}{3}~\Omega\) | 2. | \(\dfrac{40}{3}~\Omega\) |
3. | \(\dfrac{64}{3}~\Omega\) | 4. | \(\dfrac{48}{3}~\Omega\) |
1. | \(\dfrac{4V_0R}{3R_0+16R}\) | 2. | \(\dfrac{4V_0R}{3R_0+R}\) |
3. | \(\dfrac{2V_0R}{4R_0+R}\) | 4. | \(\dfrac{2V_0R}{2R_0+3R}\) |
In a potentiometer circuit, a cell of emf \(1.5~\text{V}\) gives a balance point at 36 cm length of wire. If another cell of emf 2.5 V replaces the first cell, then at what length of the wire, the balance point occur?
1. 64 cm
2. 62 cm
3. 60 cm
4. 21.6 cm
A resistance wire connected in the left gap of a meter bridge balances a \(10~\Omega\) resistance in the right gap at a point which divides the bridge wire in the ratio \(3:2\). lf the length of the resistance wire is \(1.5~\text{m}\), then the length of \(1~\Omega\) of the resistance wire will be:
1. \(1.0\times 10^{-1}~\text{m}\)
2. \(1.5\times 10^{-1}~\text{m}\)
3. \(1.5\times 10^{-2}~\text{m}\)
4. \(1.0\times 10^{-2}~\text{m}\)
The metre bridge shown is in a balanced position with \(\frac{P}{Q} = \frac{l_1}{l_2}\). If we now interchange the position of the galvanometer and the cell, will the bridge work? If yes, what will be the balanced condition?
1. Yes, \(\frac{P}{Q}=\frac{l_1-l_2}{l_1+l_2}\)
2. No, no null point
3. Yes, \(\frac{P}{Q}= \frac{l_2}{l_1}\)
4. Yes, \(\frac{P}{Q}= \frac{l_1}{l_2}\)
A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves:
1. | the potential gradients. |
2. | a condition of no current flow through the galvanometer. |
3. | a condition of cells, galvanometer, and resistances. |
4. | the cells. |
A potentiometer wire of length \(L\) and a resistance \(r\) are connected in series with a battery of EMF \(E_{0 }\) and resistance \(r_{1}\). An unknown EMF is balanced at a length l of the potentiometer wire. The EMF \(E\) will be given by:
1. \(\frac{L E_{0} r}{l r_{1}}\)
2. \(\frac{E_{0} r}{\left(\right. r + r_{1} \left.\right)} \cdot \frac{l}{L}\)
3. \(\frac{E_{0} l}{L}\)
4. \(\frac{L E_{0} r}{\left(\right. r + r_{1} \left.\right) l}\)
A potentiometer wire has a length of \(4~\text{m}\) and resistance \(8~\Omega.\) The resistance that must be connected in series with the wire and an energy source of emf \(2~\text{V}\), so as to get a potential gradient of \(1~\text{mV}\) per cm on the wire is:
1. \(32~\Omega\)
2. \(40~\Omega\)
3. \(44~\Omega\)
4. \(48~\Omega\)