301. A balance point is obtained at \(50\,\text{cm}\) when resistances \(X\) and \(R\) are placed in the two gaps. This indicates that
ⓐ. \(X=2R\)
ⓑ. \(X=\frac{R}{2}\)
ⓒ. \(X=0\)
ⓓ. \(X=R\)
Correct Answer: \(X=R\)
Explanation: At balance, the ratio of the gap resistances equals the ratio of the two wire lengths. If the balance point is at \(50\,\text{cm}\), the two wire segments are \(50\,\text{cm}\) and \(50\,\text{cm}\). Their resistance ratio is therefore \(1:1\) for a uniform wire. Hence the two gap resistances must be equal. This is why a balance near the middle is often preferred for reducing percentage error.
302. During a meter-bridge experiment, the balance point for \(X\) in the left gap and \(R\) in the right gap is very close to the left end. A better experimental adjustment is to
ⓐ. remove the galvanometer permanently
ⓑ. make the bridge wire non-uniform
ⓒ. interchange \(X\) and \(R\)
ⓓ. press the jockey harder to increase contact resistance
Correct Answer: interchange \(X\) and \(R\)
Explanation: A balance point very near one end can give larger percentage error because small length-reading errors become significant. A meter bridge gives best sensitivity when the balance point lies near the middle of the wire. Interchanging the two gap resistances shifts the balance point to the complementary length. Choosing a known resistance closer to the unknown can also move the balance point toward \(50\,\text{cm}\). Pressing the jockey harder may damage the wire and does not correct the resistance ratio.
303. When a meter bridge balances at \(60\,\text{cm}\) from the left end with \(X\) in the left gap and \(R\) in the right gap, after interchanging \(X\) and \(R\), the ideal new balance length from the left end is
ⓐ. \(20\,\text{cm}\)
ⓑ. \(60\,\text{cm}\)
ⓒ. \(40\,\text{cm}\)
ⓓ. \(80\,\text{cm}\)
Correct Answer: \(40\,\text{cm}\)
Explanation: \( \textbf{Initial balance length:} \) \(l=60\,\text{cm}\).
\( \textbf{Initial relation:} \)
\[
\frac{X}{R}=\frac{60}{40}
\]
When \(X\) and \(R\) are interchanged, the resistance ratio becomes reciprocal.
\( \textbf{New balance length \(l'\):} \)
\[
\frac{R}{X}=\frac{l'}{100-l'}
\]
Since the ratio is reciprocal, the two wire lengths must also interchange.
\( \textbf{New balance length:} \)
\[
l'=100-60=40\,\text{cm}
\]
This complementary shift is expected for an ideal uniform bridge wire.
\( \textbf{Final answer:} \) The new balance length is \(40\,\text{cm}\).
304. The meter-bridge balance point is at \(75\,\text{cm}\) from the left end with unknown \(X\) in the left gap and \(R=4\,\Omega\) in the right gap. The value of \(X\) is
ⓐ. \(\frac{4}{3}\,\Omega\)
ⓑ. \(3\,\Omega\)
ⓒ. \(16\,\Omega\)
ⓓ. \(12\,\Omega\)
Correct Answer: \(12\,\Omega\)
Explanation: \( \textbf{Known resistance:} \) \(R=4\,\Omega\).
\( \textbf{Balance length:} \)
\[
l=75\,\text{cm}
\]
\( \textbf{Other segment length:} \)
\[
100-l=25\,\text{cm}
\]
\( \textbf{Meter bridge formula:} \)
\[
\frac{X}{R}=\frac{l}{100-l}
\]
\( \textbf{Substitution:} \)
\[
\frac{X}{4}=\frac{75}{25}
\]
\( \textbf{Ratio:} \)
\[
\frac{75}{25}=3
\]
\( \textbf{Unknown resistance:} \)
\[
X=4(3)=12\,\Omega
\]
A longer balance length on the side of \(X\) means \(X\) is larger than \(R\).
\( \textbf{Final answer:} \) The value of \(X\) is \(12\,\Omega\).
305. A student uses the meter bridge formula but writes \(\frac{X}{R}=\frac{100-l}{l}\) while \(X\) is actually in the left gap and \(R\) is in the right gap. This mistake would
ⓐ. make the bridge wire non-uniform
ⓑ. remove the need for balance
ⓒ. reverse the resistance ratio
ⓓ. make \(X\) always equal to \(R\)
Correct Answer: reverse the resistance ratio
Explanation: In the stated arrangement, \(X\) corresponds to the left wire segment of length \(l\). The known resistance \(R\) corresponds to the right wire segment of length \(100-l\). Therefore, the correct relation is \(\frac{X}{R}=\frac{l}{100-l}\). Writing the reciprocal relation assigns the wire segments to the wrong gap resistances. The error is a placement-ratio error, not a failure of the Wheatstone bridge principle.
306. The meter bridge is most directly an application of
ⓐ. Kirchhoff's magnetic law
ⓑ. Wheatstone bridge principle
ⓒ. Faraday's law of induction
ⓓ. Coulomb's inverse-square law
Correct Answer: Wheatstone bridge principle
Explanation: A meter bridge is a practical form of a Wheatstone bridge. The two known bridge arms are replaced partly by the two segments of a uniform \(1\,\text{m}\) wire. At balance, the galvanometer current is zero, just as in a Wheatstone bridge. The ratio of the two wire resistances is replaced by the ratio of their lengths. This makes it possible to compare an unknown resistance with a known resistance using a length measurement.
307. For a meter bridge wire of resistance \(1.0\,\Omega\) and length \(100\,\text{cm}\), at balance, the jockey is at \(30\,\text{cm}\) from the left end. The resistance of the left segment of the wire is
ⓐ. \(0.30\,\Omega\)
ⓑ. \(0.70\,\Omega\)
ⓒ. \(1.0\,\Omega\)
ⓓ. \(3.0\,\Omega\)
Correct Answer: \(0.30\,\Omega\)
Explanation: \( \textbf{Total wire resistance:} \) \(R_{\text{wire}}=1.0\,\Omega\).
\( \textbf{Total wire length:} \) \(100\,\text{cm}\).
\( \textbf{Left segment length:} \) \(30\,\text{cm}\).
For a uniform wire, resistance is proportional to length.
\( \textbf{Resistance per centimetre:} \)
\[
\frac{1.0\,\Omega}{100\,\text{cm}}=0.010\,\Omega\,\text{cm}^{-1}
\]
\( \textbf{Left segment resistance:} \)
\[
R_L=(0.010)(30)=0.30\,\Omega
\]
The meter bridge formula uses length ratio because both segments share the same wire material and cross-section.
\( \textbf{Final answer:} \) The left segment resistance is \(0.30\,\Omega\).
308. When a null point is obtained in a meter bridge experiment, the jockey should be touched lightly on the wire because pressing hard can
ⓐ. make the unknown resistance exactly zero at contact
ⓑ. make the galvanometer insensitive to small current
ⓒ. convert the bridge into a simple series circuit
ⓓ. alter the contact length or damage the bridge wire
Correct Answer: alter the contact length or damage the bridge wire
Explanation: The jockey provides a temporary contact with the bridge wire. It should touch the wire lightly at the balance point. Pressing hard can deform or damage the wire and may change the contact condition. Since the meter bridge depends on accurate length measurement along a uniform wire, poor contact or damage can introduce error. The balance principle requires a clean contact, not a large mechanical force.
309. With \(X\) in the left gap of a meter bridge and \(R\) in the right gap, the balance point is at \(80\,\text{cm}\) from the left end. The ratio \(X:R\) is
ⓐ. \(1:4\)
ⓑ. \(2:5\)
ⓒ. \(5:2\)
ⓓ. \(4:1\)
Correct Answer: \(4:1\)
Explanation: At balance, the ratio of the gap resistances equals the ratio of the corresponding bridge-wire lengths. Since \(X\) is in the left gap, it corresponds to the left segment of \(80\,\text{cm}\). The right gap resistance \(R\) corresponds to the remaining segment \(100-80=20\,\text{cm}\). Therefore, \(X:R=80:20=4:1\). A balance point near the right end means the left-gap resistance is much larger than the right-gap resistance.
310. At balance, a meter bridge has the jockey at \(45\,\text{cm}\) from the left end. If the bridge wire is uniform, the ratio of the resistance of the left segment to the resistance of the right segment is
ⓐ. \(11:9\)
ⓑ. \(45:100\)
ⓒ. \(9:11\)
ⓓ. \(55:100\)
Correct Answer: \(9:11\)
Explanation: In a uniform bridge wire, resistance is proportional to length. The left segment has length \(45\,\text{cm}\). The right segment has length \(100-45=55\,\text{cm}\). Therefore, the resistance ratio is \(45:55\). Dividing both terms by \(5\) gives \(9:11\). The total wire length is not used as the second term of the ratio; the two segment lengths must be compared.
311. When the meter bridge reading gives \(X=R\frac{l}{100-l}\), a value of \(l\) very close to \(100\,\text{cm}\), the calculated \(X\) becomes very large mainly because
ⓐ. the bridge wire resistance becomes zero
ⓑ. the denominator \(100-l\) becomes very small
ⓒ. the known resistance becomes infinite
ⓓ. the galvanometer current must be maximum
Correct Answer: the denominator \(100-l\) becomes very small
Explanation: The formula \(X=R\frac{l}{100-l}\) contains the right-side wire length in the denominator. If the balance point is close to \(100\,\text{cm}\), then \(100-l\) is very small. This makes the ratio \(\frac{l}{100-l}\) large, so \(X\) is calculated as much larger than \(R\). Such end-region balance points are less desirable because small reading errors can strongly affect the result. A balance near the middle gives a more reliable comparison.
312. After the unknown and known resistances are interchanged in a meter bridge, the two balance lengths are \(l_1\) and \(l_2\). In an ideal meter bridge without end error, these lengths should satisfy
ⓐ. \(l_1+l_2=50\,\text{cm}\)
ⓑ. \(l_1=l_2=100\,\text{cm}\)
ⓒ. \(l_1l_2=100\,\text{cm}^2\)
ⓓ. \(l_1+l_2=100\,\text{cm}\)
Correct Answer: \(l_1+l_2=100\,\text{cm}\)
Explanation: If \(X\) and \(R\) are interchanged in an ideal meter bridge, the resistance ratio becomes reciprocal. The corresponding wire-length ratio must also become reciprocal. This means the new balance point is the complement of the original one. Therefore, if the first balance length is \(l_1\), the second should be \(l_2=100-l_1\). Hence \(l_1+l_2=100\,\text{cm}\). Deviation from this ideal result may indicate end resistance or other experimental errors.
313. If the resistance of meter-bridge end connections is not negligible, this mainly affects the result because
ⓐ. the galvanometer always carries maximum current at balance
ⓑ. effective wire-arm lengths may differ from measured lengths
ⓒ. the known resistance becomes independent of temperature
ⓓ. the bridge wire stops obeying Ohm's law everywhere
Correct Answer: effective wire-arm lengths may differ from measured lengths
Explanation: A meter bridge formula assumes that the resistance of each wire segment is proportional only to its measured length. If the end connections and contact resistances are not negligible, extra resistance is added near the ends of the bridge wire. Then the effective resistance corresponding to a segment is not exactly represented by the scale length alone. This can shift the balance point and introduce an end error. Interchanging the gap resistances and taking suitable corrections helps reduce the effect of these end-region resistances.
314. The meter bridge balances at \(25\,\text{cm}\) from the left end with \(X\) in the left gap and \(R=9\,\Omega\) in the right gap. The value of \(X\) is
ⓐ. \(1\,\Omega\)
ⓑ. \(3\,\Omega\)
ⓒ. \(6\,\Omega\)
ⓓ. \(27\,\Omega\)
Correct Answer: \(3\,\Omega\)
Explanation: \( \textbf{Known resistance:} \) \(R=9\,\Omega\).
\( \textbf{Balance length from left:} \)
\[
l=25\,\text{cm}
\]
\( \textbf{Right segment length:} \)
\[
100-l=75\,\text{cm}
\]
\( \textbf{Meter bridge relation:} \)
\[
\frac{X}{R}=\frac{l}{100-l}
\]
\( \textbf{Substitution:} \)
\[
\frac{X}{9}=\frac{25}{75}
\]
\( \textbf{Simplify length ratio:} \)
\[
\frac{25}{75}=\frac{1}{3}
\]
\( \textbf{Solve for \(X\):} \)
\[
X=9\left(\frac{1}{3}\right)=3\,\Omega
\]
A balance point closer to the left end means the left-gap resistance is smaller than the right-gap resistance.
\( \textbf{Final answer:} \) The unknown resistance is \(3\,\Omega\).
315. A student obtains a meter bridge balance at \(5\,\text{cm}\) from one end. The reading is considered less reliable mainly because
ⓐ. small length errors become large percentage errors
ⓑ. bridge wire has no resistance near the end region
ⓒ. the known resistance becomes zero at small length
ⓓ. the null point cannot exist near an end
Correct Answer: small length errors become large percentage errors
Explanation: In a meter bridge, the unknown resistance is calculated from a ratio of two wire lengths. When one length is very small, even a small uncertainty in locating the balance point becomes a large percentage error for that segment. End resistances and contact effects also become more significant near the ends. A balance near the middle of the wire gives better reliability because both lengths are reasonably large. The null point can exist near an end, but it is not the most accurate working condition.
316. If a meter bridge wire is accidentally thicker in its left half than in its right half, the usual formula using only \(l\) and \(100-l\) becomes unreliable because
ⓐ. the length of the wire becomes less than \(100\,\text{cm}\)
ⓑ. resistance per unit length is non-uniform
ⓒ. the galvanometer cannot show zero deflection
ⓓ. the known resistance loses its ohmic nature
Correct Answer: resistance per unit length is non-uniform
Explanation: The meter bridge formula depends on the bridge wire being uniform. For a uniform wire, resistance is proportional to length because \(\rho\) and \(A\) are constant. If the wire is thicker in one part, its cross-sectional area is not constant. Then equal lengths of the wire need not have equal resistances. The length ratio will no longer represent the resistance ratio accurately, so the usual balance formula becomes unreliable.
317. The working principle of a potentiometer is that, for a uniform wire carrying steady current, the potential drop along the wire is
ⓐ. directly proportional to its length
ⓑ. inversely proportional to its length
ⓒ. independent of length
ⓓ. proportional to the square of the length
Correct Answer: directly proportional to its length
Explanation: In a potentiometer, a steady current is maintained through a long uniform wire. For such a wire, resistance is proportional to length. Since the same current flows through every part of the wire, the potential drop across a length is proportional to the resistance of that length. Therefore, the potential drop is directly proportional to length. This gives the relation \(V=kl\), where \(k\) is the potential gradient of the wire.
318. In a potentiometer, the potential gradient \(k\) is defined as
ⓐ. current through the galvanometer per unit length
ⓑ. resistance of the unknown cell per unit current
ⓒ. potential drop per unit length of the wire
ⓓ. emf of the primary cell per unit resistance
Correct Answer: potential drop per unit length of the wire
Explanation: Potential gradient describes how much potential difference is available per unit length of the potentiometer wire. If a length \(l\) of the wire has potential drop \(V\), then \(k=\frac{V}{l}\). Its SI unit may be written as \(\text{V m}^{-1}\). A smaller potential gradient allows a longer balance length for the same emf, which usually improves measurement sensitivity. The galvanometer is only a null detector and does not define the potential gradient.
319. A \(4.0\,\text{m}\) potentiometer wire has a potential difference of \(2.0\,\text{V}\) across it
ⓐ. \(0.50\,\text{V m}^{-1}\)
ⓑ. \(0.25\,\text{V m}^{-1}\)
ⓒ. \(2.0\,\text{V m}^{-1}\)
ⓓ. \(8.0\,\text{V m}^{-1}\)
Correct Answer: \(0.50\,\text{V m}^{-1}\)
Explanation: \( \textbf{Total potential difference:} \) \(V=2.0\,\text{V}\).
\( \textbf{Wire length:} \) \(L=4.0\,\text{m}\).
\( \textbf{Required quantity:} \) Potential gradient \(k\).
\( \textbf{Definition:} \)
\[
k=\frac{V}{L}
\]
\( \textbf{Substitution:} \)
\[
k=\frac{2.0}{4.0}
\]
\( \textbf{Calculation:} \)
\[
k=0.50\,\text{V m}^{-1}
\]
This means each metre of the potentiometer wire has a potential drop of \(0.50\,\text{V}\).
\( \textbf{Final answer:} \) The potential gradient is \(0.50\,\text{V m}^{-1}\).
320. A cell is balanced against a potentiometer wire at length \(l\). At balance, the galvanometer shows zero deflection because
ⓐ. cell emf equals the drop along balancing length
ⓑ. potentiometer wire carries no current in primary circuit
ⓒ. the primary cell has zero emf
ⓓ. test cell is short-circuited through the galvanometer
Correct Answer: cell emf equals the drop along balancing length
Explanation: At the balance point, the potential drop along the selected length of the potentiometer wire exactly balances the emf of the test cell. There is then no potential difference across the galvanometer branch. Since no potential difference exists across the galvanometer, no current flows through it. This is a null method, so the test cell is not required to supply current at balance. The potentiometer wire still carries current from the primary circuit.
