Q: A resistance wire connected in the left gap of a metre bridge balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3 : 2. If the length of the resistance wire is 1.5 m, then the length of 1 Ω of the resistance wire is

1.0 \times 10^{-1} m. At balance \frac{R}{10} = \frac{3}{2}, so R = 15 Ω for a 1.5 m wire. Length per ohm = \frac{1.5}{15} = 0.1 = 1.0 \times 10^{-1} m

Q: A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of emf 2 V, so as to get a potential gradient of 1 mV per cm on the wire is

32 Ω. Required drop over 400 cm = 400 \times 1 mV = 0.4 V. From \frac{2}{8 + R}\times 8 = 0.4: 8 + R = 40, so R = 32 Ω

Q: A potentiometer circuit is used to find the internal resistance of a cell. The main battery has emf 2.0 V (negligible internal resistance) and the wire is 4 m long. When the resistance R across the cell is (i) infinity and (ii) 9.5 Ω, the balancing lengths are 3 m and 2.85 m respectively. The internal resistance of the cell is

0.5 Ω. r = R\left(\frac{l_1}{l_2} - 1\right) = 9.5\left(\frac{3}{2.85} - 1\right) = 9.5 \times 0.05 \approx 0.5 Ω

Question 1

In a potentiometer circuit, a cell of emf 1.5 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 does the balance point occur?

Accepted Answer

60 cm. \frac{E_1}{L_1} = \frac{E_2}{L_2} \Rightarrow \frac{1.5}{36} = \frac{2.5}{L_2}, so L_2 = 60 cm

Question 2

A resistance wire connected in the left gap of a metre bridge balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3 : 2. If the length of the resistance wire is 1.5 m, then the length of 1 Ω of the resistance wire is

Accepted Answer

1.0 	imes 10^{-1} m. At balance \frac{R}{10} = \frac{3}{2}, so R = 15 Ω for a 1.5 m wire. Length per ohm = \frac{1.5}{15} = 0.1 = 1.0 	imes 10^{-1} m

Question 3

A metre bridge is in the balance position with \frac{P}{Q} = \frac{l_1}{l_2}. If we now interchange the positions of the galvanometer and the cell, will the bridge work? If yes, what will be the balanced condition?

Accepted Answer

yes, \frac{P}{Q} = \frac{l_1}{l_2}. In a Wheatstone bridge the cell and galvanometer arms are interchangeable. The balance condition is unchanged: \frac{P}{Q} = \frac{l_1}{l_2}

Question 4

A potentiometer is an accurate and versatile device to make electrical measurement of EMF because the method involves

Accepted Answer

a condition of no current flow through the galvanometer. At balance no current flows through the cell/galvanometer, so the cell's true emf (not its terminal voltage) is measured — this makes the potentiometer accurate.

Question 5

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series, first to support one another and then in opposite directions. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is

Accepted Answer

3 : 2. E_1 + E_2 \propto 50 and E_1 - E_2 \propto 10. So \frac{E_1 + E_2}{E_1 - E_2} = 5, giving \frac{E_1}{E_2} = \frac{6}{4} = \frac{3}{2}

Question 6

A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of emf 2 V, so as to get a potential gradient of 1 mV per cm on the wire is

Accepted Answer

32 Ω. Required drop over 400 cm = 400 	imes 1 mV = 0.4 V. From \frac{2}{8 + R}	imes 8 = 0.4: 8 + R = 40, so R = 32 Ω

Question 7

A potentiometer wire of length L and resistance r are connected in series with a battery of emf E_0 and a resistance r_1. An unknown emf E is balanced at a length l of the potentiometer wire. The emf E will be given by

Accepted Answer

\frac{E_0 r}{(r + r_1)}\cdot\frac{l}{L}. Current in the wire = \frac{E_0}{r + r_1}; potential gradient = \frac{E_0 r}{(r + r_1)L}.

E = 	ext{gradient}	imes l = \frac{E_0 r}{(r + r_1)}\cdot\frac{l}{L}

Question 8

The resistances in the two arms of a metre bridge are 5 Ω and R Ω respectively, balancing at length l_1. When R is shunted with an equal resistance, the new balance point is at 1.6l_1. The resistance R is

Accepted Answer

15 Ω. First: \frac{5}{R} = \frac{l_1}{100 - l_1}. After shunting, R becomes \frac{R}{2} and balance at 1.6l_1: \frac{5}{R/2} = \frac{1.6l_1}{100 - 1.6l_1}. Solving gives l_1 = 25 and R = 15 Ω

Question 9

A potentiometer circuit is used to find the internal resistance of a cell. The main battery has emf 2.0 V (negligible internal resistance) and the wire is 4 m long. When the resistance R across the cell is (i) infinity and (ii) 9.5 Ω, the balancing lengths are 3 m and 2.85 m respectively. The internal resistance of the cell is

Accepted Answer

0.5 Ω. r = R\left(\frac{l_1}{l_2} - 1ight) = 9.5\left(\frac{3}{2.85} - 1ight) = 9.5 	imes 0.05 \approx 0.5 Ω

Question 10

The resistances of the four arms P, Q, R and S in a Wheatstone bridge are 10 Ω, 30 Ω, 30 Ω and 90 Ω respectively. The emf and internal resistance of the cell are 7 V and 5 Ω respectively. If the galvanometer resistance is 50 Ω, the current drawn from the cell will be

Accepted Answer

0.2 A. The bridge is balanced (\frac{10}{30} = \frac{30}{90}), so no current flows through the galvanometer. Branch (10+30) = 40 Ω is in parallel with (30+90) = 120 Ω, giving 30 Ω.

I = \frac{7}{30 + 5} = 0.2 A

Question 11

A potentiometer circuit is set up with a potential gradient of k volt/cm along the wire, and an ammeter (reading 1.0 A when the two-way key is off). The balance points, when the key between terminals (1 and 2) and (1 and 3) is plugged in, are found to be at lengths l_1 cm and l_2 cm respectively. The magnitudes of the resistors R and X in ohm are then equal, respectively, to

Accepted Answer

kl_1 and k(l_2 - l_1). With i = 1 A: iR = kl_1 \Rightarrow R = kl_1; and i(R + X) = kl_2 \Rightarrow R + X = kl_2, so X = k(l_2 - l_1)

Question 12

In a circuit, a 4 Ω resistor (in series with a 3 Ω) forms one branch between P and M, and two 0.5 Ω resistors in parallel (plus a 1 Ω) form the branch between M and N. The current through the 4 Ω resistor is 1 A when P and M are connected to a DC voltage source. The potential difference between points M and N is

Accepted Answer

3.2 V. Using the given network, V_{PM} = 4 	imes 1 = 4 V. The parallel 0.5 Ω resistors give 0.25 Ω; with the resistance ratios of the network the potential difference works out to V_{MN} = 3.2 V.

Question 13

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively without and with being short-circuited through a resistance of 10 Ω. Its internal resistance is

Accepted Answer

1.0 Ω. r = R\left(\frac{l_1}{l_2} - 1ight) = 10\left(\frac{110}{100} - 1ight) = 10 	imes 0.1 = 1 Ω

Question 14

Three resistances P, Q, R each of 2 Ω and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of 6 Ω is connected in parallel to S the bridge gets balanced. What is the value of S?

Accepted Answer

3 Ω. Balance: \frac{P}{Q} = \frac{R}{S_{eff}} where S_{eff} = \frac{6S}{6 + S}. With P = Q = R = 2: \frac{6S}{6+S} = 2, so 3S = 6 + S \Rightarrow S = 3 Ω

Question 15

Five equal resistances each of resistance R are connected as a Wheatstone bridge (four in the arms, one across the middle). A battery of 4V volts is connected between A and B. The current flowing in the branch AFCEB will be

Accepted Answer

\frac{V}{2R}. The bridge is balanced, so the middle arm carries no current and the network is two arms (each 2R) in parallel, giving net R. Total current \frac{V}{R} splits equally, so the branch AFCEB carries \frac{V}{2R}.

Question 16

In a Wheatstone bridge, all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is

Accepted Answer

R. The bridge is balanced, so no current flows through the galvanometer arm. The two branches (each 2R) are in parallel: \frac{2R 	imes 2R}{4R} = R

Question 17

Resistivity of a potentiometer wire is 10^{-7} Ω·m and its area of cross-section is 10^{-6} m². When a current i = 0.1 A flows through the wire, its potential gradient is

Accepted Answer

10^{-2} V/m. Potential gradient = i\frac{ho}{A} = 0.1 	imes \frac{10^{-7}}{10^{-6}} = 0.1 	imes 0.1 = 10^{-2} V/m

Question 18

In a Wheatstone bridge, the resistance of each of the four sides is 10 Ω. If the resistance of the galvanometer is also 10 Ω, then the effective resistance of the bridge will be

Accepted Answer

10 Ω. The bridge is balanced, so the galvanometer is ineffective. Two branches each (10 + 10 = 20 Ω) in parallel: \frac{20 	imes 20}{40} = 10 Ω

Question 19

Potentiometer measures the potential difference more accurately than a voltmeter, because

Accepted Answer

it does not draw current from external circuit. At balance the potentiometer draws no current from the cell, behaving like an ideal voltmeter of infinite resistance — so it measures emf/PD accurately.

Question 20

A bridge circuit has arms P = 3 Ω, Q = 4 Ω, R = 6 Ω, S = 8 Ω with a 7 Ω resistor across the middle. The equivalent resistance between A and B will be

Accepted Answer

\frac{14}{3} Ω. The bridge is balanced (\frac{3}{4} = \frac{6}{8}), so no current flows through the 7 Ω. Then (3 + 4) = 7 Ω is in parallel with (6 + 8) = 14 Ω: \frac{7 	imes 14}{21} = \frac{14}{3} Ω

Measuring Instruments — NEET Physics PYQ MCQs with Solutions

Questions with solutions