Instruments — NEET Physics MCQs with Solutions
Free NEET Physics Instruments MCQs with step-by-step solutions (40 questions). Part of Current Electricity. Practise online on Prepizo — no login needed.
▶ Practise Instruments online (free)
Questions with solutions
Q1 — Instruments · easy · theory
$10^{-3}\ amp$ is flowing through a resistance of $1000\ \Omega$. To measure the correct potential difference, the voltmeter is to be used of which the resistance should be
A. $500\ \Omega$
B. $\gg 1000\ \Omega$ ✓ Correct
C. $1000\ \Omega$
D. $0\ \Omega$
Solution: A voltmeter must draw almost no current, so its resistance has to be very much larger than the resistance across which it is connected.
Q2 — Instruments · easy · numerical
$50\ \Omega$ and $100\ \Omega$ resistors are connected in series. This connection is connected with a battery of 2.4 volts. When a voltmeter of $100\ \Omega$ resistance is connected across $100\ \Omega$ resistor, then the reading of the voltmeter will be
A. 2.0 V
B. 1.6 V
C. 1.0 V
D. 1.2 V ✓ Correct
Solution: The voltmeter and the $100\ \Omega$ resistor in parallel give $50\ \Omega$, so the total resistance is $100\ \Omega$ and $i = 2.4/100 = 0.024\ A$. The reading is $0.024 \times 50 = 1.2\ V$.
Q3 — Instruments · easy · numerical
A 2 volt battery, a $15\ \Omega$ resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance of potentiometer wire is $5\ \Omega$, then the potential gradient of the potentiometer wire is
A. 0.2 V/cm
B. 0.02 V/cm
C. 0.05 V/cm
D. 0.005 V/cm ✓ Correct
Solution: Current $i = 2/(15 + 5) = 0.1\ A$, so the p.d. across the wire is $0.1 \times 5 = 0.5\ V$ over 100 cm, i.e. 0.005 V/cm.
Q4 — Instruments · easy · theory
By ammeter, which of the following can be measured
A. Electric potential
B. Potential difference
C. Resistance
D. Current ✓ Correct
Solution: An ammeter is a low-resistance instrument connected in series to measure the current in a branch.
Q5 — Instruments · easy · numerical
When a $12\ \Omega$ resistor is connected with a moving coil galvanometer then its deflection reduces from 50 divisions to 10 divisions. The resistance of the galvanometer is
A. $60\ \Omega$
B. $24\ \Omega$
C. $36\ \Omega$
D. $48\ \Omega$ ✓ Correct
Solution: With the shunt, $\dfrac{i_g}{i} = \dfrac{S}{S + G}$, so $\dfrac{10}{50} = \dfrac{12}{12 + G}$, giving $12 + G = 60$ and $G = 48\ \Omega$.
Q6 — Instruments · easy · numerical
In Wheatstone's bridge $P = 9\ ohm$, $Q = 11\ ohm$, $R = 4\ ohm$ and $S = 6\ ohm$. How much resistance must be put in parallel to the resistance $S$ to balance the bridge
A. 18.7 ohm
B. 26.4 ohm ✓ Correct
C. 24 ohm
D. $\dfrac{44}{9}$ ohm
Solution: Balance needs $\dfrac{P}{Q} = \dfrac{R}{S'}$, so $S' = \dfrac{44}{9}\ \Omega$. From $\dfrac{1}{S'} = \dfrac{1}{6} + \dfrac{1}{X}$ we get $X = 26.4\ \Omega$.
Q7 — Instruments · easy · numerical
A Daniel cell is balanced on 125 cm length of a potentiometer wire. Now the cell is short-circuited by a resistance 2 ohm and the balance is obtained at 100 cm. The internal resistance of the Daniel cell is
A. 4/5 ohm
B. 0.5 ohm ✓ Correct
C. 1.5 ohm
D. 1.25 ohm
Solution: Using $r = R\left(\dfrac{l_1 - l_2}{l_2}\right) = 2 \times \dfrac{125 - 100}{100} = 0.5\ \Omega$.
Q8 — Instruments · easy · theory
A potentiometer is an ideal device of measuring potential difference because
A. It has a long wire hence heat developed is quickly radiated
B. It is an elaborate arrangement
C. It uses a sensitive galvanometer
D. It does not disturb the potential difference it measures ✓ Correct
Solution: At the null point no current is drawn from the source being measured, so the potentiometer behaves like a voltmeter of infinite resistance.
Q9 — Instruments · easy · numerical
A potentiometer is used for the comparison of e.m.f. of two cells $E_1$ and $E_2$. For cell $E_1$ the no deflection point is obtained at 20 cm and for $E_2$ the no deflection point is obtained at 30 cm. The ratio of their e.m.f.'s will be
A. 1/2
B. 2
C. 1
D. 2/3 ✓ Correct
Solution: On a potentiometer $E \propto l$, so $\dfrac{E_1}{E_2} = \dfrac{20}{30} = \dfrac{2}{3}$.
Q10 — Instruments · easy · theory
If in the experiment of Wheatstone's bridge, the positions of cells and galvanometer are interchanged, then balance points will
A. Change
B. None of these
C. Depend on the internal resistance of cell and resistance of galvanometer
D. Remain unchanged ✓ Correct
Solution: The balance condition $P/Q = R/S$ does not involve the cell or the galvanometer, so interchanging them leaves the balance point unchanged.
Q11 — Instruments · easy · numerical
The resistance of a galvanometer is 90 ohms. If only 10 percent of the main current may flow through the galvanometer, in which way and of what value, a resistor is to be used
A. 10 ohms in series
B. 10 ohms in parallel ✓ Correct
C. 810 ohms in series
D. 810 ohms in parallel
Solution: A shunt is needed: $\dfrac{i_g}{i} = \dfrac{S}{S + G}$, so $0.1 = \dfrac{S}{S + 90}$, giving $S = 10\ \Omega$ in parallel.
Q12 — Instruments · easy · numerical
A galvanometer of resistance $25\ \Omega$ gives full scale deflection for a current of 10 milliampere, is to be changed into a voltmeter of range 100 V by connecting a resistance of '$R$' in series with galvanometer. The value of resistance $R$ in $\Omega$ is
A. 10025
B. 9975 ✓ Correct
C. 10000
D. 975
Solution: $R = \dfrac{V}{i_g} - G = \dfrac{100}{0.01} - 25 = 10000 - 25 = 9975\ \Omega$.
Q13 — Instruments · easy · numerical
In a potentiometer circuit there is a cell of e.m.f. 2 volt, a resistance of 5 ohm and a wire of uniform thickness of length 1000 cm and resistance 15 ohm. The potential gradient in the wire is
A. $\dfrac{3}{2000}\ V/cm$ ✓ Correct
B. $\dfrac{3}{5000}\ V/cm$
C. $\dfrac{1}{500}\ V/cm$
D. $\dfrac{1}{1000}\ V/cm$
Solution: Current $i = 2/(5 + 15) = 0.1\ A$, so the p.d. across the wire is $1.5\ V$ over 1000 cm, i.e. $\dfrac{1.5}{1000} = \dfrac{3}{2000}\ V/cm$.
Q14 — Instruments · easy · theory
Which is a wrong statement
A. Kirchhoff's first law (for currents meeting at a junction in an electric circuit) expresses the conservation of charge
B. The rheostat can be used as a potential divider
C. The Wheatstone bridge is most sensitive when all the four resistances are of the same order
D. In a balanced Wheatstone bridge, interchanging the positions of galvanometer and cell affects the balance of the bridge ✓ Correct
Solution: Interchanging the cell and galvanometer does NOT disturb a balanced bridge, since the balance condition $P/Q = R/S$ involves neither — so that statement is the wrong one.
Q15 — Instruments · easy · numerical
A voltmeter having a resistance of 998 ohms is connected to a cell of e.m.f. 2 volt and internal resistance 2 ohm. The error in the measurement of e.m.f. will be
A. $2 \times 10^{-1}\ volt$
B. $4 \times 10^{-1}\ volt$
C. $2 \times 10^{-3}\ volt$
D. $4 \times 10^{-3}\ volt$ ✓ Correct
Solution: Current $i = 2/(998 + 2) = 2 \times 10^{-3}\ A$, so the voltmeter reads $2 \times 10^{-3} \times 998 = 1.996\ V$. The error is $2 - 1.996 = 4 \times 10^{-3}\ V$.
Q16 — Instruments · easy · theory
For comparing the e.m.f.'s of two cells with a potentiometer, a standard cell is used to develop a potential gradient along the wires. Which of the following possibilities would make the experiment unsuccessful
A. The diameter of the wires is the same and uniform throughout
B. The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells ✓ Correct
C. The e.m.f. of the standard cell is larger than the e.m.f.'s of the two cells
D. The number of wires is ten
Solution: If the driving cell has a smaller e.m.f. than the cells being compared, no balance (null) point can be found anywhere on the wire, so the experiment fails.
Q17 — Instruments · easy · theory
If only 2% of the main current is to be passed through a galvanometer of resistance $G$, then the resistance of shunt will be
A. $\dfrac{G}{50}$
B. $49\,G$
C. $\dfrac{G}{49}$ ✓ Correct
D. $50\,G$
Solution: From $\dfrac{i_g}{i} = \dfrac{S}{S + G} = 0.02$ we get $S + G = 50S$, so $G = 49S$ and $S = G/49$.
Q18 — Instruments · easy · theory
Constantan wire is used in making standard resistances because its
A. Specific resistance is low
B. Melting point is high
C. Density is high
D. Temperature coefficient of resistance is negligible ✓ Correct
Solution: A standard resistance must keep its value as the temperature changes, and constantan has an almost zero temperature coefficient of resistance.
Q19 — Instruments · easy · theory
The net resistance of a voltmeter should be large to ensure that
A. It does not appreciably change the potential difference to be measured ✓ Correct
B. It does not draw excessive current
C. It does not get overheated
D. It can measure large potential difference
Solution: A voltmeter is connected in parallel; a high resistance keeps the current it draws negligible so the circuit — and the p.d. being measured — is barely disturbed.
Q20 — Instruments · easy · numerical
A galvanometer having a resistance of 8 ohm is shunted by a wire of resistance 2 ohm. If the total current is 1 amp, the part of it passing through the shunt will be
A. 0.2 amp
B. 0.25 amp
C. 0.5 amp
D. 0.8 amp ✓ Correct
Solution: The current divides in inverse ratio of resistance: $i_s = i \times \dfrac{G}{G + S} = 1 \times \dfrac{8}{10} = 0.8\ A$.
Q21 — Instruments · easy · numerical
An ammeter whose resistance is $180\ \Omega$ gives full scale deflection when current is 2 mA. The shunt required to convert it into an ammeter reading 20 mA (in ohms) is
A. 18
B. 10
C. 0.1
D. 20 ✓ Correct
Solution: $S = \dfrac{i_g G}{i - i_g} = \dfrac{2 \times 180}{20 - 2} = \dfrac{360}{18} = 20\ \Omega$.
Q22 — Instruments · easy · numerical
The resistance of 10 metre long potentiometer wire is 1 ohm/meter. A cell of e.m.f. 2.2 volts and a high resistance box are connected in series to this wire. The value of resistance taken from resistance box for getting potential gradient of 2.2 millivolt/metre will be
A. $1000\ \Omega$
B. $790\ \Omega$
C. $810\ \Omega$
D. $990\ \Omega$ ✓ Correct
Solution: The wire is $10\ \Omega$ and needs a p.d. of $2.2 \times 10^{-3} \times 10 = 0.022\ V$, so $i = 0.0022\ A$. Then the total resistance is $2.2/0.0022 = 1000\ \Omega$, so the box must supply $1000 - 10 = 990\ \Omega$.
Q23 — Instruments · easy · numerical
Voltmeters $V_1$ and $V_2$ are connected in series across a D.C. line. $V_1$ reads 80 volts and has a per volt resistance of 200 ohms. $V_2$ has a total resistance of 32 kilo ohms. The line voltage is
A. 240 volts ✓ Correct
B. 160 volts
C. 220 volts
D. 120 volts
Solution: The resistance of $V_1$ is $80 \times 200 = 16\ k\Omega$, so $i = 80/16000 = 5\ mA$. Then $V_2$ reads $5 \times 10^{-3} \times 32000 = 160\ V$, and the line voltage is $80 + 160 = 240\ V$.
Q24 — Instruments · easy · numerical
An ammeter gives full deflection when a current of 2 amp. flows through it. The resistance of ammeter is 12 ohms. If the same ammeter is to be used for measuring a maximum current of 5 amp., then the ammeter must be connected with a resistance of
A. 8 ohms in parallel ✓ Correct
B. 18 ohms in parallel
C. 8 ohms in series
D. 18 ohms in series
Solution: $S = \dfrac{i_g G}{i - i_g} = \dfrac{2 \times 12}{5 - 2} = 8\ \Omega$, connected in parallel as a shunt.
Q25 — Instruments · easy · numerical
A $36\ \Omega$ galvanometer is shunted by resistance of $4\ \Omega$. The percentage of the total current, which passes through the galvanometer is
A. 10 % ✓ Correct
B. 9 %
C. 8 %
D. 91 %
Solution: $\dfrac{i_g}{i} = \dfrac{S}{S + G} = \dfrac{4}{40} = 0.1$, i.e. 10%.
Q26 — Instruments · easy · numerical
A galvanometer of resistance $20\ \Omega$ is to be converted into an ammeter of range 1 A. If a current of 1 mA produces full scale deflection, the shunt required for the purpose is
A. $0.01\ \Omega$
B. $0.02\ \Omega$ ✓ Correct
C. $0.05\ \Omega$
D. $0.04\ \Omega$
Solution: $S = \dfrac{i_g G}{i - i_g} = \dfrac{10^{-3} \times 20}{1 - 10^{-3}} \approx 0.02\ \Omega$.
Q27 — Instruments · easy · theory
If an ammeter is to be used in place of a voltmeter then we must connect with the ammeter a
A. Low resistance in parallel
B. High resistance in parallel
C. Low resistance in series
D. High resistance in series ✓ Correct
Solution: A voltmeter must have a high resistance, so a large resistance is joined in series with the ammeter.
Q28 — Instruments · easy · numerical
A 10 m long wire of $20\ \Omega$ resistance is connected with a battery of 3 volt e.m.f. (negligible internal resistance) and a $10\ \Omega$ resistance is joined to it is series. Potential gradient along wire in volt per meter is
A. 0.02
B. 1.3
C. 0.2 ✓ Correct
D. 0.3
Solution: Current $i = 3/(20 + 10) = 0.1\ A$, so the p.d. across the wire is $0.1 \times 20 = 2\ V$ over 10 m, i.e. 0.2 V/m.
Q29 — Instruments · easy · theory
The material of wire of potentiometer is
A. Copper
B. Manganin ✓ Correct
C. Steel
D. Aluminium
Solution: Manganin has a high resistivity and a very low temperature coefficient, so the potential gradient along the wire stays steady.
Q30 — Instruments · easy · theory
To convert a galvanometer into a voltmeter, one should connect a
A. Low resistance in series with galvanometer
B. Low resistance in parallel with galvanometer
C. High resistance in series with galvanometer ✓ Correct
D. High resistance in parallel with galvanometer
Solution: A large resistance in series limits the current and makes the total resistance high, as a voltmeter requires.