NEET Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ) MCQs & PYQs

Practice NEET Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ) Questions

Question 21:

easy

A potentiometer wire is $100\text{ cm}$ 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 direction. The balance points are obtained at $50\text{ cm}$ and $10\text{ cm}$ from the positive end of the wire in the two cases. The ratio of emf’s is:

(2016 – I)

When cells support each other, $E_1 + E_2 = k \cdot 50$. When they oppose each other, $E_1 - E_2 = k \cdot 10$.
Taking the ratio gives $\frac{E_1 + E_2}{E_1 - E_2} = 5$, which simplifies to $6 E_2 = 4 E_1$.
Thus, the ratio of their e.m.f.'s is $\frac{E_1}{E_2} = \frac{3}{2}$.

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Question 22:

easy

A potentiometer wire has length $4\text{ m}$ and resistance $8\text{ }\Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\text{ V}$, so as to get a potential gradient $1\text{ mV}$ per $\text{cm}$ on the wire is:

(2015)

Potential drop across the wire $V_w = k \cdot L = 10^{-3}\text{ V/cm} \cdot 400\text{ cm} = 0.4\text{ V}$.
Current required $I = \frac{V_w}{R_w} = \frac{0.4}{8} = 0.05\text{ A}$.
Total resistance of circuit $R_{\text{total}} = \frac{E}{I} = \frac{2}{0.05} = 40\text{ }\Omega$, so series resistance $R = 40 - 8 = 32\text{ }\Omega$.

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Question 23:

easy

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

(2015 Re)

Current through the potentiometer wire is $I = \frac{E_0}{r + r_1}$.
Potential gradient is $k = \frac{I r}{L} = \frac{E_0 r}{(r + r_1) L}$.
The unknown e.m.f. $E = k \cdot l = \frac{E_0 r}{(r + r_1)} \frac{l}{L}$.

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Question 24:

easy

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $2.0\text{ V}$ and a negligible internal resistance. The potentiometer wire itself is $4\text{ m}$ long. When the resistance, $R$, connected across the given cell, has values of (i) Infinity, (ii) $9.5\text{ }\Omega$
The ‘balancing lengths’, on the potentiometer wire are found to be $3\text{ m}$ and $2.85\text{ m}$, respectively. The value of internal resistance of the cell is:

(2014)

The internal resistance formula for a potentiometer is $r = R \left(\frac{l_1}{l_2} - 1\right)$.
Substituting $l_1 = 3\text{ m}$, $l_2 = 2.85\text{ m}$, and $R = 9.5\text{ }\Omega$:
$r = 9.5 \cdot \left(\frac{3}{2.85} - 1\right) = 9.5 \cdot \frac{0.15}{2.85} = 0.5\text{ }\Omega$.

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Question 25:

easy

The potentiometer is best for measuring voltage, as:

(2000)

At the balancing point, no current is drawn from the secondary circuit by the potentiometer.
Thus, it acts as an ideal voltmeter with infinite resistance.
It measures potential difference across open circuit conditions without disturbing the circuit.

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Question 26:

easy

A galvanometer having a coil resistance of $60 \Omega$ shows full scale deflection when a current of $1.0 \text{ A}$ passes through it. It can be converted into an ammeter to read currents up to $5.0 \text{ A}$ by :

(2009)

Shunt resistance is connected in parallel and given by $S = \frac{I_g G}{I - I_g} = \frac{1.0 \times 60}{5.0 - 1.0} = 15 \Omega$.

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Question 27:

easy

A galvanometer of $50\text{ }\Omega$ resistance has $25$ divisions. A current of $4 \times 10^{-4}\text{ A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25\text{ V}$, it should be connected with a resistance of: (2004)

Full scale current $I_g = 25 \times 4 \times 10^{-4} = 0.01\text{ A}$. Required series resistance $R = \frac{V}{I_g} - G = \frac{25}{0.01} - 50 = 2450\text{ }\Omega$.

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Question 28:

easy

To convert a galvanometer into a voltmeter one should connect a:

(2002)

A voltmeter is formed by connecting a very high resistance in series with a galvanometer to measure large potential differences.

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Question 29:

easy

Resistance of a galvanometer coil is $8\text{ }\Omega$ and $2\text{ }\Omega$ shunt resistance is connected with it. If main current is $1\text{ A}$ then the current flowing through $2\text{ }\Omega$ resistance will be:

(1998)

Current through shunt resistance $I_s = I \times \frac{G}{G + S} = 1 \times \frac{8}{8 + 2} = 0.8\text{ A}$.

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Question 30:

easy

To convert a galvanometer into an ammeter, one needs to connect a:

(1992)

An ammeter is constructed by connecting a low resistance shunt in parallel with the galvanometer to bypass most of the current.

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