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51.	Figure 6.36 shows a potentiometer with a cell of emf `2.0 V` and internal resistance `0.4 Omega` maintaining a potential drop across the resistor wire `AB`. A standard cell that maintains a constant emf of `1.02 V` (for very moderate current up to emf `mu A`) gives a balance point at `67.3 cm` length of the wire. To ensure very low current is drawn the standard cell, a very high resistance of `600 k Omega` is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf `epsilon` and the balance point found, similary, turns out to be at `82.3 cm` length of the wire. a. What is the value of `epsilon`? b.What purpose does the high resistance of `600 k Omega` have? c. Is the balance point affected by this high resistance? d. Is the balance point affected by internal resistance of the driver cell? e. Would the method work in the above situation if the driver cell of the potentiometer had an emf of `1.0 V` instead of `2.0 V`? f. Would the circuit work well for determining an extermely small emf, say of the order of a few `mV` (such as the typical emf of a thermocouple)? If not, how will you modify the circuit ?
Answer» Correct Answer - (a) `1.2474 V` (b) To reduced current (c ) No (d) Yes (e) No (f) No, suitable resistance is put in series with wire `AB`. (a) By formula, `(epsilon)/(epsilon_(standard)) = (l)/(l_(standard))` or `epsilon= (l xx epsilon_(standard))/(l_(standard)) = (82.3 xx 1.02)/(67.3) = 1.2474 V` (b) To reduce the current through the galvanometer when the movable constant is far from the balance point. (c ) No. (d) Yes. (e) No, if `epsilon` as it is, would be unsuited, because the balance point (for `e` of the order of a few `mV`) will be very close to the end `A` and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistor `r` in series with the wire `AB` so that potential drop across `AB` is only slightly greater than the emf to be measured. Then the balance point will be at larger length of the wire and the percentage error will be much smaller.

52.	Statement I: The resistance of an ideal voltmeter should be infinite. Statement II: Lower resistance of voltmeters gives a reading lower than the actual potential difference across the terminals.A. Statement `1` is true, Statement `2`is True , Statement `2` is correct explanation for Statement `1`.B. Statement `1` is True, Statement `2` is True, Statement `2` is NOT a correct explantion for Statement `1`.C. Statement `1` is True, Statement `2` is False.D. Statement `1` is False, Satement `2` is True.
Answer» Correct Answer - A If resistance of a voltmeter is not infinite, it will draw some current from the circuit and finally the reading will be less than actual.

53.	The reading of the ideal voltmeter in the adjoining diagram will be A. 4 VB. 8 VC. 12 VD. 14 V
Answer» Correct Answer - B

54.	In the adjoining circuit, the e.m.f. of the cell is 2 volt and the internal resistance is negligible. The resistance of the voltmeter is `80 ohm`. The reading of the voltmeter will be A. 0.80 VB. 1.60 VC. 1.33 VD. 2.00 V
Answer» Correct Answer - C

55.	A galvenometer has a resistance of `50 Omega` and its full-scale deflection current is `50 mu A`. What resistance should be added to it so that it can have a range of `0-5 V`?
Answer» Given `I_(G) = 50 muA`. Maximum voltage to be measured is `V = 5 V`. The galvanometer resistance `G = 50 Omega`. Now , `R = (V)/(I_(G)) - G = (5)/(50 xx 10^(-6)) - 50 = 100 K Omega`

56.	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 isA. 120 VB. 160 VC. 220 VD. 240 V
Answer» Correct Answer - D