Potentiometer - comparing emf and finding internal resistance — practice questions
23 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.
In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf is found to be . If this cell is replaced by another cell of emf E, the length-of null point increases by . The value of is . The value of is ____________.
In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ___________ cm.
In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3 : 2, the difference in the balancing length of the potentiometer wire in above two cases will be ___________ cm.
In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E 1 and E 2 as well. For key K 1 (closed), the jockey is adjusted to touch the wire at point J 1 so that there is no deflection in the galvanometer. Now the first battery (E 1 ) is replaced by second battery (E 2 ) for working by making K 1 open and K 2 closed. The galvanometer gives then null deflection at J 2 . The value of is , where a = _________.
With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is (A) directly proportional to the length of the potentiometer wire (B) directly proportional to the potential gradient of the wire (C) inversely proportional to the potential gradient of the wire (D) inversely proportional to the length of the potentiometer wire Choose the correct option for the above statements :
- A. A only
- B. B and D only
- C. A and C only
- D. inversely C only
A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5. When a resistance of 15 is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________.
As shown in the figure, a potentiometer wire of resistance and length is connected with resistance box (R.B.) and a standard cell of emf . For a resistance '' of resistance box introduced into the circuit, the null point for a cell of is found to be . The value of '' is ___________
A potentiometer wire of length is connected in series with a resistance 780 and a standard cell of emf . A constant current flows through potentiometer wire. The length of the null point for cell of emf is found to be . The resistance of the potentiometer wire is ____________ .
The circuit diagram of potentiometer used to measure the internal resistance of a cell (E) is shown in figure. The key 'K' is kept closed so as to send constant current through potentiometer wire. When key 'K 1 ' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K 1 ' is closed the null point is shifted at 80 cm at the potentiometer wire. The internal resistance of the given cell is _____________ .
A cell, shunted by a 8 resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 resistance. The value of internal resistance of the cell will be ____________ .
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