CBSE Class 12 Physics Practical Questions

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PRACTICAL BASED QUESTION

1. OHM’S LAW

1. A constant potential difference is maintained across two resistors of resistances R₁ and R₂ (R₂1). In which conductor less current will flow and why?

Ans; Resistor R₁ ; since current is inversely proportional to resistance.

2. Give examples of ohmic and non ohmic conductors.

Ans. Ohmicconductor :- magnet at constant temperature,Non ohmicconductor :- electric bulb, diodes

3.In the Ohm’s law experiment, if the position of ammeter and volt meter are interchanged, what happens to the measurement readings?

Ans. If both are ideal instruments, they will show any reading.

4. Two resistors of resistance R and 2R connected in parallel and the combination are connected in series with another resistor of resistance 4R, (a) find the ratio of current in parallel combination.

(b) If a supply of V volt is connected across the free terminals, what is the ratio of potential difference between the junctions?

Ans:- (a) 2 : 1 (b) 1 : 6

5. In the Ohm’s law experiment copper wire and nichrome wire are used for the same length and area of cross section. Of the two wires which has higher resistivity? If nichrome wire and the

copper wire are interchanged? Is Ohm’s law still valid?

Ans :-Nichrome wire has higher resistivity. 

6. What is resistivity ?

Ans : It is defined as the resistance offered by a wire of unit length and unit area of cross section.

7. Write the dimension of resistance and resistivity.

Ans :- [ M L² T^-3A^-2 ] and [ M L³ T^-3A^-2 ]

8. What is the S I unit of conductivity?

Ans :- Ohm^-1 m^-1.

9. Draw the graph between conductivity and resistivity of a metallic conductor

Ans :-

1. Meter Bridge & Potentiometer Worksheet 1 Name of The Chapters-Meter Bridge & Potentiometer 1 In a meter Bridge, the balance point is found to be at 39.5 cm from its zero end, when the resistor R is 12.5Ω. Determine the value of resistance X.

2. Why are the connection between resistors in a meter Bridge made of thick copper strips?

3. What happens if the galvanometer & cell are interchanged at the balance point of the bridge? Would the galvanometer will show any current?

4. In a meter bridge exp., balance point was observed at distance ‘l’ from its zero end.

(i) The values of known resistors R and X were doubled and then interchanged. What would be the new position of new balance point?

(ii) If the galvanometer and battery are interchanged at the balance point, how will the balance point get affected?

5. In a meter bridge exp., the balance point is found to at 40 cm from its zero end. When the resistor Y is of value 20 Ω is in the right gap of the meter bridge & unknown resistor X is in the left gap. Calculate the balancing length if X and Y are interchanged. 

6. In a meter bridge, two unknown resistors R and S are connected in the left and right gap respectively and the null point is found at a distance of 40 cm from the zero end. If a resistance of 12 Ω is connected in parallel with S, the null point occurs at 50 cm from zero end. Determine the values of R and S.

7. What do you mean by potential gradient?

8. A student uses a cell of emf 2 V and found the balance point at distance 350 cm. When a cell of unknown emf is attached and balance point shift 150 cm ahead than previous. Find the unknown emf.

9. The balance points in a potentiometer experiment is found to be at 600 cm and 400 cm respectively by using a known resistance 50 Ω and a cell of emf 6 V. Find the internal resistance of the cell.

10. In a potentiometer arrangement , a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm. What is the emf of the second cell? Worksheet 1 Name of The Chapters- Meter Bridge & Potentiometer ANSWER

1. R/X= 39.5/100-39.5 =39.5/60.5 =0.65 , so X= R/0.65= 12.5/0.65= 1.9 Ω

2. To minimize resistance of the connection which are not accounted for in the bridge formula.

3. The galvanometer will show no current.

4. (i) new balance point= 100-l

(ii) No change in the position of balance point.

5. _{l= 60 cm.}

6. _{R/S= 40/60 =2/3}

Now, R/S

_eq = 50/50 = 1

R(12 +S)/12 S= 1 , on solving we get S= 6 Ω & R= 4 Ω

7. Potential drop per unit length is called potential gradient.

8. ^ε1 / ^ε2=l₁/l₂₌350/500= 7/10 so, ^ε2 = 10 x 2/7 =20/7 V

9. r= (l₁/l₂-1) R=(600/400-1) x 50 = 25 Ω

10. ε₂ = 1.25 x 63/35 =2.25 V

3. CONVERTION OF GALVANOMETER INTO AMMETER AND VOLTMETER.

1. What is the resistance of an ideal ammeter?

2. What is the resistance of an ideal voltmeter?

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