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To determine the resistance of a galvanometer by the half-deflection method and to find its figure of merit | CBSE class 12th physics Practical

Objective:

To determine the resistance of a galvanometer by half deflection method and to find its figure of merit.

Apparatus Required:

A Weston type galvanometer, a voltmeter, a battery or
battery eliminator, two(10000 ohm and 1000 ohm) resistance boxes, two one-way keys, a rheostat, a screw gauge, a metre scale, an ammeter if given range,
connecting wires and a piece of sand paper.

Formula Used:

a) To find the resistance of galvanometer:

Where,

G is the resistance of the galvanometer

R is the resistance of the resistance box.

S is the shunt resistance.

b) To find the figure of merit:

The figure of merit is current per unit deflection.

Where,

I is the current in the circuit.

k is the figure of merit.

θ is the deflection.

E is the e.m.f. of the cell or battery eliminator.

Circuit Diagram:

Observations :

(a) Resistance of galvanometer by half deflection method: 
(b) The figure of merit:

Least count of voltmeter = 0.01 V
Range of voltmeter = 0-20 V
Zero error of voltmeter = nil

Calculations :

a) Resistance of Galvanometer

b) Figure of merit

Result:

Resistance of given Galvanometer, G = 55.43 Ω

Figure of merit of Given Galvanometer, k= 6.25 x 10¯5 A/div

Precautions:

1. All connections must be neat and tight.
2. Key Kl should be closed after taking out a high resistance from the
resistance box R.
3. The deflection of galvanometer should be large.
4. The emf of cell or battery should be constant.
5. All the plugs in resistance boxes should be tight.

Sources Of Error:

1. The screws of the instruments may be loose.
2. The plugs of resistance boxes may not be clean.
3. The emf of battery may not be constant.
4. The galvanometer divisions may not be of equal size.

Viva Questions

Q.1. Name two devices used for measuring potential difference between two points in an electric field.

Ans. Potentiometer and voltmeter.

Q.2. How do you determine the resistance of the galvanometer?

Ans. By half deflection method.

Q.3. What do you understand by resistance of a galvanometer and can it be determined?

Ans. It is the resistance offered by the coil of the galvanometer to the flow of current through it. It can be found by half deflection method.

Q.4. Which method is being used by The half deflection method. you to determine the resistance of galvanometer?

Ans. The half deflection method.

Q.5. Define current sensitivity of a galvanometer.

Ans. The deflection produced in the galvanometer, when unit current is passed through it, is called current sensitivity of the galvanometer.

Q.6. Why galvanometer is called a moving coil galvanometer?

Ans. In this galvanometer, the coil moves, while the magnet remains fixed.

Q.7. Which part of galvanometer offers resistance?

Ans. Coil of the galvanometer.

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To verify the laws of combination (parallel) of resistances using a metre bridge | Cbse Class 12th physics Practical

Objective

To verify the laws of combination (parallel) of resistances using a metre bridge.

Apparatus Required

A metre bridge, a Leclanche cell or battery eliminator, a galvanometer, a resistance box, a jockey, two resistance wires or two resistance coils known resistances, a set square, sand paper and connecting wires.

Formula Used

Diagram

Here R1 = 1 Ω and R2 = 2 Ω

Calculation

Result

Hence, laws of combination (parallel) of resistance is verified using a metre bridge.

Precautions

  • All connections should be neat and tight.
  • All the plugs in the resistance box should be tight.
  • Clean the connecting wires and the connecting points of metre bridge properly with sandpaper.
  • The jockey should be held perpendicular to wire of metre bridge.
  • Try to obtain the balance point between 40 cm and 60 cm.
  • Move the jockey gently on the wire and do not keep the jockey and the wire in contact for a long time.

Sources of errors

  • The keys of the resistance box may not be clean and tight.
  • The wire may not be uniformly thick throughout.
  • The screws of the instrument may be loose.
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To verify the laws of combination (series) of resistances using a metre bridge | Cbse Class 12th Physics Practics

Objective : To verify the laws of combination (series) of resistances by using a meter bridge.

Required Apparatus:

A metre bridge, a Leclanche cell (battery eliminator), a galvanometer, a resistance box, a jockey, two resistance wires or two resistance coils of known resistances, a set square, connecting wires, a piece of sandpaper.

Formula Used: 

(i) The resistance r of a resistance wire or coil is given by

Where, R is known resistance placed in the left gap and unknown resistance r in the right gap of the metre bridge. I cm is the length of a metre bridge wire from zero end up to the balance point.

(ii) When r1 and r2 are connected in series, then their combined resistance.

Circuit Diagram:

Observations Table:

Table for unknown resistance

Calculations:

The experimental value of Rs = 6.74 Ω

Theoretical value of Rs = r1 + r2 = 1.91 Ω + 4.79 Ω = 6.70 Ω

Difference ( if any ) = 6.74 Ω – 6.70 Ω = 0.04 Ω

Result:

Within the limits of experimental error, experimental and theoretical values of R are the same. Hence, the law of resistance in series is verified.

Precautions:

1. The connections should be neat, tight and clean.

2. Plugs should be tightly connected in the resistance box.

3. The movement of the jockey should be gentle and it shouldn’t be rubbed.

4. The key K should be inserted only when the observations are to be taken.

5. The null point should be between 30 cm and 70 cm.

6. To avoid the error of parallax, the set square should be used to note the null point.

7. There shouldn’t be any loops in the wire.

Sources Of Error:

1. The screws of the instrument might be loose.

2. The plugs may not be clean.

3. The wire might be of non-uniform diameter.

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To find the resistance of a given wire / standard resistor using a metre bridge | CBSE Class 12 physics practical

Objective:

To find the resistance of a given wire/standard resistor using a metre bridge.

Required Apparatus:

A meter bridge, a battery eliminator, a galvanometer meter, a resistance box, a jockey, a one-way key, a resistance wire, and connecting wires.

Formula Used:

Where,
R is known resistance placed in the left gap of the metre bridge.

X is the unknown resistance placed in the right gap of the metre bridge.

L (cm) is the length of the metre bridge wire from zero end up to the balance point.

Circuit Diagram:

Observation Table:

Calculations:

Result

  • The value of unknown resistance X is 5.4056 ohm.

Precautions:

  • The balance point should lie between 40 cm and 60 cm.
  • All connections should be clean and tight.
  • Hold the jockey perpendicular to the wire of the metre bridge.
  • Move the jockey gently on the wire and do not keep the jockey and wire in contact for a long time.

Sources of Error:

  • The key of the resistance box may not be clean and tight.
  • The wire may not be uniformly thick throughout.
  • The screw of the instruments may be loose.

Viva Question-Answers

Q.1. What is metre bridge? Explain in brief.

Ans. The metre bridge was originally developed by Charles Wheatstone to measure unknown resistance values and as a means of calibrating measuring instruments, voltmeters, ammeters, etc. by the use of a long resistive slide wire.

Q.2. Explain the structure of metre bridge.
Ans. The metre bridge consists of a one metre long wire of uniform cross sectional area fixed on a wooden block. A scale is attached to the block. Two gaps are formed on it by using thick metal strips in order to make the Wheatstone’s bridge. The terminal B between the gaps is used to connect galvanometer and jockey. The metre bridge operates under Wheatstone’s principle.

Q.3. Why is the metre bridge so called?

Ans. It is so because it uses a one metre long wire.

Q.4.When is a Wheatstone bridge most sensitive?

Ans A bridge is most sensitive when the resistances in the four arms of the bridge are of the same order of magnitude.

Q.5 How do you define resistivity or specific resistance of material?

Ans . It is defined as the resistance of a metre long wire of the material and having an area of cross-section 1 m².

Q.6 Does resistance depend upon length and area of cross-section of the material?

Ans. Yes

Q.7. What happens to the resistivity of a semiconductor, if its temperature is increased?

Ans. The resistivity of a semiconductor decreases with the rise of temperature.

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To determine the resistivity of two / three wires by plotting a graph for potential difference versus current | CBSE Class 12th Physics Practical

Experiment – 1

Objective:

To determine the resistivity of two wires by plotting a graph for potential difference versus current.

Required Apparatus:

Two resistance wires, a voltmeter (0-3)V and an ammeter (0-3 se) A of appropriate range, a battery/battery eliminator, a rheostat, a meter scale, a one-way key, connecting wires, and a screw gauge.

Formula Used:

By Ohm’s law

Where R is the constant of proportionality, it is known as the resistance of the conductor. R depends on the nature of the material, temperature and dimension of the conductor.

 

Specific resistance (ρ) of the material is given by

Where L is the length and D is the diameter of the given wire

Circuit Diagram:

Observations:

  • Range of ammeter = (0 – 3) A
  • The least count of ammeter = 0.05 A
  • Range of voltmeter = (0 -3) V
  • The least count of voltmeter = 0.05 V
  • The least count of metre-scale (L.C.) =0.1 cm

Zero error, e=0 mm,   Zero correction, c = 0 mm

 

For 1st wire:

Length of the given wire, L= 15 cm = 0.15 m

Observation Table for Resistance

Mean value of resistance, R=

.Table for Diameter of wire:

Calculation for Specific Resistance:

Graph – Potential difference versus Current

For 2nd wire:

Length of the given wire, L= 29cm = 0.29cm

Observation table for Resistance:

Table for Diameter of wire:

Calculation for Specific Resistance:

Graph – Potential difference versus Current:

Result:

  • Resistance of first wire from table = 1.04 Ω and from graph 1.05 Ω
  • Resistivity of 1st wire=48.98×10–⁸ Ω-m
  • Percentage error = 0.04%.
  • Resistance of 2nd wire from table = 2.06 Ω and from graph = 2.00 Ω
  • Resistivity of 2nd wire     =      50.19×10–⁸ Ω-m
  • Percentage error = 2.43%.

Precautions:

  1. The connection should be neat clean and tight.
  2. Thick connections wire should be used for the connections.
  3. The voltmeter and ammeter should be of proper range.
  4. A low-resistance rehosteat should be used.
  5. The key should be inserted only while taking observation to avoid heating of resistance.
  6. At one place, the diameter of the wire should be measured in two mutually perpendicular directions.
  7. The wire should not make a loop.

Source of Error:

  1. The instrument crew may be lose.
  2. Thick connection wires may not be available.
  3. Rehosteat may have high resistance.
  4. The wire may not have uniform thickness.
  5. The screw gauge may have faults like a ‘back lash’ error and a wrong pitch.
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Chapter 15 Communication System assertation & reasoning Questions Class 12th Physics .

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

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Chapter 14 Semiconductor Electronic: Material, Devices And Simple Circuits  assertation & reasoning Questions Class 12th Physics

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

Q.1. Assertion : A pure semiconductor has negative temperature coefficient of resistance.
Reason : In a semiconductor on raising the temperature, more charge carriers are released, conductance increases and resistance decreases.

Answer(a) In semiconductors, by increasing temperature, covalent bond breaks and conduction hole and electrons increase.

Q.2. Assertion : If the temperature of a semiconductor is increased then its resistance decreases.
Reason : The energy gap between conduction band and valence band is very small.

Answer(a) In semiconductors the energy gap between conduction band and valence band is small (1 eV). Due to temperature rise, electron in the valence band gain thermal energy and may jumpy across the small energy gap, (to the conduction band). Thus conductivity increases and hence resistance decreases.

Q.3. Assertion : In semiconductors, thermal collisions are responsible for taking a valence electron to the conduction band.
Reason : The number of conduction electrons go on increasing with time as thermal collisions continuously take place.

Answerc

Q.4. Assertion : A p-type semiconductors is a positive type crystal.
Reason : A p- type semiconductor is an uncharged crystal.

Answer(d) There is no charge on P-type semiconductor, because each atom of semiconductor is itself neutral.

Q.5. Assertion : Silicon is preferred over germanium for making semiconductor devices.
Reason : The energy gap in germanium is more than the energy gap in silicon.

Answer(c) Silicon is cheaper than germanium, so it is preferred over germanium. But energy gap in germanium is smaller than silicon.

Q.6. Assertion : Electron has higher mobility than hole in a semiconductor.
Reason : The mass of electron is less than the mass of the hole.

Answer-a

Q.7. Assertion : The number of electrons in a p-type silicon semiconductor is less than the number of electrons in a pure silicon semiconductor at room temperature.
Reason : It is due to law of mass action.

Answer-a

Q.8. Assertion : When two semi conductor of p and n type are brought in contact, they form p-n junction which act like a rectifier.
Reason : A rectifier is used to convent alternating current into direct current.

Answer(b) Study of junction diode characteristics shows that the junction diode offers a low resistance path, when forward biased and high resistance path when reverse biased. This feature of the junction diode enables it to be used as a rectifier.

Q.9. Assertion : Diode lasers are used as optical sources in optical communication.
Reason : Diode lasers consume less energy.

Answer(c) Statement – 1 is True, Statement- 2 is False

Q.10. Assertion : The diffusion current in a p-n junction is from the p-side to the n-side.
Reason : The diffusion current in a p-n junction is greater than the drift current when the junction is in forward biased.

Answer(b) Diffusion current is due to the migration of holes and electrons into opposite regions, so it will be from p-side to n-side. Also in forward bias it will increases.

Q.11. Assertion : The drift current in a p-n junction is from the n-side to the p-side.
Reason : It is due to free electrons only.

Answer-a

Q.12. Assertion : A p-n junction with reverse bias can be used as a photo-diode to measure light intensity.
Reason : In a reverse bias condition the current is small but it is more sensitive to changes in incident light intensity.

Answer-a

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Chapter 13 Nuclei assertation & reasoning Questions Class 12th Physics

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

Q.1. Assertion : Density of all the nuclei is same.
Reason : Radius of nucleus is directly proportional to the cube root of mass number.

Answer-a

Q.2. Assertion : Neutrons penetrate matter more readily as compared to protons.
Reason : Neutrons are slightly more massive than protons.

Answer(b) Both statements are separately correct.

Q.3. Assertion : The mass number of a nucleus is always less than its atomic number.
Reason : Mass number of a nucleus may be equal to its atomic number.

Answer(d) In case of hydrogen atom mass number and atomic number are equal.

Q.4. Assertion : The binding energy per nucleon, for nuclei with atomic mass number A > 100, decrease with A.
Reason : The forces are weak for heavier nuclei.

Answer(c) Nuclear force is nearly same for all nucleus.

Q.5. Assertion : Radioactivity of 108 undecayed radioactive nuclei of half life of 50 days is equal to that of 1.2 × 108 number of undecayed nuclei of some other material with half life of 60 days.
Reason : Radioactivity is proportional to half-life.

Answer-c

Q.6. Assertion : The ionising power of β-particle is less compared to β-particles but their penetrating power is more.
Reason : The mass of β-particle is less than the mass of α- particle.

Answer(b) β-particles, being emitted with very high speed compared to α-particles, pass for very little time near the atoms of the medium. So the probability of the atoms being ionised is comparatively less. But due to this reason, their loss of energy is very slow and they can penetrate the medium through a sufficient depth.

Q.7. Assertion : Radioactive nuclei emit β–1 particles.
Reason : Electrons exist inside the nucleus.

Answer(c) Electrons are not inside nucleus.

Q.8. Assertion : ZXA undergoes 2α, 2β- particles and 2γ-rays, the daughter product is Z-2YA – 8.
Reason : In α- decay the mass number decreases by 4 and atomic number decreases by 2. In β-decay the mass number remains unchanged, but atomic number increases by 1.

Answer-a

Q.9. Assertion : The heavier nuclei tend to have larger N/Z ratio because neutron does not exert electric force.
Reason : Coulomb forces have longer range compared to the nuclear force.

Answer-a

Q.10. Assertion : A free neutron decays to a proton but a free proton does not decay to a neutron. This is because neutron is an uncharged particle and proton is a charged particle.
Reason : Neutron has larger rest mass than the proton.

Answer-d

Q.11. Assertion : Cobalt-60 is useful in cancer therapy.
Reason : Cobalt -60 is source of γ- radiations capable of killing cancerous cells.

Answer-d

Q.12. Assertion : It is not possible to use 35Cl as the fuel for fusion energy.
Reason : The binding energy of 35Cl is to small.

Answer-c

Q.13. Assertion : Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.
Reason : For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z.

Answer(d) We know that energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Therefore statement (1) is correct.
The second statement is false because for heavy nuclei the binding energy per nucleon decreases with increasing Z and for light nuclei, B.E/nucleon increases with increasing Z.

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Chapter 12 Atoms assertation & reasoning Questions Class 12th Physics

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

Q.1. Assertion : The force of repulsion between atomic nucleus and α-particle varies with distance according to inverse square law.
Reason : Rutherford did α-particle scattering experiment.

Answer(b) Rutherford confirmed that the repulsive force of α- particle due to nucleus varies with distance according to inverse square law and that the positive charges are concentrated at the centre and not distributed throughout the atom.

Q.2. Assertion : According to classical theory the proposed path of an electron in Rutherford atom model will be parabolic.
Reason : According to electromagnetic theory an accelerated particle continuously emits radiation.

Answer(d) According to classical electromagnetic theory, an accelerated charged particle continuously emits radiation. As electrons revolving in circular paths are constantly experiencing centripetal acceleration, hence they will be losing their energy continuously and the orbital radius will go on decreasing, form spiral and finally the electron will fall in the nucleus.

Q.3. Assertion : Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate.
Reason: According to classical physics all moving electrons radiate.

Answer(b) Bohr postulated that electrons in stationary orbits around the nucleus do not radiate. This is the one of Bohr’s postulate, According to this the moving electrons radiates only when they go from one orbit to the next lower orbit.

Q.4. Assertion : Electrons in the atom are held due to coulomb forces.
Reason : The atom is stable only because the centripetal force due to Coulomb’s law is balanced by the centrifugal force.

Answer(c) According to postulates of Bohr’s atom model the electron revolves around the nucleus in fixed orbit of definite radii. As long as the electron is in a certain orbit it does not radiate any energy.

Q.5. Assertion : Hydrogen atom consists of only one electron but its emission spectrum has many lines.
Reason : Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission
spectrum, all the series are found.

Answer(b) When the atom gets appropriate energy from outside, then this electron rises to some higher energy level. Now it can return either directly to the lower energy level or come to the lowest energy level after passing through other lower energy levels hence all possible transitions take place in the source and many lines are seen in the spectrum.

Q.6. Assertion : Between any two given energy levels, the number of absorption transitions is always less than the number of emission transitions.
Reason : Absorption transitions start from the lowest energy level only and may end at any higher energy level. But emission transitions may start from any higher energy level and end at any energy level below it.

Answer(a)

Q.7. Assertion : In Lyman series, the ratio of minimum and maximum wavelength is 3/4
Reason : Lyman series constitute spectral lines corresponding to transition from higher energy to ground
state of hydrogen atom.

Answer(b)

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Chapter 11 Dual Nature of Radiation and Matter  assertation & reasoning Questions Class 12th Physics

Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

Q.1. Assertion : In process of photoelectric emission, all emitted electrons do not have same kinetic energy.
Reason : If radiation falling on photosensitive surface of a metal consists of different wavelength then energy acquired by electrons absorbing photons of different wavelengths shall be different.

Answer(b) Both statement I and II are true; but even it radiation of single wavelength is incident on photosensitive surface, electrons of different KE will be emitted.

Q.2. Assertion : Though light of a single frequency (monochromatic) is incident on a metal, the energies of
emitted photoelectrons are different.
Reason : The energy of electrons emitted from inside the metal surface, is lost in collision with the other atoms in the metal.

Answer(a) When a light of single frequency falls on the electrons of inner layer of metal, then this electron comes out of the metal surface after a large number of collisions with atom of it’s upper layer.

Q.3. Assertion : The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies.
Reason : The work function of the metal is its characteristics property.

Answer(b) The kinetic energy of emitted photoelectrons varies from zero to a maximum value. Work function depends on metal used.

Q.4. Assertion : Photoelectric saturation current increases with the increase in frequency of incident light.
Reason : Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.

Answer(d) Photoelectric saturation current is independent of frequency. It only depends on intensity of light.

Q.5. Assertion : Photosensitivity of a metal is high if its work function is small.
Reason : Work function = hf0 where f0 is the threshold frequency.

Answer(b) Less work function means less energy is required for ejecting out the electrons.

Q.6. Assertion : The photon behaves like a particle.
Reason : If E and P are the energy and momentum of the photon, then p = E / c .

Answer-a

Q.7. Assertion : In an experiment on photoelectric effect, a photon is incident on an electron from one direction and the photoelectron is emitted almost in the opposite direction. It violate the principle of conservation of linear momentum.
Reason : It does not violate the principle of conservation of linear momentum.Answerd

Q.9. Assertion : Two sources of equal intensity always emit equal number of photons in any time interval.
Reason : Two sources of equal intensity may emit equal number of photons in any time interval.

Answer(d) Total number of emitted photons depends on energy of each photon. The energy of photons of two sources may be different.

Q.10. Assertion : Two photons of equal wavelength must have equal linear momentum.
Reason : Two photons of equal linear momentum will have equal wavelength.

Answer(d) To photons of equal wavelength will have equal momentum (magnitude), but direction of momentum may be different.

Q.11. Assertion : The kinetic energy of photoelectrons emitted from metal surface does not depend on the intensity of incident photon.
Reason : The ejection of electrons from metallic surface is not possible with frequency of incident photons below the threshold frequency.

Answer-b

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