Students must start practicing the questions from RBSE 12th Physics Model Papers Set 9 with Answers in English Medium provided here.

## RBSE Class 12 Physics Model Paper Set 9 with Answers in English

Time: 2:45 Hours

Maximum Marks: 56

General Instruction to the Examinees:

- Candidate must write first his/her Roll No. on the question paper compulsorily.
- All the questions are compulasory.
- Write the answer to each question in the given answer book only.
- For questions having more than one part the answers to those parts are to be written

together in continuity.

Section – A

Question 1.

Write the Correct Answer from multiple choice question 1 (i to ix) and write in given answer book:

(i) Identify the wrong statement in the following. Coulomb’s law correctly describes the electric force that: [1]

(a) binds the electrons of an atom to its nucleus.

(b) binds the protons and neutrons in the nucleus of an atom

(c) binds atoms together to form molecules.

(d) binds atoms and molecules together to form solids.

Answer:

(b) binds the protons and neutrons in the nucleus of an atom

(ii) Which of the following statement is not true? [1]

(a) Electrostatic force is a conservative force.

(b) Potential at a point is the work done per unit charge in bringing a charge from infinity to that point.

(c) An equipotential surface is a surface over which potential has a constant value.

(d) Inside a conductor, electrostatic field is zero.

Answer:

(b) Potential at a point is the work done per unit charge in bringing a charge from infinity to that point.

(iii) An electric heater is connected to the voltage supply. After few seconds, current gets its steady value then its initial current will be: [1]

(a) equal to its steady current

(b) slightly higher than its steady current

(c) slightly less than its steady current

(d) zero.

Answer:

(b) slightly higher than its steady current

(iv) An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T.

The radius of path is (q = 1.6 x 10^{-19} C, m_{c} = 9.1 x 10^{31} kg) [1]

(a) 2.585 x 10^{-4} m

(b) 3.58 x 10^{-4} m

(c) 2.58 x 10^{-3} m

(d) 3.58 x 10^{-3} m

Answer:

(d) 3.58 x 10^{-3} m

(v) The magnetic flux linked with a coil of N turns of area of cross-section A held with its plane parallel to the field B is : [1]

(a) \(\frac{N A B}{2}\)

(b) NAB

(c) \(\frac{N A B}{4}\)

(d) Zero

Answer:

(d) Zero

(vi) The minimum energy required to remove an electron is called: [1]

(a) stopping potential

(b) Kinetic energy

(c) work function

(d) none of these.

Answer:

(c) work function

(vii) The quantity which is not conserved in a nuclear reaction is: [1]

(a) momentum

(b) charge

(c) mass

(d) none of these.

Answer:

(c) mass

(viii) Bonds in a semiconductor: [1]

(a) trivalent

(b) covalent

(c) bivalent

(d) monovalent.

Answer:

(b) covalent

(ix) truth table entries are necessary for a four input circuit:

(a) 4

(b) 8

(c) 12

(d) 16

Answer:

(a) 4

Question 2.

Fill in the blanks :

(i) The capacitor which has air in between its plates is called ……………………………… . [1]

Answer:

air capacitor

(ii) SI unit of electrical conductivity is ……………………….. . [1]

Answer:

mho per meter

(iii) ……………………….. produced due to change in electric field. [1]

Answer:

Displacement current

(iv) Boolean expression of OR gate is ……………………………. . [1]

Answer:

Y=A +B.

Question 3.

Give the answer of the following questions in one line :

(i) If the electron drift is so small, how can we obtain large amount of current in a conductor? [1]

Answer:

It is on account of the fact that the number density of free electrons in a conductor is very large, e.g.,8 x 10^{28}

electrons/m^{3}.

(ii) Write the underlying principle of a Moving coil galvanometer.[1]

Answer:

The principle of moving coil galvanometer is based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque.

(iii) What is the direction of induced currents in metal rings 1 and 2,img when current I in the wire is increasing steadily? [1]

Answer:

(iv) Show on a plot the nature of variation of photoelectric current with the intensity of radiation incident on a photosensitive surface. [1]

Answer:

Graph of variation of photoelectric current with the intensity of incident radiation on a photosensitive surface is given as alongside.

(v) A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why? [1]

Answer:

de-Brogue wavelength,

λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m K}}\)

where, K = KE

For given KE, λ ∝ \(\frac{1}{\sqrt{m}}\)

Electron have smaller mass,

∴ λ_{e} > λ_{p} [m_{e} < m_{p} ]

Forgiven kinetic energy, electrons have greater wavelength as it has smaller mass.

(vi) A general impression exists that mass-energy interconversion takes place only in nuclear reaction and never in chemical reaction. This is strictly speaking, incorrect. Explain. [1]

Answer:

The mass defects are also involved in a chemical reaction, but they are almost a million times smaller than those in nuclear reaction. That is why a general impression exists that mass-energy interconversion takes place only in nuclear reaction.

(vii) Obtain approximately the ratio of the nuclear radii of the gold isotope _{79197}Au silver isotope _{47107}Ag.

Answer:

R ∝ A ^{1/3}

∴ \(\frac{R_{\mathrm{Au}}}{R_{\mathrm{Ag}}}=\left(\frac{197}{107}\right)^{1 / 3}\)

= 1.23

(viii) Write the truth table for a NAND gate as shown in the figure.[1]

Answer:

A | B | Y |

0 | 0 | 1 |

0 | 1 | 1 |

1 | 0 | 1 |

1 | 1 | 0 |

Section-B

Question 4.

Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by \(\left(\vec{E}_{2}-\vec{E}_{1}\right) \cdot \hat{n}=\sigma / \varepsilon_{0}\) where \(\hat{n} \) is a unit vector normal to the surface at a point and o is the surface charge density at that point.

(The direction of \(\hat{n} \) is from side 1 to side 2). Hence show that just outside a conductor, the electric field is a \(\sigma \hat{\boldsymbol{n}} / \varepsilon_{0} \) . [1½ ]

Answer:

Proceeding as in normal component of electric field intensity due to a thin infinite plane sheet of charge,

\(\vec{E}_{1}=-\frac{\sigma}{2 \varepsilon_{0}} \hat{n}\) and on right side,

\(\vec{E}_{2}=\frac{\sigma}{2 \varepsilon_{0}} \hat{n}\)

Discontinuity in the normal component from one side to the other is

Inside a closed condcutor, \(\vec{E}_{1}\) = o

∴ E = \(\vec{E}_{2}=\frac{\sigma}{\varepsilon_{0}} \hat{n}\)

Question 5.

A slab of material of dielectric constant K has the shape area as the plates of a parallel plate capacitor but has a thickness (3/4) d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates? [1½ ]

Answer:

Given:t= \(\frac{3}{4}\) d

Capacitance of air capacitor;

C_{0} = \(\frac{A \varepsilon_{0}}{d}\) ……………………… (i)

Capacitance of capacitor partially field with dielectric medium.

C = \(\frac{A \varepsilon_{0}}{(d-t)+\frac{t}{K}}=\frac{A \varepsilon_{0}}{\left(d-\frac{3 d}{4}\right)+\frac{3 d}{4 K}}\)

Question 6.

(a) The electron drift speed is estimated to be only a few mm/s for currents in the range of a few amperes. How then is current established almost the instant a circuit is closed?

(b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed?

(c) If the electron drift is so small, how can we obtain large amount of current in a conductor? [1½]

Answer:

(a) As soon as a circuit is closed, an electric field is established throughout the circuit with the speed of light. This results in local electron drift at all points in the circuit. For the current to be established. It is not necessary for the electrons at one end to reach the other end of the circuit. But in order to attain its steady value, the current does take some time.

(b) Each free-electron accelerates due to which its drift speed increases but decreases on colliding with a positive ion in the metal. After the collision, the electron starts accelerating again and the above process is repeated time and again throughout its motion in the conductor. Thus, on the average electrons acquire only a steady average drift speed.

(c) It is on account of the fact that the number density of free electrons in a conductor is very large, e.g., 8 x 10^{28}

electrons/m^{3}.

Question 7.

In following two circuits, galvanometer gives no deflection. Determine the ratio of R_{1} and R_{2} in both circuits. [1½ ]

Answer:

In the circuit-1, Wheatstone bridge is in balance condition, therefore,

\(\frac{4}{R_{1}}=\frac{6}{9}\) ⇒ R_{1} = \(\frac{4 \times 9}{6}\) = 6 Ω

In the circuit-2, The positions of battery and galvanometer have been changed, therefore there will be no effect on balance condition. Hence,

\(\frac{6}{12}=\frac{R_{2}}{8}\) ⇒ R_{2} = \(\frac{6 \times 8}{12}\) = 4 Ω

∴ \(\frac{R_{1}}{R_{2}}=\frac{6}{4}=\frac{3}{2}\) or R_{1} :R_{2} = 3:2

Question 8.

(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid.

(b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor? [1½]

Answer:

(a) The magnetic energy is

U_{B} = \(\frac{1}{2} L I^{2}=\frac{1}{2} L\left(\frac{B}{\mu_{0} n}\right)^{2}\)

(since B = u_{0}nI, for a solenoid)

= \(\frac{1}{2}\left(\mu_{0} n^{2} A l\right)\left(\frac{B}{\mu_{0} n}\right)^{2}=\frac{1}{2 \mu_{0}} B^{2} A l\) ………………. (1)

(b) The magnetic energy per unit

volume is, u_{B} = \(\frac{U_{B}}{V}\)

where V is volume that contains

= \(\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\) [Using (i)]

We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor,

u_{E} = \(\frac{1}{2} \varepsilon_{0} E^{2}\) ………………. (ii) 10

In both the cases energy is proportional to the square of the field strength.

Question 9.

A coil whose inductance is 2H and resistance is 10 Ω, is connected to a battery of 100 V and negligible internal resistance. Determine :

(i) time constant of the circuit.

(ii) the stable current in the circuit.

(iii) the energy stored to obtain stable current in the coil. [1½ ]

Answer:

(i) Time constant of circuit

= \(\frac{L}{R}=\frac{2}{10}\) = 0.2 s

(ii) Stable current in the circuit,

I_{0} = \(\frac{\varepsilon}{R}=\frac{100}{10}\) = 10 A

(iii) Stored energy in the coil,

U = \(\frac{1}{2} L I_{0}^{2}=\frac{1}{2}\) x 2 x 10 x 10 = 100 J

Question 10.

Write the conditions for observing a rainbow. Show by drawing suitable diagram how one understands the formation of a rainbow. [1½ ]

Answer:

To see the rainbow important conditions are:

(i) The Sun should shine in a part of sky while in opposite part of sky there should be rain.

(ii) The observer should stand with his back towards the Sun.

Question 11.

What is meant by power of a lens? Explain briefly. [1½ ]

Answer:

“Power of a lens can be defined as the deviation produced by the lens in the rays coming parallel to principal axis at a unit distance from the principal axis.”

It is shown, that the deviation δ is produced in a ray coming at a height of h from principal axis and parallel to this. Therefore, from the figure.

tanδ= \(\frac{h}{f}\)

if δ is small, then tan δ =δ

∴ δ=\(\frac{h}{f}\)

If h = 1, then δ = P (As per the definition of the power of lens).

∴ P= \(\frac{1}{f}\)

“Therefore the reciprocal of the focal length is known as the power of the lens.”

Question 12.

A convex lens is used to obtain a magnified image of an object on a screen 10 cm from the lens.

If the magnification is 19, find the focal length of the lens. [1½ ]

Answer:

Since, real and inverted image of an object can be taken on the screen.

Given, v=+10cm

and magnification, u = — ve (for real image)

∴ m = \(\frac{I}{O}=\frac{v}{u}\) ⇒ -19 = \(\frac{v}{u}\) ⇒ u = -19 u

⇒ u = \(-\frac{v}{19}\)

Using lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

∴ f = \(\frac{1}{2}\) cm

⇒ f = 0.5cm

Question 13.

What do you mean by persistence of vision? [1½ ]

Answer:

Image formed on retina is unstable yet after removal of an object its image persists on retina up to 1/10 vision. Due to persistence of vision luminous ring appears when illuminated object is whirled and time interval between two scenes in cinema hail is on the same principle.

Question 14.

Write any two characteristic properties of nuclear force. [1½ ]

Answer:

Two characteristics of nuclear force are given as below:

- These are short-range forces.
- These are strongest forces of attractive nature up to certain distance.

Question 15.

A radioactive isotope has a half-life of T years. How long will it take the activity for reduce to

(a) 3.125%

(b) 10% of its original value? [1½ ]

Answer:

(a) \(\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{2}\)

But, \(\frac{N}{N_{0}}=\frac{3.125}{100}=\frac{1}{32}\)

∴ \(\left(\frac{1}{2}\right)^{2}=\frac{1}{32}\)

⇒ n = 5 year

∴ t=nT

t = 5T years

(b) \(\frac{N}{N_{0}}=\frac{-1}{100}\)

t = \(\frac{2.303}{\lambda} \log \frac{N_{0}}{N}=\frac{2.303 \mathrm{~T}}{0.693} \log \left(\frac{100}{1}\right)\)

= \(\frac{2.303 \mathrm{~T} \times 2}{0.693} \cong 6.65\) T years

Section-C

Question 16.

How is moving coil galvanometer converted into a voltmeter? Explain giving the necessary circuit diagram and the required mathematical relation used. [3]

Or

(i) State the underlying principle of a moving coil galvanometer.

(ii) Give two reasons to explain why a galvanometer cannot as such be used to measure the value of the current in given circuit.

(iii) Define the terms (a) voltage sensitivity and (b) current sensitivity of a galvanometer. [3]

Answer:

Conversion of galvanometer in voltmeter:

As seen from the construction side it is also basically a pivoted coil galvanometer. If in a pivoted coil galvanometer a very high resistance known as “series resistance” is connected in series then the galvanometer is converted to a very high range Voltmeter shows a voltmeter.

A voltmeter is connected in parallel to a circuit to calculate the potential difference between two points in a circuit. To calculate the actual value of the potential difference, it is necessary that while measuring the potential difference the voltmeter takes no current from the circuit. This is only possible when the resistance of the voltmeter will be high. An ideal voltmeter’s resistance should be infinite.

If a G resistance galvanometer is to be converted in a V range voltmeter, then the high resistance R_{H} from the figure

V= I_{g}(R_{H} + G)

or R_{H}+G = \(\frac{V}{I_{g}}\)

Or R_{H} = \(\frac{V}{I_{g}}\) – G

R_{v} =G+R_{H} …(16)

since, R_{H} >> G

Therefore, R RH, meaning the effective resistance of the voltmeter is approximately equal to the high resistance of the galvanometer connected in series.

Question 17.

(i) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.

(ii) Obtain the mirror formula and write the expression for the linear magnification. [3]

Or

Under what conditions is the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium. [3]

Answer:

(i)

(ii) Mirror Formula for Concave Mirror

M_{1}M_{2} is a concave mirror and AB is an object placed in front of it. A’B’ is

the image.

In Δ ABCand ΔCA’B’

∠BAC = ∠CA’B’= 90^{0}

and ∠BCA ∠A’CH

(Because these are vertically opposite angles). Therefore Δ ABC and ΔCA’B’ are similar triangles. In

these similar triangles,

\(\frac{A B}{A^{\prime} B^{\prime}}=\frac{A C}{C A^{\prime}}\) ……………….. (1)

Now, In similar ΔA’B’F and FDN

\(\frac{D N}{A^{\prime} B^{\prime}}=\frac{F N}{F A^{\prime}}\) ……………………. (2)

DN = AB

∴ \(\frac{A B}{A^{\prime} B^{\prime}}=\frac{F N}{F A^{\prime}}\) ……………… (3)

On comparing equations (1) and (3)

\(\frac{A C}{C A^{\prime}}=\frac{F N}{F A^{\prime}} \) …………………….. (4)

If aperture of the mirror is small then P and N will be very close together.

∴ FP≈FN

Therefore from equation (4)

Formula For Magnification

In ∆ABP and ∆A’B’P

∠BAP= ∠B’A’P= 90°

∠BPA= ∠A’ Pff

From the laws of reflection,

∠i = ∠r

∴ From similar ∆ABP and ∆A’B’P

\(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} P}{A P}\)

AB = + O; A’B’ = -I

A’P = – ν;

AP = – u

∴ From equation (1),

\(\frac{-I}{+O}=\frac{-v}{-u}\)

or,

\(\frac{I}{O}=-\frac{v}{u}\)

Thus, magnification

m = \(\frac{I}{O}=-\frac{v}{u}\)

Question 18.

The work function of Cesium metal is 2.14 eV. When light of frequency 6 x 10^{14} Hz is incident on the metal surface, photoemission of electrons occurs.

What is the:

(a) Maximum kinetic energy of the emitted electrons,

(b) Stopping potential and

(c ) Maximum speed of the emitted photoelectrons? [3]

Or

Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies u_{A} > u_{B}.

(i) In which case, is the stopping potential more and why?

(ii) Does the slope of the graph depend on the nature of the material used? Explain. [3]

Answer:

Given, work function

W_{0} =2.14eV

Frequency u = 6x 10^{14} Hz

(a) Maximum kinetic energy of electrons K_{max} = hu – w_{0}

= 6.63 x 10^{-34} x 6x 10^{14} J- 2.14eV

= \(\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}\) eV – 2.14 eV

= 0.34 eV

(b) Stopping potential

V_{0} = \(\frac{K_{\max }}{e}=\frac{0.34 \mathrm{eV}}{e}\) =0.34V

(c) Maximum kinetic energy

K_{max} =\(\frac{1}{2}\) mv^{2}_{max} =0.34eV

Question 19.

A charge is distributed uniformly over a ring of radius a. Obtain an expression for the electric field intensity E at a point on the axis of the ring. Hence, show that for points at large distances from the ring, it behaves like a point charge.[4]

(i) Define electric field.

(ii) Why do the electric field lines never cross each other?

(iii) Why must electrostatic field at the surface of a charged conductor be perpendicular to every point on it?

(iv) Why do the electrostatic field lines not form closed loop? [4]

Answer:

According to the question, suppose that the ring is placed with its plane perpendicular to the X-axis as shown in the figure. Consider small element dl of the ring.

As the total charge q is uniformly distributed so, the charge dq on element dl is dq = \(\frac{q}{2 \pi a}\) dl

= \(\frac{k q}{2 \pi a} \frac{d l}{r^{2}}\) cos θ = dE cos θ

(where, cosθ = \(\frac{x}{r}\) )

Since only the axial component gives the net E at point P due to charge on ring.

So, ∫_{0E} dE = ∫_{02πd} dE cosθ

Now, for points at large distances from the ring x >> a.

∴ E=\(\frac{k q}{x^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{x^{2}} \)

This is same as the field due to a point charge indicating that for far-off axial point, the charged ring behaves as a point charge.

Question 20.

Write the boolean expression, symbol and truth table of the following logic gates:

(a) OR gate

(b) And gate. [4]

Or

(i) Draw the circuit diagram of a p-n junction diode in (a) forward bias. (b) reverse bias.

How are these circuits used to study the VI characteristics of a silicon diode? Draw the typical V-I characteristics.

(ii) What is a Light Emitting Diode (LED)? Mention two important advantages of LEDs over conventional lamps. [4]

Answer:

OR gate

Boolean expression Y = A + B.

Symbol:

Truth table

A | B | Y =A+B |

0 | 0 | 0 |

1 | 0 | 1 |

0 | 1 | 1 |

1 | 1 | 1 |

(b) AND gate Boolean Expression

Y=A-B

Symbol

Truth Table

A | B | Y=A.B |

0 | 0 | 0 |

1 | 0 | 0 |

0 | 1 | 0 |

1 | 1 | 1 |

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