Bihar Board 12th Physics Model Question Paper 1 in English Medium

Instructions for the candidates :

Candidates are required to give answers in their own words as far as practicable.
Figures in the Right-hand margin indicates full marks.
While answering the questions, candidates should adhere to the words limit as far as practicable.
15 Minutes of extra time has been allotted to read the questions carefully.
This question paper is divided into two sections : Section -A and Section-B
In Section A there are 35 Objective type questions which are compulsory, each carrying 1 mark. Darken the circle with Blue/Black ball pen against the correct option on OMR Sheet provided to you. Do not use Whitener/Liquid/ Blade/Nail on OMR Sheet, otherwise, your result will be treated as invalid.
In section-B, there are 18 Short answer type questions (each carrying 2 marks), out of which only 10 (ten) questions are to be answered Apart from this there are 06 Long Answer type questions each carrying 5 marks, out of which 3 questions are to be answered.
Use of any electronic device is prohibited.

Time : 3 Hours 15 Minute
Total Marks : 70

Objective Type Questions

In the following questions from no. 1 to 35, there is only one correct answer against each question. For each question, mark (darken) the correct answer on the OMR Sheet provided to you. (1 x 35 = 35)

Question 1.
The force acting between two electrons separated by distance r is proportional to
(a) r (b) r² (c) r^-2 (d) r^-1
Answer:
(c) r^-2

Question 2.
The dimensional formula for electric flux is
(a) [ML²T^-3l^-1] (b) [ML³T^-3l^-1] (c) [ML²T⁺³l^-1] (d) [ML³T^-3l]
Answer:
(b) [ML³T^-3l^-1]

Question 3.
The Torque experienced by an electric dipole of electric dipole moment \(\overrightarrow{\mathrm{P}}\) in a uniform field \(\overrightarrow{\mathrm{E}}\) is

Answer:
\(\text { (a) } \overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}\)

Question 4.
The work done in moving a unit positive charge through a distance x on an equipotential surface is
(a) xj (b) \(\frac{1}{x} j\) (c) zero (d) x²j
Answer:
(c) zero

Question 5.
The electric field intensity inside a charged hollow spherical conductor is
\((a) \epsilon_{0}
(b) E o \sigma
(c) zero
(d) \frac{\epsilon \circ}{\sigma}\)
Answer:
(c) zero

Question 6.
A conductor of capacity 20 mF is charged to 1000 V. The potential energy of the conductor will be
(a) 20 x 10⁴ J (b) 10⁴ J (c) 20 x 10³ J (d) 10³ J
Answer:
(d) 10³ J

Question 7.
The emf of a cell is measured by
(a) Voltmeter (b) Potentiometer (c) Ammeter (d) Voltameter
Answer:
(b) Potentiometer

Question 8.
Kirchoff’s current rule is related to Principle of conservation of
(a) Momentum (b) Energy (c) Charge (d) Angular momentum
Answer:
(c) Charge

Question 9.
Electron volt (eV) is a measure of
(a) Potential difference (b) Charge (c) Energy (d) Current
Answer:
(c) Energy

Question 10.
The Torque experienced by current carrying loop of magnetic moment \(\overrightarrow{\mathbf{M}}\) placed in a magnetic field \(\overrightarrow{\mathbf{B}}\) is

Answer:
(a)

Question 11.
Resistance of Galvanometer is G. Shunt used to allow 1 % of main current through the Galvanometer is
\((a) \frac{\mathrm{G}}{99}
(b) \frac{\mathrm{G}}{90}
(c) \frac{\mathrm{G}}{100}
(d) \frac{99 \mathrm{G}}{100}\)
Answer:
\(\text { (a) } \frac{\mathrm{G}}{99}\)

Question 12.
A Current is flowing in a circular coil. If the radius of the coil is doubled, then the value of magnetic field intensity at its centre will be
(a) same (b) half (c) double (d) four times
Answer:
(b) half

Question 13.
The working of a Dynamo is based on
(a) heating effect of current (b) electro-magnetic induction (c) induced magnetism (d) induced electricity
Answer:
(b) electro-magnetic induction

Question 14.
Unit of self inductance is
(a) Weber (Wb) (b) Ohm (Ω) (c) Henry (d) Gauss
Answer:
(c) Henry

Question 15.
If N₁ and N₂ are number of turns in primary and secondary coil of a transformer, then
(a) N₁ > N₂ (b) N₂ > N₁ (c) N₁ = N₂ (d) N₁= 0
Answer:
(b) N₂ > N₁

Question 16.
Power factor of L – R circuit is

Answer:
(b)

Question 17.
The ratio of value of magnetic field at same distance from mid point of a small magnet at its axial and equatorial position is
(a) 2′: 1 \(\text { (b) } 1: \sqrt{2}\) \(\text { (c) } \sqrt{2}: 1\) (d) 1 : 2
Answer:
(a) 2 : 1

Question 18.
The angle of dip at magnetic pole of earth is
(a) 0° (b) 90° (c) 45° (d) 180°
Answer:
(b) 90°

Question 19.
The direction of transmission of the electromagnetic wave is
(a) parallel to \(\overrightarrow{\mathrm{E}}\) (b) parallel to \(\overrightarrow{\mathrm{B}}\) (c) parallel to \(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{E}}\) (d) parallel to \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)
Answer:
(d) parallel to \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

Question 20.
An air bubble inside water shines due to
(a) refraction (b) reflection (c) total internal reflection (d) diffraction
Answer:
(c) total internal reflection

Question 21.
The equivalent focal length of co-axial contact combination of two lenses of power -15D and +5D will be
(a) -10 cm (b) +10cm (c) -20 cm (d) +20 cm
Answer:
(a) -10 cm

Question 22.
The final image in astronomical telescope is
(a) Real and inverted (b) Virtual and erect (c) Virtual and inverted (d) Real and erect
Answer:
(b) Virtual and erect

Question 23.
Cylindrical lenses are used to correct defect of eye is called
(a) Myopia (b) Hypermetropia (c) Astigmatism (d) Presbyopia
Answer:
(c) Astigmatism

Question 24.
In Rayleigh scattering of light, the intensity of scattered light is proportional to the wave-length X in the form of
\((a) \frac{1}{\lambda^{2}}
(b) \frac{1}{\lambda}
(c) \frac{1}{\lambda^{3}}
(d) \frac{1}{\lambda^{4}}\)
Answer:
\((d) \frac{1}{\lambda^{4}}\)

Question 25.
The colour of a thin film is due to
(a) interference (b) diffråction (c) scattering (d) dispersion
Answer:
(a) interference

Question 26.
Ilrcwstcrs Jaw is
(a) μ = cos i_p
(c) μ = sin i_p
(b) μ = tan i_p
(d) μ = cot i_p
Answer:
(a) μ = cos i_p

Question 27.
The energy of a photon of wave-length λ is
(a) \(\frac{h c}{\lambda}\) (b) \(\frac{h \lambda}{C}\) (c) hcλ (d) \(\frac{\lambda}{\mathrm{hc}}\)
Answer:
(b) \(\frac{h \lambda}{C}\)

Question 28.
Which of the spectral line of hydrogen spectrum lies in visible region?
(a) 50 Hz (b) 25 Hz (c) 100 Hz (d) 200 Hz
Answer:
(a) 50 Hz

Question 29.
Which one of the following is a chargeless particle?
(a) α-partcle (b) β-partcle (c) proton (d) photon
Answer:
(a) α-partcle

Question 30.
The time during which the quantity of any radio¬active material becomes half of its original value is called
(a) Average age (b) Half age (c) Decay constant (d) Time period
Answer:
(b) Half age

Question 31.
The majority charge carrier in /i-type semi¬conductor is
(a) a-partcle (b) electron (c) proton (d) hole
Answer:
(b) electron

Question 32.
In full wave rectification, if input frequency is 50 Hz, then output frequency will be
(a) 50 Hz (b) 25 Hz (c) 100 Hz (d) 200 Hz .
Answer:
(a) 50 Hz

Question 33.
The Boolean – expression for NAND gate is
\(\text { (a) } \overline{A+B}=Y\) (b) A + B = Y \(\text { (c) } \overline{A.B}=Y\) (d) A.B = Y
Answer:
\(\text { (c) } \overline{A.B}=Y\)

Question 34.
Result of conversion of decimal number 15 into Binary will be
(a) (1100)₂ (b) (1001)₂ (c) (1111)₂ (d) (11001)₂
Answer:
(c) (1111)₂

Question 35.
Which frequency range is used in TV transmission?
(a) 30 Hz – 300 Hz (b) 30 KHz – 300 KHz (c) 30 MHz – 300 MHz (d) 30 GHZ – 300 GHz
Answer:
(c) 30 MHz – 300 MHz

Section – II

Non-Objective Type Questions
Short Answer Type Questions

Question nos. from 1 to 18 are short answer type questions. Each question of this category carries 2 marks. Answer any ten (10) questions.

Question 1.
What is Electrostatic Shielding? Give its practical application.
Answer:
Electrostatic Shielding:
The electrostatic shielding is the best way to protect some area in vacuum or in alir from the influence of externd electric field.

The technique of electrostatic shielding is applied basically in designing the cable of television.

Question 2.
Give two factors which affect capacitance of a capacitor.
Answer:
The capacity of a capacitor depends upon the following two factors.
(a) On the surface area of the plate of the capacitor.
(b) Distance between the two plates of the capacitor.

Question 3.
Define current density \(\overrightarrow{\mathrm{J}}\) and write expression for Current density in terms of Drift velocity.
Answer:
Current density The current per unit area of conductor is called Current density. It is denoted by J.
\(\mathrm{J}=\frac{1}{\mathrm{A}}\)

It is a Vector quantity.
S.I. unit Ampere/m²
We know that I = ηAeV_d; \(\mathrm{V}_{\mathrm{d}}=\frac{\mathrm{I}}{\eta \mathrm{Ae}} ; \quad \mathrm{V}_{\mathrm{d}}=\frac{1}{\mathrm{ne}}\)

Question 4.
A wire of 12 Ω resistance is stretched to double its length. Find out the new resistance of the wire.
Answer:
In this position volume remains constant

Question 5.
Explain Ampere’s circuital law.
Answer:
Ampere’s circuital law : It states that the line integral of magnetic field \(\overrightarrow{(B)}\) around any closed path or circuit is equal to μ₀ times of total current (I) threading the closed circuit.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} 1}=\mu_{0} \mathrm{I}\)
where μ₀ = permeability of vacuum.

Question 6.
What is Shunt? Explain its uses.
Answer:
Shunt :
It is wire of low resistance, joined in parallel with circuit. The current capacity of pivoted type galvanometer is not very large. When large current passes through the cirucit then its coil may burn. To save them from damage, a low resistance wire is joined in parallel and it is called Shunt.

Uses of Shunt :

In the protection of coil of Galvanometer
In the conversion of Galvanometer into Ammeter.

Question 7.
Lenz’s law is the law of conservation of energy. Explain.
Answer:
Lenz’s Law : It states that the induced current produced in a circuit always flows in such a direction that it opposes the change or the cause that produces it.

According to Lenz’s law, the induced emf. opposes the change that produces it. It is this opposition against which we perform mechanical work in causing the change in magnetic flux. So, it is the mechanical energy which is converted into electrical energy. Thus, Lenz’s law is in accordance with the law of conservation of energy.

Question 8.
State two differences betwen the magnetic properties of soft iron and steel.
Answer:
Magnetic properties of steel and soft iron :
Steel have high value of retentivity and high value of coercivity. The area of hystersis loop is large for steel. So, it is used to make permanent magnet.

The soft iron have high retentivity and low coercivity. The area of hysteresis loop is small for soft iron, so it is used to make electro-magnet.

Question 9.
Explain Wattless current.
Answer:
Three properties of electro-magenetic waves :

The electro-magnetic waves are produced by accelerated charges.
Electro-magnetic waves do not require any material medium for their propagation.
The electro-magnetic waves obey the principle of Superposition.

Question 10.
State two main characteristics of electro-magnetic waves.
Answer:

Electromagnetic waves are emitted by electrically charged particles undergoing acceleration.
Electromagnetic waves of different frequency are called by different names. Since they have different source and effects no matter.

Question 11.
Two thin convex lenses of power 4D and 6D are placed co-axially at a distance of 20 cm apart. Find the focal length and power of the combination of lenses.
Answer:

Question 12.
Write necessary conditions for Interference?
Answer:
Necessary conditions for Interference of light.

The two sources should emit the light waves continuously.
The light waves should be of same wave-length.
The light waves emitted should preferably be of the same amplitude.
The two light sources should be very narrow.
The two sources of light must be very close to each other.

Question13.
Write Truth table and Boolean algebraic expression of OR gate and AND gate.
Answer:
Boolean expressiosn of OR gate A + B = Y
Truth table for OR gate

Boolean expression for AND gate A • B = Y Truth table for AND gate

Question 14.
Write the de-Broglie wave-length of an electron of Kinetic energy E.
Answer:
de-Broglie wave length of an electron :
Let m is mass of electron, e is the charge and it is accelerated with potential v.
Energy of electron E = ev …………………….. (i)

K.E. of electron

where h = Plank” constants.

Question 15.
What is Amplitude Modulation? Explain.
Answer:
Amplitude Modulation When the modulating wave is superimposed on a high-frequency carrier wave in a manner that the frequency of the modulated wave is same as that of the carrier wave but its amplitude is modified in accordance with that of the modulating wave the process is called amplitude modulation.

Question 16.
What will be the value of flux of electric field on a closed surface ABC placed near charges +q₁ +q₂ and -q₃?

Answer:
Total charge inside the closed surface = + q₁ – q₃
total electric flux charge q₂ is outside Φ = \(\frac{q_{1}-q_{3}}{\epsilon_{0}}\) the closed surface.

Question 17.
In going from water to glass will speed of light increase or decrease? Explain.
Answer:
When the light ray passes from water to glass its speed decreases because glass is denser medium in comparision to water. When light ray passes from water to glass its wave length decreases. So, speed of light also decreases.

Question 18.
A particle of mass m and charge q is projected with velocity \(\overrightarrow{\mathrm{v}}\) perpendicular to a uniform magnetic field \(\overrightarrow{\mathrm{B}}\). The particle is set in a circular motion. What will be the radius of this circular path?
Answer:
When the charge particle moves perpendicular to magnetic field then its experiences the force F = qBVsinθ
Here θ = 90°; F = qBV ……………………..(i)
when it moves on aircular path then centripetal force
\(F=\frac{m V^{2}}{r}\) …………… (ii)

From equation (i) and (ii)
\(\frac{m v^{2}}{r}=q B V ; m v=q B r ; \quad r=\frac{m v}{q B}\)
This is the expression for radius of circular path.

Long Answer Type Questions

Question nos. from 19 to 24 are long answer type questions. Each question of this category carries 5 marks. Answer any three questions. (3 x 5 = 15)

Question 19.
State Gauss’s law. Applying this law obtain electric field due to an infinitely long thin uniform charged straight wire.
Answer:
Gauss’s Theorem :
It states that the total electric flux through a closed surface enclosing a charge is equal \(\frac{1}{\epsilon_{0}}\) to times the magnitude of charge enclosed.
\(\phi=\frac{q}{\epsilon_{0}}\)

where ε₀= permitivity of vacuum or air Expression for electric intensity due to infinitely long charged st. wire.

Consider a thin infinitely long st. line charge having a linear charge density λ placed along yy’.

To find electric field due to line charged at point P distance r from it, draw a cylindrical surface of radius r and length l about the line charge as its axis. The cylindrical surface may be treated as the Gaussian surface for the line charge.

If E is the magnitude of electric field at point P, then electric flux through the Gausssian surface is given by
Φ = E x Area of curved surface of cylinder
Φ = E x 2πrl …………………… (i)

According to Gauss’s theorem

Question 20.
Establish the formula for a lens
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{\mathbf{R}_{1}}-\frac{1}{\mathbf{R}_{2}}\right)\)
Answer:
Lens Maker’s Formula :
Suppose L is a thin lens. The refractive index of the material of lens is n₂ and it is placed in a medium of refractive index nx. The optical centre or lens is C and X. X is principal axis. The radii of curvature of the surfaces of the lens are Rj and R₂ and their poles are P₁ and P₂. The thickness of lens is t, which is very small. O is a point object on the principal axis of the lens. The distance of O from pole P₁ is u. The first refracting surface forms the image of O at V at a distance v’ from P₁. From the refraction formula at spherical

The image I acts as a virtual object for second surface and after refraction at second surface, the final image is formed at I. The distance of I from pole P₂ of second surface is v. The distance of virtual object (E) from pole P₂ is (v -1).

For refraction at second surface, the ray is going from second medium (refractive index n₁) to first medium (refractive index (n₂), therefore from refraction formula at spherical surface.
\(\frac{n_{1}}{v}-\frac{n_{2}}{\left(v^{\prime}-1\right)}=\frac{n_{1}-n_{2}}{\mathrm{R}_{2}}\) …………………….. (ii)

For a thin lens t is negligible as compared to v’, therefore from (ii)

where \(\mathrm{l} n_{2}=\frac{n_{2}}{n_{1}}\) is refractive index of second medium (i.e., medium of lens) with respect to first medium.

If the object O is at infinity, the image will be formed at second focus i.e.,

if u = ∞, v = f₂ = f , Therefore from equation (iv)

This is the formula of refraction for a thin lens. This is also called Len’s maker’s formula.

Question 21.
Define Interference of light. Derive an expression for Fringe-Width in Young’s double slits experiment.
Answer:
Interference of Light :
The phenomenon of non-uniform distribution of energy in the medium due to superposition of two light waves is called interference of light.

Expression for Fringe Width :
Consider that two coherent source of light S₁ and S₂ are placed at a distance apart and screen is placed at a distance D from the plane of two source. Let P be a point on the screen at distance y from the centre of screen. If .v is the path difference between light waves reaching point P from the source S₁ and S₂ then x = S₂P – S₁P

Fringe Width :
The spacing between any two consecutive fright fringes is equal to the width of dark frine therefore width of dark fringe.

Thus, in Young’s double slit experiment, the bright and dark fringes are of same width. It is denoted by β.
\(\beta=\frac{D \lambda}{d}\)

Question 22.
State KirchofTs laws. Applying these laws obtain the condition for balanced Wheatstone Bridge.
Answer:
Kirchoff’s law : First law :
In the network of conductor the algebraic sum of current meeting at a point is always zero. It also causes junction rule.

Second law :
It states that in any closed part of an electrical circuit, the algebraic sum of the emf is equal to the algebraic sum of the products of the resistance and current flowing through them. It is also called loop rule. This law is based on the law of conservation of energy.

Wheatstone bridge :
It is one of the accurate arrangements for measuring an unknown resistance. It is based on the principle of Kirchoff s law.

Four resistances P, Q, R and S are connected to form a quadrilateral of galvanometer and a tapping key k₁ (called galvanometer) are connected between points B and D as shown in the figure.

Applying Kirchoff s second law, for closed part ABDA in the circuit I₁ P + I_g g – (I – I₁) S = 0
where g is the resistance of galvanometer In the balanced state I_g = 0

Applying Kirchoff s law in closed part BDCB

From equation (i) and (ii)
\(\frac{P}{S}=\frac{Q}{R}\left[S=\frac{P}{Q} \cdot R\right]\)

Thus unknown resistance can be calculated.

Question 23.
State Biot-Savart law. Applying this law obtain expression of magnetic Held at a point on the axis of a circular coil carrying electric current.
Answer:
Biot-Savart law :
This law establish the relation between magnetic field and current.

Let xy is aonductor in which current I is flowing from v to y. P is the point at the distance r and at this point magnetic field \(\overrightarrow{d B}\) depends upon following factor :

where μ₀ is permeability of vacuum or air.

Magnetic field at the axis of a circular loop :
Consider a circular loop of radius R carrying current I, with its plane perpendicular to the plane of paper. Let P be a point of observation on the axis of this circular loop at a distance x from its centre O. Consider a small element of length dl of the coil at point A. The magnitude of the magnetic induction \(\overrightarrow{\mathrm{dB}}\) at point P due to this element is given by

The direction of \(\overrightarrow{\mathrm{dB}}\) is perpendicular to the plane containing \(\overrightarrow{\mathrm{dl}}\) and \(\overrightarrow{\mathrm{r}}\) is given by Right hand screw rule. As the angle between I \(\overrightarrow{\mathrm{dl}}\) and \(\overrightarrow{\mathrm{r}}\) is 90°, the magnitude of the magnetic induction \(\overrightarrow{\mathrm{dB}}\) is given by,
\(\overrightarrow{d B}=\frac{\mu_{0} I}{4 \pi} \frac{d l \sin 90^{\circ}}{r^{2}}=\frac{\mu_{0} I d l}{4 \pi r^{2}}\)

If we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to the axis will be added up. Thus the resultant magnetic induction \(\overrightarrow{\mathrm{B}}\) at axial point P is along the axis and may be evaluated as follows :

The component of \(\overrightarrow{\mathrm{dB}}\) along the axis,

Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by

Question 24.
What is Zener diode? Explain its use in a Voltage regulator.
Answer:
Zener diode :
The specially designed Junction diode, which can operate in the reverse breakdown voltage region continuously without being damaged are called Zener diode.

Symbol of Zener diode

Zener diode as a voltage regulator :
The use of Zener diode is based on the fact that in reverse break down region, a very small change in voltage across the Zener diode produces a very large change in the current through the circuit.

If voltage is increased beyond Zener voltage, the resistance of the Zener diode drops considerably. Consider that the Zener diode and a resistance R, called dropping resistor are connected to fluctuating d.c. supply, such that the Zener diode is reverse biased. Whenever voltage across the diode tends to increase, the current through the diode rises out of proportion and causes a sufficient increase in voltage drop across the dropping resistor. As a result, the output voltage lowers back to the normal value. Similarly, when the voltage across the diode tends to decrease, the current through the diode goes down out of proportion, so that the voltage drop across the dropping resistor is much less and now the output voltage is raised to normal.