SOLUTION (XII BOARD PAPER PT-2) PHYSICS

KS/IIT/SSB1719/PT-2/20180914 BOARD EXAM [1]

KALRASHUKLA CLASSES – C IV IL L INES, KAKADEO, SOUTH KANPUR – 7617800800

SOLUTION (XII BOARD PAPER PT-2) PHYSICS

1. Biot-Savart’s law : The strength of magnetic field or magnetic flux density at a point P (dB)

due to current element dl depends on,

(i) dB I

(ii) dB dl

(iii) dB sin

(iv) dB 2

1

r.

Combining, (i) to (iv) dB 2

Id sin

r

l

dB = k 2

Id sin

r

l [k = Proportionality constant]

In S.I. units, k = 0

4

where 0 is called permeability of free

space.

0 = 4 × 10–7

TA–1

m

dB = 0

2

Id sin

4 r

l

2. (a) Magnetic field due to straight part

0

3

Id rB

4 r

lB

For point O, dl and r for each element of the straight segments AB and DE are parallel.

Therefore, dl × r = 0. Hence, magnetic field due to straight segments is zero.

(b) Magnetic field at the centre due to circular point

Magnetic field at the centre of circular coil

2

0 0I i1

B [ Here, coil is half]2 2r 4r

7

0

2

I 4 10 12B

4r 4 2 10

= 6 × 10–5

T

3. The radius of atom whose principal quantum number is n, is given by

r = n2r0

r0 = 5.3 × 10–11

m

For second excited state, n = 3

r = 32 × r0 = 9 × 5.3 × 10

–11

or, r = 4.77 × 10–10

m

4. Total magnetic field at the centre of the loop will be zero because direction of magnetic field

due to semicircle APB and semicircle AQB are opposite and magnitude is equal.



5. I = 5A, dl = 1 cm = 0.01m, r = 1 m

= 45° [ Direction is North – East)

By using Biot Savart’s Law,

0

2

Id sindB

4 r

l

7 9

2

5 0.01 sin 4510 3.54 10 T

(1)

Its direction is vertically downwards.

6. According to Rutherford’s model, the electrons revolve around the nucleus in any orbit and

radiate waves of all frequencies, so the spectrum is continuous. On the other hand in Bohr’s

model, the electrons revolver around the nucleus in some definite orbits and emit only waves of

some definite frequencies and so they give line spectrum.

7. The total charge of the capacitor remain conserved and the capacitance of capacitor increases to

K times of original values on introduction of dielectric slab.

CV = C'V'

CV = (KC)V', V' = V

K

New electric field

E' = V' V / K V 1

d d d K

E' = E

K

8. Two postulates of Bohr’s theory of hydrogen atom.

(i) Every atom consists of small and massive central core, known as nucleus around which

electron revolve and necessary centripetal force prevailed by electrostatic force of

attraction between positively charged nucleus and negatively charged electrons.

(ii) The electrons are revolved around the nucleus in only those circular orbits which satisfy

the quantum condition that the angular momentum of electrons is equal to integral

multiple of h

2, where, h is Planck’s constant.

i.e., nh

mvr2

where, n = 1, 2, 3, …

9. Limitation of Bohr’s model

(1) It could not explain the spectra of atoms containing more than one electron.

(2) There was no theoretical basis for selecting mvr to be an integral multiple of h/2. (3) It involved the orbit concept which could not be checked experimentally.

(4) It could not explain Zeeman and Stark effect and fine lines of spectra.

(5) It was against de-Broglie concept and uncertainly principle.



10. After disconnection from battery and doubling the separation between two plates.

(i) Charge on capacitor remains same.

Q = Q'

i.e., CV = C'V'

C

CV V'2

V' = 2V Electric field between the plates

V' 2V

E 'd' 2d

V

E ' Ed

Therefore, Electric field between the two plates remains same.

(ii) Capacitance reduces to half of original value as

1

Cd

0A C

C'2d 2

(iii) Now, energy stored in capacitor after disconnection from battery

2 2 2

2

q q qU

C2(C') C2

2

2 2

2 1

q qU 2 2U U

2C 2C

Energy stored in capacitor gets doubled to its initial value.

11. Effect of filling dielectric with battery connected

When there is no dielectric

Capacitance 0

0

AC

d

Potential difference between the plates V

Charge on a plate Q = CV

Energy 2

0 0

1E C V

2

Electric field 0

VE

d

When dielectric is inserted

0

0

KAC KC

d

0 0

Q KC V KQ

2

0 0

1U KC V KU

2

0

VE E

d



12. From Biot-Savart’s law, the magnetic field due to an element of length 2l of the current

carrying conductor

0

2

Id sindB

4 r

l

Here = 90° ( dl is tangential, angle between the radius and the tangent is 90°)

0

2

IddB .

4 r

l

Total magnetic field at the centre O due to the whole circular coil

0 0 0

2 2

I I IB dB d .2 r

4 r 4 r 2r

l =

For a coil of n number of turns, 0nI

B2r

It is perpendicular to the plane of the coil and directed inwards.

13. (i) Capacitance of the parallel plate capacitor is given by,

0A

Cd

0A

Xd

r 0 0A 4 A

Yd d

or, Y = 4X

Now, X and Y are in series combination and their equivalent capacitance is 4F.

1 1 1 1 1

4 X Y X 4X

or, 1 5

4 4X

or, X = 5F.

Y = 4 × = 20 F. (ii) Total charge, Q = CV

= 4 × 10–6

× 12

= 48 × 10–6

C

Potential difference between the plates of capacitor

6

6

Q 48 10X 9.6V.

C 5 10

Potential difference between the plates of capacitor

6

6

Q 48 10Y 2.4V.

C 20 10

(iii) Energy stored 21CV

2

The ratio of electrostatic energy stored in X and Y.

26 21 1

6 22

2 2

1C V

(5 10 )(9.6)21 (20 10 )(2.4)

C V2

=

2

1 4 16 44 :1.

4 1 4 1



14. (i) 20 F and C are in series

1 1 1

4 20 C

1 1 1 5 1 1 4 1

4 20 C 20 C 20 C

C = 5 F.

(ii) q = CV, q = 4 × 10–6

× 12 = 48 × 10–6

48 C.

(iii) 6

1 6

1

q 48 10V 2.4V

C 20 10

6

2 6

2

q 48 10V 9.6V

C 5 10

Potential drop across 20 F capacitor = 2.4V

And potential drop across 5 F capacitor = 9.6V.

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SOLUTION (XII BOARD PAPER PT-2) PHYSICS