OR Initial value of potential energy,

U_i = 12C V^{1 2}

On joining C₁ with C₂, after being disconnected from the battery, common potential,

′ = +

V C V

C C₁¹ ₂

( )

Hence, final value of potential energy U_f =1 C C V+ ′

10. Electric field at a point distant x from the ring of charge q,

Force on the electron, F = eE Acceleration of the electron,

d x

11. The capacitance of a parallel plate capacitor of plate area A and plate separation d with vacuum between its plates is given by

C A

0=e0d

Suppose initially the charges on the capacitor plates are ±Q. Then the uniform electric field set up between the capacitor plates is

E Q

0 A

0 0

= σ =

e e

When a dielectric slab of thickness t < d is placed between the plates, the field E₀ polarises the dielectric.

This induces charge –Q_p on the upper surface and +Q_p on the lower surface of the dielectric.

These induced charges set up a field E_p inside the dielectric in the opposite direction of 

E₀. The induced field is given by

E P Q

p p A P

=σ = P = =

e e σ

0 0[ ,polarisation density] The net electric field inside the dielectric is

physics for you|june ‘15 between the capacitor plates, the field E exists over a distance t and field E₀ exists over the remaining distance (d–t). Hence the potential difference between the capacitor plates is

V E d t Et E d t EK t E

The capacitance of the capacitor on introduction of dielectric slab becomes

C QV A

d t tK

= = − +

e₀ .

12. Suppose the three charges are placed as shown in the figure. Let the charge q be positive.

For the equilibrium of charge +q, we must have, Force of repulsion F₁ between + 4e and +q only x = 2a/3 is possible. Hence for equilibrium, the charge q must be placed at a distance 2a/3 from the charge + 4e.

We have considered the charge q to be positive. If we displace it slightly towards charge e, from the equilibrium position, then F₁ will decrease and F₂ will increase and a net force (F₂ – F₁)will act on q towards left i.e., towards the equilibrium position.

Hence the equilibrium of positive q is stable.

Now if we take charge q to be negative, the force F₁ and F₂ will be attractive, as shown in figure.

The charge –q will still be in equilibrium at x = 2a/3. However, if we displace charge –q slightly towards right, then F₁ will decrease and F₂ will increase. A net force (F₂ – F₁) will act on –q towards right i.e., away from the equilibrium position. So the equilibrium of the negative q will be unstable.

13. Consider a small sphere of radius r placed inside a large spherical shell of radius R. Let the spheres carry charges q and Q, respectively.

Total potential on the outer sphere, V_R = Potential due to its own charge Q

+ potential due to the charge q on the inner sphere

=  +

Potential on the inner sphere due to its own charge is

V q sphere is the same as that on its surface, so potential on the inner sphere due to charge Q on outer sphere is

V Q

2 R

0

41

= pe ⋅

\ Total potential on inner sphere

V q

Hence the potential difference is

V V q sphere will always be higher than that of the outer sphere. Now if the two spheres are connected by a conducting wire, the charge q will flow to the outer sphere, irrespective of the charge Q already present on the outer sphere. In fact this is true for conductors of any shape.

14. As the two spheres are connected to each other by a wire, so they have same electric potential i.e.,

V_a = V_b

physics for you |june ‘15 57

So, the ratio of electric fields at the surfaces of the two spheres is produces more electric field on its surface. Hence, the charge density on the sharp and pointed ends of conductor is higher than on its flatter portions.

15. Consider an electric dipole consisting of charges +q and –q and of length 2a placed in a uniform electric field  uniform electric field is zero. But the two equal and opposite forces act at different points of the dipole.

They form a couple which exerts a torque.

Torque = Either force × Perpendicular distance between the two forces t = qE × 2a sinq = (q × 2a)E sinq

or t = pE sinq (p = 2aq)

As the direction of torque t is perpendicular to both   pand , so we can write  t = ×E p E

The direction of vector t is that in which a right handed screw would advance when rotated from p

to 

E. As shown in figure, the direction of vectort is perpendicular to both pand E, and points into the plane of paper.

When the dipole is released, the torque t tends to align the dipole with the field

E i.e., tends to reduce angle q to 0. When the dipole gets aligned with the torque t becomes zero. E ,

Clearly, the torque on the dipole will be maximum when the dipole is held perpendicular to

E .Thus t_max = pE sin90° = pE.

Dipole moment:

As t = pEsinq

If E = 1 unit, q = 90°,then t = p

Hence dipole moment may be defined as the torque acting on an electric dipole, placed perpendicular to a uniform electric field of unit strength.

As shown in figure, suppose the charges + q and –q OR are located at points A and B distance d apart.

Let P be a point such that AP = r₁ and BP = r₂. Electric field at P due to charge +q is

E q

Electric field at P due to charge –q is angles made by AP and BP with AB, then

Component of E₁ along PR = E₁ cosa opposite or zero. Hence

(i) When E₁ sina = E₂ sinb

Thus the resultant intensity will be parallel to the

physics for you|june ‘15

58

line joining A and B if (i) P lies on the perpendicular bisector of AB, or (ii) P lies on either side of AB i.e., a = b = 0° or 180°

16. (a) q₁ = 7 mC = 7 × 10^–6 C, q₂ = –2 ×10^–6 C, r = 18 cm = 0.18 m

Electrostatic potential energy of the two charge system is

(b) Work required to separate two charges infinitely away from each other,

W = U₂ – U₁ = 0 – U = –(–0.7) = 0.7 J

(c) Energy of the two charges in the external electric field = Energy of interaction of two charges with the external electric field

+ Mutual interaction energy of the two charges

=q V r +q V r + q q

17. The velocity of the particle, normal to the direction of field

v_x = 1000 m s^–1, is constant.

The velocity of the particle, along the direction of field, after 10 s, is given by

v_y = u_y + a_yt

The net velocity after 10 s,

v= v_x²+v²_y = (1000) (²+ 2000)² =1000 5m s⁻¹ Displacement, along the x-axis, after 10 s,

x = 1000 × 10 m = 10000 m

Displacement along y-axis (in the direction of field) after 10 s, On connecting charged capacitor to uncharged capacitor, the common potential V across the capacitors is

V C V C V= C C+

Energy stored in capacitors before connection is U_i = 12C V + = ×_{1 1}² 0 12 600 10× ⁻¹²×200² or U_i = 12 mJ

Energy stored in capacitors after connection is

U_f = 12 1

2 600 600 10 100

1 2 2 12 2

(C C V+ ) = ( + )× ⁻ × or U_f = 6 mJ

Hence the energy lost in the process is DU= U_i – U_f = (12 – 6) mJ or DU= 6 mJ.

19. Here, r₁ = 13 cm, r₂ = 12 cm, K = 32, Q = 2.5 mC (a) Capacitance of capacitor is

C =4 1 32

(b) Electric potential of inner sphere is V_B = V_BB + V_BA

(c) Capacitance of isolated sphere of radius 12 cm is C₀ = 4pe₀r= 1 near it, negative charge induced on it decreases the electric potential on A, hence more charge can be stored on A.

physics for you |june ‘15 59 Since C₄ and C′′ are in series, so net capacitance of

the network is

1 1 1 1

200 1

100 1 2

4 200

C C= ′′ + = + =C + or C = 2003 pF = 66.7 pF

Net charge stored on the combination is Q = CV = 2003 ×10⁻¹²×300=2 × 10^–8 C As C′′ and C₄ are in series, so

Q′′ = Q₄ = Q

or Q′′ = Q₄ = 2 × 10^–8 C and hence V′′= ′′

Q′′

C = 2 10 200 10

8 12

×

×

−

−C

F= 100 V and V₄ = Q

C₄⁴

8 12

2 10 100 10

= ×

×

−

−C

F = 200 V C₁ and C′ are in parallel, so

V₁ = V′ = V′′

or V₁ = V′ = 100 V

Hence, Q₁ = C₁V₁ = 100 × 10^–12 × 100 = 1 × 10^–8 C and Q′ = C′V′ = 100 × 10^–12 × 100 = 1 × 10^–8 C C₂ and C₃ are in series, so

Q₂ = Q₃ = Q′ = 1 × 10^–8 C Hence,V₂ = QC²₂

8

1 10 12

200 10

= ×

×

−

−C

F= 50 V and V₃ = QC³₃

8 12

1 10 200 10

= ×

×

−

−C

F= 50 V.

21. Given situation is shown with DABC.

In D ABD, cos 30° = ADAB AD

= l or AD = l cos 30° = 32l As AO = 23 2

3 3

AD= × 2l or AO = l

3 ....(i) Similarly BO = CO = l

Forces on charge +Q at O due to charges +q at A, B 3 and C are

F_OA = F_OB = F_OC = 14 3

41 3

0 2 0 2

pe Qq pe

l

Qq

 l

 



=

Horizontal component of net force on +Q charge at O is

F_x = F_OB cos 30° – F_OC cos 30° = 0

or F_x = 0 ...(ii)

and vertical component of net force on +Q charge at O is

F_y = F_OB sin 30° + F_OC sin 30° – F_OA

or F_y = F_OA [2sin30° – 1] [ F_OB = F_OC = F_OA] or F_y = F_OA 2 1

× −2 1







= F_OA [1–1]

or F_y = 0 ...(iii)

So, net force on charge +Q at O is F_net = F_x²+F_y²

or F_net = 0

22. Consider an infinite line of charge with uniform linear charge density l, as shown in figure. We wish to calculate its electric field at any point P at a distance y from it. The charge on small element dx of the line charge will be

dq = ldx

The electric field at the point P due to the charge element dq will be

dE dq

r

dx y x

= =

+

41 1

0 2 4 0 2 2

pe . pe . l

physics for you|june ‘15

60

The field dE has two components : dE_x = –dE sinq and dE_y = dE cosq

The negative sign in x-component indicates that dE_xacts in the negative x-direction. Every charge element on the right has a corresponding charge element on the left. The x-components of two such charge elements will be equal and opposite and hence cancel out. The resultant field

E gets contributions only from y-components and is given by

E E_y dE_y dE

23. (i) Inquisitiveness and sharing.

(ii) The picture on the screen is produced by striking accelerated electrons on the positively charged screen, that is why Devansh observed charging on the screen. The charged screen also attracts airborne infected particles floating around. As many of the particles collected on the screen surface carry bacteria, the screen becomes contaminated with bacteria.

When a surgeon puts his finger with gloves near the screen the bacteria may shift to the gloves. To avoid the risk, surgeons are suggested not to bring finger near a video monitor.

24. Refer point 1.4 (2, 3, 4), page-6(MTG Excel in

Now battery remains connected and dielectric of constant K = 10 introduced, the potential will

physics for you|june ‘15

62

Silvering of Lens

An object is kept infront of a lens whose one face has been mirrored and location of the image is asked, as below.

To solve such questions, we need to understand first of all that when thin optical devices (lens or mirror) are kept in contact, the optical power follows algebraic addition rule which simply means, a converging (lens or mirror) + another converging (lens or mirror) increases the converging nature whereas converging + diverging will be less of converging.

Conventionally, the ability to converge is taken to be positive power and ability to diverge is negative power.

Note that for a lens, P and f have same sign whereas for mirror they follow opposite sign.

Clearly following the definition of power stated above,

For mirror, Power, P^m= − 1f_m ,

where f_m = focal length of mirror For lens,

Power, P^l = 1 , where ff_l l = focal length of lens

Hence, if two thin lenses of focal lengths f₁ and f₂ are kept in contact

we can replace the combination with a single lens of equivalent power.

f₁ f₂ P_eq = P₁ + P₂ ⇒ 1 = 1 + 1

1 2

f_eq f f

Here, we need to follow the sign convention of the respective lenses.

Now, what happens if a lens is silvered?

Any ray which is incident on the set up, has to pass through lens → mirror → lens, in the same order as indicated.

 F_eq

Hence, the incident ray passes through lens twice and once through mirror.

\ Equivalent power of this combination, whose overall nature (behaviour) is that of a mirror can be written as

P_eq = P_l + P_m + P_l = 2P_l + P_m

⇒ − 1 = 2 − 1 ⇒ 1 = 1 − 2

f_eq f_l f_m f_eq f_m f_l

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

physics for you |june ‘15 63 This formula can be used as a standard result for zfinding

out equivalent focal length of the combination and then we can use mirror formula to trace the image formed.

1 1 1

v u f+ = _eq

Note : If focal length of lens has not been given, instead the radius of curvature of the two surfaces and refractive index of the material is given, first find out the focal length of the lens using lens makers formula

1 1 1 1

1 2

f_l = − R −R

 

 (m )

and if the second surface of radius of curvature R₂ has been silvered, the focal length of the mirror becomes

1 2² f

R

m = Apparent depth

Assumption : Until and unless specified, we would assume near normal incidence of incident rays, i.e., angle of incidence would be small, such that we can assume sini ≈ i.

This is a very common phenomenon, in which objects kept in denser medium when viewed from rarer medium appear to be shifted closer.

Let us see the physics involved behind this.

Object O, is kept in medium of refractive index (R.I.) m₂ and is viewed from a medium of refractive index m₁.

R is the real depth of the object whereas A is the apparent depth.

\ m2sini = m₁sinr ⇒ m₂i ≈ m₁r

⇒ m₂ x =m₁ R

x

A ⇒ A R= = R m m₂ m

1 21

where, m m

21 m2

= 1 = refractive index of 2^nd medium with respect to 1^st medium.

On similar lines, if the object was kept in rarer medium and viewed from denser medium, it would

appear to be shifted even more far than the real depth as below :

Again since the object is in medium of refractive index m₁,

A= R = R

m m

12 m1 2

Clearly from both the results, we note that, depth increases when viewed from denser medium and decreases when viewed from rarer medium.

Note : If the rarer medium is air, m₁ = 1 whereas the other medium might be water, glass etc. whose refractive index, m₂ = m (say).

\ =







A R

R m m

when depth decreases when depth increases Let us see some applications :

1. Two thin similar convex glass pieces are joined together front to front with its rear portion silvered such that a sharp image is formed 20 cm from the mirror. When the air between the glass pieces is replaced by water (m = 4/3), then find the location of image.

Soln.: The first thing to note here is, what would happen with the thin convex glass pieces? Do they have any role to play?

Imagine a thick slab,

The ray gets laterally displaced, inclination does not change since radius of curvature is identical.

physics for you|june ‘15

64

So, if the glass piece is said to be thin, it means lateral displacement will be negligible.

So, basically, these glass pieces are used only to hold water, nothing else.

In the first situation (without water), the entire set up is equivalent to only a concave mirror on which a ray travelling from infinity is incident. Hence they would be focussed at f R=

2 .

\ R= ⇒ R

2 20 cm = 40 cm

Now, we have a lens made of water.

 =4 3

R1= +R R2= –R

\ = −

 

 −

−







=

1 4

3 1 1 1 2

3

f_l R R R

\ 1 = 1 − 2

f_eq f_m f_l =

−2 − 4 3 ( )R R

= − = −

× = −

10 3

10 3 40

1 12 R

\ feq = –12 cm, i.e., the set up behaves like a concave mirror of focal length 12 cm.

Hence, rays get focussed at 12 cm from combination.

2. A glass slab of thickness t and refractive index m is kept in between a point object and observer.

By what distance the object appears shifted?

Soln.:

Let O, I′ and I represent the object, image formed after 1^st refraction, and final image respectively.

Remember the rule, as many times there is a change in medium for a plane surface we can apply the concept of real and apparent depth, where we apply mR for increment in depth, R

m for decrement in depth and that can be understood by taking a ray and see how it bends. If it bends towards normal, depth increases. If it bends away from normal, depth decreases.

For 1^st refraction (interface AB), R = x, A = mx

2^nd refraction (interface CD), R = mx + t, A=mx t+

m

\ Shift in location of image, OI= x x t= + − x t+





∆ ( ) m 

m

∆x t=  −



 1 1

m

3. A point object is kept in front of a glass slab of thickness 10 cm, at a distance of 16 cm. The rear face of slab is silvered and hence starts behaving as a mirror. The image of the object is formed 12 cm behind mirrored face. Find the refractive index of glass slab.

Soln.:

To the object the mirror appears to be shifted by,

∆x =  −



 10 1 1

m

\ From this shifted mirror, object distance = image distance

⇒ + = +  −



 16 10 12 10 1 1

m m

⇒ 20 6= ⇒ =10 3

m m

nn

Physics for you|June ‘15 65

Category I (Q1 to Q30)

Each question has one correct option and carries 1 mark, for each wrong answer 1/4 mark will be deducted.

1. An object is located 4 m from the first of two thin converging lenses of focal lengths 2 m and 1 m respectively. The lenses are separated by 3 m. The final image formed by the second lens is located from the source at a distance of

(a) 8.0 m (b) 7.5 m (c) 6.0 m (d) 6.5 m

2. A simple pendulum of length L swings in a vertical plane. The tension of the string when it makes an angle q with the vertical and the bob of mass m moves with a speed v is (g is the gravitational acceleration)

(a) mv²/L (b) mgcosq + mv²/L (c) mgcosq – mv²/L (d) mgcosq

3. The length of a metal wire is L₁ when the tension is T₁ and L₂ when the tension is T₂. The unstretched length of the wire is

(a) L L^{1 2} 2

+ (b) L L1 2

(c) T L T L

2 1T T1 2

2 1

−

− (d) T L T L

2 1T T1 2

2 1

+ +

4. A hollow sphere of external radius R and thickness t (<<R) is made of a metal of density r. The sphere will float in water if

(a) t≤R

r (b) t≤ R

3r (c) t R≤

2r (d) t R≥

3r

5. A metal wire of circular cross-section has a resistance R₁. The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes R₂ then R₂ : R₁ is

(a) 1 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4

6. Assume that each diode shown in the figure has a forward bias resistance of 50 W and an infinite reverse bias resistance. The current through the resistance 150 W is

(a) 0.66 A (b) 0.05 A (c) zero (d) 0.04 A

7. A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T. The velocity of the conductor is 15 m/s and is directed perpendicular to the field.

The e.m.f. induced between the two ends of the conductor is

(a) 0.10 V (b) 0.15 V (c) 1.50 V (d) 15.00 V

8. A ray of light is incident at an angle i on a glass slab of refractive index m. The angle between reflected and refracted light is 90°. Then the relationship between i and m is

(a) i = 

 

− 

tan ¹ 1

m (b) tani = m (c) sini = m (d) cosi = m

9. Two particles A and B are moving as shown in the figure. Their total angular momentum about the point O is

Physics for you|June ‘15

66

(a) 9.8 kg m²/s (b) zero

(c) 52.7 kg m²/s (d) 37.9 kg m²/s 10. Particle A moves along X-axis

with a uniform velocity of magnitude 10 m/s. Particle B moves with uniform velocity 20 m/s along a direction making an angle of 60° with the positive direction of X-axis as shown in the figure.

The relative velocity of B with respect to that of A is (a) 10 m/s along X-axis

(b) 10 3 m/s along Y-axis (perpendicular to X-axis) (c) 10 5 m/s along the bisection of the velocities

of A and B

(d) 30 m/s along negative X-axis

11. When light is refracted from a surface, which of its following physical parameters does not change?

(a) velocity (b) amplitude (c) frequency (d) wavelength

12. A solid maintained at t₁ °C is kept in an evacuated chamber at temperature t₂ °C (t₂ >> t₁). The rate of heat absorbed by the body is proportional to (a) t₂⁴ – t₁⁴ (b) (t₂⁴ + 273) – (t₁⁴ + 273) (c) t₂ – t₁ (d) t₂² – t₁²

13. The work function of metals is in the range of 2 eV to 5 eV. Find which of the following wavelength of light cannot be used for photoelectric effect.

(Consider, Planck constant = 4 × 10^–15 eVs, velocity of light = 3 × 10⁸ m/s)

(a) 510 nm (b) 650 nm (c) 400 nm (d) 570 nm

14. A thin plastic sheet of refractive index 1.6 is used to cover one of the slits of a double slit arrangement.

The central point on the screen is now occupied by what would have been the 7^th bright fringe before the plastic was used. If the wavelength of light is 600 nm, what is the thickness (in mm) of the plastic?

(a) 7 (b) 4

(c) 8 (d) 6

15. The length of an open organ pipe is twice the length of another closed organ pipe. The fundamental frequency of the open pipe is100 Hz. The frequency of the third harmonic of the closed pipe is

(a) 100 Hz (b) 200 Hz (c) 300 Hz (d) 150 Hz

16. A 5 mF capacitor is connected in series with a 10 mF capacitor. When a 300 volt potential difference is applied across this combination, the total energy stored in the capacitors is

(a) 15 J (b) 1.5 J (c) 0.15 J (d) 0.10 J

17. Two particles of mass m₁ and m₂, approach each other due to their mutual gravitational attraction only. Then

(a) accelerations of both the particles are equal.

(b) acceleration of the particle of mass m₁ is proportional to m₁.

(c) acceleration of the particle of mass m₁ is proportional to m₂.

(d) acceleration of the particle of mass m₁ is inversely proportional to m₁.

18. Three bodies of the same material and having masses m, m and 3m are at temperatures 40°C, 50°C and 60°C respectively. If the bodies are brought in thermal contact, the final temperature will be

(a) 45°C (b) 54°C

(c) 52°C (d) 48°C

19. A satellite has kinetic energy K, potential energy V and total energy E. Which of the following statements is true?

(a) K = –V/2 (b) K = V/2 (c) E = K/2 (d) E = – K/2 20. The line AA′ is on a charged infinite conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge s and B is a ball of mass m with a like charge of magnitude q. B is connected by a string from a point of the line AA′. The tangent of the angle (q) formed between the line AA′ and the string is

(a) q mg s e

2 ₀ (b) q

mg s 4pe₀ (c) q

mg s pe

2 ₀ (d) q

mg s e₀

Physics for you|June ‘15 67 21. The current I in the circuit shown is

(a) 1.33 A (b) zero (c) 2.00 A (d) 1.00 A 22. The r.m.s. speed of oxygen is v at a particular

temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the r.m.s. speed becomes

(a) v (b) 2v (c) 2v (d) 4v

23. Two particles, A and B, having equal charges, after being accelerated through the same potential difference enter a region of uniform magnetic field and the particles describe circular paths of radii R₁

23. Two particles, A and B, having equal charges, after being accelerated through the same potential difference enter a region of uniform magnetic field and the particles describe circular paths of radii R₁

In document Physics for You June 2015 (Page 48-68)