Solutions of Physics JEE Main—2016

1. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (Take g = 10 ms–2)

(a) 2 s (b) 2 2 s

(c) 2 s (d) 2p 2 s

2. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2

(a) 6.45 × 10–3 kg (b) 9.89 × 10–3 kg (c) 12.89 × 10–3 kg (d) 2.45 × 10–3 kg 3. A point particle of mass m, moves along the

uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals m. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ and QR. The values of the coefficient of

friction m and the distance x (= QR), are, respectively close to:

(a) 0.2 and 3.5 m (b) 0.29 and 3.5 m (c) 0.29 and 6.5 m (d) 0.2 and 6.5 m 4. Two identical wires A and B, each of length ‘l’ carry

the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of magnetic field at the centres of the circle and square respectively, then

the ratio A

B

BB

is:

(a) 2

16 2

p (b)

2

16

p

(c) 2

8 2

p (d)

2

8

p

5. A galvanometer having a coil resistance of 100 W gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is:

(a) 2 W (b) 0.1 W (c) 3 W (d) 0.01 W 6. An observer looks at a distant tree of height 10 m

with a telescope of magnifying power of 20. To the observer the tree appears:

(a) 10 times nearer (b) 20 times taller (c) 20 times nearer (d) 10 times taller 7. The temperature dependence of resistances of Cu

and undoped Si in the temperature range 300-400K, is best described by:

(a) Linear increase for Cu, exponential increase for Si

(b) Linear increase for Cu, exponential decrease for Si

(a) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal

(b) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal

(c) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal

(d) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal

9. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially the samples have equal number of nuclei. After 80 minutes, the ratio of decayed number of A and B nuclei will be:

(a) 4 : 1 (b) 1 : 4 (c) 5 : 4 (d) 1 : 16 10. ‘n’ mole of an ideal gas undergoes a process AÆB

as shown in the figure. The maximum temperature of the gas during the process will be:

(a) 0 03

2

P VnR

(b) 0 09

2

P VnR

(c) 0 09P VnR

(d) 0 09

4

P VnR

11. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

(a) 0.08 H (b) 0.044 H (c) 0.065 H (d) 80 H 12. A pipe open at both ends has a fundamental frequency

f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column now is

(a) 3f/4 (b) 2f

(c) f (d) f/2

13. The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength l the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when:

(a) a = Ll and bmin =

22

Ll

(b) a = Ll and bmin = 4 Ll

(c) a = 2

Ll

and bmin = 4 Ll

(d) a = 2

Ll

and bmin =

22

Ll

14. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 mF and 9 mF capacitors), at a point distant 30 m from it, would equal:

(a) 360 N/C (b) 420 N/C (c) 480 N/C (d) 240 N/C 15. Arrange the following electromagnetic radiations

per quantum in the order of increasing energy: A : Blue light B : Yellow light C : X-ray D : Radiowave (a) A, B, D, C (b) C, A, B, D (c) B, A, D, C (d) D, B, A, C 16. Hysteresis loops for two magnetic materials A and

B are given below:

These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:

(a) A for electromagnets and B for electric generators. (b) A for transformers and B for electric generators. (c) B for electromagnets and transformers. (d) A for electric generators and transformers. 17. A pendulum clock loses 12 s a day if the temperature

is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (a) of the metal of the pendulum shaft are respectively:

(a) 60°C; a = 1.85 × 10–4/°C (b) 30°C; a = 1.85 × 10–3/°C (c) 55°C; a = 1.85 × 10–2/°C (d) 25°C; a = 1.85 × 10–5/°C 18. The region between two concentric spheres of radii

‘a’ and ‘b’, respectively (see figure), has volume

charge density r = Ar

, where A is a constant and r

is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is:

(a) −2 22 ( )

Qb ap

(b) −2 2

2

( )

Qb ap

(c) 2

2Qap

(d) 22

Qap

19. In an experiment for determination of refractive index of glass of a prism by i – d, plot, it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In the case which of the following is closest to the maximum possible value of the refractive index?

(a) 1.6 (b) 1.7 (c) 1.8 (d) 1.5 20. A student measures the time period of 100 oscillations

of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

(a) 92 ± 5.0 s (b) 92 ± 1.8 s (c) 92 ± 3 s (d) 92 ± 2 s 21. Identify the semiconductor devices whose

characteristics are given below, in the order (a), (b), (c), (d)

(a) Zener diode, Simple diode, Light dependent resistance, Solar cell

(b) Solar cell, Light dependent resistance, Zener diode, Simple diode

(c) Zener diode, solar cell, Simple diode, Light dependent resistance

(d) Simple diode, Zener diode, Solar cell, Light dependent resistance

22. Radiation of wavelength l, is incident on a photocell. The fastest emitted electron has speed v

If the wavelength is changed to 3

4

l , the speed of the

fastest emitted electron will be:

(a) <

1

24

3v (b) =

1

24

3v

(c)

1

23

4= v (d)

1

23

4> v

23. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it

is at a distance 2

3

A from equilibrium position. The

new amplitude of the motion is:

(a) 3A (b) A 3

(c) 7

3

A (d) 413

A

24. A particle of mass m is moving along the side of square of side ‘a’ with a uniform speed v in the x-y plane as shown in the figure:

Which of the following statements is false for the angular momentum L about the origin?

(a) when the particle is moving

from C to D.

(b

from B to C.

when the particle is moving from D

when the particle is moving from

A to B. 25. An ideal gas undergoes a quasistatic, reversible

process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PV n = constant, then n is given by (Here Cp and Cv are molar specific heat at constant pressure and constant volume, respectively):

(a) n = −−

PC CC Cv

(b) n = −

−PC C

C Cv

(c) n = −− P

C CC C

v (d) n = PCCv

26. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminum. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

(a) 0.80 mm (b) 0.70 mm (c) 0.50 mm (d) 0.75 mm

27. A roller is made by joining, together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:

(a) turn right. (b) go straight. (c) turn left and right alternately. (d) turn left. 28. If a, b, c, d are inputs to a gate and x is its output,

then, as per the following time graph, the gate is:

(a) AND (b) OR (c) NAND (d) NOT 29. For a common emitter configuration, if a and b

have their usual meaning, the incorrect relationship between a and b is

(a) a = −1

bb

(b) a = +1

bb

(c) a = +

2

21

bb

(d) = +1 11

a b

30. A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius or earth R; h<< R). The minimum increase in its orbit velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere)

(a) gR (b) 2gR

(c) −( 2 1)gR (d) 2gR

Answers

Solutions

1. Since the string is very long (L = 20 m), it has a finite mass. Hence the tension in the string is not the same at all points. Consider a point P at a distance y from end A. Since the string is uniform, the tension T at P is proportional to y and is given by (see figure)

T = Mg

yL

; M = mass of the string

Speed of pulse is v = Tm

; m = Mass per unit length of the string

Now v = dydt

\ dydt

= Mg y

gyLm

=

fi dy

y = g dt

Integrating, we have.

L

o∫ (y–1/2 dy = g

t

o∫ dt

fi 0

2L

y = g t

fi 20

02 y = 10 t

fi 2 20 = 10 t fi t = 2 2 s

So the correct choice is (b). 2. Potential energy needed to lift the mass m = 10 kg to

a height h = 1 m, 1000 times is P.E. = mgh × 1000 = 10 × 9.8 × 1 × 1000 = 9.8 × 104 J (1) Let M kg be the mass of the fat lost, the energy

required is

E = 20% of M × 3.8 × 107

= 20

100 × M × 3.8 × 107

= (7.6 × 106 × M) J (2)

Equating (1) and (2)

(7.6 × 106) M = 9.8 × 104

fi M = 1.289 × 10–2 kg

So the correct choice is (c).

3. Frictional force acting on the particle as it moves from P to Q is

f1 = m mg cos q, where q = 30°

Potential energy lost by the particle in this displacement is

E1 = work done against f1

= f1 × PQ

= m mg cos q × sin

hq

= tan

mghmq

= 2

tan 30

mgm ×°

fi E1 = 2 3 m mg (1) Frictional force acting on the particle as it moves

from Q to R is

f2 = m mg

Potential energy lost by the particle in this displacement is

E2 = f2 × QR = m mg x (2)

Given E1 = E2. Equating (1) and (2), we get x = 2 3 m 3.5 m

Total P.E. lost is E = E1 + E2 = 2 3 m mg + m mg x

= m mg ( 2 3 + x)

= m mg ( 2 3 + 2 3 )

= 4 3 m mg

Total P.E. lost is also equal to

E¢ = mgh = mg × 2 = 2 mg

Since E = E¢, we get

4 3 m mg = 2 mg fi m = 1

2 3 0.29

So the correct choice is (b).

4. 2pR = l fi R = 2

lp

The magnetic field at the centre of the circular coil A of radius R is

BA = 0 0

2

I IR l

m m p= (1)

Refer to the following figure.

Magnetic field at O due to current I in PQ is

BPQ = 0

4

Ia

mp

(sin 45° + sin 45°)

where a = OT = PT = 8

l ( PT = OT tan 45° = OT)

BPQ = 04

2

I

l

mp

Therefore, the magnetic field at O due to the complete sqs loop PQRS is

BB = 4 × 0 04 8 2

2

I Ill

m mpp

= (2)

Dividing (1) by (2)we get

A

B

BB

= 2

8 2

p

So the correct choice is (c).

5.

IgG = (I – Ig)S

S = g

g

I G

I I−

Given I = 10 A, Ig = 1 mA = 0.001 A and G = 100 W.

S = 0.001 100

(1 0.001)

×−

= 0.01

0.999

0.01 W So the choice is (d). 6. A person looking through a telescope observes

angular magnification (which is called magnifying power). If a distant tree subtends an angle a at his unaided eye, then the angle subtended at his eye (when he looks through the telescope) by the image will be 20 a. Hence the image will appear 20 times taller, so the correct choice is (b).

7. In the given temperature range, the resistance of a metal (such as copper) increases linearly with temperature but the resistance of a semi-conductor (such as silicon) decreases exponentially with temperature. So the correct choice is (b).

8. The correct choice is (d). 9. Let No be the radioactive nuclei present is each

sample at t = 0.

For sample A: No. of half lives in 80 minutes = 80

20

= 4

Number of nuclei decayed in 4 half lives = 42 16o oN N

=

Number of nuclei left undecayed is

NA = No – 15

16 16o oN N

=

For sample B: No. of half lives in 80 minutes = 80

20

= 2

Number nuclei decayed in 2 half lives = 22 4o oN N

=

Number of nuclei left undecayed is

NB = No – 3

4 4o oN N

=

A

B

NN

= 15 4 5

16 3 4o

o

NN

× =

So the correct choice is (c). 10. The equation of straight line AB is

P = mV + C (1)

where m is the slope and C is the intercept. From the given figure,

Slope m = 0 0 0

0 0 0

2

2

P P PV V V

−= −

− (2)

To find C, we have for point A,

2 P0 = mV0 + C = – 0

0

PV

× V0 + C

fi C = 3 P0 (3)

Using (2) and (3) in (1) we have

P = – 0

0

PV

V + 3 P0 (4)

Equation of state is PV = nRT

T = PVnR

(5)

Using (4) in (5) we have

T = 200

0

13

PV P V

nR V

− +

(6)

T will be maximum if dTdV

= 0 and 2

2

d TdV

< 0. From

(6) we find that

dTdV

= 00

0

213

PV P

nR V

− × +

Setting dTdV

= 0, we have

0 = 00

0

213

P VP

nR V

− ×

fi V = 03

2

V

Putting this value in (6), we get

Tmax = 2

0 0 00

0

3 313

2 2

P V VP

nR V

− × + ×

= 0 09

4

P VnR

It is easy to check that for V = 2

02

3,

2

V d TdV

is negative.

Hence the correct choice is (d). 11. Resistance of the lamp is

R = 80 V

10 A = 8 W

Given Irms = 10 A, Vrms = 220 V and w = 2pn = 100p rad s–1.

Irms = rms

2 2L

V

R X+

fi 10 = 2 2

220

8 LX+

fi 64 + X2L =

2220

10

= 484

fi XL = 420 = 20.5 W

fi wL = 20.5

fi L = 20.5

2 50p ×

12.

Fundamental frequency of open pipe is

and hence l = 2L.

f = 2Ll

=v v

Fundamental frequency of the closed pipe is

f ¢ = 2Ll

=v v = f

So the correct choice is (c).

13. Refer to the following figure.

Linear width of the circular spot for undiffracted beam is 2a.

Half angular width of the first maximum is

q1 = al

Angular width of the first maximum is

2q1 = 2

al

\ Linear width of the first maximum = 2 Lal

Sum of two linear widths is

b = 2a + 2 L

al

(1)

b will be minimum if db

da = 0 and

2

2

d b

da > 0.

Now db

da =

dda

22

La

al +

= 2 – 2

2 Lal

Putting db

da = 0 we set

0 = 2 – 2

2 La L

al l⇒ =

Putting this value of a in (1) we have

bmin = 2

2L

LL

lll

+

= 2 2 4L L Ll l l+ =

It is easy to check that when a = 2

2,

d bL

dal is

positive. So the correct choice is (b).

14. The given circuit can be redrawn as follows.

Potential difference across the series combination of C1 and C ¢ is 8 V. If V1 and V ¢ are the potential differences across C1

and C¢ respectively, then

C1V1 = C ¢V ¢

fi 4 × 10– 6 × V1 = 12 × 10– 6 × V¢

fi V1 = 3 V¢ (1)

Also V1 + V ¢ = 8 (2)

Equations (1) and (2) give V1 = 6 V and V ¢ = 2 V. Therefore, charges on 4 mF capacitor and 9 mF capacitor is

Q1 = (4 × 10– 6) × 6 = 24 × 10– 6 C

and Q2 = (9 × 10– 6) × 2 = 18 × 10– 6 C

Total charge Q = Q1 + Q2 = 42 × 10– 6 C The electric field at r = 30 m is

E = –6 9

20

42 10 9 10

4 30 30

Qrp

× × ×=∈ ×

= 420 NC –1, which is option (c).

15. Energy E = hn = hcl

. The wavelength of radiowaves

is the longest and of X-rays is the shortest, In fact lradio > lyellow > lblue > lx-rays.

Hence nD < nB < nA < nC

So ED < EB < EA < EC

Thus the correct choice is (d). 16. The area enclosed by the hysteresis loop gives the

energy dissipated. It is clear that material B dissipates less energy than material A. Hence, if the core of a transformer is made of material B, it will have a higher efficiency. Also the core of an electromagnet

should be made of a material which has a small retentivity and a small coercivity. It follows from the given loops that material B satisfies both these conditions. Hence the correct choice is (c).

17. T = 2Lg

p . Therefore,

1 1 1

2 2 2

T L LT L LD D D Da q a q= = × =

\ Loss of time per second = 1

2Da q

Loss of time per day = 1

(24 60 60)2

Da q × × × s.

If q is the temperature at which the clock gives correct time,

1

2a (40 – q) × (24 × 60 × 60) = 12 (1)

and 1

2a (q – 20) × (24 × 60 × 60) = 4 (2)

Dividing (1) and (2) we have

40 –3

– 20

qq

= fi q = 25°C

Using q = 25°C in (1) or (2) we get a = 1.85 × 10– 5 per °C so the correct choice is (d).

18. Divide the region between the two spheres into a large number of concentric spherical elements each of a very small width dr. Consider one such element at a distance r from the centre O of the spheres as shown in the figure.

Volume of element = 4pr2 dr. Volume charge density

is r = Ar

(given). Therefore, charge on the element is

dq = 4pr2 dr × Ar

= 4p A rdr

Total charge in the region between r = a and r = r is

q = 4r r

a a

dq A rdrp=∫ ∫ = 2p A (r2 – a2) This charge can be assumed to be concentrated at the

centre O. So the total charge at O is

Q¢ = Q + q = Q + 2p A (r2 – a2)

The electric field at a distance r from Q¢ is

E = 2

04

Qrp

′∈

= 20

1

4 rp∈ [Q + 2p A (r2 – a2)]

= 2 20 0

2

4 4

Q Ar r

pp p

+∈ ∈

(r2 – a2)

fi E = 2

2 20 0 0

–2 4 2

A Q Aar rp

+∈ ∈ ∈

E will be constant (i.e., E will be independent of r) if the last two terms in the above equation cancel each other, i.e., if

2

2 20 04 2

Q Aar rp

=∈ ∈

fi A = 22

Qap

So the correct choice is (d). 19. Given i = 35° , d = 40° and e = 79°. Now

d = i + e – A

fi A = i + e – d = 35° + 79° – 40° = 74°

If dm is the angle of minimum deviation, the refractive index of the prism is given by

m = sin

2

sin2

mA

A

d+

(1)

fi m sin sin2 2 2

mA A d = +

= sin cos cos sin2 2 2 2

m mA Ad d +

Partial differentiation of this equation gives (since m and A are constants),

0 = 1 1

– sin sin( ) cos cos( )2 2 2 2m m

A Ad d +

fi tan dm = 1 1

tan 37tan2

A=

° ( A = 74°)

Since tan 37° is slightly greater than tan 30° 1

which is3

, It follows that dm is slightly less

than 60°. The actual calculation (using trignometric tables given dm = 53°. This angle is greater than the given value of deviation (which is d = 40°). Now, with dm = 53° and A = 74°,

m =

74 53sin

sin 63.5274 sin 37

sin2

° + ° °=

° °

Using tables m turns out to be nearly equal to 1.5. So the closest option is (d).

20. The mean value of the four measurements of time for 100 oscillations is

t = 1 2 3 4

4

t t t t+ + +

= 90 91 95 92

4

+ + + = 92 s

Deviations (or errors) of values of t from the mean value are

| t – t1| = |92 – 90| = 2s

| t – t2| = |92 – 91| = 1s

| t – t3| = |92 – 95| = 3s

| t – t4| = |92 – 92| = 0s

Average error = 2 1 3 0

4

+ + + = 1.5s

Since the least count of the clock is 1 s, average error rounded off to appropriate significant figure is either 1s or 2s. So the reported mean time should be either (92 ± 1) s or (92 ± 2) s. So the correct choice is (d).

21. It is a question based on the experimental observation as written in NCERT book which states that the correct option is (d).

22. hn = 1

2 mv2 + w0. Since n =

c

l

v = 0

2–

hcw

m l

(1)

For wavelength 3

4

l , we have

v¢ = 0

2 4–

3

hcw

m l

(2)

Dividing (2) by (1)

1/2

0

0

4–

3

–

hcw

hcw

l

l

′ =

vv

or 2

0

0

4–

3

–

hcw

hcw

l

l

′ = vv

0 0

0 0

4–

3

–

hc hcw w

hc hcw w

l l

l l

+ + =

+ +

41 73

2 6

+= =

fi 71.08

6

′ = =vv

. So choices (b) and (c) are in-

correct. Choice (a) gives 4

3

′ <vv

or 1.15′ <v

v and

choice (d) gives 3

4

′ >vv

or 0.87′ >v

v. But 1.08

′ >vv

= 1.08,

i.e. v¢ > v. So choice (a) also wrong. Hence the correct choice is (d).

23. v = w 2 2–A x (1)

At x = 2

3

A,

2 22 4 5 5

–9 9 3

A AA Aw w w′ = = = ×v

Given v¢ = 3v. Hence

v¢ = 5 w A

Let An be the new amplitude. Then

v¢ = w (An2 – x2)1/2

fi 5 wA = w (An2 – x2)1/2

fi 5A2 = An2 – x2

fi An2 = 5A2 + x2 = 5A2 +

24

9

A

= 249

9

A

fi An = 7

3

A

So the correct choice is (c). 24. The angular momentum of a particle moving with

momentum about origin O is defined as

Where is the position vector of the particle from O. is

L = mrv sin q = mv r sin q = mv r^

where r^ = perpendicular of direction of motion of the particle from O.

The direction is obtained from the right hand screw rule. Now, let us first examine option (a). Refer to the following figure.

The magnitude of the position vector of C from O is

OC = r = OA + AC = R + 2 a

Now r^ = OE = OC sin 45°

= (R + 2 a) × 1

2

= 2

Ra

+

\ L = mv 2

Ra

+

is along position z-axis. Hence

So choice (a) is false. Using the above method, it is easy to check that choices (b), (c) and (d) are all true. So the correct answer is (a).

25. The molar heat capacity is given by

C = Cv + (1 – )

R

n

fi C – Cv = (1 – )

R

n =

–

(1 – )p vC C

n

fi 1 – n = –

–p v

v

C C

C C

fi n = 1 – –

–p v

v

C C

C C

= – –

–v p v

v

C C C C

C C

+

= –

–p

v

C C

C C, which is choice (a)

26. Least count of screw gauge = 0.5 mm

50 = 0.01 mm.

Zero error = – 0.5 + 45 × 0.01 = – 0.05 mm Measured value = 0.5 + 25 × 0.01 = 0.75 mm Corrected value = measured value – zero error = 0.75 – (– 0.05) = 0.80 mm So the correct option is (a). 27. As the roller moves with its centre O in a direction

parallel to CD, the radius of the wheel decreases because the rails AB and CD are not parallel to each other and the distance between them decreases as O moves upward. Hence, for the same number of rotations, the distance moved AB becomes less. This can happen if the roller will tend to turn left. So the correct option is (d).

28. It follows from the given graph, x = 1 if a = 1 or b = 1 or c = 1 or d = 1. Hence the gate is an OR gate, which is choice (b).

29. We know that IE = IC + IB and two d.c. current gains

of a transistor are defined as a = C

E

I

I and b = C

B

I

I.

a = /

/ 1 1C C B

C B C B

I I I

I I I I

bb

= =+ + +

. So option (a) is

incorrect. Option (b) is correct.

2

1 ( 1)bb ba

b b= =

+ +. So option (c) is also incorrect.

Now 1 1

1 1C BE B

C C C

I II I

I I Ia b+

= = = + = + . So option

(d) also correct. So the incorrect options are (a) and (c).

30. Orbital speed vO = GM

R h+

Escape speed ve = 2GM

R h+

Minimum increase required

= ve – vO

( )2 – 1GM

R h

= +

( )1/2

2 – 1GM

R =

( )2 – 1 gR=

So the correct choice is (c).