KVPY-SX 2018 (PHYSICS) - Byju's

KVPY-SX_2018 (PHYSICS)

KVPY-SX_2018 (PHYSICS) Page | 1

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PART-I

1. A table has a heavy circular top of radius 1 m and mass 20 kg, placed on four light

(considered massless) legs placed symmetrically on its circumference. The maximum

mass that can be kept anywhere on the table without toppling it is close to

(A) 20 kg (B) 34 kg (C) 47 kg (D) 59 kg

2. Air (density ) is being blown of a soap film (surface tension T) by a pipe of radius R

with its opening right next to the film. The film is deformed and a bubble detaches

from the film when the shape of the deformed surface is a hemisphere. Given that the

dynamic pressure on the film due to the air blown at speed v is 1

2v2, the speed at

which the bubble is formed is

(A) T

R (B)

2T

R (C)

4T

R (D)

8T

R

3. For an ideal gas the internal energy is given by U = 5PV/2 + C, where C is a constant.

The equation of the adiabats in the PV plane will be:

(A) P5V7 = constant (B) P7V5 = constant

(C) P3V5 = constant (D) P5V2 = constant

4. An ideal gas undergoes change in its state from the initial state I to the final state F

via two possible paths as shown. Then

(A) there is no change in internal energy along path 1

(B) heat is not absorbed by the gas in both paths

(C) the temperature of the gas first increases and then decreases for path 2

(D) work done by the gas is larger in path 1

P

I F

2

1

V




5. A thermally insulated rigid container of one litre volume contains a diatomic ideal

gas at room temperature. A small paddle installed inside the container is rotated

from the outside such that the pressure rises by 105 Pa. The change in internal

energy is close to

(A) 0 J (B) 67 J (C)150 J (D) 250 J

6. In a Young’s double slit experiment the amplitudes of the two waves incident on the

two slits are A and 2A. If I0 is the maximum intensity, then the intensity at a spot on

the screen where the phase difference between the two interfering waves is .

(A) I0cos2(/2) (B) 0I

3sin2(/ 2)

(C) 0I

9(5 + 4cos()) (D) 0I

9(5 + 8cos())

7. Figures below show water flowing through a horizontal pipe from left to right. Note

that the pipe in the middle is narrower. Choose the most appropriate depiction of

water levels in the vertical pipes.

(A) (B)

(C) (D)

8. A plank is moving in a horizontal direction with a constant acceleration ˆai . A

uniform rough cubical block of side rests on the plank, and is at rest relative to the

plank.

Let the center of mass of the block be at (0, /2) at a given instant. If a = g/10, then

the normal reaction exerted by the plank on the block at that instant acts at

(A) (0, 0) (B) (–/20, 0) (C) (–/10, 0) (D) (/10, 0)

Plank

Block

y

x a




9. Using the Heisenberg uncertainty principle, arrange the following particles in the

order of increasing lowest energy possible.

(I) an electron in H2 molecule

(II) a H atom in a H2 molecule

(III) a proton in the carbon nucleus

(IV) a H2 molecule within a nanotube

(A) (I) < (III) < (II) < (IV) (B) (IV) < (II) < (I) < (III)

(C) (II) < (IV) < (III) < (I) (D) (IV) < (I) < (II) < (III)

10. The current is flowing along the path abcd of a cube (shown to the left) produces a

magnetic field at the centre of cube of magnitude B. Dashed line depicts the non-

conducting part of the cube.

Consider a cubical shape shown to the right which is identical in size and shape to the

left. If the same current now flows in along the path daefgcd, then the magnitude of

magnetic field at the centre will be:

(A) zero (B) 2 B (C) 3 B (D) B

11. A thin metallic disc is rotating with constant angular velocity about a vertical axis

that is perpendicular to its plane and passes through its centre. The rotation causes

the free electrons in the disc to redistribute. Assume that there is no external electric

or magnetic field. Then

(A) a point on the rim of the disc is at a higher potential than the centre.

(B) a point on the rim of the disc is at a lower potential than the centre.

(C) a point on the rim of the disc is at the same potential as the centre.

(D) the potential in the material has an extremum between center and the rim.

a

b

d

c

d

c

a

f

g

e




12. One mole of a monatomic gas and one mole of a diatomic gas are initially in the same

state. Both gases are expanded isothermally and then adiabatically such that they

acquire the same final state. Choose the correct statement.

(A) Work done by diatomic gas is more than that by monatomic gas

(B) Work done by monatomic gas is more than that by diatomic gas

(C) Work done by both the gases are equal

(D) Change in internal energies of both the gases are equal

13. An ideal gas is made to undergo the cyclic process shown in the figure below. Let W

depict the work done,U be the change in internal energy of the gas and Q be the

heat added to the gas. Sign of each of these three quantities for the whole cycle will

be (0 refers to no change)

(A) –, 0, – (B) +, 0, + (C) 0, 0, 0 (D) +, +, +

14. Two balls of mass M and 2M are thrown horizontally with the same initial velocity v0

from top of a tall tower and experience a drag force of –kv(k > 0), where v is the

instantaneous velocity. Then

(A) The heavier ball will hit the ground further away than the lighter ball

(B) The heavier ball will hit the ground closer than the lighter ball

(C) Both balls will hit the ground at the same point

(D) Both balls will hit the ground at the same time

v0

Pre

ssu

re

Volume




15. Consider a glass cube slab of dielectric bound by the planes x = 0, x = a; y = 0, y = b;

z = 0, z = c; with b > a > c. The slab is placed in air and has a refractive index of n. The

minimum value for n such that all rays entering the dielectric at y = 0 reach y = b is

(A) 1 (B) 2 (C) 3 (D) 2

16. The graph shows the log of activity (log R) of a radioactive material as a function of

time t in minutes.

The half-life (in minutes) for the decay is closest to

(A) 2.1 (B) 3.0 (C) 3.9 (D) 4.4

17. The magnetic field is uniform for y > 0 and points into the plane. The magnetic field is

uniform and points out of the plane for y < 0. A proton denoted by filled circle leaves

y = 0 in the y-direction with some speed as shown below.

Which of the following best denotes the trajectory of the proton.

(A) (B)

(C) (D)

time (min) 0 5 10 15 20 25

2

4

6

8

10 lo

g R




18. The Hitomi satellite recently observed the Lyman alpha emission line (n = 2 to n = 1)

of Hydrogen-like iron ion (atomic number of iron is 26) from the Perseus galaxy

cluster. The wavelength of the line is closest to

(A) 2Å (B) 1Å (C) 50Å (D) 10Å

19. Assume that the drag force on a football depends only on the density of the air,

velocity of the ball and the cross-sectional area of the ball. Balls of different sizes but

the same density are dropped in an air column. The terminal velocity reached by

balls of masses 250 g and 125 g are in the ratio:

(A) 21/6 (B) 21/3 (C) 21/2 (D) 22/3

20. An electrostatic field line leaves at an angle from point charge q1 and connects with

point charge –q2 at an angle (q1 and q2 are positive) (see figure below). If q2 = 3

2q1

and = 30°, then

(A) 0° << 30° (B) = 30° (C) 30° < 60° (D) 60° < 90°

PART-II

21. In steady state heat conduction, the equations that determine the heat current J(r)

[heat flowing per unit time per unit area] and temperature T(r) in space are exactly

the same as those governing the electric field E(r) and electrostatic potential V(r)

with the equivalence given in the Table below:

Heat flow Electrostatics

T( r )

j(r)

V( r )

E(r)

We exploit this equivalence to predict the rate Q of total heat flowing by conduction

from the surface of spheres of varying radii, all maintained at the same temperature.

If Q Rn, here R is the radius, then the value of n is

(A) 2 (B) 1 (C) –1 (D) –2

+q1 –q2




22. An arrangement of spring, strings, pulley and masses is shown in the figure. The

pulley and the strings are massless and M > m. The spring is light with spring

constant k. If the string connecting m to the ground is detached, then immediately

after detachment

(A) the magnitude of the acceleration of m is zero and that of M is g

(B) the magnitude of the acceleration of m is (M – m)g/m and that of M is zero.

(C) the accelerations of both masses are same

(D) the elongation in the spring is (M – m)g/k.

23. The potential due to an electrostatic charge distribution is

V(r) = – r

o

qe

4 r

Where is positive. The net charge within a sphere centered at the origin and of

radius 1/ is

(A) 2q/e (B) (1 – 1/e)q (C) q/e (D) (1 + 1/e)q

24. A wheel of radius R is trapped in a mud pit and spinning. As the wheel is spinning, it

splashes mud blobs with initial speed u from various points on its circumference. The

maximum height from the centre of the wheel, to which a mud blob can reach is

(A) 2u

2g (B)

2 2

2

u gR

2g 2u+ (C)

2u

2g (D)

2uR

2g+

m M




25. Two rods of copper and iron with the same cross sectional area are joined at S and a

steady current I flows through the rods as shown in the figure.

Choose the most appropriate representation of charges accumulated near the

junction S.

(A) (B)

(C) (D)

26. Graphs below show the entropy vs energy (U) of two system I and II at constant

volume. The initial energies of the systems are indicated by UI,i and UII,i, respectively.

Graphs are drawn to the same scale. The system are then brought into thermal

contact with each other. Assume that at all times the combined energy of the two

systems remains constant. Choose the most appropriate option indicating the

energies of the two systems and the total entropy after they achieve the equilibrium.

(A) UI increases and UII decreases and the total entropy remains the same

(B) UI decreases and UII increases and the total entropy remains the same

(C) UI increases and UII decreases and the total entropy increases

(D) UI decreases and UII increases and the total entropy increases

En

tro

py

UI,i UI

En

tro

py

UII,i UII

Fe I I Cu

S

– – –

– – –

Fe I I Cu

S

Fe I I Cu

S

+ + +

+ + +

Fe I I Cu

S

+ + +

– – –

Fe I I Cu

S




27. The image of an object O due to reflection from the surface of a lake is elongated due

to the ripples on the water surface caused by a light breeze. This is because the

ripples act as tilted mirrors as shown. Consider the case where O and the observer E

are at the same height above the surface of the lake. If the maximum angle that the

ripples make with the horizontal is α, the angular extent of the image will be

(A) 2

(B) (C) 2 (D) 4

28. A spiral galaxy can be approximated as an infinitesimally thin disk of a uniform

surface mass density (mass per unit area) located at z = 0. Two stars A and B start

from rest from heights 2z0 and z0 (z0<< radial extent of the disk), respectively, and

fall towards the disk, cross over to the other side, and execute periodic oscillations.

The ratio of time periods of A and B is

(A) 2–1/2 (B) 2 (C) 1 (D) 21/2

29. Two mutually perpendicular infinitely long straight conductors carrying uniformly

distributed charges of linear densities 1 and 2 are positioned at a distance r from

each other.

Force between the conductors depends on r as

(A) l/r (B) l/r2 (C) r (D) r0

B C

O

E




30. The graph below shows the variation of a force (F) with time (t) on a body which is

moving in a straight line. Dependence of force on time is F tn. Initially body is at rest

If the speed of the object is 2 m/s at 3 s, the speed at 4 s will be approximately (in

m/s)

(A) 2.5 (B) 6.5 (C) 7.8 (D) 3.1

t (s) 0 1 2 3 4 5

10

20

30

F(N

)




ANSWER KEY

1. (C) 2. (D) 3. (A) 4. (C) 5. (D)

6. (C) 7. (A) 8. (B) 9. (B) 10. (C)

11. (B) 12. (B) 13. (A) 14. (A) 15. (B)

16. (B) 17. (D) 18. (A) 19. (A) 20. (A)

21. (B) 22. (B) 23. (A) 24. (B) 25. (B)

26. (C) 27. (C) 28. (D) 29. (D) 30. (B)

SOLUTIONS

PART-I

1. (C)

Given: Radius of circular table = 1 m

Mass of the table = 20 kg

Applying Rotational-equilibrium about reference-axis AB:

20gR

2

= MgR1

1–2

M[ 2 – 1] = 20

M = 48.28 kg

Hence, correct option is (C).

A

45° 45° M

R

R

C.O.M.

R

2

R–R

2

B

Maximum mass

that can be

placed on

circumference




2. (D) The rate of formation of hemi-spherical bubble is same as air-entering speed inside

bubble.

For bubble formation:

Excess-pressure [P] = 4T

R

Where, T = Tension in bubble, R = Radius of hemispherical bubble This excess pressure provided by Dynamic-pressure on the film.

1

2v2 =

4T

R

v = 8T

R

Hence, correct option is (D). 3. (A) The equation of adiabatic process in PV plane is: PV = Constant (C) (P = Pressure, V=Volume, =ratio of specific heat)

Given: U = 5PV

2 + C

For an ideal gas: PV = nRT Where, R= Universal gas constant, T=Temperature, n = Number of moles

U = 5

2nRT + C

Change in internal energy:

U = 5

2nRT …(1)

We know:

U = nfR T

2

…(2)

From eq.(1) & (2): Degree of freedom, f = 5 Now,

= 1 + 2

f = 1 +

2

5 =

7

5

So, PV7/5 = C [PV7/5]5 = C5 P5V7 = Constant

R Pipe

Hemi-spherical bubble




Hence, correct option is (A). 4. (C)

• Internal-energy is a state function. Hence, there is a change in U along-path 1.

• Area under curve represent work-done on PV axis. Hence, option (D) is wrong.

• For an ideal gas

PV = nRT

T = PV

nR

• Firstly PV increases in path 2 hence temperature increases and then PV decreases

in path 2 then temperature decreases.

Hence, option (C) is correct.

5. (D)

For an adiabatic process:

dQ = 0

dU = dW

U = Work done in adiabatic process

U = 1 1 2 2P V –P V

–1

U = 2 1–[P –P ]V

–1

U = –PV

–1

{given: V=1litre, Pressure rises= 105Pa}

= –5 –3 310 10 m

7–1

5

(for diatomic gas =7/5)

U = –250 J [Negative sign indicates decrease in U]

Hence, correct option is (D).

Motor

Paddle wheel

V = 1 litre




6. (C) Intensity (I) (area)2 I A2 I1 = kA2 I2 = 4kA2 I2 = 4I1

2

1 2I I +

= Imax

I0 = 9I1

I1 = 0I

9

I2 = 04I

9

Resultant-Intensity:

I = 2

1I

+ 2

2I

+ 2 1I 2I cos

I = I1 + I2 + 2 1 2I I cos

I = 0I

9 + 04I

9 +

2016I

81cos

= 05I

9 + 04I

9cos

= 0I

9[5 + 4cos]

Hence, correct option is (C). 7. (A) Recall Bernoulli’s equation:

P

g +

2V

2g+ Z = constant

Where, P

g= pressure head,

2V

2g = kinetic head

Z = datum/reference level For flow to be possible: A1V1 = A2V2 (From equation of continuity)

V 1

A

• As area decreases velocity increases at narrow-section. • As velocity increases; pressure decreases.




Hence, correct option is (A).

8. (B)

Given: a = g

10

Observing block with respect to plank:

For rotational equilibrium:

Net torque about O

o = 0

Mg2

= Ma.2

+ N – x2

N = Mg

Mg2

= Mg

10 2 + Mg – x

2

x = 20

The normal reaction act at:

– , 020

Hence, correct option is (B).

Mg

[0,0]

fS

ma x 2

–x

N

O

–y

[0,0]

+y

+x –x

Block

Plank




9. (B)

Heisenberguncertainty principle

(x)(P) h

4

E = 2P

2m (The relation between kinetic energy and momentum)

• For option (III): Uncertainty in position (x) is quite small hence (P) is quite

large hence increase in lowest energy possible (E) is maximum for this condition.

• For option (IV): Uncertainty of position of H2 molecule in a nano tube is more

hence uncertainty in energy (E) is quite small.

• Similarly for option (I) & (II):

Uncertainty of position (x) of electron (e–)in H2 molecule

<Uncertainty of position (x) of H atom in H2 molecule

Hence,

Uncertainty of momentum/increase in lowest energy of electron in H2 molecule

>Uncertainty of momentum/increase in lowest energy of H atom is H2 molecule.

Hence, using above concept

IV < II < I < III

Hence, option (B) is best possible answer.




10. (C)

The magnitude of magnetic-field creates at the centre of cube by one face carrying

current i: OB = B

The magnetic field by face dabc at O:

The magnetic field by face bfgc at O:

The magnetic field by face acfb at O:

Net magnetic field at O:

netB = 3 B


B

B

B

B z

O

B

O B

d

c

a

f

g

e

O

d

c

a

f

g

e

=

a

b

d

c O




11. (B)

• Due to centrifugal force electrons have a tendency to go towards the

circumference.

• As electron gather on circumference it will generate negative potential on

circumference and deficiency of electron at centre it will generate higher potential at

the centre.


12. (B)

• Recall the concept of slope in isothermal and adiabatic process.

dP P

–dV V

=

diatomic monoatomic

7 3;

5 2

= =

• Area under curve of diatomic process is less, hence work done is less.


V

P

1

2

3

Monoatomic

Diatomic

r e–

F = m2r

Axis of rotation passing

through centre

disc (m=mass of the disc, = angular velocity of disc r = radius of disc)




13. (A)

Cycle is anti-clockwise. Hence

Work done W = –ive

For a cycle initial & final state are same. Hence

Change in internal energy U = 0

For a cycle:

Q = W

Change in heat Q = –ive


14. (A)

As in question it is asked about range.

Both heavier and lighter ball has ux = v0

Now retardation in x-direction

ax = xF

m (Fx = force in x-direction, m=mass of the ball)

ax = xKv

m

ax1

m

So, heavier block will hit the ground further away than the lighter ball.


v0 Kv Kvy

Kvx mg

Pre

ssu

re

Volume




15. (B)

• Recall TIR [Total internal reflection]

c

90 – cc

c 45°

sinc sin45°

1

n

1

2

n 2

Minimum value of refractive index n should be 2 .


16. (B)

time (min) 0 5 10 15 20 25

2

4

6

8

10

log

R

x x=a y=0 90°

Origin c

90°–c

y

y=b




From radioactive decay

R = Roe–t

Taking log:

log R = log Ro – t [y = mx + c]

Graph of log R vs time is straight line and its slope is . [Decay-constant]

– = 8–6

16– 8

= 1

4

T1/2 = n2

= 0.693

(1/ 4)

= 2.772 minute

Closest value is 3 minutes.


17. (D)

Concept: Charge entering into a uniform magnetic field.

When charge (q) enter into a uniform magnetic field (B) with velocity (V) then force

on the charge.

F = q[ V × B]

Correct option is (D).

q

•




18. (A)

Line of emission OR emission-spectrum depends on the energy radiated by photon.

Energy of a photon:

E = hc

=

12400Å

…(1)

E = 13.6Z22 2

1 2

1 1–

n n

…(2)

Z = atomic number

From eq.(1) & (2)

12400

= 13.6[26]2

1 1–

1 4

= 12400Å 4

13.6 676 3

= 1.79 2Å


19. (A)

Given, drag force FD depends on air density , velocity of ball v and area of cross-

section of ball A.

FDxvyAz

Writing the dimensional-formula for r, v, A & FD:

[FD] = [M1L1T–2]

[] = [M1L–3]

[v] = [L1T–1]

[A] = [L2]

[M1L1T–2] [M1L–3]x[L1T–1]y[L2]z

[M1L1T–2] [M]x[L]–3x+y+2z[T]–y

On comparing:

x = 1

y = 2

z = 3x – y

2 = 1

FD v2A

FD = Kv2A

Where K is proportionality constant which has a value of 1

2.




At terminal velocity ball must be in Equilibrium. So FD = mg

KAv2 = mg As density is same:

v m

A

1

2

v

v = 1 2

2 1

m A

m A =

21 2

22 1

m 4 R

m 4 R

=

21 2

22 1

m R

m R

Mass of ball

m = v = × 4

3R3

1

2

m

m =

3

1

2

R

R

1

2

R

R

=

1/3

1

2

m

m

2

2

1

R

R

=

2/3

2

1

m

m

Substituting the value of

2

2

1

R

R

, we get

1

2

v

v =

2/3

1 2

2 1

m m

m m

1

2

v

v =

1/6

1

2

m

m

=

1/6250 g

125 g

= 21/6 Hence, correct option is (A).

FD

m

mg

Air-column




20. (A)

Now EF field lines emerges from q1 and terminate at –q2.

Number of electric-field lines Magnitude of charge Electric flux

The relation between plane angle and solid angle.

= 2[1 - cos]

Electric flux coming out of q1= Electric flux entering at q

2

1

o

q

42[1–cos] = 2

o

q

42[1 – cos]

1

2

q

q =

[1–cos ]

[1–cos ]

1

2

q

q

=

2

2

2sin2

2sin2

2

3

×sin215° = sin2

2

24.40°


R

+q1 –q2

+q1 +q2




PART-II 21. (B)

Heat flow Electrostatics

T( r )

j(r)

V( r )

E(r)

• Rate of heat-flowing by conduction from the surface of sphere:

Q = –KAdT

dr

Q = –K4r2dT

dr

2

Q

K(4 r )dr = –dt …(1)

• As temperature is co-related with Electric-potential. We know for reference: r = v() = 0 Similarly from given analogy r = ; T() = 0 r = r ; V(r) = V r = r ; T(r) = T Applying limits to equation (1):

2

R

Qdr

4 Kr

= 0

T

– dT

R

Q –1

4 K r

= 0

R–T

Q 1

4 K R

= [T]

Q = 4KRT

Q Rn (given)

n = 1 Hence, correct option is (B).

r Sphere

Q Q

Q

Q




22. (B)

System is in equilibrium now,

Just after cutting:

• Spring is non-impulse force hence it will not suddenly wipe out as you cut the

string

For block M

Ma = NETF

M = sMg – F

M = 0

• As tension in the string is a self-adjusting force. As Fs is there T will be there in

string.

For block m:

ma = T – mg

m =

Mg – mg

m =

(M– m)g

m


m

T = Fs = Mg

mg

[Given: M > m]

M

Fs = Mg

Mg

m M

T

T

Fs

Fs

Mg

M

Fs

Mg

Fs = Mg

• Free body diagram of

Block of mass M before cutting




23. (A)

Given: V(r) = – r

o

qe

4 r

E = –dV

dr

E = – rKqe

r

+

– r

2

Kqe

r

…………. (1) [Here o

1

4 = K]

Electric-field at a distance [r = 1

] From equation (1)

E = Kq[2e–1 + 2e–1]

E = 22Kq

e

We know from Gauss theorem:

= in

o

Q

E dA = in

o

Q

(for spherical distribution; E dA = EdAcos0° = EdA and E = uniform)

EA = in

o

Q

Substituting the values of E and A, we get

Qin = 22Kq

e

4

21

o

Qin = 2q

e


C [0,0]

r=1




24. (B)

Let a mud blob is detached from the circumference of wheel with initial speed u at an

angle as shown below.

Estimating problem for a particular angle .

H = H1 + R sin

H1 = Hmaxfor projectile leaving at angle (90-)

H = 2 2u sin 90

Rsin2g

− +

H = Rsin + 2 2u cos

2g

For maximum Height:-

dH

0d

=

R cos + 2u

2sin cos 02g

− =

R cos = 2u

sin cosg

2

Rgsin

u =

22

max 2 2

Rg u RgH R 1

u 2g u

= + −

2 2

max 2

u gRH

2g 2u= +

Hence, option (B) is correct.

90°–

R

C

Rsin

H1

u




25. (B)

From the definition of current density i.e. current per unit area is given by

J = I

S = same

Where, J = current density, I = current, S = cross-sectional area

The relation between current density and thermal conductivity is given by

J = E = E

Where, = conductivity, = resistivity, E = electric field intensity

E = J

Fe|E | > Cu|E |

Net flux through Gaussian-surface:

= EFeS – ECuS [> 0 as Fe|E | > Cu|E | ]

[Flux through curved surface is zero as E A⊥ ]

= in

o

Q

= (+ive)

Qin = +ive [Accumulated charge on x-section should be positive]


I I

Cu Fe

CuE FeE

Gaussian-Surface

Cu

<Fe




26. (C)

System-I System-II

Utotal = iIU +

iIIU = constant

[S]system = [S]1 + [S]2

system initial[ S] =S1 + S2

As total energy of the system remain same

iI| U | =

iiI| U |

|S|I> |S|II (from graph)

Checking option (C); now entropy after thermal equilibrium:

By second law of thermodynamics, entropy of a system either increases or remains

constant (only in reversible quasistatic processes)

[S]system = [S1 + S1] + [S2 – S2]

As, S1>S2

[S]system = [S1 + S2] + 1 2

ive

[ S – S ]

+


En

tro

py

UIi UI

+UIi

S1 S1

En

tro

py

UIIi UII

S2

–UIIi

S2




27. (C)

We can consider ripples as plane mirrors reflecting light to form image with angular

extent .

It is given O and E are at same level, so OEBC is a trapezium with OE||BC.

Now, different angles of trapezium OEBC are shown below.

1 = 90° – –

2 = 90° – –

OPB = EPC

and OPE = BOC

90° – – = 90° – –

= …(i)

As, OPC = 180°

180° = + + 2 + 180° – 2 –

2 + – – = 0

= 2 [ = r from eq.(i)]

B C

O

E

2 1

180–1–2

1 2 P

B C

O

E

h




28. (D)

Gravitational field due to disc:

EG = 2G2 2

z1–

R z

+

As z <<< R

EG 2G (constant)

Acceleration of star B aB = B G

B

m E

m = EG

Acceleration of star A aA = A

A

F

m = A G

A

m E

m = EG

For Star-A

From 2ndequation of motion (s = ut + 1

2at2)

2A

1at

2 = 2z0

tA = 04z

a

For Star-B

From 2ndequation of motion (s = ut + 1

2at2)

2B

1at

2 = z0

B

A

B mB

A mA

z0

2z0

2z0

z0

tB

tA

Thin disk of

uniform mass

density




tB = 02z

a

Time-period:

TA = 4tA

TB = 4tB

A

B

T

T = A

B

t

t =

0

0

4z

a

2z

a

= 2


29. (D)

tan = y

r

y = r tan

Differentiating with respect to

dy = rsec2d

dFnet = 2dFcos

= 2(dqE)cos

Electric field for infinitely long straight wire

r

d

d

1

2

dy

dFsin

E

dFcos

dq=2dy

y

dFcos

E

dF dFsin




E = 12K

rsec

Where, K = constant, = linear charge density, r = distance from the straight wire)

dFnet = 22dyEcos

= 22rsec2d 12K

rsec

cos

netdF = /2

1 2

0

4K d

Fnet = 4K122

netF r0


30. (B)

Given:

Dependence of force on time:

F tn

F = Ktn …(a)

Where K is constant.

From graph:

t = 2 sec; F = 2N

2 = K[2]n …(1)

t = 4 sec; F = 16N

16 = K[4]n …(2)

Dividing eq.(2)/eq.(1)

[2]n = 8

n = 3

Putting n value in eq.(a)

F = Kt3




We know

F = P

t

( P =change in momentum&t = change in time)

P = F t m[v – 0] = Fdt

mv = F t

v = 31Kt dt

m

v = 4Kt

4m

4

3

v

v

= 4

4

3

v4 = 4

4

3

[2]

v4 = 6.32 m/s


KVPY-SX 2018 (PHYSICS) - Byju's

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Transcript of KVPY-SX 2018 (PHYSICS) - Byju's