JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS · JEE ADVANCED-PART TEST-1 (PHYSICS) SOLUTIONS OF...

JEE ADVANCED-PART TEST-1 (PHYSICS)

SOLUTIONS OF MOCK TEST

CONDUCTED ON DOUBTNUT APP

Download Doubtnut Today

Ques No. Question & Solution

Q No. - 1

JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS

Consider a solid sphere of density and radius 4R. Centre

of the sphere is at origin. Two spherical cavities centered at

(2R, 0) and (�2R, 0) are created in sphere. Radii of both

cavities is R. In left cavity material of density is filled

while second cavity is kept empty. What is gravitational field

at origin.

ρ

2ρ

https://doubtnut.app.link/52ujms05cV



(a).

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

Above distribution can be represented as shown in figure.

Gravitational field due to sphere of radius R at a distance

GρπR

32GρπR

34GρπR

33GρπR

2

3R

so net field at center will be

Q No. - 2


The lens shown is equiconvex having refractive Index. 1.5.

In the situation shown the final image of object coincides

with the object. The region between lens and mirror is now

filled with a liquid of Rrefractive Index 2. Then find the

separation between O & image formed by convex mirror.

(a). 33 cm

(b). 66 cm

(c). 16 cm

(d). 32 cm

CORRECT OPTION: B

SOLUTION

Case-1:

Eg = =Gρ πR34

3

4R2

GρπR

3

2Fg =2GρπR

3

Radius of curvature of lens is 20 cm

Image of fomred by convex lens should be at centre of

curvature of mirror

Radius curvature of mirror should be 40 m.

Case-II

So for convex mirror

Separation between object and this image

Q No. - 3


The electric field at the centre of a uniformly charged

hemispherical shell is . Now two portions of the

hemisphere are cut from either side and remaining portion is

shown in figure. If , then electric field intensity

at centre due to remaining portion is

+ =1

V

1

30

1

20

= − ⇒ V = 60cm1

V

1

20

1

30

+ = +2

V1

1

30

1.5 − 1

20

2 − 1.5

−20⇒ V = − 60

u = − 80

− =1

V

1

80

1

20V = 16cm

O = 66cm

E0

α = β =π

3

(a).

(b).

(c).

(d).

CORRECT OPTION: C

E0

3E0

6E0

2E0

SOLUTION

consider the whole hemispere as three portion if electric

field due to one portion is then

Q No. - 4


A thin converging lens forms a real image of an object

located far away from the lens as shown in the figure. The

image is located a distance and has height h. A diverging

E1 2E1 sin 30 + E1 = E0

2E1 = E0

⇒ E1 =E0

2

L1

4l

lens of focal length is placed 2l from lens . Another

converging lens of focal length 2l is placed 3l from lens .

The height of final image thus formed is (Both diverging and

converging lenses are placed at right side of -

(a).

(b).

(c).

(d).

CORRECT OPTION: D

SOLUTION

From lens or

From lens or

l L1

L1

L1

h

h

24h

2h

2nd − =1

9v

1

2l

1

− lv = − 2l

m1 = − 1

3rd − =1

v

1

−3l

1

2lv = 6l

m2 = − 2

Q No. - 5


A point object P is moving towards left with speed 5 mm/sec

parallel to optical axis of a concave mirror of focal length f =

20 cm. The seperation between object and optical axis is 1

cm. Find velocity of image of object in vector form when foot

of perpendicular from object on the optical axis is at a

distance 30 cm from pole.

(a).

(b).

(c).

(d).

CORRECT OPTION: B

hi(m1 × m2)h0

= 2h

→V i = 20 i + 4jmm/sec→V i = 20 i + jmm/sec→V i = 20 i − j /sec→V i = 20 imm/sec

SOLUTION

for for

+ =1

v

1

−30

1

−20v = − 60

m = =yi

yo

v

u

yi = − 2cm

−→––– (v)1,→v 1

→v 1 = − (→

v p)v2

u2

→v 2

→v 2 ⇒ =

yi

yo

v

u

yiu = − yov

Q No. - 6


Monochromatic light rays parallel to x-axis strike a convex

lens AB. If the lens oscillates such that AB tilts upto a small

angle (in radian) on either side of y-axis, then find the

distance between extreme positions of oscillating image (f =

focal length of the lens) :

(u) + yi = − y0dyidt

dy

dt

dv

dt

( − 30) + ( − 2)( − 5) = − (20)dyidt

= 1mm/secdyidtVi = 20 i + jmm/sec

θ

(a).

(b).

(c).

(d). the image will not move

CORRECT OPTION: C

SOLUTION

When the lens is tited by , the image is formed at the

intersection (Q) of focal plane of lens in tited position and x-

axis.

As the lens ocillates. The image shifts on x-axis in between

2f(sec θ − 1)

f sec2 θ

f(sec θ − 1)

θ

P and Q

Distance between two exterme position of the image

Q No. - 7


Two point charges having charge +Q, �q and mass M, m

respectively are separated by a distance L. They are

released from rest in a uniform electric field E. The electric

field is parallel to line joining both the charges and is

directed from negative to positive charge. For the separation

between particles to remain constant, the value of L is

(a).

(b).

(c).

(d).

CORRECT OPTION: A

∴

= PQ = − f = f(sec θ − 1)f

cos θ

(K = )1

4π ∈0

√ (M + m)KQq

E(qM + Qm)

√ (M + m)KQq

E(qm + QM)

√ mMKQq

E(qM + Qm)

√mMKQq

E(QM + qm)

SOLUTION

In order to maintain constant separation the particles must

have the same acceleration assuming the system of both

charges to accelerate towards left. Applying newton\'s

second law.

.(i)

Undergiven condition the acceleration of both charges

should be same and should also be equal to acceleration of

centre of mass of both the charges.

QE = = MaKQq

L2

Hence from equation (1) and (2) we get

Q No. - 8


At distance from a point charge the ratio where is

energy density and V is potential) is best reprsented by

(a).

(b).

L = √((M + m) )KQq

E(qM + Qm)

rU

V 2U



(c).

(d).

CORRECT OPTION: B

SOLUTION

Because

so the correct option is B.

Q No. - 9


U = ε0E2 =

1

2

1

2

ε0K2Q2

r4

V =KQ

r

= =U

V 2

ε0K21

2Q2

r4

K2Q2

r2

1

2

ε0

r2

∝U

V 2

1

r2

A cylindrical portion of radius r is removed from a solid

sphere of radius R and uniform volume charge density in

such a way that the axis of the hollow cylinder coincides

with one of the diameters of the sphere. (r is negligible

compared to R). Then the electric field intensity at point A is

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

Field at A

Due at A due the cylinder of length 2R (which can assumed

to be inifinite, since )

ρ

iρr

3ε0

− iρr

3ε0

iρr

6ε0

− iρr

6ε0

r < < R

net field

Q No. - 10


Two satellites revolve around the �Sun� as shown in the

figure. First satellite revolves in a circular orbit of radius R

with speed . Second satellite revolves in elliptical orbit,

for which minimum and maximum distance from the sun are

and respectively. Velocities at these positions are

and respectively. The correct order of speeds is

(a).

(b).

(c).

(d).

CORRECT OPTION: C

E2 = ( − i) = − ri2K(ρπr2)

r

ρ

2ε0

∴ E = E1 + E2 = iρr

6ε0

v1

R

3

5R

3v2 v3

v2 > v3 > v1

v3 < v2 < v1

v2 > v1 > v3

v2 > v3 = v1

SOLUTION

(orbital velocity in circular path)

For elliptical orbit

conservation of angular momentum

conservation energy

Solving and

Q No. - 11


A small area is removed from a uniform spherical shell of

mass M and radius R. Then the gravitatioinal field intensity

near the hollow portion is

(a).

(b).

(c).

(d). zero

CORRECT OPTION: B

V1 = √GM

R

mV2 = mV3R

3

5R

3

− + mV 22 = + mV 2

3

GMm

R/3

1

2

−GMm

5R/3

1

2

V2 = √ 5GM

RV3 = √GM

5R

GM

R2

GM

2R2

3GM

2R2

SOLUTION

Consider a small area (shaded strip)

here gravitational field due to this strip

and gravitational field due to the rest of spherical

shell.

gravitational field just inside the strip due to whole

shell.

gravitaional field just outside the strip due to whole

shell.

Eself =

Eext =

Ein =

Eout =

Ein = Eext − Eself = 0

After the shaded area has been removed there is no

and no and only hence

Q No. - 12


A meteorite approaching a planet of mass M (in the straight

line passing through the centre of the planet) collides with

an automatic space station orbiting the planet in a circular

trajectory of radius R. The mass of the station is ten times

as large as the mass of the meteorite. As a result of the

collision, the meteorite sticks in the station which goes over

to a new orbit with the minimum distance R/2 from the

planet. Speed of the meteorite just before it collides with the

planet is : .

(a).

(b).

(c).

(d).

CORRECT OPTION: A

⇒ Eext = Eself

Eout = Eext + Eself = ⇒ Eext =GM

R2

GM

2R2

Eself

Eself Eext Enet = Eext =GM

2R2

√ 58GM

R

√ 38GM

R

√ 28GM

R

√ 18GM

R

SOLUTION

As the space station is moving in circular orbit,

..(i)

Let u be the velocity of meteorite.

Velocity of the space station after collision can be obtained

from momentum conservation

=GM(10m)

R2

(10m)v20

R

⇒ v0 = √GM

R

μ = (10m + m)v1 ⇒ v1 =u

11

Let v be the velocity space stationa at closest distance

from angu7lar momentum conservation

from energy conservation

Ans

Q No. - 13


10m, v0 = (10m + m)v2 ⇒ v2 = v010

11

10mv0 × R = 11mv ⇒ v =R

2

20v0

11

× (11m)(v21 + v2

2) − =

× (11m)v2 −

1

2

GM(11m)

R

1

2GM.11m

R/2

⇒ ( )2

+ ( )2

− = ( )2

−

u

11

10v0

11

2GM

R

20v0

11

4GM

R

⇒ = − −u2

112

400v20

112

100v20

112

2GM

R

⇒ u2 = (400 − 100 − 242) = 58GM

R

GM

R

u = √ 58GM

R

Two converging lenses have focal length and

). The optical axis of the two lenses coincide.

This lens system is used to from an image of real object. It

is observed that final magnification of the image does not

depend on the distance x. Whole arrangement is shown in

figure. Final magnification is :

(a).

(b).

(c).

(d).

CORRECT OPTION: B

f1 f2

(f1 > > f2

−f1

f2

−f2

f1

−f1

f1 + f2

−f2

f1 − f2

SOLUTION

Image-1

Image -2

− =1

v1

1

−x

1

f1

v1 =xf1

x − f1

m1 = = = − ( )v1

u1

v1

x

f1

x − f1

u2 = − (d − v1)

Since m is

independent of x

Q No. - 14


In the figure shown an infinitely long wire of uniform linear

charge density is kept perpendicular to the plane of figure

such that it extends upto infinity on both sides of the paper.

− =1

v2

1

−(d − v1)

1

f2

v2 =(d − v1)f2

d − v1 − f2

m2 = = − ( )v2

−(d − v1)

f2

d − v1 − f2

⇒ (d − f1 − f2) = 0 ⇒ d = f1 + f2

⇒ m = −f2

f1

λ

Find the electrostatic force on a semicircular ring kept such

that its geometrical axis coincides with the wire. The

semicircular ring has a uniform linear charge density .

(a).

(b).

(c).

(d).

CORRECT OPTION: D

SOLUTION

The electrostatic field intensity at a point on the ring is

the force on the elementary charge dq is

The sine component of dF will get cancelled and cosine

component will get added. Net force on the ring

λ'

λλ'

πε0Rλλ'

4πε0λλ'

2πε0

(λλ' )πε0

E =λ

2πε0

1

R

dF = dqE = (λ'Rdthη).λ

2πε0

1

R

F = 2∫+π / 2

0dF cos θ = 2∫

+π / 2

0dthη. cos θ

=

λ. λ'

2πε0

λλ'

πε0

ans

Q No. - 15


Consider a spherical planet rotating about its axis. The

velocity of a point at equator is v. The angular velocity of this

planet is such that it makes apparent value of �g� at the

equator half of value of �g� at the pole. The escape speed

for a polar particle on the planet expressed as multiple of v

is :

(a). v

λλ'

πε0

(b). 2v

(c). 3v

(d). 4v

CORRECT OPTION: B

SOLUTION

According to question (At equator)

using conservation of energy

Q No. - 16


Mg − = ⇒ v2 = =Mv2

R

Mg

2

Rg

2

GM

2R

− + mv2e = 0 ⇒ v2

e = = 4v2GMm

R

1

2

2GM

R



The figure shows two equal, positive charges, each of

magnitude , fixed at points (3, 0) m and (�3, 0)m

respectively. A charge , moving along negative

y�axis has a kinetic energy of 4J at the instant it crosses

point (0,4)m. Determine the position of this charge where

the direction of its motion reverses for the first time after

crossing this point (neglect gravity).

(a).

(b).

(c).

(d).

CORRECT OPTION: B

50μC

� 50μC

(0m, − 7√2)(0, − 6√2m)(0, − 5√2)(0m, − 4√2)

SOLUTION

The charge will move in straight line along y-axis

as it does not experience any force in x-direction let B be

the locatio where the charge comes to rest momentarily and

them return. Total energy of the system, remai constant.

= 50μC

∴ KE + PE

= 4 + × 21

4πε0

(50 × 10− 6)( − 50 × 10− 6)5

= 0 + × 21

4πε0

(50 × 10− 6)( − 50 × 10− 6)

√32 + y2

Q No. - 17


Orbital velocity of a satellite in its orbit (around earth) of

radius r is v. It collides with another body in its orbit and

comes to rest just after the collision. Taking the radius of

earth as R, the speed with which it will fall on the surface of

earth will be :

(a).

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

.. (i)

..(ii)

from (i) and (ii) we have

Q No. - 18


v√( − 1)r

R

v√2( − 1)r

Rv

√2( + 1)rR

v√2( + 1)r

R

v = √GM

r

− = + mv2GMm

r

GMm

R

1

2

v' = √2( − 1)r

R

A solid spherical planet of mass 2m and radius \'R\' has a

very small tunnel along its diameter. A small cosmic particle

of mass m is at a distance 2R from the centre of the planet

as shown. Both are initially at rest, and due to gravitational

attraction, both start moving toward each other. After some

time, the cosmic particle passes through the centre of the

planet. (Assume the planet and the cosmic particle are

isolated from other planets)

(a).

(b). velocity of the cosmic particle at that instant is

(c). Total work done by the gravitational force on both the

particle is

4R

3

√ 8Gm

3R

−2Gm2

R

(d). Total work done by the gravitational force on both the

particle is

CORRECT OPTION: A,B,C

SOLUTION

Applying momentum conservation

..(i) ,br> From energy conservation

−2Gm2

R

0 = mv1 − 2mv2

⇒ v2 =v1

2Ki + Ui = kf + Uf

..(ii)

Solving eqs. (i) and (ii) get

(A) COM will be fixed so

(B).

(D).

Q No. - 19


In the figure shown A & B are two charged particles having

charges q and � q respectively are placed on a

0 + ( − )m = mv21 + (2m)v2

2

+ ( − )(m)

G(2m)

2R

1

2

1

2

3

2

G(2m)

R

Scm =m1s1 + m2s2

m1 + m2

0 = ⇒ x =(m)(x) + (2m)( − (2R − x))

m + 2m

4R

3Fnet = 0 ⇒ a = 0

Wgr = U⏐⏐ ⏐ ⏐↓

⇒ Wgr = ( − )m

− ( − )m

G(2m)

2R

3

2

G(2m)

R

nonconducting fixed horizontal smooth plane. B is fixed and

A is attached to a non conducting massless spring of spring

constant k. The other end of the spring is fixed. Mass of A is

m, A and B are in equilibrium when the distance between

them is r. Choose the correct options

(a). time period of small oscillation of block A about is

means position

(b). time period of small oscillation of block A about is

means position

(c). To perform SHM, K must be greater than

(d). to perform SHM, K must be greater than

CORRECT OPTION: A,C

SOLUTION

Let extension in the spring when A is in equilibrium

then

.(i)

Now let A be shifted be a small distance x towards B. Then

the resultant force towards A is,

= 2π√m

k − q2

2π∈0 r3

= 2π√m

k + q2

2π∈0 r3

( )q2

2πε0r3

( )2q2

πε0r3

x0 =

kx0 =1

4π ∈0

q2

r2

: :

binomial expansion

, using (i)

with

For reat T,

Q No. - 20


Fres = k(x0 + x) − = k(x0 + x)

− (1 − )− 2

q2

4π ∈0 (r − x)2

q2

4π ∈0 r2

x2

r

= k(x0 + x) − (1 + )q2

4π ∈0 r2

2x

rx < < r

= kx − xq2

2π ∈0 r3

Fres = (k − )xq2

2π ∈0 r3

∵ sFαx ∴ SHM T = 2π√m

k − q2

2π∈0 r3

k > ∴ kmin =q2

2π ∈0 r3

q2

2π ∈0 r3

A light ray enters into a medium whose refractive index

varies along the x-axis as .

The medium is bounded by the planes

if the ray enters at the origin at an

angle with x-axis. where n0 = 1. The medium is

bounded by the planes x = 0, x = 1 & y = 0. If the ray enters

at the origin at an angle 30� with x�axi

(a). equation of trajectory of the light ray is

(b). equation of trajectory of the light ray is

(c). the coordinate the point at which light ray comes out

from the medium is

n(x) = n0√1 +x

4n0 = 1

x = 0, x = 1&y = 0

30 ∘

y = √x + 3 − √3

y = 2[√x + 3 − √3]

[1, 2(2 − √3)]

(d). the coordinate the point at which light ray comes out

from the medium is

CORRECT OPTION: B,C

SOLUTION

[0, 2(2 − √3)]

(a).

position at which ray comes out of the medium is

1 × sin 30 ∘ = n sin i

sin i =1

2n

tan i =1

√4n2 − 1

=dy

dx

1

√x + 3

∫y

0dy = ∫

x

0(x + 3)− 1 / 2

dx

y = 2(√1 + 3 − √3) y = 2(2 − √3)∴

(1, 2(2 − √3))

Q No. - 21


A charge is placed on the diagonal of a cube at a

distance from the point A. Choose the correct option.

(a). the sum of electric flux passing through the surfaces

and is

(b). the sum of electric flux passing through the surfaces

and is

(c). the flux through both the surfaces and

are same

(d). the flux through the surfaces is larger than the

' q' AP

AP

3

ABCD PQRSq

3ε0

ABCD PQRSq

8ε0

ABCD PQRS

ABCD

flux through surfaces .

CORRECT OPTION: A,D

SOLUTION

(a) We can easily see that charge q is placed symetrically to

surface ABCD, ABSR and ADQR. Charge q is also placed

symetrically to rest of the surfaces. If the flux through the

surface ABCD is x and through RSPQ is y then the total flux

will be 3x + 3y Now by Gauss law Now by Gauss

(b). Flux through two surfaces are not same flux via ABCD

is larger.

Q No. - 22


Two infinite, parallel, non-conducting thin sheets carry equal

positive charge denisity . One is placed in the plane

and the other at . Take potential at .

Choose the correct statements

(a). For , potential .

(b). For , potential

(c). For , potential

PQRS

= ϕqin

ε0

⇒ 3x + 3y =q

ε0

⇒ x + y =q

3ε0

σ yz

x = a V = 0 x = 0

0 < x < a V = 0

x > a V = − (x − a)σ

∈0

x > a V = (x − a)σ

∈0

(d). For potential

CORRECT OPTION: A,B,D

SOLUTION

(as

)

Q No. - 23


In the figure shown there is a hollow hemisphere of radius

\'R\'. It has a uniform mass distribution having total mass m.

The gravitational potential at points A, D and B are ,

and respectively. Distance of D and B from centre C are

R/2 and 2R respectively. The points C, D and B are lying on

radial line of the hollow hemisphere

x < 0 V = xσ

∈0

0 < x < a :V = [ − ∫x

0Exdx] + V ( 0 ) = 0

Ex = 0

x > a, V = − ∫x

a

Exdx + V (a )

= [ − ∫x

a

dx] + V (a ) = − (x − a)σ

∈0

σ

∈0

x < 0, V = − ∫x

0Exdx + V ( 0 ) =

− ( − . x) + V ( 0 ) = . xσ

∈0

σ

∈0

VA VD

VB

(a).

(b).

(c).

(d).

CORRECT OPTION: A,B,C,D

VA = −Gm

R

VD = −Gm

R

VB = −Gm

2RVA = VD < VB

SOLUTION

Consider another identical hemisphere to complete a hollow

spherical shell the potential at a point D due to half shell

potantial due to complete shell at D (due to

symmetry)

potential due to complete shell at

potential due to complete shell ab B (again

VD = ×1

2

= × ( − ) = −1

2

G.2m

R

Gm

R

VA = ×1

2

A = × ) = −1

2

−G.2m

R

Gm

R

VB = ×1

2

due to symmetry)

Q No. - 24


A cavity of radius r is present inside a fixed solid dielectric

sphere of radius R, having a volume charge density of .

The distance between the centres of the sphere and the

cavity is a. An electron is released inside the cavity at an

angle as shown. The electron (of mass m and

charge �e) will take time to touch the

sphere again. Neglect gravity. find the value of P:

= × − = −1

2

G × 2m

2R

Gm

2R

VA = VD = − , VB = −Gm

R

Gm

2R

ρ

θ = 450

( )1 / 2

P√2mrε0

eaρ



(a). 2

(b). 4

(c). 6

(d). 8

CORRECT OPTION: C

SOLUTION

Electric field inside the cavity [here along

line joining centres of sphere and cavity]

froce on the electron inside the cavity (e) Now for

distance

by

=ρ

→a

3ε0

→a =

=ρ

→a

3ε0

d = √r2 + r2 = √2r

S = ut + 1/2at2, √2r = × t2 ⇒ t

= ( )1 / 2

1

2

ρae

3mε0

6√2rmε0

eaρ

Q No. - 25


A planet revolves around the sun in elliptical orbit of

semimajor axis m. The areal velocity of the planet

when it is nearest to the sun is . The least

distance between planet and the sun is m. Find

the minimum speed of the planet in km/s.

(a). 32

(b). 40

(c). 44

(d). 52

CORRECT OPTION: B

SOLUTION

Area covered by line joining planet and sun in time dt is

area velocity

where distance between planet and sun

and angular speed of planet about sum

From keplers second law areal velocity of planet is constant

At farthest position

2 � 1012

4.4 � 1016 sm2

/

1.8 � 1012

dS = x2dthη1

2

= dS/dt = x2dthη/dt = x2ω1

2

1

2x =

ω =

Download Doubtnut to Ask Any Math Question By just a click

Get A Video Solution For Free in Seconds

Doubtnut Has More Than 1 Lakh Video Solutions

Free Video Solutions of NCERT, RD Sharma, RS Aggarwal, Cengage (G.Tewani), ResonanceDPP, Allen, Bansal, FIITJEE, Akash, Narayana, VidyaMandir

Download Doubtnut Today

A = dS/dt = (2R − r3)ω

= (2R − r)[(2R − r)ω] = (2R − r)VB

1

21

2

1

2

VB = 40km/s



JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS · JEE ADVANCED-PART TEST-1 (PHYSICS) SOLUTIONS OF...

Documents

Transcript of JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS · JEE ADVANCED-PART TEST-1 (PHYSICS) SOLUTIONS OF...