JEE ADVANCED-PART TEST-1 (PHYSICS)
SOLUTIONS OF MOCK TEST
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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
(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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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
JEE ADVANCED-PART TEST-1 (PHYSICS) - PHYSICS
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 =
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A = dS/dt = (2R − r3)ω
= (2R − r)[(2R − r)ω] = (2R − r)VB
1
21
2
1
2
VB = 40km/s