PHYSICS Class
II IIT-JEE Achiever 2016-17 Intensive Revision Program Solution to Worksheet - 16
Topics
Gravitation
Date:
22-12-2016
Multiple choice questions with one correct alternative
1. If radius of the earth is increased by a factor, 2 by what factor would its density have to be changed to keep the g the same? 1 (A) 8
(B) 4
(C)
1 2
(D)
1 4
Ans (C) GM G 4 4 3 g = = π R ρ = πG R ρ R2 R 2 3 3 Thus, if R increases by a factor 2, ρ is to be decreased by a factor
1 2
.
2. A simple pendulum has a time period T 1 when on the earth’s surface, and T 2 when taken to a height R T above the earth surface, where R is the radius of earth. The value 2 is T1 (A) 1
(B)
2
(C) 4
(D) 2
Ans (D) Period of the pendulum, T α
∴
T2 T1
g
=
=
gh
GM
4R 2
R2
GM
1 g
=2
3. Three uniform spheres, of mass M and radius R each, are kept in such a way that each touches the other two. The magnitude of the gravitational force on any of the spheres due to the other two is (A) (C)
3 GM 2
(B)
(D)
R2
4
3 GM 2 R2
3 GM 2 R2
2
3 GM 2 2
R2
Ans (A) Fnet = 2
GM 2 ( 2R ) 2
o
cos 30 =
3 GM 2 4
R2
4. A particle on earths surface is projected with escape velocity. It’s mechanical energy will be (A) negative
(B) positive
(C) zero
(D) infinite
Ans (C) Ans (C) 5. Two earth-satellites are revolving in the same circular orbit round the centre of the earth. They must have the same (A) mass
(B) angular momentum
(C) kinetic energy
(D) linear speed
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Ans (D) Ans (D) 6. A hole is drilled from the surface of earth to its centre. A particle is dropped from rest at the surface of earth. The speed of the particle when it reaches the centre of the earth in terms of its escape speed on the surface of earth ve is v (A) e 2
(B) ve
(C)
2 ve
(D)
ve 2
Ans (D) Low of conservation of mechanical energy, − v=
GM R
ve
=
2
GMm R
+0=−
3GMm 2R
+
1 2
mv 2
.
7. Density of a solid sphere is ρ. Radius of the sphere is R. The gravitational field at a distance r from the centre of the sphere inside it is (A) (C)
4ρGπr 3 4ρGR 3r 2
4ρGr 2
(B)
ρGR 3 (D) πr
3
3
Ans (A) Ans (A) Gravitational field inside the sphere GM Gr 4 4πρGr E = 3 r = 3 πR 3ρ = 3 R 3 R 8. The percentage increase in the energy of a satellite when it is shifted from an orbit of radius r to an higher orbit of radius 2r is (A) 20
(B) 25
(C) 50
(D) 60
Ans (C) Initial energy = − Final energy = −
GMm
2r GMm 4r
− Percentage increase =
GMm
+
GMm
4r 2r GMm
100 = 50 .
2r 9. A planet of mass m describes an elliptical orbit around the sun. The areal velocity of the planet is proportional to −1
(A) m
(B) m 1
0
(C) m
(D) m 2
Ans (C) Ans (C) Areal velocity is a constant dA L = = constant dt 2m
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10. The potential at the surface of a solid sphere of mass M and radius R assumed to be zero. Choose the most appropriate option. GM (A) The potential at infinity is R (B) The potential at the centre of sphere is −
GM 2R
(C) Both (A) and (B) are correct (D) Both (A) and (B) are wrong Ans (C) Ans (C) Position
Potential with infinity as the
Potential with surface as the
Centre
reference 3GM
reference GM
−
Surface
−
Infinity
−
2R GM
2R 0
R GM
0
R 11. The gravitational field due to certain mass distribution is given by E =
A x
2
constants. If gravitational potential is zero at infinity, potential at x is A B (A) + x 2x 2 A B (B) − x 2x 2 A B + 2 (C) x x A B (D) − 2 x x Ans (B) Ans (B) x
A
∫
V = − Edx = − −
x
∞
B B A = − 2x 2 x 2 x 2
− −
12. A spherical portion of a solid sphere of mass M and radius R is evacuated as shown in the figure. The gravitational field at the centre of the hollow portion is (A) zero (C)
GM 2R 2
(B)
Ans (C) Ans (C)
(D)
GM 8R 2 GM R2
E = E remainder + E cavity
E = E remainder + 0
∴ E remainder =
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GM R3
r=
GM 2R 2
R ∵ r = 2
3
−
B x3
, where A and B are
Multiple choice questions with one or more than one correct alternative/s
13. For a circular orbit (A) orbital speed is
GM r
2
(B) time period, T ∝ r 3 GMm (C) kinetic energy is 2r GMm (D) potential energy is − 2r Ans (A), Ans (A), (B) and (C) Read the passage given below and answer questions 1 to 1! b" choosing the correct alternative
Spherical portion of a uniform solid sphere of mass M and radius R is evacuated as shown in the figure. Y B
O
C
X
A
15. The force on a particle of mass m, placed at A(2R, 0, 0) is 3GMm 5GMm 7GMm 11GMm (A) (B) (C) (D) 36R 2 36R 2 2R 2 8R 2 3R 16. The force on a particle of mass m kept at B 0, ,0 has its X-component 2 (A)
GMm 20 10 R
2
(B)
GMm 10 R
2
17. The force on a particle of mass m kept at C is GMm GMm (A) (B) R2 2R 2
(C) −
(C)
GMm 20 10 R
GMm 3R 2
2
(D) −
(D)
GMm 10 R
2
GMm 4R 2
Solution to passage
15. (A)
A
M m GMm GMm 1 1 7Mm 8 − = − = F= . 2 2 ( 2R ) R 2 4 18 36R 2 3R 2 G
θ 10 R 3R
16. (C)
O
Hollow portion can be treated as the negative mass. 2IIT1617PPWS16S
2
2
C
R 2
4
Force due to original sphere is along Y axis and hence does not contributes to X-component. M G m 8 . sin θ FX = − R2 10 4 GMm R 2 GMm
=−
20R 2 2
=−
10 R
20 10 R
2
17. (B) F=
GMm R
3
r + 0 =
GMm 2R 2
Choose the appropriate entr"/entries #rom column II to match each o# the entries o# the column I$
18. On the surface of earth acceleration due to gravity is g and gravitational potential is V. Column − I (i) (ii) (iii) (iv)
At height h = R, value of modulus of g R
Column − II (P)
, value of modulus of g
(Q)
At height h = R, value of modulus of V
(R)
At depth h =
At depth h =
2
R 2
, value of modulus of V
1
decreases by a factor decreases by a factor increases by a factor
4 1 2 11 8
(S)
increases by a factor 2
(T)
none
Ans (i) → (P), (ii) → (Q), (iii) → (Q), (iv) → (T) 19. Let V denote the gravitational potential and E denote the gravitational field. R is the radius of the sphere/shell and r is the distance from the centre of the sphere. Then, choose the appropriate graph/graphs from column II to match each of the entries of the column I. Column − I (i)
Column − II
The plot of E against r due to a spherical shell
(P)
R
O
r
V
(ii)
The plot of V against r for a spherical shell
(Q)
R
O
r
E
(iii)
The plot of E against r for a solid sphere.
(R)
V 2IIT1617PPWS16S
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r
O
−
3GM 2R
(iv)
The plot of V against r for a solid sphere.
(S)
−E
r
R
O
R
(T)
V
Ans (i) Ans (i) → Q (ii) → (P)
(iii) → (S)
−
r
3GM 2R
(iv) → (T)
For a spherical shell GM V=− for r < R R GM =− for r > R r E = 0 for r < R GM E = − 2 for r > R r For Solid sphere V=− = −
GM(3R 2 − r 2 ) 2R 3
for r < R
GM
for r > R r GMr E=− for r < R R3 GM = − 2 for r > R r Sub%ective questions
20. A cord of length 64 m is used to connect a m = 100 kg astronaut to a space ship whose mass M is much larger than that of the astronaut. Estimate the value of the tension in the cord. Assume that the space ship is orbiting near the surface of the earth. Assume also that the space-ship and astronaut fall on a straight line from the earth’s centre. The radius of the earth is 6400 km. km. Solution M >>m. So, the centre of mass of the astronaut spaceship system is close to the space-ship. Gravitation force provides centripetal force GM e (M + m) GM e g GM e = ( M + m ) Rω 2 ⇒ Rω 2 = = g ⇒ ω2 = 2 = 3 2 2 R R R R 2IIT1617PPWS16S
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T+
GM e m r2
rg R 2g r R2 2 GM e = mrω ⇒ T = m rω − 2 = m − 2 = mg − 2 r r R R r h h 2 3mgh R + h R2 = mg − = mg 1 + − 1 + = R (R + h ) 2 R R R 2
* * * DPP Multiple choice questions with one correct alternative
1. The ratio of acceleration due to gravity at a depth h below the surface of earth and at a height h above the surface of earth for h << R (where R is radius of earth) (A) is a constant
(B) increases linearly with h
(C) increases parabolically with h
(D) decreases
Ans (B)
Acceleration at a depth h, g1 = g1 −
h
R
g
Acceleration at a height h, g 2 =
2h 1 + R g h h 2h The ratio, 1 = 1 − 1 + ≈ 1 + g 2 R R R ∴It increases linearly with h 2. Mass of a planet is eight times mass of earth. Density of the planet is eight times density of earth. Then, if acceleration due to gravity on the surface of earth is g, the acceleration due to gravity on the surface of planet is (A) 2g
(B) 4g
(C) 8g
(D) 16g
Ans (C) We have, g =
GM R
2
and M =
4 3
πR 3 ρ
R =
3
3M 4πρ
2
4π ρ 3 3M
∴ g = GM
1
2
3
= k M ρ 3 If gp is the acceleration due to gravity on the surface of the planet 1
gp g
2
3
=
(8M ) (8ρ) 3 1
2
= 8 ⇒ gp = 8g
M3 ρ3
3. Three particles A, B, and C each of mass m, are placed at the vertices of an equiangular triangle of side 3 m. The particle at A is released at t = 0. The ratio of its acceleration soon after release and when it reaches centeroid is 2IIT1617PPWS16S
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(A) 3
1
(B)
3
(C)
3
(D)
1 3
Ans (D) Gm 2
At A, FA = 2 aA =
3
2
cos 30 o =
1 3 3
Gm 2
Gm 3 3
At the centroid,
Fo = 2 a o =
∴
aA ao
Gm 2
0
( 3) 2
cos 60 =
Gm 2 3
Gm 3
=
1 3
.
4. A particle of mass M is placed at the centre of a uniform spherical shell of mass 2M and radius R. The gravitational potential on the surface of the shell is GM 3GM (A) − (B) − R R
(C) −
2GM R
(D) zero
Ans (B) Ans (B) Potential is additive. GM G 2M 3GM V=− − =− R R R 5. A satellite is launched into a circular orbit of radius r while a second satellite is launched into an orbit of radius 1.02r. The percentage difference in the time periods of the two satellites is (A) 0.7
(B) 1.0
(C) 1.5
(D) 3.0
Ans (D) Ans (D) T 2 ∝ r 3 ⇒ T12 ∝ r 3 ; T22 ∝ (1.02 r ) 3
1.02r T2 = r
3 / 2
T1 = (1 + 0.02)
3 / 2
T1 = 1 +
3 2
× 0.02 T1
= 1.03 T 1
T − T1 ∆T × 100 = 0.03 × 100 = 3.0% × 100% = 2 T T 1 6. A solid sphere of mass M from which a spherical cavity of radius R/2 is
Y
removed as shown in the figure. The gravitational field strength at centre C of the cavity is (A) (3GM / 4 R 2 ) (ˆi )
(B) − (GM / 2 R 2 ) ˆi
(C) (3GM / 2 R 2 ) ˆi
(D) (3GM / 4 R 2 )( − ˆi )
X
Ans (B) Ans (B) 7. A particle is projected upward from the surface of earth (radius = R) with a speed equal to the orbital speed of a satellite near the earth’s surface. The height to which it would rise is
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8
(A)
2R
(B)
R
(C) R
2
(D) 2R
Ans (C) Ans (C) Law of conservation of mechanical energy, GMm 1 GMm GMm − + =− +0 R 2 R R+h 2R = R + h h = R. 8. A ring has a total mass M but non-uniformly distributed over its circumference. The radius of the ring is R. A point mass m is placed at the centre of the ring. Work done in taking away this point mass from centre to infinity GMm (A) is R (C) is greater than
(B) is less than
GMm
GMm R
(D) can not be calculated
R
Ans (A) Ans (A)
GMm GMm . = R R
W = 0 − −
9. The time period of revolution of a satellite close to earth is 90 min. The time period of another satellite in an orbit at a distance of three times the radius of earth from its surface will be (A) 90 8 min
(B) 360 min
(C) 270 min
(D) 720 min
Ans (D) Ans (D) 3
T2 T1
r2 2 r 1
=
3
T2 = (4) 2 (90) = 720 min . 10. A double star consists of two stars having masses m and 2m separated by a distance r. Of the following the correct option is (A) Radius of circular path traced by 2m is
2r 3
.
(B) Kinetic energy of heavier star is double than that of lighter star. (C) Time period of revolution of both are not same. (D) Angular momentum of lighter star is more. Ans (D) Ans (D) The stars describe concentric circles centred at their centre of mass with equal angular speeds. 2r Radius of lighter star r1 = 3 r Radius of heavier star r2 = 3
2rω Kinetic energy of lighter star K 1 = m 2 3 1
2
rω Kinetic energy of heavier star K 2 = ( 2m) 2 3 1
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2
9
∴
K1
=2
K2
2rω 2 r× 3 3 rω 1 Angular momentum of heavier star L 2 = m r × 3 3 Angular momentum of lighter star L1 = m
∴
L1
=4.
L2
11. A particle lies on the axis of a ring of mass M and radius R, at a distance R from the centre of the ring. The particle starts from rest under the gravitational attraction of the ring. It’s speed at the centre of the ring is (A)
2GM R
(B)
2GM 1 1 − (C) R 2
2GM R
(
)
2 −1
(D) zero.
Ans (B) Ans (B) Law of conservation of mechanical energy GMm GMm 1 − +0=− + mv 2 R 2 2R v=
2GM 1 1 − R 2
(
)
12. The gravitational field in a region is given by E = 3ˆi + 2 jˆ N kg −1 . Work done by this field is zero when the particle is moved along the line (B) 2y − 3x = 6
(A) 3y + 2x = 5
(C) 6y + 5x = 4
(D) 6y + 9x = 2
Ans (D) Ans (D) When displacement is perpendicular to the field, work done is zero. 13. Two bodies of mass m1 and m2 are initially placed infinite distance apart. They are then released to move under mutual gravitational attraction. Their relative velocity when they are r distance apart is (A)
2G ( m 1 + m 2 ) r
(B)
2Gm 1 m 2 (m1 + m 2 )r
(C)
G (m1 + m 2 ) r
(D)
Gm1 m 2 (m 1 + m 2 )r
Ans (A) Ans (A) Law of conservation of mechanical energy
−
Gm1 m 2
vr =
r
+
1 2
µv 2r = 0
2G ( m 1 + m 2 ) r
m1 m 2 µ = m1 + m 2
.
Multiple choice questions with one or more than one correct alternatives
14. Two objects of masses m and 4m are at rest at infinite separation. They move towards each other under mutual gravitational attraction. Then, at separation r 1
(A) mechanical energy of the system is zero.
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10 Gm 2
(B) their relative speed is
10
r
.
(C) the kinetic energy of the system is
4Gm 2 r
.
(D) angular momentum of the system is zero.
Ans (A), Ans (A), (B), (C) and (D)
15. For a planet in elliptical orbit (A) angular momentum about the centre of the sun is constant (B) potential energy is constant (C) kinetic energy is constant (D) total mechanical energy is constant Ans (A) Ans (A) and (D) Read the passage given below and answer questions 1& to 1' b" choosing the correct alternative
A body of mass m is kept on a sensitive spring balance in a stationary ship at the equator. Normal force exerted on the body is FN. Angular speed of the earth is ω. The observations are made by an observer in the ship. 16. Number of forces exerted on the body are (A) three.
(B) two.
(C) four.
(D) five.
17. If the ship travels due east with speed v the normal force exerted on the body becomes F ′N. Then F′ F′ 1 (A) 2mωv = N (B) 2mωv ∝ N = . FN FN 2mωv (C) FN′ = FN − 2mωv .
(D) FN′ = FN − mωv .
18. The normal force exerted on the body becomes FN′′ when the ship moves with the speed 2v in the northsouth direction. Then (A) FN′′ − FN′ = mωv .
(B) FN′′ − FN′ = 0 .
(C) F′′ − F′ = 0.3 mωv nearly .
(D) F′′ − F′ = 3.4 mv nearly .
Solution to passage
16. (A) Forces exerted on the body are gravitational force, spring force and the pseudo force since the observer also rotates with the earth. 17. (C) At the equator, latitude λ = 0 f p= mω2R C
mg
m
FN
Net force along the equator is zero.
∴ mg = FN + FP ⇒ FN = mg – m ω2R
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11
When the ship moves due east, the angular speed ω becomes ω + v FN′ = mg − mR ω + R
v
R
2
2 v 2 2ωv = mg − mR ω + 2 + R R 2
= mg − mω R −
mv 2 R
−
2ωv R
mv2 ∵ is very small R
= FN − 2 mωv 18. (D)
Replace v by v cos θ and ω by ω −
∴ FN′′ = FN +
2mωv 2
v cos θ
R
= FN + 2 mωv
FN′′ − FN′ = ( 2 + 2)mωv = 3.4 mωv Choose the appropriate entr"/entries #rom column II to match each o# the entries o# the column I$
19. Two concentric spherical shells are shown in figure. Match the following. B A D C
Column − I
Column − II
(i)
Gravitational potential at A
(P)
greater than that at B
(ii)
Gravitational field at A
(Q)
less than that at B
(iii)
As one moves from C to D
(R)
potential remains constant
(iv)
As one moves from D to A
(S)
gravitational field decreases.
(T)
none
Ans (i) Ans (i) → (Q), (ii) → (T), (iii) → (R), (iv) → (S) Sub%ective questions
20. What would be the period of rotation of the earth about its axis if its rotation speed increased until an object at the equator became weightless? Solution At the equator, mg e − FN =
mv 2 R
; v=
2πR T
Where ge → free fall acceleration at the equator. v → speed of the body of mass m. 2IIT1617PPWS16S
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R → equatorial radius of the earth. T → period of the earth.
4π 2 R ∴ FN = mg e − = m g e − 2 R T mv 2
where g → acceleration as measured at the location FN is a measure of the apparent weight of the body. The body becomes weightless when F N = 0. Thus, ge =
4π 2 R T2
⇒ T = 2π
R ge
= 2π
6.4 × 10 6 10
= 2π × 800 = 160 π s =
4π 9
***
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hour
