Radius of moon = R
C
A vA
h
R
100
R
B
Mass= M
Moon
vB
Speed of particle at B = vB
Escape velocity on
surface of moon = ve
∴ v GM
e R
=^2
At highest point vA = 0
Mechanical energy is conserved in the process.
Decrease in kinetic energy =^1
2
mv^2
B
Increase in PE = UA – UB
VAGM
A Rh
= Potential at =
( + )
−
VBGM
R
B =Potential at = 1.5^ RR 0.5 R
(^3100)
2
2
−− −
= 3
2
1
2
99
(^3100)
2 2
−−
GM
R
RR
−− GM −
R
RGM
(^3) R
2
×
2
[3 0.98]= × 2.02
2
∴−Increase in PE =
× (1.01)
GMm
Rh
GMm
R
Equate KE and PE.
1
2
( + )
mv^2 GMm +^ × 1.01
Rh
GMm
B R
−
GM
R
GM
Rh R
= 1
− + 1.01
or^1 =^1
+
+ 1.01
RRhR
−
or^1
+
= 0.01 = 1
Rh RR (^100)
or 100 R = R + h or h = 99R
- (a) : Let R be the radius of circular layer of water.
en pR^2 dρ = m
d
A
T
Pressure at AP
T
d
=− 0
2
(meniscus is cylindrical in
shape)
Pressure between the plates is less than the
atmospheric pressure and so the plates are pressed
together.
F = Force of attraction = 'ρ × area
F
T
d
=×R
(^22)
π
=×^22 T = 2
d
m
d
Tm
.ρ dρ
=
×××
××
=
−
−
20210007
00110 1000
28
3
(^24)
..
.
.N
- (c)
- (b) : Variation of acceleration due to gravity with
altitude
gg h
h Re
=−
(^12) ; ∆g gh
Re
=
2
Variation of l with temperature = 'l
Linear expansivity = ∆
∆
l
l()θ
T l
gg
l
g
g
g
T g
h g
=
−
=−
=+
−
2211
2
12
ππ
∆
∆∆/
T ll
g
T l
θ l
21 π= + =+
2
∆∆
e clock shows correct time if Th = Tq , ∆
l ∆
l
g
22 g
=
Linear expansivity = ∆
l ∆
l
g
g
h
R
h
(^1010) eeR
2
10 5
===
National Talent Search Examination
The Schedule of NTSE-2018-19
Stage Area Tentative Dates
Stage-I
(State)
Last Date for Submission of
Application Form
To be notified by the
respective State/UT. May
vary from state to state.
Examination in Mizoram,
Meghalaya, Nagaland and
Andaman and Nicobar Islands
03 th November, 2018
(Saturday)
Examination in All other
States and Union Territories
04 th November, 2018
(Sunday)
West Bengal^18
th November, 2018
(Sunday)
Stage-II
(National)
Examination in All States and
Union Territories^12
th May, 2019 (Sunday)