Physics Times – July 2019

2.Sol:

So increased by^0

V

R



At t 0

 1 2 

dN

t

dt

   

on integrating

^12 

0

loge

N

t

N

 

 

   

 

Given that

0 99

N N

N



  N 0  100 N

10

2.3 log 10 100 5 10

N

t

N

     

  

 

(^) t 9.2 10^9 Year
3.Sol: dQ Cdt
where C is the heat capacity per unit mass
.
dQ dT
C
dt dt

0 3/
1
4
P C T     t
3/
0
4
.
P
t C
T 
 
Now T T T t  0 0 1/
^
3
3/4 0
0
T T
t
T
  
 
 
(^) 
3
0
4 4
0
4 (P T T)
C
T


4.Sol: Let M is the mass of the sphere having
radius r
2
2
GMm mv
r r
^2
2 1
2
mv
r
  
 
 
^2
GMm K 2 2 Kr
M
r r Gm
  
^
dM^2 Kdr 4 r dr^22 Kdr
Gm Gm
   
2 2
K
Gmr


 
 (^) 2 2
2
K
m Gm r



SECTION-
1.Sol: Given that [ ] [M Mass M LT ] [ 0 0^0 ]
[J] = [Angular momentum] [ML T^2 ^1 ]
[ ] [Length]L 
Now; [M L T0 2 ^1 ] [Dimensionless quantity]
^ [ ] [ ]L T^2 
Power [P] = [M LT LT^0 ^2. ^1 ]
[M L T0 2 ^3 ]
[ ] [ ]P L ^4
Energy = [M LT L^0 ^2  ]
[L L^2 ^4 ][ ]L^2
Force [F]  [ ] [F M LT^0 ^2 ] [. ] [ ] L L^4  L^3
Linear momentum [ ] [p M LT^0 ^1 ] [. ] L L^2
[ ] [ ]p L ^1
2.Sol:
2.Sol:
3.Sol:
4.Sol:
SECTION-
1.Sol:
2.Sol:
(^) V 100 10^3 V 10 ^1 V
V I R R g( g V)