TIFR Question With Solution 2017

TIFR-2017 Physics by fiziks
[SOLUTION]
TIFR-2017 (Mathematical Physics Question and Solution)

SECTION A-(For both Int. Ph.D. and Ph.D. candidates)
Q1. Denote the commutator of two matrices A and B by [ A, B=] AB − BA and the
anticommutator by { A, B
=} AB + BA .
If { A, B} = 0 , we can write [ ABC ] =
(a) − B [ A, C ] (b) B { A, C} (c) [ A, C ] B (d) − B { A, C}
Ans. : (d)
Solution: [A, BC] = ABC –
BCA = ABC − BCA + BAC − BAC = ABC + BAC − ( BCA + BAC )
=( AB + BA ) C − B ( AC=
+ CA ) { A, B} C − B { A, C} = − B { A, C}
Hence 'd' is the correct answer.

Q2. Consider the waveform x ( t ) shown in the diagram below.
x
1
−2T −T t
0 T 2T 3T
−1
The Fourier series for x ( t ) which gives closest approximation to this waveform is
2 πt 1 4π t 1 3π t 
(a) x ( t ) =  cos − cos + cos + ...
π T 2 T 3 T 
2 2π t 1 4π t 1 6π t 
(b) x ( t ) =−
 cos + cos − cos + ...
π T 2 T 3 T 
2  πt 1 4π t 1 3π t 
(c) x ( t ) =  sin − sin + sin + ...
π T 2 T 3 T 
2 πt 1 2π t 1 3π t 
(d) x ( t ) =  − sin + sin − sin + ...
π T 2 T 3 T 
Ans. : (c)
T 
Solution: The given wave form is from 0 to  ,1
1 2 
T
Let's find equation of the line with end points (T,0)
(0,0)
x2 − x1
( 0, 0 ) and  ,1 ; x=
T
− x1 ( t − t1 )
2  t2 − t1
T 
 , −1
2 
Revised Edition-2022 1
H.No. 40-D, G.F., Jia Sarai, Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com, Email: fiziks.physics@gmail.com
TIFR-2017
[SOLUTION]
Physics by fiziks
1− 0 2 T
=
x−0
T
( t − 0 ) ;=x t, 0 < t <
−0 T 2
2
2 −T
=x t, < t < 0 as well
T 2
T /2
4 T 2 T 2 
T /2 T /2
2 2 2t 4 t2
x ( t ) dt
T −T∫/ 2 T −T∫/ 2 T
a0 = = = dt =  − = 0
T2 2 −T / 2
T8 8 
−2t
As x(−t ) = =− x(t ) , function is odd, So an = 0
T
4  −t cos nωt T /2 
T /2 T /2
2 2t 1
T −T∫/ 2 T
bn = = sin nω t dt  + sin nω t −T / 2 
T 2  nω −T / 2 n 2ω 2 
4  1 T nωT T nωT  1  nωT nωT  

= 2 
−  cos + cos  + 2 2 sin 2 + sin 2  
T  nω  2 2 2 2  nω 
nωT 2nπ
sin = sin(
= sin nπ ) 0 . Therefore second term vanishes
=
2 2
 −2 
−4T nωT −4 −4 −2  nπ , for even n 
Thus, bn = 2 = cos = cos nπ = cos nπ cos nπ =  
T nω 2 nωT n 2π nπ  2 , for odd n 
 nπ 
a0 + ∑ an cos nωt + ∑ bn sin nωt

Thus, x ( t ) =
2nπ t 2π t 4π t 6π t
x ( t ) = ∑ bn sin = b1 sin + b2 sin + b3 sin
T T T T
2 −2 2 −2
=b1 =, b2 = , b3 = , b4
π 2π 3π 4π
2 2π t 2 4π t 2 6π t
x ( t ) = sin − sin + sin + .....
π T 2π T 3π T
2  2π t 1 4π t 1 6π t 1 8π t 
x (t )=  sin − sin + sin − sin + ......
π T 2 T 3 T 4 T 
Q3. The matrix
100 2 x 0 
 
 −x 0 −x 
 
 0 x 100 2 
where x > 0 , is known to have two equal eigenvalues. Find the value of x.
Ans. : 50
2 Revised Edition-2022
[SOLUTION]
Solution:
Given that the matrix has two eigen values
Let's find the eigen values
100 2 x 0  1 0 0 100 2 − λ x 0
     
 −x 0 −x  − λ  0 1 0  =
0 ⇒ −x −λ −x
= 0
  0 0 1  
 0 x 100 2     0 x 100 2 − λ 
−λ −x −x −x
⇒ 100 2 − λ −x =
0
x 100 2 − λ 0 100 2 − λ
( )
⇒ 100 2 − λ λ 2 − 100 2λ + x 2  + x 2 100 2 − λ  =
0
⇒ (100 2 − λ ) λ 2
− 100 2λ + x 2  =
0
One of the eigen values = 100 2

The other two eigen values are given by solution of
λ 2 − 100 2λ + 2 x 2 =
0 ...(1)
Roots of (1) will be equal if b 2 − 4ac =
0
( )
2
⇒ 100 2 − 4 × 2 × x2 =0 ⇒ 8 x 2 = 100 ×100 × 2
⇒ x=
2
2500 ⇒ =
x 50
100 2 ± 0
will be λ
The two eigen values= = 50 2, 50 2
2
value of x = 50
SECTION B- (only for Int.-Ph.D. candidates)
iπ H
Q4. A unitary matrix U is expanded in terms of a Hermitian matrix H , such that U = e 2
 1 3
 0 
 2 2 
if we know that H =  0 1 0  , then U must be
 
 3 1
0 − 

 2 2
 i i 3
 0   1 0 3
 2 2 
 
(a)  0 i 0  (b)  0 2 0 
   
i 3 i   3 0 −1 
 0 − 
 2 2
TIFR-2017
[SOLUTION]
Physics by fiziks
 1 3
 i   3
 2 2   2i 1 
 1 1   2 
(c)  i (d)  1 2i 0 
2 2   
 
   3
2i 
3 1  0
 i   2 
 2 2 
Ans. : (a)
Solution:
If a matrix is unitary, uu † = I
Let’s test this relation for every possible solution.
 i 3   −i −i 3   1 3 3 3
 0 i  0   + 0 − 
 2 2  2 2   4 4 4 4  1 0 0 
−i 0 =  
(a)  0 i 0  0  0 1 0  = 0 1 0  =I
    
i 3 −i   −i 3 i   3 3 3 1  0 0 1 
0 0 − 0 +
 2 2   2  
2   4 4 4 4 
(b)
1 0 3  1 0 3  1+ 0 + 3 0+0+0 3 + 0 − 3 4 0 0
      
0 2 0  0 2 0=   0+0+0 0+4+0 0=+0+0  0 4 0 ≠ I
 −1   3 0 −1   3 + 0 + 1   0 0 4 
 3 0  3 +0− 3 0+0+0
 1 3  1 3  1 3 i i 3 3 1 3 
 i  −i   1+ + − + i+ − i
 4 4 2 2 4 2 4 2 
 2 2  2 2 
 1 1  1 1   −i i 3 1 1 3 1 1 
(c)  i −i =  + + +1+ + − 
2 2  2 2   2 2 4 4 4 4 2 2 
  
 3 1  3 1  − 3 1 3 3 i i 3 1 
 i  −i   i+ + i − + + +1 
 2 2  2 2   2 4 2 4 2 2 4 4 
 3 1 
 2 
 4 4 
 3 3
=  2 ≠I
 4 4 
 1 3 
 2 
 4 4 
[SOLUTION]
 3  3
 2i 1  −2i 1 
 2  2 
(d)  1 2i 0  1 −2i 0 
  
 3  3
0 −2i 
 0 2i 
 2  2 
 3   23 
 4 +1+ 4 2i − 2i + 0 3i + 0 − 3i  
4
0 0 
   
 3   3
=  −2i + 2i + 0 1+ 4 + 0 + 0 − 0  = 0 5 ≠I
2 2 
   
 3 3   3 19 
 − 3i + 3i +0+0 +0+4   0
4 
 2 4   2
Hence (a) is the correct answer.
Q5. Evaluate the expression
xn xn−1 xn−2 x3 x2 x1
n ! ∫ dxn −1 ∫ dx ∫ dx n−2 n −3... ∫ dx ∫ dx ∫ dx

2 1 0
0 0 0 0 0 0
Ans. : (a)
Solution:
x1
x1
∫ dx= x0 0= x=
1x
0 1
0
1!
x2 x1 x2 x2
x12 x22 x23
∫ dx ∫ dx= ∫ x dx=
0
1
0
0
0
1 1 = =
2 0 1.2 2!
x3 x2 x1 x3 x3
x22 dx2 1 x33
∫0 dx 2 ∫0 dx=
1 ∫ dx0 ∫0= 1 ⋅ 2 ⋅ 3 ∫0 0
= 3
x2
0
1⋅ 2 3 !
x4 x3 x2 x1 4 x4
x33 1 4 x4 x4−
∫0 dx3 ∫0 dx2 ∫0 dx=
1 ∫ dx0
0
∫0 =
3!
dx3 = x3
4! 0 4!
x5
x44 x5
∫
0
4!
dx4 = 9
5!
An
Similarly, if we repeat the procedure, we will get I =n !× =An
n!
SECTION B-(Only for Ph.D. candidates)
∞ dx
Q6. The value of the integral ∫ 0 x +4
4
is
π π π
(a) (b) (c) (d) π
8 4 2
TIFR-2017
[SOLUTION]
Physics by fiziks
Ans.: (a)
Solution:
∞
dx
∫x
Consider and C1
4
+4 √ √
−∞
–1+i 1+i
1
f ( z) = –R
1–i R
z +4
4 –1–i
× ×
Poles are given by z 4 + 4 =0
z 4 =−4 =4eπ eπ =−1 ;
( 2 n +1)π
1
z = 4 4e 4
 2 n +1 
i π
=z =
2e 4  {n 0,1, 2,3}
iπ
 π π  1 i 
z1 = 2 cos + i sin  = 2  +  = 1 + i = 2 e 4
 4 4   2 2
2π
 3π 3π   1 i 
z2 = 2 cos + i sin  = 2  − +  =−1 + i = 2e
4
 4 4   2 2 
z3 =−1 − i, z4 =1 − i
z3 and z4 are located in the lower half complex plane. Hence, they do not contribute.
Let's calculate residue at z = z1 and z = z2 .
z1 : Residue
1 1 1 1 − 34iπ 1  3π 3π 
= = 3iπ
=
= e cos 4 − i sin 4 
( ) ( 2) e
3
d 4 4z 3 8 2 8 2
z +4 z = z1 4 4
dz z = z1
1  −1 i  −1
=  − = (1 + i )
8 2 2 2  16
1 1 1 − i49π 1  1 i  1
=
z2 : Residue = 3
e =  −  = [1 − i ]
4z3  i 3π
 8 2 8 2 2 2  16
z = z2
4  2e 4 
 
R
dz dx
Thus, ∫C z 4 += ∫
2π i Σ Residue = + ∫ f ( z ) dz =
2π i Σ Residue
4 −R
x 4
+ 4 C1
[SOLUTION]
Take limit R → ∞ and note that, limit R → ∞ ∫ f ( z ) dz = 0

C1
∞
dx 1 1 
+∫ = + 0 2π i  ( −1 − i ) + (1 − i ) 
−∞
x +4
4
16 16 
∞
dx 2π i −4π i 2 π
∫ x 4 + 4 16 
−∞
=  −1 − i + =
1 − i 
 =16 4
∞ ∞
dx 1 dx 1π π
Now ∫0 x 4 += ∫ = =
4 2 −∞ x + 4 2 4 8
4
Hence the answer is (a)
Q7. Write down x ( t ) , where x ( t ) is the solution of the following differential equation
d  d 
 + 2  + 1 x =
1,
 dt  dt 
with the boundary conditions
dx 1
= 0, x ( t ) t =0 = −
dt t =0 2
Ans.:
Solution:
d   dx  d  dx   dx  d 2x dx
 + 2  + x  =
1 ⇒  + x  + 2  + x  =
1 ⇒ + 1 + 2 + 2x =
1
 dt   dt  dt  dt   dt 
2
dt dt
d 2x dx
⇒ 2 +2 +2= 0
dt dt
Auxiliary equation is given by D 2 + 2 D + 2 =0
−2 ± 4 − 8 −2 ± 2i
D= = =−1 ± i
2 2
=
Thus, x ( t ) e − t [ A cos t + B sin t ]
1 1 1
x ( 0) =− =e −0 [ A cos 0 + B ⋅ 0] =− =1[ A + 0] ⇒ A =−
2 2 2
−e − t [ A cos t + B sin t ] + e − t [ − A sin t + B cos t ]
x ' (t ) =
x ' ( t ) t =0 =
0=−e −0 [ A cos 0 + B sin 0] + e −0 [ − A sin 0 + B cos 0]
1
0= e −0 [ − A + B ] ⇒ B =A⇒ A=B=−
2
1
− e − t [ cos t + sin t ]
Thus x ( t ) =
2
TIFR-2017
[SOLUTION]
Physics by fiziks
TIFR-2017 (EMT Question and Solution)

Q1. Two long hollow conducting cylinders, each of height h , are placed
concentrically on the ground, as shown in the figure (top view). The outer
cylinder is grounded, while the inner cylinder is insulated. A positive
V =0
charge (the black dot in the figure) is placed between the cylinders at a
h
height from the ground. Which of the following figures gives the most accurate
2
representation (top view) of the lines of force?
(a) (b)
V =0
V =0
(c) (d)
V =0
V =0
Ans. : (b)
Solution. :
Electric field can not penetrate the conductor.
Q2. A common model for the distribution of charge in a hydrogen atom has a point-like
proton of charge + q0 at the centre and an electron with a static charge density
q0 −a2 r 
distribution ρ ( r ) = − e where a is a constant. The electric field E at r = a due
πa 3
to this system of charges will be

5q0 5q0
(a) − rˆ (b) − rˆ
4π ∈0 e 2 a 2 4π ∈0 e a 2
3q0 5q0
(c) rˆ (d) rˆ
4π ∈0 e 2 a 2 4π ∈0 e 2 a 2
Ans. : (d)
[SOLUTION]
Solution. :
1   q0 −a2 r  
→ → → a
1
∫ E .d a = ε ⇒ E × 4π a=  q0 + ∫  − 3 e  4π r dr 
2 2
Qenc
S 0 ε 0  0
πa  
1  4q0 2 −a2 r  1  4q0 2 −α r 

→ a a
2
3 ∫ 3 ∫
⇒ E × 4π a=
2
 0
q − r e =
dr   0
q − r e dr  where α = .
ε0  a 0  ε0  a 0  a
n  ( αa) 
a i
n
 ∫ r e=dr n −α r
1 − e ∑
−α a

0
α n +1  i =0 i 

2 ( αa)  2   ( α a ) 
i 2
−α a 
a 2
Thus ∫ r e 2 −α r
dr= 1 − e ∑
−α a
= 1 − e 1 + α a + 
α 3  i =0 i  α 3
  2 
0 
a
a3  − a 
2
2 4 a 2  a 3
⇒ ∫ r 2 e −α r dr =×
2 1 − e a
1 + a + 2  = 1 − 5e 
−2
0
8   a a 2  4
→ 1  4q0  a 3  1
⇒ E × 4π a=
2
 q0 − 3 
ε0 
(
1 − 5e −2 =
)  (  ε0 
)
 1 5q0 e −2 
 q0 − q0 1 − 5e −2 =

a 4  ε 0
→ 5q
⇒ E = 02 2 rˆ
4πε 0 e a
Q3. A rectangular metallic loop with sides L1 and L2 is

z
placed in the vertical plane, making an angle φ with
respect to the x -axis, as shown in the figure, and a L2
 L1
spatially uniform magnetic field B = Byˆ is applied. The 
B
loop is free to rotate about the ẑ axis (shown in the
y
figure with a double line). The magnetic field changes
with time at a constant rate
φ
x
dB
=κ
dt
If the resistance of the loop is R , the torque τ required to prevent the loop from rotating
will be
(L L ) (L L )
2 2
(a) κ B 1 2 sin φ zˆ (b) −κ B 1 2 sin 2φ zˆ

2R 2R
(L L ) (L L )
2 2
(c) κ B 1 2 sin φ cos φ zˆ (d) −κ B 1 2 sin φ zˆ

R R
TIFR-2017
[SOLUTION]
Physics by fiziks
Ans. : (b)
Solution. : The torque τ required to prevent the loop
from rotating will ẑ
  
be τ= m × B . L2 n̂
Magnetic flux through the loop is L1
 
φm ∫=
= B.d a ( Byˆ ) . ( L=
1 L2 n )
ˆ L1 L2 B cos φ
S
φ
 ŷ
Induced emf in the loop B
φ
dφm dB x̂
ε
= = L1 L2 cos φ = L1 L2 cos φκ
dt dt
ε L1 L2 cos φκ
=
Induced current in the loop I ind = .
R R
 ( L1 L2 ) cos φκ
2
Magnetic dipole moment of the current loop is m =I ind × ( L1 L2 ) nˆ =

− nˆ .
R
( L1L2 )
 cos φκ (L L ) cos φκ (L L )
2 2 2
τ= nˆ × ( Byˆ ) =
− 1 2 ( B sin φ zˆ ) −κ B 1 2
= sin 2φ zˆ
R R 2R
Q4. An electromagnetic wave in free space is described by

E ( x, =
1
(
y, z , t ) zˆ E0 cos kx − 3ky − 2ωt
2
)
The Poynting vector associated with this wave is along the direction
(a) xˆ + 3 yˆ (b) 3xˆ + yˆ (c) xˆ − 3 yˆ (d) − 3xˆ + yˆ
Ans. : (c)
 
Solution. : E= ( x, y, z, t ) zˆ E0 cos k.r − ωt ( )
  k
E ( x, =
1
(
y, z , t ) zˆ E0 cos kx − 3ky − 2ωt ⇒ k=
2 2
xˆ − 3 yˆ ) ( )

Poynting vector is in the direction of k .
[SOLUTION]
Q5. Consider the following situations.

A B
d q d q
In situation A , two semi-infinite earthed conducting planes meet at right-angles. A point

charge q , is placed at a distance d from each plane, as shown in the figure A . The
magnitude of the force exerted on the charge q is denoted FA .
In situation B , the same charge q is kept at the same distance d from an infinite earthed
conducting plane, as shown in the figure B . The magnitude of the force exerted on the
FA
charge q is denoted FB . Find the numerical ratio
FB
Ans. :
  q2  q2 q2
=
Solution. : F =
F k = and F k= k F3
( )
1 2 3 2
4d 2 8d 2 d d
2 2d −q q
F1
q2 F2 d
Resultant of F 1 , F 2 is F12 = F12 + F22 = 2k .
4d 2
 d
=
Net force
q2 
FA k 2  2=
4d 
1
− 
q2
2  32πε 0 d 2
2 2 −1 ( ) +q −q
2d
(towards the corner)
1 q2 q2
FB =
4πε 0 4d 2 16πε 0 d 2
• •
16πε 0 d 2 (2 =
2 − 1)
−q d d +q
=
FA q2
FB 32πε 0 d 2
2 2 − 1(× =
q2
) 2
0.9142
Q6. Consider a spherical shell with radius R such that the potential on
the surface of the shell in spherical coordinates is given by,
V ( r R=
= , θ , φ ) V0 cos 2 θ
P
where the angle θ is shown in the figure. There are no charges R θ
except for those on the shell. The potential outside the shell at the C
point P a distance 2R away from its center C (see figure) is
TIFR-2017
[SOLUTION]
Physics by fiziks
(a)=
V
V0
8
(1 + 2 cos 2 θ ) (b)=
V
V0
4
(1 − cos 2 θ )
(c)=
V
V0
8
(1 + cos2 θ ) (d) V =0 ( −2 cos θ + cos3 θ )
V
2
Ans. : (c)
Solution. :
∞
Bl
V ( R, θ )
= ∑=
R
P ( cos θ )
l =0
l +1 l
V0 (θ )
B0 B B
⇒ P0 ( cos θ ) + 12 P1 ( cos θ ) + 23 P2 ( cos θ ) =
V0 (θ )
R R R
B0 B1 B2  3cos 2 θ − 1 
+ 2 cos θ + 3  = V0 cos 2 θ
R R R  2 
B B  B 3 B2
⇒  0 − 23  + 12 cos θ + cos 2 θ = V0 cos 2 θ
 R 2R  R
3
2R
3 B2 2 B B  B 1
B1 =0, =V0 ⇒ B2 = V0 R 3 ,  0 − 23  =0 ⇒ B0 = 22 = V0 R
 R 2R 
3
2R 3 2R 3
1 2
=
Thus B0 V0 R=
, B1 0,=
B2 V0 R 3
3 3
Potential outside r > R is
∞
Bl B0 B B
V ( r ,θ ) = ∑r
l =0
P ( cos θ ) =
l +1 l
r
P0 ( cos θ ) + 21 P1 ( cos θ ) + 32 P2 ( cos θ )
r r
3  3cos θ − 1 
2
1 1 1 2
V ( 2 R, θ )= × V0 R ×1 + 0 + × V R ×  
( 2R ) 3
3 0
2R 3  2 
1 2 3cos 2 θ 2 1 1 1 1
⇒ V ( 2 R, θ ) = V0 + V0 × − V0 × ⇒ V ( 2 R, θ ) =+V0 V0 cos 2 θ − V0
6 24 2 24 2 6 8 24
V0
⇒ V ( 2 R, θ ) =
8
(1 + cos 2 θ )
[SOLUTION]
TIFR-2017 (Quantum Mechanics Question and Solution)

Q1. Denote the commutator of two matrices A and B by [ A, B=] AB − BA and the
anticommutator by { A, B
=} AB + BA .
If { A, B} = 0 , we can write [ ABC ] =
(a) − B [ A, C ] (b) B { A, C} (c) [ A, C ] B (d) − B { A, C}
Ans. : (d)
Solution.: [=
A, BC ] B [ A, C ] + [ A, B ] C
= B [ A, C ] + ( AB − BA ) C
= B ( AC − CA ) + ( AB + BA − 2 BA ) C= B ( AC − CA ) + ({ A, B} − 2 BAC )
− B ( AC + CA ) = − B { A, C}
B ( AC − CA − 2 AC ) =
=
Q2. Consider the 1 − D asymmetric double well potential V ( x ) as sketched below.
V ( x)
The probability distribution p ( x ) of a particle in the ground state of this potential is best
represented by
(a) (b)
p ( x) p ( x)
x x
(c) (d)
p ( x) p ( x)
x x
Ans. : (c)
TIFR-2017
[SOLUTION]
Physics by fiziks
Solution. :
This problem is solved through elimination method.
(a) Since the potential of the particle is not symmetrical, so the probability density of
the particle is also not symmetrical. This eliminates option ( a ) & ( d ) .
(b) The depth of well in the region x < 0 is lower than the depth of well in the region
x > 0, it implies that the probability of finding the particle in the region x < 0 is
lower than finding the particle in region x > 0 .
(c) This is correctly depicted in the option (c).
Q3. The normalized wave function of a particle can be written as
n
∞
 1 
ψ ( x) = N ∑   ϕn ( x )
n =0  7
where ϕn ( x ) are the normalized energy eigenfunctions of a given Hamiltonian. The
value of N is
(a)
(6 − 2 7 ) (b)
1
(c)
3
(d)
6
7 7 7 7
Ans. : (d)
Solution. :
The normalized wave function of a particle can be written as
n
∞
 1   1 1 
ψ ( x) = N ∑   φn ( x ) = N  φ0 ( x ) + φ1 ( x ) + φ2 ( x ) + ... 
n =0  7  7 7 
Applying the condition of normalization, we have
 1 1 
( x ) ψ ( x ) N 2  φ0 ( x ) φ0 ( x ) + φ1 ( x ) φ1 ( x ) + 2 φ2 ( x ) φ2=
ψ= ( x ) + ......  1
 7 7 
 1 1   1  27 6
N 2 1 + + 2 + ..  =1 ⇒ N 2   =1 ⇒ N   =1 ⇒ N =
 7 7   1 − 1/ 7  6 7
The value of N is 6 / 7 .
Q4. A quantum mechanical system consists of a one-dimensional infinite box, as indicated in
the figures below.
E0
[SOLUTION]
3 (three) identical non-interacting spin-1/ 2 particles; are first placed in the box, and the
ground state energy of the system is found to be E0 = 18 eV . If 7 (seven) such identical
particles are placed in the box, what will be the ground state energy, in units of eV ?
Ans. :
Solution. :
2π 2  2 22 π 2  2 6π 2  2 π 22
=
E2' 2 E1 + 2 E2 = + = =18 ⇒ =
3eV
2ma 2 2ma 2 2ma 2 2ma 2
Now the seven particles are placed in the box, then the
n=4 ε4
distribution of fermion in ground state is shown below
n=3 ε3
i.e.,
The ground state energy in this state is by n=2 ε2
n =1 ε1
E11 = 2 E1 + 2 E 2 + 2 E3 + E4
2π 2  2 2.4π 2  2 29π 2  2 16π 2  2 44π 2  2

= + + + =
2ma 2 2ma 2 2ma 2 2ma 2 2ma 2
π 22
E11 =44 =44 × 3 eV =132 eV
2ma 2
Q5. Electrons in a given system of hydrogen atoms are described by the wave function
υ/ ( r=
, θ , ϕ ) 0.8ψ 100 + 0.6 ei π / 3ψ 311
(
where the ψ nm denote normalized energy eigenstates. If Lˆx , Lˆ y , Lˆz are the components )
of the orbital angular momentum operator, the expectation value of Lˆ2x in this system is
(a) 1.5  2 (b) 0.18 2 (c) 0.36  2 (d) Zero

Ans. : (b)
Solution. :
The wave function of electron in a given system of hydrogen atom is defined as
ψ ( r ,θ , φ ) =
0.8 ψ 100 + 0.6eiπ /3ψ 3,1 =
0.8 1, 0, 0 + 0.6eiπ /3 3,1,1
1 2 2
The expectation value of Lˆ2x is given by=
Lˆ2x  L − L2 
2
Let us determine the expectation value of L2
=L2 ψ ( eiπ / 3 311

L2 0.8 1, 0, 0 + 0.6= )
0.8 L2 1, 0, 0 + 0.6eiπ / 3 L2 311
= 0.8 2 0 ( 0 + 1) 1.0, 0 + 0.8eiπ / 3 2 2 311 = 0.6eiπ / 3 2 2 311
TIFR-2017
[SOLUTION]
Physics by fiziks
or ψ L2 ψ =
036 × 2 2 311 311 =
0.72 2
Similarly the expectation value of L2z is
=Lz ψ ( eiπ /3 Lz 3,1,1

Lz 0.8 1, 0, 0 + 0.6= )
0.8 ( 0. ) 1, 0, 0 + 0.6eiπ / 3  311
L2z ψ = 0.6eiπ /3  2 311
or , ψ L22 ψ
= =
0.36  2 311 311 0.36 2
The expectation value of L2x is given by
=L2x ψ
2
(
1 2 2
L − Lz ψ
=
1
2
ψ L2 ψ − (ψ ) L2z ψ( ) = 12 ( 0.72 2
)
− 0.36 2 = 0.18 2

Q6. A one-dimensional quantum harmonic oscillator of natural frequency ω is in thermal
equilibrium with a heat bath at temperature T . The mean value E of the energy of the
oscillator can be written as
ω  ω  ω  ω 
(a) coth   (b) csch  
2  2 k BT  2  2 k BT 
ω  ω  ω  ω 
(c) sech   (d) tanh  
2  2 k BT  2  2 k BT 
Ans. : (a)
Solution. :
Consider a one-dimensional quantum harmonic oscillator of natural frequency ω is in
thermal equilibrium with a heat bath at temperature T . The energy of the oscillator
N
 1
=E ∑ ω  n + 2 
i =1
i
The single oscillator partition function is

∞  1 ∞ − β ω
e − β ω  n + 
2
− β ω 1 1
=Z1 ∑= e e ∑ = e =  2 2 − β ω n
β ω
=
= n 0= 1 − e − β ω β ω 2 − 2sinh hβ  / 2
n 0
e −e 2
The partition function for the system is

N
 1 
Z N (=
Z1 )
N
=  
 2sinh β ω / 2 
This yields for the partition function
[SOLUTION]
 1 β ω 
=
ln Z N N=
ln Z1 N  ln − ln sinh 
 2 2 
Which enables us to determine the internal energy
∂ ∂  1  β ω  
E = − ln Z N ⇒ E =
− N  ln − ln sinh  
∂β ∂β  2  2 
 ω 
cos h  β 
ω  2  N ω cot h  β ω 
=⇒ E N =  
2  ω  2  2 
sin h  β 
 2 
The mean value E / N of the energy of the oscillator can be written as
E ω  ω 
= cot h  
N 2  2 k BT 
Q7. A quantum mechanical system which has stationary states 1 , 2 and 3 , corresponding
to energy levels 0 eV ,1eV and 2 eV respectively, is perturbed by a potential of the form
=V ε 1 3 + ε 3 1
where, in eV , 0 < ε  1 .
The new ground state, correct to order ε , is approximately.
 ε ε ε
(a) 1 −  1 + 3 (b) 1 + 2 −ε 3
 2 2 2
ε ε
(c) 1 − 3 (d) 1 + 3
2 2
Ans. : (c)
Solution. :
nvk
The correction for wave function is given by n = n0 + E k 0 ( 0)
new k +n E n − Ek( 0)
1v 2 1v 3
The correction in the ground state is 1 new =
1 + 2 ( 0) ( 0)
+ 3 ( 0)
E1 − E2 E1 − E3( 0)
Let us evaluate the value of 1 v 2 i.e.
=
1v 2 1 (3 1 3 + e 3 1 ) 2 =  1 1 3 2 +  1 3 1 2 = 0
Similarly, Let us evaluate the value of 1 v 3
=
1v 3 1 (  1 3 + =
3 1 ) 3  11 3 3 + 1 3 1 3 =
Substituting the value of 1 v 2 and 1 v 3 in the equation for wave function we get
2 .0  
1 new =
1 + + 3 ⇒ 1 new =−
1 3
( 0 − 1) 0−2 2
TIFR-2017
[SOLUTION]
Physics by fiziks
Q8. A particle of mass m , confined to one dimension x , is in the ground state of a harmonic
oscillator potential with a normalized wave function
1
 2a  4
ψ 0 ( x ) =   e − ax
2
π 
mω
where a = . Find the expectation value of x8 in terms of the parameter a
2
105
Ans. :
256a 4
Solution. :
The ground state wave function for a particle of must in confined in one dimensional
1
 2a  4
harmonic potential is given by ψ 0 ( x ) =   e − ax
2
π 
1
2 ∞
 2a 
ψ x ψ ∫xe
8 −2 ax 2
The expectation value of x is given =
8
by x 8
= 8
  dx
π  −∞
1 12 1 1 − 12
Introducing new variable, 2ax 2 =t ⇒ x = t ⇒ dx = t dt
2a 2a 2
Substituting the value of x in above equation, we get
1 ∞ 1 4 ∞
 2a  2  2a  2  1  1 7
 ∫xe .2.∫ t 2 e − t dt
8 −2 ax 2
=x =
8
dx   .   .
 π  −∞  π   2a  2 ( 2a ) 2 0
1
∞
1 1 9 1 1 7 5 3 1 105
4 ∫
=
= t 2 −t
e dt = . . . π
π ( 2a ) 0 π ( 2a ) 2 2 2 2
4
2 5a 46
TIFR-2017 (Electronics Question and Solution)

Q1. A current source produces a square wave I ( t ) of 1.0V peak-to-peak voltage and is used
to drive the RC circuit shown below.

I (t )
R
V (t )
C
Which of the following represents the correct voltage across the capacitor C ?
(a) (b)
[SOLUTION]
(c) (d)
Ans. : (b)
Solution. :
RC circuit behaves as an integrator circuit.
Q2. The output (Y ) of the following circuit will be
A A B B C C
(a) C (b) B (c) A (d) A + B + C

Ans. : (a)
A A B B C C
Solution. : Y1
Y = Y1.Y2 .Y3 .Y4 = Y 1 + Y 2 + Y 3 + Y 4
Y2
= =
Y1 ABC =
, Y2 ABC =
, Y3 ABC , Y4 ABC Y
Y3
⇒ Y= ABC + ABC + ABC + ABC
Y4
⇒ Y= ABC + ABC + ABC + ABC
⇒Y = ( A + A) BC + ( A + A) BC
⇒Y = BC + BC = ( B + B ) C = C
Q3. An AC voltage source has an internal resistance of 50 Ω and is specified to deliver an

rms voltage of 50V to a matched load. If you connect this AC source to a cathode- ray
oscilloscope with 1M Ω input setting, what will be the peak-to-peak voltage you
observe?
Ans. : 283 50 Ω A
• +
Solution. : +
If R= 50 Ω , v0 = 50 V rms , then vin = 100 V rms . vin ~ RL v0 ( t )
L
−
• −
B
TIFR-2017
[SOLUTION]
Physics by fiziks
106 Ω
If R= 1M Ω , v0= ×100 V rms= 100 V rms
50 Ω + 106 Ω
L
Peak voltage of input is V0 m = 100 2 V .
Thus peak to peak voltage of input= =

is 2V0 m 200 2 V 283 V .

Q4. For exact calculation and minimum complexity, two four-digit binary numbers can be
added with
(a) 3 full adders and 1 half-adder (b) 2 full adders and 2 half-adders
(c) 1 full adder and 3 half-adders (d) 4 full adders
Ans. : (a)
Solution. :
Q5. Which digital logic gate is mimicked by the following silicon diode and silicon transistor
circuit?
+5V +5V
R1
RC 2
RC1
A
Vout
R3
B R2
Solution:
AND Gate
[SOLUTION]

Q6. The following circuit is fed with an input sine wave of frequency 50 Hz .
10 pF
10kΩ
Vin 10kΩ
−
Vout
+
Which of the following graphs (solid line is input and dashed line is output) best
represents the correct situation?
(a) (b)
(c) (d)
Ans. : (c)
Solution. :
v0 X  RF X C RF RF RF
=
− C =
− =
− =
−
vin R1 R1 ( X C + RF ) R1 (1 + RF / X C ) R1 (1 + jωCF RF )
=
v0
=
( RF / R1 ) e jπ ( RF / R1 ) e( )
j π −θ
where θ = tan −1 ( 2π fCF RF ) .
vin 1 + ( 2π fCF RF ) e jθ 1 + ( 2π fCF RF )
2 2
Let us calculate 2π fCF RF =2 × 3.14 × 50 × (10 ×10−12 F ) × (10 × 103 Ω ) =314 × 10−7 → 0 .
v0
⇒=
( RF / R1 ) = (1)= 1 ⇒ v0 =
vin
vin 1 + ( 2π fCF RF ) 1 + ( 0)
2 2
and θ tan −1 ( 2π =
= fCF RF ) tan
= −1
(0) 0 ⇒ φ =
π
TIFR-2017
[SOLUTION]
Physics by fiziks
TIFR-2017 (Atomic- Molecular Physics Questions and Solution)

Q1. The separation between neighbouring absorption lines in a pure rotational spectrum of
the hydrogen bromide ( HBr ) molecule is 2.23meV . If this molecule is considered as a
rigid rotor and the atomic mass number of Br is 80 , the corresponding absorption line
separation in deuterium bromide ( DBr ) molecule, in units of meV ,would be
(a) 2.234 (b) 1.115 (c) 4.461 (d) 1.128
Ans.: (d)
Solution:
ε J BJ ( J + 1) m −1 , ε J +1 − ε=
= J 2 B ( J + 1) m −1
EJ 2 Bhc ( J + 1) Joule
EJ +1 −=
h h
=B =
8π Ie 8π µ r 2 e
2 2
mH mBr 1× 80
µ HBr = = ×1.66 ×10−27 kg
mH + mBr 1 + 80
mD mBr 2 × 80
µ DBr= = ×1.66 ×10−27 kg
mD + mBr 2 + 80
( EJ +1 − EJ )DBr µ HBr 80 82 41
= = × =
( EJ +1 − EJ )HBr µ DBr 81 160 81
41
( EJ +1 − EJ )DBr = × 2.23 =1.128 meV
81
Q2. The energy of an electron in the ground state of the He atom is −79 eV . Considering the
Bohr model of the atom, what would be 10 times the first ionization potential for a He +
ion, in units of eV ?
Ans.: 246
Solution:
Ground state energy of He atom Gs
EHe = −79 eV
He - atom becomes He + ion after ionization.

Ground state energy of He + is
Z2
E GS
He+
=
−13.6 2 =
−13.6 × 4 ⇒ EHe
GS
+ =
−54.4 eV
n
First ionization potential ( IP ) of He
= + GS
EHe + − E He
GS
=−54.4 − ( −79 ) = 24.6 eV
So, ten times of first I.P= 10 × 24.6 = 246 eV
[SOLUTION]

Q3. Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average
kinetic energy of 1eV . The ratio of the number of hydrogen atoms in the 2nd excited
state ( n = 3) to the number in the ground state ( n = 1) is
(a) 5.62 ×10−6 (b) 3.16 ×10−11

(c) 3.16 ×10−8 (d) 1.33 ×10−8
Ans.: (d)
Solution:
3 2
=
Average kinetic energy = kT 1eV ⇒ kT = eV
2 3
Energy of ground state E1 = −13.6 eV
13.6 13.6
Energy of second excited state ( n = 3) ; E3 =
− 2 = − eV
3 9
 13.6 
N3 + −13.6  / ( 2 / 3) 3 8
− × ×13.6 N3
= e ( 3 1 ) = e
− E − E / kT 9
= e=

2 9
e −54.4 / 3 ⇒ = 1.33 ×10−8
N1 N1
TIFR-2017 (Solid State Physics Questions and Solution)

Q1. Consider a 2 − D square lattice. The ratio of the kinetic energy of a free electron at a
corner of the first Brillouin zone ( Ec ) to that of an electron at the midpoint of a side
Ec
face-of the same zone ( Em ) is =
Em
1
(a) 2 (b) 2 (c) 1 (d)
2
Ans: (a)
Solution:
2k 2
Kinetic energy of the free electron is written as a function of k is E =
2m
Kinetic energy of the free electron at a corner of the first Brillouin zone ( Ec ) is
 2 kc2  2 2π 2
=
Ec =
2m 2m a 2
TIFR-2017
[SOLUTION]
Physics by fiziks
Kinetic energy of the free electron at the midpoint of a side face-of the same zone ( Em )
is
 2 kc2  2 π 2
=
Ec =
2m 2m a 2
The ratio is
 2 2π 2
Ec 2m a 2
= = 2
Em 2 π 2
2m a 2
Thus, the correction answer is
option (a)

Q2. In two dimensions, two metals A and B , ha the number density of free electrons in the
ratio
nA : nB = 1: 2
The ratio of their Fermi energies is
(a) 2 : 3 (b) 1: 8 (c) 1: 2 (d) 1: 4
Ans: (c)
Solution:
A d-dimensional metal with volume Ld contain N electrons and can be written as
2
2 (
EF = Nζ d )
2/d
2mL
π
=
For ζd
d 1,= = ζ d π ; and
; d 2,= = ζ d 3π 2
d 3,=
2
For two dimensions, the Fermi energy is
2 2 Nπ π 2
= ( = π ) =
2/2
EF N n
2mL2 2m L2 2m
The ratio of the Fermi energy of two metals A and B is
EFA nA 1
= =
EFB nB 2
Thus, the correct answer is option (c).
[SOLUTION]

Q3. Electrons in a metal are scattered by both impurities and phonons. The impurity
scattering time is 8 ×10−12 s and the phonon scattering time is 2 ×10−12 s . Taking the
density of electrons to be 3 ×1014 m −3 , find the conductivity of the metal in units of
AV −1m −1 . [Assume that the effective mass of the electrons is the same as that of a free
electron.]
Ans: 1.4 ×10−5
Solution:
1
Conductivity ( σ ) and hence the resistivity ( ρ = )of the metal is
σ
ne 2τ 1 m*
σ= , ρ
= = ,
m* σ ne 2τ
Where, n is the electron concentration, m* is the effective mass of the electron that is
equal to the rest mass of the electron and τ is the collision time.
According to the Matthiessen rule the total resistivity ρ total can be approximated by
adding up several different terms:
m m m  1 1 
ρ = ρ phonons + ρimpurity = + = 2 
+ 
ne τ phonon
2
ne τ impurity
2
ne  τ phonon τ impurity 
Given, τ phonon = 8 × 10−12 s , τ impurity = 2 × 10−12 s , n= 3 × 1014 m −3 and m
= 9.1 × 10−31 kg
9.1 × 10−31  1 1 
ρ
=  + = 7.4 × 104 Ω − m
−12 
3 × 10 × (1.6 × 10 )
−12
14 −19 2
 8 × 10 2 × 10 
The electrical conductivity is

1 1
σ== 0.135 × 10−4 AV −1m −1 =
= 1.4 × 10−5 AV −1m −1
ρ 7.4 × 10 4
TIFR-2017 (Nuclear Physics Questions and Solution)

Q1. Cosmic ray muons, which decay spontaneously with proper lifetime 2.2 µ s , are produced
in the atmosphere, at a height of 5km above sea level. These move straight downwards at
N 
98% of the speed of light. Find the percent ratio 100 ×  A  of the number of muons
 NB 
measured at the top of two mountains A and B , which are at heights 4,848 m and
2, 682 m respectively above mean sea level.
TIFR-2017
[SOLUTION]
Physics by fiziks
Ans.: 52
Solution.:
Proper life time τ=
0 2.2 ×10−6 s
τ0 2.2 ×10−6
=
Life time measured from the earth τ = = 11.06 ×10−6 s
v2 1 − ( 0.98 )
2
1− 2
c
=
Time taken by muon to reach at the top of mountain A1 , t A
(=
5000 − 48 48 ) 152
0.98c 0.98c
152 152
− −
− λt A
=
N A N=
0e N 0 e − t A /τ ⇒ N A =
N0e 0.98 c×11.06×10−6
⇒ NA =
N0e 3251
2318 2318
− −
Similarly, N B = N0e 0.98 c×11.06×10−6
⇒ NB =
N0e 3251
152
−
N N 0 e 3251 NA N 
⇒ A = 2318
⇒ = e0.67 = 1.95 ⇒ 100 ×  A  =195
NB −
N 0 e 3251
NB  NB 
NB 1 N 
⇒ = = 0.52 ⇒ 100 ×  B  = 52
N A 1.95  NA 
Q2. A deuteron of mass M and binding energy B is struck by a gamma ray photon of energy
Eγ and is observed to disintegrate into a neutron and a proton. If B  Mc 2 , the minimum
value of Eγ must be
B2 1 B2 
(a) 2 B + (b)  3B + 
2 Mc 2 2 Mc 2 
B2 1 B2 
(c) B + (d)  2 B + 
Mc 2 2 Mc 2 
Ans.: (d)
Solution: D + γ → p + n ;
Eγ E
Momentum of photon = Combined momentum of ( p + n ) =γ
,
c c
Q -value of the nuclear reaction,
Q=Mc 2 − ( m p + mn ) c 2 =−
0 B ⇒ B= (m p + mn ) c 2 − Mc 2
2
E 
(m + mn )
2
Apply the conservation of energy Eγ + Mc = 2
p c +  γ  c2
4
 c 
Eγ2 + M 2 c 4 + 2 Eγ .Mc 2 =
B 2 + 2 BMc 2 + M 2 c 4 + Eγ2 ⇒ 2 Eγ Mc 2 =
B 2 + 2 BMc 2
B2 1  B2 
⇒ E=
γ 2
+ B ⇒ =
Eγ  2
+ 2B 
2 Mc 2  Mc 
[SOLUTION]

Q2. A subatomic particle ψ and its excited state ψ * have rest masses 3.1GeV / c 2 and
3.7 GeV / c 2 respectively. A table of its assigned quantum numbers is given below.
Angular Momentum Parity C-Parity Isospin Electric charge
J =1 P = −1 C = −1 I =0 Q=0
If π 0* is an excited state of π 0 with a mass of about 1.3 GeV / c 2 , which of the following
reactions is possible when the above quantum numbers are conserved?
(a) ψ * → γγ (b) ψ * → π 0π 0
(c) ψ * → ψ π +π − (d) ψ * → ψ π 0*
Ans.: (c)
Solution:
Bosons and their antiparticles have same parity.
ψ →ψ + π + + π −
S 1 1 0 0 conserved
parity −1 −1 −1 −1 conserved
Isospin 0 0 0 0 conserved
Q 0 0 +1 −1 conserved
Q3. In a theoretical model of the nucleus, the binding energy per nucleon was predicted as
shown in the figure below
Binding energy per nucleon (BE/A) (MeV)
9
8
7
6
5
4
3
2
1
0
0 50 100 150 200 250
Mass Number (A)
If a nucleus of mass number A = 240 undergoes a symmetric fission to two daughter

nuclei each of mass number A = 120 , write down the amount of energy released in this
process, in units of MeV , using this theoretical model.
TIFR-2017
[SOLUTION]
Physics by fiziks
Ans.: 240
Solution: 240
X →120 Y +120 Y
E 2 BEY − BE X =( 2 ×120 × 8.5 ) − ( 240 × 7.5 ) = 240 × ( 8.5 − 7.5 ) = 240 MeV
∆=
General Physics

Q3. A solid tetrahedron (solid with four plane sides) has the following projections (drawn to
scale) when seen from three different sides:
Left Top Right

When viewed from the front, its projection will be
(a) (b)
(c) (d)
Ans: (b)
Solution:
The projection seen from front will have one of its side non-modulated and will be equal
in length to one of sides of the left or right view. Only option which fits into this scheme
is choice (b). Hence choice is (b)
Left Right Front

TIFR Question With Solution 2017

Uploaded by

Copyright:

Available Formats

TIFR Question With Solution 2017

Uploaded by

Document Information

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

TIFR Question With Solution 2017

Uploaded by

Copyright:

Available Formats

TIFR-2017 Physics by fiziks

TIFR-2017 (Mathematical Physics Question and Solution)

(a) − B [ A, C ] (b) B { A, C} (c) [ A, C ] B (d) − B { A, C}

Hence 'd' is the correct answer.

4  1 T nωT T nωT  1  nωT nωT  

a0 + ∑ an cos nωt + ∑ bn sin nωt

One of the eigen values = 100 2

n ! ∫ dxn −1 ∫ dx ∫ dx n−2 n −3... ∫ dx ∫ dx ∫ dx

Let's calculate residue at z = z1 and z = z2 .

Take limit R → ∞ and note that, limit R → ∞ ∫ f ( z ) dz = 0

Hence the answer is (a)

TIFR-2017 (EMT Question and Solution)

to this system of charges will be

1  4q0 2 −a2 r  1  4q0 2 −α r 

Q3. A rectangular metallic loop with sides L1 and L2 is

(a) κ B 1 2 sin φ zˆ (b) −κ B 1 2 sin 2φ zˆ

(c) κ B 1 2 sin φ cos φ zˆ (d) −κ B 1 2 sin φ zˆ

Magnetic dipole moment of the current loop is m =I ind × ( L1 L2 ) nˆ =

Q5. Consider the following situations.

In situation A , two semi-infinite earthed conducting planes meet at right-angles. A point

TIFR-2017 (Quantum Mechanics Question and Solution)

where ϕn ( x ) are the normalized energy eigenfunctions of a given Hamiltonian. The

2π 2  2 2.4π 2  2 29π 2  2 16π 2  2 44π 2  2

(a) 1.5  2 (b) 0.18 2 (c) 0.36  2 (d) Zero

=L2 ψ ( eiπ / 3 311

= 0.8 2 0 ( 0 + 1) 1.0, 0 + 0.8eiπ / 3 2 2 311 = 0.6eiπ / 3 2 2 311

Similarly the expectation value of L2z is

=Lz ψ ( eiπ /3 Lz 3,1,1

L2z ψ = 0.6eiπ /3  2 311

The expectation value of L2x is given by

SECTION B-(Only for Ph.D. candidates)

oscillator can be written as

The single oscillator partition function is

The partition function for the system is

Similarly, Let us evaluate the value of 1 v 3

TIFR-2017 (Electronics Question and Solution)

to drive the RC circuit shown below.

(a) C (b) B (c) A (d) A + B + C

Q3. An AC voltage source has an internal resistance of 50 Ω and is specified to deliver an

Peak voltage of input is V0 m = 100 2 V .

Thus peak to peak voltage of input= =

SECTION B- (only for Int.-Ph.D. candidates)

SECTION B-(Only for Ph.D. candidates)

TIFR-2017 (Atomic- Molecular Physics Questions and Solution)

He - atom becomes He + ion after ionization.

SECTION B-(Only for Ph.D. candidates)

state ( n = 3) to the number in the ground state ( n = 1) is

(a) 5.62 ×10−6 (b) 3.16 ×10−11

SECTION A-(For both Int. Ph.D. and Ph.D. candidates)

SECTION B- (only for Int.-Ph.D. candidates)

SECTION B-(Only for Ph.D. candidates)

density of electrons to be 3 ×1014 m −3 , find the conductivity of the metal in units of

The electrical conductivity is

TIFR-2017 (Nuclear Physics Questions and Solution)

SECTION B-(Only for Ph.D. candidates)

If a nucleus of mass number A = 240 undergoes a symmetric fission to two daughter

SECTION A-(For both Int. Ph.D. and Ph.D. candidates)

Left Top Right

Left Right Front

You might also like