JEE Main 2021 Physics March 18 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


 
 
18
th
 March 2021 | Shift 2
SECTION – A 
 
1. The decay of a proton to neutron is :  
 (1) Not possible as proton mass is less than the neutron mass    
 (2) Always possible as it is associated only with ?
+
 decay   
 (3) Possible only inside the nucleus    
 (4) Not possible but neutron to proton conversion is possible   
Sol. (3) 
 Positron emission or Beta plus decay is a subtype of radioactive decay called Beta decay, in 
which a proton inside a radionuclide nucleus is converted into a neutron while releasing a 
positron and an electron neutrino. 
 So, decay of a proton to neutron is possible only inside the nucleus. 
 
 
 
2. An object of mass m
1
 collides with another object of mass m
2
, which is at rest. After the 
collision the objects move with equal speeds in opposite direction. The ratio of the masses   
 m
2 
: m
1 
 is :  
(1) 2 : 1   (2) 1 : 1   (3) 1 : 2   (4) 3 : 1   
Sol. (4) 
 
 
m
1
u
1 
m
2
u
2
=0 
Before Collision
m
1
v
1
=v
m
2 
After Collision
v
2
=v
 
 From conservation of linear momentum; 
 P
i
 = P
f
  
 
1 1 2 1 2
m u m (0) m ( v) m v ? ? ? ? ? 
 
1 1 2 1
m u v(m m ) ? ? ? …(i) 
 
sep
app 1 1
V
v ( v) 2v
e 1
V u u
? ?
? ? ? ? ? 
 
1
u 2V ? ? …(ii) 
 from (i) & (ii) 
 
1 2 1
m (2v) v(m m ) ? ? 
 
1 2 1
2m m m ? ? ? 
 
1 2
3m m ? ? 
 
2
1
m 3
m 1
? ?
 
 
 
 
3. A plane electromagnetic wave propagating along y-direction can have the following pair of 
electric field (E)
?
and magnetic field (B)
?
 
components. 
 (1) E
x
, B
z
 or E
z
, B
x
  (2) E
y
, B
x
 or E
x
, B
y
  
 (3) E
x
, B
y
 or E
y
, B
x
  (4) E
y
, B
y
 or E
z
, B
z
  
 
Page 2


 
 
18
th
 March 2021 | Shift 2
SECTION – A 
 
1. The decay of a proton to neutron is :  
 (1) Not possible as proton mass is less than the neutron mass    
 (2) Always possible as it is associated only with ?
+
 decay   
 (3) Possible only inside the nucleus    
 (4) Not possible but neutron to proton conversion is possible   
Sol. (3) 
 Positron emission or Beta plus decay is a subtype of radioactive decay called Beta decay, in 
which a proton inside a radionuclide nucleus is converted into a neutron while releasing a 
positron and an electron neutrino. 
 So, decay of a proton to neutron is possible only inside the nucleus. 
 
 
 
2. An object of mass m
1
 collides with another object of mass m
2
, which is at rest. After the 
collision the objects move with equal speeds in opposite direction. The ratio of the masses   
 m
2 
: m
1 
 is :  
(1) 2 : 1   (2) 1 : 1   (3) 1 : 2   (4) 3 : 1   
Sol. (4) 
 
 
m
1
u
1 
m
2
u
2
=0 
Before Collision
m
1
v
1
=v
m
2 
After Collision
v
2
=v
 
 From conservation of linear momentum; 
 P
i
 = P
f
  
 
1 1 2 1 2
m u m (0) m ( v) m v ? ? ? ? ? 
 
1 1 2 1
m u v(m m ) ? ? ? …(i) 
 
sep
app 1 1
V
v ( v) 2v
e 1
V u u
? ?
? ? ? ? ? 
 
1
u 2V ? ? …(ii) 
 from (i) & (ii) 
 
1 2 1
m (2v) v(m m ) ? ? 
 
1 2 1
2m m m ? ? ? 
 
1 2
3m m ? ? 
 
2
1
m 3
m 1
? ?
 
 
 
 
3. A plane electromagnetic wave propagating along y-direction can have the following pair of 
electric field (E)
?
and magnetic field (B)
?
 
components. 
 (1) E
x
, B
z
 or E
z
, B
x
  (2) E
y
, B
x
 or E
x
, B
y
  
 (3) E
x
, B
y
 or E
y
, B
x
  (4) E
y
, B
y
 or E
z
, B
z
  
 
 
 
Sol. (1) 
 
 Y 
v
?
Z
X
 
 
ˆ ˆ ˆ
E B C ? ? ? for electromagnetic waves. 
 
ˆ ˆ
E B ? ? should point in the  
 direction of propagation of wave (y-direction here)  
 ? ?possible combinations are  
 (E
x 
, B
z
) or (E
z 
, B
x
) 
 
4. A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough 
inclined plane as shown in the figure. The frictional force acting between the cylinder and the 
inclined plane is :  
 
60°
 
 
 [The coefficient of static friction, ?
s
, is 0.4] 
 (1) 
7
mg
2
  (2) 0   
 (3) 
mg
5
   (4) 5 mg   
Sol. (3) 
 
 
T
N
f
mg cos60°
mg
60°
60°
mg sin60° 
 
 
 Let's assume equilibrium condition of cylinder 
Page 3


 
 
18
th
 March 2021 | Shift 2
SECTION – A 
 
1. The decay of a proton to neutron is :  
 (1) Not possible as proton mass is less than the neutron mass    
 (2) Always possible as it is associated only with ?
+
 decay   
 (3) Possible only inside the nucleus    
 (4) Not possible but neutron to proton conversion is possible   
Sol. (3) 
 Positron emission or Beta plus decay is a subtype of radioactive decay called Beta decay, in 
which a proton inside a radionuclide nucleus is converted into a neutron while releasing a 
positron and an electron neutrino. 
 So, decay of a proton to neutron is possible only inside the nucleus. 
 
 
 
2. An object of mass m
1
 collides with another object of mass m
2
, which is at rest. After the 
collision the objects move with equal speeds in opposite direction. The ratio of the masses   
 m
2 
: m
1 
 is :  
(1) 2 : 1   (2) 1 : 1   (3) 1 : 2   (4) 3 : 1   
Sol. (4) 
 
 
m
1
u
1 
m
2
u
2
=0 
Before Collision
m
1
v
1
=v
m
2 
After Collision
v
2
=v
 
 From conservation of linear momentum; 
 P
i
 = P
f
  
 
1 1 2 1 2
m u m (0) m ( v) m v ? ? ? ? ? 
 
1 1 2 1
m u v(m m ) ? ? ? …(i) 
 
sep
app 1 1
V
v ( v) 2v
e 1
V u u
? ?
? ? ? ? ? 
 
1
u 2V ? ? …(ii) 
 from (i) & (ii) 
 
1 2 1
m (2v) v(m m ) ? ? 
 
1 2 1
2m m m ? ? ? 
 
1 2
3m m ? ? 
 
2
1
m 3
m 1
? ?
 
 
 
 
3. A plane electromagnetic wave propagating along y-direction can have the following pair of 
electric field (E)
?
and magnetic field (B)
?
 
components. 
 (1) E
x
, B
z
 or E
z
, B
x
  (2) E
y
, B
x
 or E
x
, B
y
  
 (3) E
x
, B
y
 or E
y
, B
x
  (4) E
y
, B
y
 or E
z
, B
z
  
 
 
 
Sol. (1) 
 
 Y 
v
?
Z
X
 
 
ˆ ˆ ˆ
E B C ? ? ? for electromagnetic waves. 
 
ˆ ˆ
E B ? ? should point in the  
 direction of propagation of wave (y-direction here)  
 ? ?possible combinations are  
 (E
x 
, B
z
) or (E
z 
, B
x
) 
 
4. A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough 
inclined plane as shown in the figure. The frictional force acting between the cylinder and the 
inclined plane is :  
 
60°
 
 
 [The coefficient of static friction, ?
s
, is 0.4] 
 (1) 
7
mg
2
  (2) 0   
 (3) 
mg
5
   (4) 5 mg   
Sol. (3) 
 
 
T
N
f
mg cos60°
mg
60°
60°
mg sin60° 
 
 
 Let's assume equilibrium condition of cylinder 
 
 
18
th
 March 2021 | Shift 2
 T f mgsin60 ? ? ? ? ….(i) 
 & TR – fR = 0  ….(ii) 
 from (i) & (ii) 
 
req
mgsin60
T f
2
?
? ? 
 But limiting friction < required friction. 
 ? ? ?mgcos60°  
mgsin60
2
?
? 
 ?cylinder  won't be in equilibrium 
 ? ?f will be kinetic 
 & f = ?
K
 N 
 
k
mgcos60 ? ? ? 
 
1 mg
0.4 mg
2 5
? ? ? ? 
 
5. An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is 
S
1
 and that of the other part is S
2
. Given that S
1
 > S
2
. If the piston is removed then the total 
entropy of the system will be :  
 (1) S
1
 + S
2
   (2) S
1
 - S
2
   (3) S
1
 × S
2
   (4) 
1
2
S
S
  
Sol. (1) 
 
 
(S
1
>S
2
)
1 2
S
2 
S
1
Piston
 
 for gas 1 , 
1 1
f
S n R
2
? 
 for gas 2, 
2 2
f
S n R
2
?  } identical gas, so f will be same. 
 after removal of piston, 
 
1 2 1 2
f
s (n n )R S S
2
? ? ? ? 
 
6. The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of 
resistance R, inductance L is connected to a battery, is : 
 (1) 
L
ln2
R
 (2) 
L
ln10
R
  (3) Infinite  (4) 
L
ln5
R
  
Sol. (1) 
 ? ?Magnetic energy, 
2
1
U LI
2
? , when current in circuit is I. 
 ? When circuit has maximum current, 
 maximum value of Magnetic energy, 
2
0 0
1
U LI
2
? 
Page 4


 
 
18
th
 March 2021 | Shift 2
SECTION – A 
 
1. The decay of a proton to neutron is :  
 (1) Not possible as proton mass is less than the neutron mass    
 (2) Always possible as it is associated only with ?
+
 decay   
 (3) Possible only inside the nucleus    
 (4) Not possible but neutron to proton conversion is possible   
Sol. (3) 
 Positron emission or Beta plus decay is a subtype of radioactive decay called Beta decay, in 
which a proton inside a radionuclide nucleus is converted into a neutron while releasing a 
positron and an electron neutrino. 
 So, decay of a proton to neutron is possible only inside the nucleus. 
 
 
 
2. An object of mass m
1
 collides with another object of mass m
2
, which is at rest. After the 
collision the objects move with equal speeds in opposite direction. The ratio of the masses   
 m
2 
: m
1 
 is :  
(1) 2 : 1   (2) 1 : 1   (3) 1 : 2   (4) 3 : 1   
Sol. (4) 
 
 
m
1
u
1 
m
2
u
2
=0 
Before Collision
m
1
v
1
=v
m
2 
After Collision
v
2
=v
 
 From conservation of linear momentum; 
 P
i
 = P
f
  
 
1 1 2 1 2
m u m (0) m ( v) m v ? ? ? ? ? 
 
1 1 2 1
m u v(m m ) ? ? ? …(i) 
 
sep
app 1 1
V
v ( v) 2v
e 1
V u u
? ?
? ? ? ? ? 
 
1
u 2V ? ? …(ii) 
 from (i) & (ii) 
 
1 2 1
m (2v) v(m m ) ? ? 
 
1 2 1
2m m m ? ? ? 
 
1 2
3m m ? ? 
 
2
1
m 3
m 1
? ?
 
 
 
 
3. A plane electromagnetic wave propagating along y-direction can have the following pair of 
electric field (E)
?
and magnetic field (B)
?
 
components. 
 (1) E
x
, B
z
 or E
z
, B
x
  (2) E
y
, B
x
 or E
x
, B
y
  
 (3) E
x
, B
y
 or E
y
, B
x
  (4) E
y
, B
y
 or E
z
, B
z
  
 
 
 
Sol. (1) 
 
 Y 
v
?
Z
X
 
 
ˆ ˆ ˆ
E B C ? ? ? for electromagnetic waves. 
 
ˆ ˆ
E B ? ? should point in the  
 direction of propagation of wave (y-direction here)  
 ? ?possible combinations are  
 (E
x 
, B
z
) or (E
z 
, B
x
) 
 
4. A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough 
inclined plane as shown in the figure. The frictional force acting between the cylinder and the 
inclined plane is :  
 
60°
 
 
 [The coefficient of static friction, ?
s
, is 0.4] 
 (1) 
7
mg
2
  (2) 0   
 (3) 
mg
5
   (4) 5 mg   
Sol. (3) 
 
 
T
N
f
mg cos60°
mg
60°
60°
mg sin60° 
 
 
 Let's assume equilibrium condition of cylinder 
 
 
18
th
 March 2021 | Shift 2
 T f mgsin60 ? ? ? ? ….(i) 
 & TR – fR = 0  ….(ii) 
 from (i) & (ii) 
 
req
mgsin60
T f
2
?
? ? 
 But limiting friction < required friction. 
 ? ? ?mgcos60°  
mgsin60
2
?
? 
 ?cylinder  won't be in equilibrium 
 ? ?f will be kinetic 
 & f = ?
K
 N 
 
k
mgcos60 ? ? ? 
 
1 mg
0.4 mg
2 5
? ? ? ? 
 
5. An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is 
S
1
 and that of the other part is S
2
. Given that S
1
 > S
2
. If the piston is removed then the total 
entropy of the system will be :  
 (1) S
1
 + S
2
   (2) S
1
 - S
2
   (3) S
1
 × S
2
   (4) 
1
2
S
S
  
Sol. (1) 
 
 
(S
1
>S
2
)
1 2
S
2 
S
1
Piston
 
 for gas 1 , 
1 1
f
S n R
2
? 
 for gas 2, 
2 2
f
S n R
2
?  } identical gas, so f will be same. 
 after removal of piston, 
 
1 2 1 2
f
s (n n )R S S
2
? ? ? ? 
 
6. The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of 
resistance R, inductance L is connected to a battery, is : 
 (1) 
L
ln2
R
 (2) 
L
ln10
R
  (3) Infinite  (4) 
L
ln5
R
  
Sol. (1) 
 ? ?Magnetic energy, 
2
1
U LI
2
? , when current in circuit is I. 
 ? When circuit has maximum current, 
 maximum value of Magnetic energy, 
2
0 0
1
U LI
2
? 
 
 
 Given : U = 25% of U
0
. 
 
2 2
0
1 1 1
LI LI
2 4 2
? ? ? 
 
2
2 0 0
I I
I I
4 2
? ? ? ? 
 ? 
t/
0
I I (1 e )
? ?
? ? 
 
t/ 0
0
I
I (1 e )
2
? ?
? ? ? 
 
t/
1
e
2
? ?
? ? 
 
t/
e 2
?
? ? 
 t n2 ? ? ? ? 
 
L
t n2
R
? ? ? 
 
7. For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to 
(where ? is the ratio of specific heats) : 
 (1) 
dV
V
? ? (2) 
dV
V
  (3) 
1 dV
V
?
?
  (4) 
V
dV
? ?  
Sol. (1) 
 for adiabatic expansion : 
 PV const.
?
? 
 nP lnv const. ? ? ? ? ? 
 ? differentiating both sides; 
 
dp dv
0
p v
? ? ? 
 
dp dv
p V
? ? ? ? 
 
8. The correct relation between ? (ratio of collector current to emitter current) and ? (ratio of 
collector current to base current) of a transistor is :  
 (1) 
1
?
? ?
? ?
 (2) 
1
?
? ?
? ?
 (3) 
1
1
? ?
? ?
 (4) 
1
?
? ?
? ?
 
Sol. (1) 
 
C
E
I
I
? ? & 
C
B
I
I
? ? 
 & I
E
 = I
B
 + I
C
  
 
C E B
C C C
I I I
I I I
? ? ? 
 
1 1 1
1
? ?
? ? ? ?
? ? ?
 
 
1
?
? ? ?
? ?
 
Page 5


 
 
18
th
 March 2021 | Shift 2
SECTION – A 
 
1. The decay of a proton to neutron is :  
 (1) Not possible as proton mass is less than the neutron mass    
 (2) Always possible as it is associated only with ?
+
 decay   
 (3) Possible only inside the nucleus    
 (4) Not possible but neutron to proton conversion is possible   
Sol. (3) 
 Positron emission or Beta plus decay is a subtype of radioactive decay called Beta decay, in 
which a proton inside a radionuclide nucleus is converted into a neutron while releasing a 
positron and an electron neutrino. 
 So, decay of a proton to neutron is possible only inside the nucleus. 
 
 
 
2. An object of mass m
1
 collides with another object of mass m
2
, which is at rest. After the 
collision the objects move with equal speeds in opposite direction. The ratio of the masses   
 m
2 
: m
1 
 is :  
(1) 2 : 1   (2) 1 : 1   (3) 1 : 2   (4) 3 : 1   
Sol. (4) 
 
 
m
1
u
1 
m
2
u
2
=0 
Before Collision
m
1
v
1
=v
m
2 
After Collision
v
2
=v
 
 From conservation of linear momentum; 
 P
i
 = P
f
  
 
1 1 2 1 2
m u m (0) m ( v) m v ? ? ? ? ? 
 
1 1 2 1
m u v(m m ) ? ? ? …(i) 
 
sep
app 1 1
V
v ( v) 2v
e 1
V u u
? ?
? ? ? ? ? 
 
1
u 2V ? ? …(ii) 
 from (i) & (ii) 
 
1 2 1
m (2v) v(m m ) ? ? 
 
1 2 1
2m m m ? ? ? 
 
1 2
3m m ? ? 
 
2
1
m 3
m 1
? ?
 
 
 
 
3. A plane electromagnetic wave propagating along y-direction can have the following pair of 
electric field (E)
?
and magnetic field (B)
?
 
components. 
 (1) E
x
, B
z
 or E
z
, B
x
  (2) E
y
, B
x
 or E
x
, B
y
  
 (3) E
x
, B
y
 or E
y
, B
x
  (4) E
y
, B
y
 or E
z
, B
z
  
 
 
 
Sol. (1) 
 
 Y 
v
?
Z
X
 
 
ˆ ˆ ˆ
E B C ? ? ? for electromagnetic waves. 
 
ˆ ˆ
E B ? ? should point in the  
 direction of propagation of wave (y-direction here)  
 ? ?possible combinations are  
 (E
x 
, B
z
) or (E
z 
, B
x
) 
 
4. A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough 
inclined plane as shown in the figure. The frictional force acting between the cylinder and the 
inclined plane is :  
 
60°
 
 
 [The coefficient of static friction, ?
s
, is 0.4] 
 (1) 
7
mg
2
  (2) 0   
 (3) 
mg
5
   (4) 5 mg   
Sol. (3) 
 
 
T
N
f
mg cos60°
mg
60°
60°
mg sin60° 
 
 
 Let's assume equilibrium condition of cylinder 
 
 
18
th
 March 2021 | Shift 2
 T f mgsin60 ? ? ? ? ….(i) 
 & TR – fR = 0  ….(ii) 
 from (i) & (ii) 
 
req
mgsin60
T f
2
?
? ? 
 But limiting friction < required friction. 
 ? ? ?mgcos60°  
mgsin60
2
?
? 
 ?cylinder  won't be in equilibrium 
 ? ?f will be kinetic 
 & f = ?
K
 N 
 
k
mgcos60 ? ? ? 
 
1 mg
0.4 mg
2 5
? ? ? ? 
 
5. An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is 
S
1
 and that of the other part is S
2
. Given that S
1
 > S
2
. If the piston is removed then the total 
entropy of the system will be :  
 (1) S
1
 + S
2
   (2) S
1
 - S
2
   (3) S
1
 × S
2
   (4) 
1
2
S
S
  
Sol. (1) 
 
 
(S
1
>S
2
)
1 2
S
2 
S
1
Piston
 
 for gas 1 , 
1 1
f
S n R
2
? 
 for gas 2, 
2 2
f
S n R
2
?  } identical gas, so f will be same. 
 after removal of piston, 
 
1 2 1 2
f
s (n n )R S S
2
? ? ? ? 
 
6. The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of 
resistance R, inductance L is connected to a battery, is : 
 (1) 
L
ln2
R
 (2) 
L
ln10
R
  (3) Infinite  (4) 
L
ln5
R
  
Sol. (1) 
 ? ?Magnetic energy, 
2
1
U LI
2
? , when current in circuit is I. 
 ? When circuit has maximum current, 
 maximum value of Magnetic energy, 
2
0 0
1
U LI
2
? 
 
 
 Given : U = 25% of U
0
. 
 
2 2
0
1 1 1
LI LI
2 4 2
? ? ? 
 
2
2 0 0
I I
I I
4 2
? ? ? ? 
 ? 
t/
0
I I (1 e )
? ?
? ? 
 
t/ 0
0
I
I (1 e )
2
? ?
? ? ? 
 
t/
1
e
2
? ?
? ? 
 
t/
e 2
?
? ? 
 t n2 ? ? ? ? 
 
L
t n2
R
? ? ? 
 
7. For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to 
(where ? is the ratio of specific heats) : 
 (1) 
dV
V
? ? (2) 
dV
V
  (3) 
1 dV
V
?
?
  (4) 
V
dV
? ?  
Sol. (1) 
 for adiabatic expansion : 
 PV const.
?
? 
 nP lnv const. ? ? ? ? ? 
 ? differentiating both sides; 
 
dp dv
0
p v
? ? ? 
 
dp dv
p V
? ? ? ? 
 
8. The correct relation between ? (ratio of collector current to emitter current) and ? (ratio of 
collector current to base current) of a transistor is :  
 (1) 
1
?
? ?
? ?
 (2) 
1
?
? ?
? ?
 (3) 
1
1
? ?
? ?
 (4) 
1
?
? ?
? ?
 
Sol. (1) 
 
C
E
I
I
? ? & 
C
B
I
I
? ? 
 & I
E
 = I
B
 + I
C
  
 
C E B
C C C
I I I
I I I
? ? ? 
 
1 1 1
1
? ?
? ? ? ?
? ? ?
 
 
1
?
? ? ?
? ?
 
 
 
18
th
 March 2021 | Shift 2
9. In a series LCR circuit, the inductive reactance (x
L
) is 10 ? and the capacitive reactance (X
C
) is 
4 ?. The resistance (R) in the circuit is 6 ? ?  
 (1) 
1
2
  (2) 
3
2
  (3) 
1
2
  (4) 
1
2 2
  
Sol. (1) 
 Given :  X
L
 = 10 ? 
              X
C
 = 4 ? 
               R = 6 ? 
 ? ?Power factor = cos ?  = 
R
Z
 
  
2 2
L C
R
R (x X )
?
? ?
 
  
2 2
6
6 (10 4)
?
? ?
 
  
6 1
6 2 2
? ? 
 
10. A proton and an ?-particle, having kinetic energies K
p
 and K
?
 repectively, enter into a magnetic 
field at right angles.  
 The ratio of the radii of trajectory of proton to that ?-particle is 2 : 1. The ratio of K
P
 : K
?
 is :  
(1) 1 : 8   (2) 1 : 4   (3) 8 : 1   (4) 4 : 1   
Sol. (4) 
 
mv p
r
qB qB
? ? ? 
 
p
m 4
&
m 1
?
? 
 
p
P
P
r
q P 2
.
r P q 1
?
? ?
? ? ? 
 
P P
P 2q 1
2 1
P q 2
? ?
? ?
? ? ? ?
? ?
? ?
 
 Now, 
2
p p
p
K P
m 4
. 1 4
K P m 1
?
? ?
? ?
? ? ? ? ? ?
? ?
? ?
 
 
 
11. The function of time representing a simple harmonic motion with a period of 
?
?
 is :  
 (1) cos( ?t) + cos(2 ?t) + cos(3 ?t) (2) 3cos 2 t
4
? ? ?
? ?
? ?
? ?
   
 (3) sin
2
( ?t)    (4) sin( ?t) + cos( ?t)  
Sol. (2) 
 for expression, 3cos 2 t
4
? ? ?
? ?
? ?
? ?