Work, Energy and Power: JEE Main Previous Year Questions (2021-2026)

 Page 1


 
JEE Main Previous Year Questions (2021-2026): 
Work Power & Energy  
 
(January 2026) 
 
 
Q1: A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 
0 and at this instant the horizontal component of its velocity is v. Another bead Q of the 
same mass as P is ejected from point A at t = 0 along the horizontal string AB, with the 
speed v, friction between the beads and the respective strings may be neglected in both 
cases. Let t
P
 and t
Q
 be the respective times taken by beads P and Q to reach the point B, 
then the relation between t
P
 and t
Q
 is 
 
(a) t
P 
< t
Q 
(b) t
P
 = t
Q 
(c) t
P 
> 1.25 t
Q 
(d) t
P
 > t
Q
  
Ans: (a) 
Sol: 
For bead Q  
Bead Q moves along the horizontal straight string AB. The length AB is 2R where R is the 
radius of semicircular path.  
Since the string is frictionless and horizontal, there is no component of gravity acting along the 
Page 2


 
JEE Main Previous Year Questions (2021-2026): 
Work Power & Energy  
 
(January 2026) 
 
 
Q1: A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 
0 and at this instant the horizontal component of its velocity is v. Another bead Q of the 
same mass as P is ejected from point A at t = 0 along the horizontal string AB, with the 
speed v, friction between the beads and the respective strings may be neglected in both 
cases. Let t
P
 and t
Q
 be the respective times taken by beads P and Q to reach the point B, 
then the relation between t
P
 and t
Q
 is 
 
(a) t
P 
< t
Q 
(b) t
P
 = t
Q 
(c) t
P 
> 1.25 t
Q 
(d) t
P
 > t
Q
  
Ans: (a) 
Sol: 
For bead Q  
Bead Q moves along the horizontal straight string AB. The length AB is 2R where R is the 
radius of semicircular path.  
Since the string is frictionless and horizontal, there is no component of gravity acting along the 
direction of motion.  
Bead Q maintains a constant horizontal velocity v throughout the motion.  
 
For bead P Bead P moves along the semi-circular path ACB under the influence of gravity. And 
along horizontal the net force is zero, i.e., the acceleration along vertical will be non-zero and 
the acceleration along horizontal will be zero.  
At point S , the horizontal component of its velocity is given as v.  
As bead P moves from S toward the bottom-most point C, the vertical component of velocity 
changes but the horizontal component remains same. 
 
When bead P reaches point B then the horizontal displacement is S'O + OB  
S'O = R cos  45
°
  
Then the time taken to reach point B is 
 
So, the ratio of time taken by Q and P to reach B is, 
 
Page 3


 
JEE Main Previous Year Questions (2021-2026): 
Work Power & Energy  
 
(January 2026) 
 
 
Q1: A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 
0 and at this instant the horizontal component of its velocity is v. Another bead Q of the 
same mass as P is ejected from point A at t = 0 along the horizontal string AB, with the 
speed v, friction between the beads and the respective strings may be neglected in both 
cases. Let t
P
 and t
Q
 be the respective times taken by beads P and Q to reach the point B, 
then the relation between t
P
 and t
Q
 is 
 
(a) t
P 
< t
Q 
(b) t
P
 = t
Q 
(c) t
P 
> 1.25 t
Q 
(d) t
P
 > t
Q
  
Ans: (a) 
Sol: 
For bead Q  
Bead Q moves along the horizontal straight string AB. The length AB is 2R where R is the 
radius of semicircular path.  
Since the string is frictionless and horizontal, there is no component of gravity acting along the 
direction of motion.  
Bead Q maintains a constant horizontal velocity v throughout the motion.  
 
For bead P Bead P moves along the semi-circular path ACB under the influence of gravity. And 
along horizontal the net force is zero, i.e., the acceleration along vertical will be non-zero and 
the acceleration along horizontal will be zero.  
At point S , the horizontal component of its velocity is given as v.  
As bead P moves from S toward the bottom-most point C, the vertical component of velocity 
changes but the horizontal component remains same. 
 
When bead P reaches point B then the horizontal displacement is S'O + OB  
S'O = R cos  45
°
  
Then the time taken to reach point B is 
 
So, the ratio of time taken by Q and P to reach B is, 
 
 
Q2: An object is projected with kinetic energy K from a point A at an angle 60
°
 with the 
horizontal. The ratio of the difference in kinetic energies at points B and C to that at point 
A (see figure), in the absence of air friction is : 
 
(a) 3 : 4 
(b) 1 : 4 
(c) 2 : 3 
(d) 1 : 2 
Ans: (a)  
Let the mass of the object be m and the initial speed be u.  
The angle of projection is ? = 60
°
.  
At point A  
The kinetic energy at point A is given as K.  
 
At Point B (Maximum Height)  
At the highest point of a projectile's path, the vertical component of velocity becomes zero (v
y
 = 
0) . The object only possesses the horizontal component of velocity (v
x
).  
The horizontal velocity remains constant throughout the motion (assuming no air resistance): 
 
Page 4


 
JEE Main Previous Year Questions (2021-2026): 
Work Power & Energy  
 
(January 2026) 
 
 
Q1: A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 
0 and at this instant the horizontal component of its velocity is v. Another bead Q of the 
same mass as P is ejected from point A at t = 0 along the horizontal string AB, with the 
speed v, friction between the beads and the respective strings may be neglected in both 
cases. Let t
P
 and t
Q
 be the respective times taken by beads P and Q to reach the point B, 
then the relation between t
P
 and t
Q
 is 
 
(a) t
P 
< t
Q 
(b) t
P
 = t
Q 
(c) t
P 
> 1.25 t
Q 
(d) t
P
 > t
Q
  
Ans: (a) 
Sol: 
For bead Q  
Bead Q moves along the horizontal straight string AB. The length AB is 2R where R is the 
radius of semicircular path.  
Since the string is frictionless and horizontal, there is no component of gravity acting along the 
direction of motion.  
Bead Q maintains a constant horizontal velocity v throughout the motion.  
 
For bead P Bead P moves along the semi-circular path ACB under the influence of gravity. And 
along horizontal the net force is zero, i.e., the acceleration along vertical will be non-zero and 
the acceleration along horizontal will be zero.  
At point S , the horizontal component of its velocity is given as v.  
As bead P moves from S toward the bottom-most point C, the vertical component of velocity 
changes but the horizontal component remains same. 
 
When bead P reaches point B then the horizontal displacement is S'O + OB  
S'O = R cos  45
°
  
Then the time taken to reach point B is 
 
So, the ratio of time taken by Q and P to reach B is, 
 
 
Q2: An object is projected with kinetic energy K from a point A at an angle 60
°
 with the 
horizontal. The ratio of the difference in kinetic energies at points B and C to that at point 
A (see figure), in the absence of air friction is : 
 
(a) 3 : 4 
(b) 1 : 4 
(c) 2 : 3 
(d) 1 : 2 
Ans: (a)  
Let the mass of the object be m and the initial speed be u.  
The angle of projection is ? = 60
°
.  
At point A  
The kinetic energy at point A is given as K.  
 
At Point B (Maximum Height)  
At the highest point of a projectile's path, the vertical component of velocity becomes zero (v
y
 = 
0) . The object only possesses the horizontal component of velocity (v
x
).  
The horizontal velocity remains constant throughout the motion (assuming no air resistance): 
 
So, the kinetic energy at B is, 
 
At Point C 
Point C is at the same horizontal level as point A . By the law of conservation of energy, the 
speed of the projectile when it hits the ground at the same level is equal to the speed of 
projection. 
v
C
 = u 
Therefore, the kinetic energy at C is equal to the kinetic energy at A : 
K
C
 = K
A
 = K 
The difference in kinetic energies at B and C : 
 
So, the ratio of the difference in kinetic energies at points B and C to that at point A is 
 
Hence, the correct option is (a). 
 
Q3: Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B 
as shown in figure. The energy stored in A is E. The energy stored in B, when spring 
constants k
A
 , k
B
 of A and B, respectively satisfy the relation 4 k
A
 = 3k
B
, is : 
Page 5


 
JEE Main Previous Year Questions (2021-2026): 
Work Power & Energy  
 
(January 2026) 
 
 
Q1: A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 
0 and at this instant the horizontal component of its velocity is v. Another bead Q of the 
same mass as P is ejected from point A at t = 0 along the horizontal string AB, with the 
speed v, friction between the beads and the respective strings may be neglected in both 
cases. Let t
P
 and t
Q
 be the respective times taken by beads P and Q to reach the point B, 
then the relation between t
P
 and t
Q
 is 
 
(a) t
P 
< t
Q 
(b) t
P
 = t
Q 
(c) t
P 
> 1.25 t
Q 
(d) t
P
 > t
Q
  
Ans: (a) 
Sol: 
For bead Q  
Bead Q moves along the horizontal straight string AB. The length AB is 2R where R is the 
radius of semicircular path.  
Since the string is frictionless and horizontal, there is no component of gravity acting along the 
direction of motion.  
Bead Q maintains a constant horizontal velocity v throughout the motion.  
 
For bead P Bead P moves along the semi-circular path ACB under the influence of gravity. And 
along horizontal the net force is zero, i.e., the acceleration along vertical will be non-zero and 
the acceleration along horizontal will be zero.  
At point S , the horizontal component of its velocity is given as v.  
As bead P moves from S toward the bottom-most point C, the vertical component of velocity 
changes but the horizontal component remains same. 
 
When bead P reaches point B then the horizontal displacement is S'O + OB  
S'O = R cos  45
°
  
Then the time taken to reach point B is 
 
So, the ratio of time taken by Q and P to reach B is, 
 
 
Q2: An object is projected with kinetic energy K from a point A at an angle 60
°
 with the 
horizontal. The ratio of the difference in kinetic energies at points B and C to that at point 
A (see figure), in the absence of air friction is : 
 
(a) 3 : 4 
(b) 1 : 4 
(c) 2 : 3 
(d) 1 : 2 
Ans: (a)  
Let the mass of the object be m and the initial speed be u.  
The angle of projection is ? = 60
°
.  
At point A  
The kinetic energy at point A is given as K.  
 
At Point B (Maximum Height)  
At the highest point of a projectile's path, the vertical component of velocity becomes zero (v
y
 = 
0) . The object only possesses the horizontal component of velocity (v
x
).  
The horizontal velocity remains constant throughout the motion (assuming no air resistance): 
 
So, the kinetic energy at B is, 
 
At Point C 
Point C is at the same horizontal level as point A . By the law of conservation of energy, the 
speed of the projectile when it hits the ground at the same level is equal to the speed of 
projection. 
v
C
 = u 
Therefore, the kinetic energy at C is equal to the kinetic energy at A : 
K
C
 = K
A
 = K 
The difference in kinetic energies at B and C : 
 
So, the ratio of the difference in kinetic energies at points B and C to that at point A is 
 
Hence, the correct option is (a). 
 
Q3: Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B 
as shown in figure. The energy stored in A is E. The energy stored in B, when spring 
constants k
A
 , k
B
 of A and B, respectively satisfy the relation 4 k
A
 = 3k
B
, is : 
 
(a) 4E 
(b) 2E 
(c) 4/3 E 
(d) 3E 
Ans: (d) 
Sol:  
When a mass m is suspended from a spring of constant k, the spring stretches.  
If the stretch in spring is x, kx = mg ? x = mg/k  
The potential energy (U) stored in the spring is given by : 
 
Mass of block A is m
A
 = 100 g. 
The spring constant is k
A
. 
So, the stored energy in spring A is E
A
 = E. 
Using the formula :