14 NEET previous-year questions on Work, Energy and Power, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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2
3
4
1
Solution
FC=mv1 2 r =2mv2 2 (r 2) =4mv2 2 r So V1=2V2
14 m sโปยน
20 m sโปยน
28 m sโปยน
10 m sโปยน
Solution
KE_f / KE_i = 1/2 โ V_f / V_i = 1/โ2. From energy conservation: โ(2gh) / โ(Vโยฒ + 2gh) = 1/โ2 โ Vโ = 20 m/s.
โgR
โ2gR
โ3gR
โ5gR
Solution
Minimum velocity required is v =โ5gR 29. When a metallic surface is illuminated with radiation of wavelength ฮป, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2ฮป, the stopping potential is V 4. The threshold wavelength for the metallic surface is: (1) 4ฮป (2) 5ฮป (3) 5 2ฮป (4) 3ฮป Solution: (4) In photo electric effects eV0 = 48โ W eV0 = hc ฮป โ W eV= hc ฮป โ W โฆ(i) e V 4 = hc 2ฮป โ W โฆ.(ii) From (i) and (ii) hc ฮป โ W = 4(hc 2ฮปโ W) hc ฮป โ W =2hc ฮป โ 4W 3W = hc ฮป โ W =hc 3ฮป hc ฮปmax = hc 3ฮป โ ฮปmax = threshold wavelength 3ฮป www.vedantu.com 20
(i) โ 10 J (ii) โ8.25 J
(i) 1.25 J (ii) โ8.25 J
(i) 100 J (ii) 8.75 J
(i) 10 J (ii) โ8.75 J
Solution
wg + wa = Kf โ Ki mgh + wa = 21 02 mv - 10โ3 ร 10 ร 10 3 + wa = 321 10 (50)2 -รร wa = โ8.75 J i.e. work done due to air resistance and work done due to gravity = 10 J www.vedantu.com 9
5 7 D
D
2 3 D
4 5 D
Solution
For a body to complete a vertical circle, the minimum kinetic energy at the bottom must be . Using conservation of energy, . However, the height must also account for the radius of the circle, so . But the correct interpretation for the problem is , which simplifies to when considering the total height from the ground, so option (a) is correct.
5 9
1 9
8 9
4 9
Solution
๐1=( ๐1โ๐2 ๐1+๐2 )๐ข1+( 2๐2 ๐1+๐2 )๐ข2 ๐ฃ2=( 2๐1 ๐1+๐2 )๐ข1+(๐2โ๐1 ๐1+๐2 )๐ข2 Here, v2 = 0 โด ๐ฃ1=( ๐1โ๐2 ๐1+๐2 )๐ฃ1;๐ฃ2=( 2๐1 ๐1+๐2 )๐ฃ1 ๐ฃ1=(4๐โ2๐ 6๐ )๐ข ๐ฃ1=2๐ 6๐๐ข=1 3 A 4m u B 2m Rest 4m v1 v2 2m Before collision After collision 3 SPACE FOR ROUGH WORK โด ๐พ๐ = 1 24๐๐ข2 and ๐พ๐ = 1 24๐ ๐ข2 9 ๐พ๐ =2๐๐ข2 ๐พ๐ = 2 9๐๐ข2 Loss = 2๐๐ข2โ 2 9๐๐ข2=2๐๐ข2ร 8 9 โด fraction = 2๐๐ข2ร8 9 2๐๐ข2 = 8 9
๐ฆ(๐ก)= 3cos( ฯ๐ก 2), where y in m
๐ฆ (๐ก)= โ3cos2ฯ๐ก, where y in m
๐ฆ (๐ก)=4sin(ฯ๐ก 2), where y in m
๐ฆ (๐ก)=3cos( 3ฯ๐ก 2), where y in m
Solution
Conceptual.
10.2 kW
8.1 kW
12.3 kW
7.0 kW
Solution
The power input to the turbine is given by , where , , and . Substituting the values, . Since 10% of the energy is lost to friction, the power generated by the turbine is . Thus, option (b) is correct.
16U
2U
4U
8U
Solution
The potential energy stored in a spring is given by . If the spring is stretched by 8 cm, the new potential energy . Therefore, option (a) is correct.
10
5
7
6
Solution
x = 2t โ 1 12 m s dxv dt โ== P = F. v = 2 ร 5 = 10 W - 4 - NEET (UG)-2024 (Code-Q1)
1 : 2
2 : 1
4 : 1
1 : 4
Solution
Before collision โ It undergoes completely inelastic collision Using conservation of linear momentum Initial momentum = Final momentum โ mv1 = mv2 + mv2 โ mv1 = 2mv2 โ 1 2 2 1 v v = - 5 - NEET (UG)-2024 (Code-Q1)
(1)
(2)
(3)
(4)
Solution
By work-energy theorem,
Solution
Sol.
At Point P, ... (1)
By conservation of mechanical energy at point
Put using (1)
21 NS
7 NS
0
84 NS
Solution
Sol.
and
Impulse
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