Home

/

Physics

/

Work, Energy and Power

Work, Energy and PowerNEET Physics ยท Class 11 ยท NCERT Chapter 5

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.

PYQ frequency ยท topic ร— year

15
16
17
18
19
21
23
24
25
Work, Energy and Power
2
1
1
1
2
1
1
2
3

Darker = more questions in our PYQ bank for that topic and year.

Want to time yourself?

Take a free 10-question chapter mock test on Work, Energy and Power โ€” no login needed for your first attempt.

Take timed test โ†’
All (14)
Work, Energy and Power (14)

A

2

B

3

C

4

D

1

Solution

FC=mv1 2 r =2mv2 2 (r 2) =4mv2 2 r So V1=2V2

A

14 m sโปยน

B

20 m sโปยน

C

28 m sโปยน

D

10 m sโปยน

Solution

KE_f / KE_i = 1/2 โ‡’ V_f / V_i = 1/โˆš2. From energy conservation: โˆš(2gh) / โˆš(Vโ‚€ยฒ + 2gh) = 1/โˆš2 โ‡’ Vโ‚€ = 20 m/s.

A

โˆšgR

B

โˆš2gR

C

โˆš3gR

D

โˆš5gR

Solution

Minimum velocity required is v =โˆš5gR 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 = 48โˆ’ W eV0 = hc ฮป โˆ’ W eV= hc ฮป โˆ’ W โ€ฆ(i) e V 4 = hc 2ฮป โˆ’ W โ€ฆ.(ii) From (i) and (ii) hc ฮป โˆ’ W = 4(hc 2ฮปโˆ’ W) hc ฮป โˆ’ W =2hc ฮป โˆ’ 4W 3W = hc ฮป โ‡’ W =hc 3ฮป hc ฮปmax = hc 3ฮป โ‡’ ฮปmax = threshold wavelength 3ฮป www.vedantu.com 20

A

(i) โ€“ 10 J (ii) โ€“8.25 J

B

(i) 1.25 J (ii) โ€“8.25 J

C

(i) 100 J (ii) 8.75 J

D

(i) 10 J (ii) โ€“8.75 J

Solution

wg + wa = Kf โ€“ Ki mgh + wa = 21 02 mv - 10โ€“3 ร— 10 ร— 10 3 + wa = 321 10 (50)2 -ร—ร— wa = โ€“8.75 J i.e. work done due to air resistance and work done due to gravity = 10 J www.vedantu.com 9

A

5 7 D

B

D

C

2 3 D

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.

A

5 9

B

1 9

C

8 9

D

4 9

Solution

๐‘‰1=( ๐‘š1โˆ’๐‘š2 ๐‘š1+๐‘š2 )๐‘ข1+( 2๐‘š2 ๐‘š1+๐‘š2 )๐‘ข2 ๐‘ฃ2=( 2๐‘š1 ๐‘š1+๐‘š2 )๐‘ข1+(๐‘š2โˆ’๐‘š1 ๐‘š1+๐‘š2 )๐‘ข2 Here, v2 = 0 โˆด ๐‘ฃ1=( ๐‘š1โˆ’๐‘š2 ๐‘š1+๐‘š2 )๐‘ฃ1;๐‘ฃ2=( 2๐‘š1 ๐‘š1+๐‘š2 )๐‘ฃ1 ๐‘ฃ1=(4๐‘šโˆ’2๐‘š 6๐‘š )๐‘ข ๐‘ฃ1=2๐‘š 6๐‘š๐‘ข=1 3 A 4m u B 2m Rest 4m v1 v2 2m Before collision After collision 3 SPACE FOR ROUGH WORK โˆด ๐พ๐‘– = 1 24๐‘š๐‘ข2 and ๐พ๐‘“ = 1 24๐‘š ๐‘ข2 9 ๐พ๐‘– =2๐‘š๐‘ข2 ๐พ๐‘“ = 2 9๐‘š๐‘ข2 Loss = 2๐‘š๐‘ข2โˆ’ 2 9๐‘š๐‘ข2=2๐‘š๐‘ข2ร— 8 9 โˆด fraction = 2๐‘š๐‘ข2ร—8 9 2๐‘š๐‘ข2 = 8 9

A

๐‘ฆ(๐‘ก)= 3cos( ฯ€๐‘ก 2), where y in m

B

๐‘ฆ (๐‘ก)= โˆ’3cos2ฯ€๐‘ก, where y in m

C

๐‘ฆ (๐‘ก)=4sin(ฯ€๐‘ก 2), where y in m

D

๐‘ฆ (๐‘ก)=3cos( 3ฯ€๐‘ก 2), where y in m

Solution

Conceptual.

A

10.2 kW

B

8.1 kW

C

12.3 kW

D

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.

A

16U

B

2U

C

4U

D

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.

A

10

B

5

C

7

D

6

Solution

x = 2t โ€“ 1 12 m s dxv dt โˆ’== P = F. v = 2 ร— 5 = 10 W - 4 - NEET (UG)-2024 (Code-Q1)

A

1 : 2

B

2 : 1

C

4 : 1

D

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)

A

(1)

B

(2)

C

(3)

D

(4)

Solution

By work-energy theorem,

A

B

C

D

Solution

Sol.

At Point P, ... (1)

By conservation of mechanical energy at point

Put using (1)

A

21 NS

B

7 NS

C

0

D

84 NS

Solution

Sol.

and

Impulse

Track Your NEET Score Across All 90 Chapters

Free 14-day trial. AI tutor, full mock tests and chapter analytics โ€” built for NEET 2027.

Free 14-day trial ยท No credit card required