Start free

Physics · Work, Power & Energy

JEE Advanced 2018 — Paper 2 — Question 17

In the List-I below, four different paths of a particle are given as functions of time. In these functions, α\alpha and β\beta are positive constants of appropriate dimensions and α≠β\alpha \neq \beta. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned: p→\overrightarrow{\mathrm{p}} is the linear momentum, L→\overrightarrow{\mathrm{L}} is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path.

List-IList-II
P. r⃗(t)=αti^+βtj^\quad \vec{r}(t)=\alpha t \hat{i}+\beta t \hat{j}1. p→\overrightarrow{\mathrm{p}}
Q. r→(t)=αcos⁡ωti^+βsin⁡ωtj^\overrightarrow{\mathrm{r}}(\mathrm{t})=\alpha \cos \omega \mathrm{t} \hat{\mathrm{i}}+\beta \sin \omega \mathrm{t} \hat{\mathrm{j}}2. L→\overrightarrow{\mathrm{L}}
R. r→(t)=α(cos⁡ωti^+sin⁡ωtj^)\overrightarrow{\mathrm{r}}(\mathrm{t})=\alpha(\cos \omega \mathrm{t} \hat{\mathrm{i}}+\sin \omega \mathrm{t} \hat{\mathrm{j}})3. K
S. r→(t)=αti^+β2t2j^\quad \overrightarrow{\mathrm{r}}(\mathrm{t})=\alpha \mathrm{t} \hat{\mathrm{i}}+\frac{\beta}{2} \mathrm{t}^{2} \hat{\mathrm{j}}4. U
5. E
  1. Option A:

    P→1,2,3,4,5;Q→2,5;R→2,3,4,5;S→5\mathrm{P} \rightarrow 1,2,3,4,5 ; \quad \mathrm{Q} \rightarrow 2,5 ; \quad \mathrm{R} \rightarrow 2,3,4,5 ; \quad \mathrm{S} \rightarrow 5

    Correct
  2. Option B:

    P→1,2,3,4,5;Q→3,5;R→2,3,4,5;S→2,5\mathrm{P} \rightarrow 1,2,3,4,5 ; \mathrm{Q} \rightarrow 3,5 ; \quad \mathrm{R} \rightarrow 2,3,4,5 ; \mathrm{S} \rightarrow 2,5

  3. Option C:

    P→2,3,4;Q→5;R→1,2,4;S→2,5\mathrm{P} \rightarrow 2,3,4 ; \quad \mathrm{Q} \rightarrow \mathbf{5} ; \quad \mathrm{R} \rightarrow \mathbf{1}, \mathbf{2}, \mathbf{4} ; \quad \mathrm{S} \rightarrow \mathbf{2}, \mathbf{5}

  4. Option D:

    P→1,2,3,5;Q→2,5;R→2,3,4,5;S→2,5\mathrm{P} \rightarrow \mathbf{1}, \mathbf{2}, \mathbf{3}, 5 ; \quad \mathrm{Q} \rightarrow 2,5 ; \quad \mathrm{R} \rightarrow 2,3,4,5 ; \quad \mathrm{S} \rightarrow 2,5

Answer: A

Step-by-step solution

' P ' is straight line with zero acceleration.

' Q ' is elliptical path. ' RR ' is circular path

' S ' is parabolic path

For 'S' ∣v∣=α2+β2t2|v|=\sqrt{\alpha^{2}+\beta^{2} t^{2}}

dU=−F→⋅dr→\mathrm{dU}=-\overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}

U=−mβ2t22U=-\frac{m \beta^{2} t^{2}}{2}

Total energy =KE+U=12mα2==\mathrm{KE}+\mathrm{U}=\frac{1}{2} m \alpha^{2}= constant

Answer key and solution verified before publishing.

Practise Work, Power & Energy

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Advanced 2018
Paper
Paper 2
Subject
Physics
Chapter
Work, Power & Energy
Topic
Conservative Forces and Potential Energy

More Work, Power & Energy questions from this paper

In the List-I below, four different paths of a particle are given as… | JEE Advanced 2018 PYQ with Solution · DhiX AI