In the List-I below, four different paths of a particle are given as functions of time. In these functions, α and β are positive constants of appropriate dimensions and α ≠β . 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→ is the linear momentum, 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 -I LIST-II A) r→t=αti^+βj^ P) p→ B) r→t=αcos⁡ωti^+βsin⁡ωtj^ Q) L→ C) r→t=αcos⁡ωti^+sin⁡ωtj^ R) K D) r→t=αti^+β2t2j^ S) U T) E

(A) a-r,s;b-s,t;c-q,r,s,t;d-q,r,s,t;

JEE Advanced · Physics · 6. Work Power Energy

In the List-I below, four different paths of a particle are given as functions of time. In these functions, $α a n d β$ are positive constants of appropriate dimensions and $α \neq β$ . 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: $\vec{p}$ is the linear momentum, $\vec{L}$ 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 -I LIST-II

A) $\vec{r} (t) = α t \hat{i} + β \hat{j}$ P) $\vec{p}$

B) $\vec{r} (t) = α \cos ω t \hat{i} + β \sin ω t \hat{j}$ Q) $\vec{L}$

C) $\vec{r} (t) = α \cos ω t \hat{i} + \sin ω t \hat{j}$ R) $K$

D) $\vec{r} (t) = α t \hat{i} + \frac{β}{2} t^{2} \hat{j}$ S) $U$

T) $E$

A a-r,s;b-s,t;c-q,r,s,t;d-q,r,s,t;
B a-p,q,r,s,t;b-q,t;c-q,r,s,t;d-t;
C a-t;b-r,s,t;c-t;d-q;
D a-r;b-r,s;c-q,r,s,t;d-t;

Verified Solution

Answer & Solution

Correct Answer

(B) a-p,q,r,s,t;b-q,t;c-q,r,s,t;d-t;

Step-by-step Solution

Detailed explanation

(i)

\vec{r} (t) = α t \hat{i} + β t \hat{j}

\vec{v} = \frac{d \vec{r} (t)}{d t} = α \hat{i} + β \hat{j}

{constant}

\vec{a} = \frac{\vec{d v}}{d t} = 0

\vec{P} = m \vec{v}

(remain constant)

k = \frac{1}{2} m v^{2}

{remain constant}

\vec{F} = - [\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{i}] = 0

\Rightarrow U \to

constant

E = K + U

\frac{d \vec{L}}{d t} = - \vec{τ} = \vec{r} \times \vec{F} = 0

\vec{L} =

constant

(i i)

\vec{r} = α \cos (ω t) \hat{i} + β \sin (ω t) \hat{j}

\vec{v} = \frac{d \vec{r}}{d t} = - α ω \sin (ω t) \hat{i} + β ω \cos (ω t) \hat{j}

\vec{a} = \frac{d \vec{v}}{d t} = - α ω^{2} \cos (ω t) \hat{i} + β ω \sin (ω t) \hat{j}

= - ω^{2} [α \cos (ω t) \hat{i} + β \sin (ω t) \hat{j}]

\vec{a} = = - α ω^{2}

\vec{τ} = \vec{r} \times \vec{F} = 0 \{\vec{r} a n d \vec{F} a r e p a r a l l e l\}

∆ U = - \int F . d r + \int_{0}^{r} m ω^{2} . r . d r

∆ U = m ω^{2} [\frac{r^{2}}{2}]

U \propto r^{2}

r = \sqrt{α^{2} \cos^{2} (ω t) + β^{2} \sin^{2} (ω t)}

R is a function of time (t)
U depends on r hence it will change with time
Total energy remain constant because force is central

(i i i)

\vec{r} (t) = α (\cos ω t \hat{i} + \sin (ω t) \hat{j})

\vec{v} (t) = \frac{d \vec{r} (t)}{d t} = α [- ω \sin (ω t) \hat{i} + ω \cos (ω t) \hat{j}]

|\vec{v}| = α ω

(Speed remains constant)

\vec{a} (t) = \frac{d \vec{v} (t)}{d t} = α [- ω^{2} \cos (ω t) \hat{i} + ω^{2} \sin (ω t) \hat{j}]

= - α ω^{2} [\cos (ω t) \hat{i} + \sin (ω t) \hat{j}]

\vec{a} (t) = - ω^{2} (\vec{r})

|\vec{r}| = α

(remain constant)
Force is central in nature and distance from fixed point is constant.
Potential energy remains constant
Kinetic energy is also constant (speed is constant)

(i v)

\vec{r} = α t \hat{i} + \frac{β}{2} t^{2} \hat{j}

\vec{v} = \frac{d \vec{r}}{d t} = α t \hat{i} + β t \hat{j}

(speed of particle depends on 't')

\vec{a} = \frac{d \vec{v}}{d t} = β \hat{j} \{c o n s t a n t\}

\vec{F} = m \vec{a} \{c o n s t a n t\}

∆ U = - \int \vec{F} . d \vec{r} = - m \int_{0}^{t} β \hat{j} . (α \hat{i} + β t \hat{j}) d t

U = \frac{- m β^{2} t^{2}}{2}

k = \frac{1}{2} m v^{2} = \frac{1}{2} m (α^{2} + β^{2} t^{2})

E = k + U = \frac{1}{2} m α^{2}

[remain constant]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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	LIST -I		LIST-II
A)	$\vec{r} (t) = α t \hat{i} + β \hat{j}$	P)	$\vec{p}$
B)	$\vec{r} (t) = α \cos ω t \hat{i} + β \sin ω t \hat{j}$	Q)	$\vec{L}$
C)	$\vec{r} (t) = α \cos ω t \hat{i} + \sin ω t \hat{j}$	R)	$K$
D)	$\vec{r} (t) = α t \hat{i} + \frac{β}{2} t^{2} \hat{j}$	S)	$U$
		T)	$E$