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IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

The acceleration of this particle for $$|x| > {X_0}$$ is

A
proportional to V0.
B
proportional to V0/mX0.
C
proportional to $$\sqrt {{V_0}/m{X_0}} $$.
D
zero.

Explanation

For $$|x| > {X_0}$$, $$V = {V_0}$$ = constant

Force $$ = - {{dV} \over {dx}} = 0$$

Hence, acceleration of the particle is zero for $$|x| > {X_0}$$.

2

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

For periodic motion of small amplitude A, the time period T of this particle is proportional to

A
$$A\sqrt {m/\alpha } $$
B
$${1 \over A}\sqrt {m/\alpha } $$
C
$$A\sqrt {\alpha /m} $$
D
$${1 \over A}\sqrt {\alpha /m} $$

Explanation

As $$V = a{x^4}$$

$$[\alpha ] = {{[V]} \over {[{x^4}]}} = {{[M{L^2}{T^{ - 2}}]} \over {[{L^4}]}} = [M{L^{ - 2}}{T^{ - 2}}]$$

By method of dimensions,

$$\left[ {{1 \over A}\sqrt {{m \over \alpha }} } \right] = {{{{[M]}^{1/2}}} \over {[L]{{[M{L^{ - 2}}{T^{ - 2}}]}^{1/2}}}} = [T]$$

Only option (b) has the dimensions of time.

3

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

If the total energy of the particle is E, it will perform periodic motion only if

A
E < 0
B
E > 0
C
V0 > E > 0
D
E > V0

Explanation

The kinetic energy of the particle cannot be negative. The total energy E is the sum of the kinetic energy K(x) and potential energy V(x) i.e.,

E = K(x) + V(x). ....... (1)

From the given figure, V(x) $$\ge$$ 0 for all x. If E $$\le$$ 0 for some x then kinetic energy K(x) = E $$-$$ V(x) $$\le$$ 0 for those x and hence the motion of the particle is not allowed for those x. On the other hand, if E $$\ge$$ V0 then K(x) = E $$-$$ V(x) $$\ge$$ 0 for all x and particle is allowed to move for all x, including infinity (in this case the particle will escape to infinity). Thus, for the particle to have a periodic motion, V0 > E > 0. In this case, particle is allowed at points where

K(x) = E $$-$$ V(x) = E $$-$$ $$\alpha$$x4 $$\ge$$ 0 i.e.,

| x | $$\le$$ (E/$$\alpha$$)1/4.

4

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)
A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y = Kt2, (K = 1 m/s2) where y is the vertical displacement. The time period now become T2. The ratio of $${{T_1^2} \over {T_2^2}}$$ is (g = 10 m/s2)
A
$${5 \over 6}$$
B
$${6 \over 5}$$
C
1
D
$${4 \over 5}$$

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