IIT-JEE 1999 Screening | JEE Advanced Year Wise Previous Years Questions

Physics

The coordinates of a particle moving in a plane are given by x(t) = a cos(pt) and y(t) = b sin(pt) where a, b (< a) and p are positive constants of approprite dimensions. Then

A spring of force-constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of

A particle free to move along the x-axis has potential energy given by $$U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]$$ for $$ - \infty \le x \le - \infty $$, where k is a positive constant of appropriate dimensions. Then

IIT-JEE 1999 Screening

MCQ (More than One Correct Answer)

-0.75

The coordinates of a particle moving in a plane are given by x(t) = a cos(pt) and y(t) = b sin(pt) where a, b (< a) and p are positive constants of approprite dimensions. Then

the path of the particle is an ellipse

the velocity and acceleration of the particle are normal to each other at t = $${\pi \over {2p}}$$

the acceleration of the particle is always directed towards a focus

the distance traveled by the particle in time interval t = 0 to t = $${\pi \over {2p}}$$ is a

IIT-JEE 1999 Screening

MCQ (Single Correct Answer)

-0.5

A spring of force-constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of

$$\left( {{2 \over 3}} \right)k$$

$$\left( {{3 \over 2}} \right)k$$

3 k

6 k

IIT-JEE 1999 Screening

MCQ (Single Correct Answer)

-0.5

at points away from the origin, the particle is in unstable equilibrium

for any finite nonzero value of x, there is a force directed away from the origin

if its total mechanical energy is $${k \over 2}$$, it has its minimum kinetic energy at the origin

for small displacement from x = 0, the motion is simple harmonic