1
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+2
-0.5
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
$${4 \over 5}$$
2
IIT-JEE 2001 Screening
MCQ (Single Correct Answer)
+2
-0.5
A particle executes simple harmonic motion between x = - A to x = + A. The time taken for it to go from 0 to $${A \over 2}$$ is T1 and to go from $${A \over 2}$$ to A is T2. Then
A
T1 < T2
B
T1 > T2
C
T1 = T2
D
T1 = 2T2
3
IIT-JEE 2000 Screening
MCQ (Single Correct Answer)
+2
-0.5
The period of oscillation of a simple pendulum of length $$L$$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $$\alpha$$, is given by
A
$$2\pi \sqrt {{L \over {g\cos \alpha }}} $$
B
$$2\pi \sqrt {{L \over {g\sin \alpha }}} $$
C
$$2\pi \sqrt {{L \over g}} $$
D
$$2\pi \sqrt {{L \over {g\tan \alpha }}} $$
4
IIT-JEE 1999 Screening
MCQ (Single Correct Answer)
+2
-0.5
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
A
at points away from the origin, the particle is in unstable equilibrium
B
for any finite nonzero value of x, there is a force directed away from the origin
C
if its total mechanical energy is $${k \over 2}$$, it has its minimum kinetic energy at the origin
D
for small displacement from x = 0, the motion is simple harmonic
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