Acceleration of a particle which is at
rest at x = 0 is a^→ = (4 − 2x) i^∧ . Select
the correct alternative(s).
(a) Maximum speed of the particle is
4 units
(b) Particle further comes to rest at
x = 4
(c) Particle oscillates about x = 2
(d) Particle will continuously accelerate
along the x-axis.
Two particles A and B start from the
same position along the circular path of
radius 0.5 m with a speed v_A = 1 ms^(−1)
and v_B = 1.2 ms^(−1) in opposite direction.
Determine the time before they collide.
If in a ΔABC, tan(A/2), tan(B/2) and tan(C/2)
are in HP, then the minimum value of
cot(B/2) is greater than
(1) 2(√3)
(2) ((√3)/2)
(3) (√3)
(4) (1/(√3))
A small bead is slipped on a horizontal
rod of length l. The rod starts moving
with a horizontal acceleration a in a
direction making an angle α with the
length of the rod. Assuming that
initially the bead is in the middle of
the rod, find the time elapsed before
the bead leaves the rod. Coefficient of
friction between the bead and the rod
is μ. (Neglect gravity).
Consider a disc rotating in the horizontal
plane with a constant angular speed ω
about its centre O. The disc has a
shaded region on one side of the diameter
and an unshaded region on the other
side as shown in the Figure. When the
disc is in the orientation as shown, two
pebbles P and Q are simultaneously
projected at an angle towards R. The
velocity of projection is in the y-z
plane and is same for both pebbles with
respect to the disc. Assume that (i) they
land back on the disc before the disc has
completed (1/8) rotation, (ii) their range
is less than half the disc radius, and
(iii) ω remains constant throughout.
Then
(a) P lands in the shaded region and Q
in the unshaded region
(b) P lands in the unshaded region and
Q in the shaded region
(c) Both P and Q land in the unshaded
region
(d) Both P and Q land in the shaded
region
A sphere is rolling without slipping on
a fixed horizontal plane surface. In the
Figure, A is a point of contact, B is the
centre of the sphere and C is its topmost
point. Then
(a) v_C ^→ − v_A ^→ = 2(v_B ^→ − v_C ^→ )
(b) v_C ^→ − v_B ^→ = v_B ^→ − v_A ^→
(c) ∣v_C ^→ − v_A ^→ ∣ = 2∣v_B ^→ − v_C ^→ ∣
(d) ∣v_C ^→ − v_A ^→ ∣ = 4∣v_B ^→ ∣