- A uniform rod of length l and mass m is swinging
freely about a horizontal axis passing through its
end. Its maximum angular speed is Z. Its centre of
mass rises to a maximum height of
(a)1
3
l^22
gω
(b)1
6
l
gω(c)1
2
l^22
gω
(d)1
6
l^22
gω- The ratio of the radii of gyration of a circular disc and
a circular ring of the same radii about a tangential
axis is
(a) 12: (b) 56: (c) 23: (d) 21: - A solid sphere of uniform density and radius R
applies a gravitational force of attraction F 1 on a
particle placed at a distance 2R from the centre of
the sphere. A spherical
cavity of radius (R/2) is
now made in the sphere
as shown in gure. The
sphere with cavity now
applies a gravitational
force F 2 on the same
particle. The ratio (F 2 /
F 1 ) is
(a) 1/2 (b) 3/4 (c) 7/8 (d) 7/9 - A capillary tube is attached horizontally to a
constant pressure head arrangement. If the radius
of the tube is increased by 10%, the rate of ow of
liquid will change nearly by
(a) +10% (b) –10% (c) +46% (d) +40% - Two rigid boxes containing dierent ideal gases
are placed on a table. Box A contains one mole of
nitrogen at temperature T 0 while box B contains
one mole of helium at temperature (7/3)T 0. e
boxes are then put into thermal contact with each
other and heat ows between them until the gases
reach a common nal temperature. (Ignore the heat
capacity of boxes). Then the final temperature of
the gases, Tf, in terms of T 0 , is
(a) TTf 5
2 0(b) TTf3
7 0
(c) TTf 7
3 0(d) TTf 3
2 0- Two vibrating tuning forks producing progressive
waves given by y 1 = 4 sin (500pt) and y 2 = 2sin (506pt)
are held near the ear of a person. The person will
hear
OR/2R R(a) 3 beats/s with intensity ratio between maxima
and minima equal to 2
(b) 3 beats/s with intensity ratio between maxima
and minima equal to 9
(c) 6 beats/s with intensity ratio between maxima
and minima equal to 2
(d) 6 beats/s with intensity ratio between maxima
and minima equal to 9- If 2 moles of an ideal monoatomic gas at
temperature T 0 is mixed with 4 moles of another
ideal monoatomic gas at temperature 2T 0 , then the
temperature of the mixture is
(a) (5/3)T 0 (b) (3/2)T 0 (c) (4/3)T 0 (d) (5/4)T 0
SOLUTIONS- (d) : Maximum acceleration = slope of BC
=40
10
cms^1
s−
= 4 cm s–2- (b) : Height through which a body falls in t seconds,
(^) hg 1 t
2
(^2) ...(i)
Height through which the body falls in
(t – 2) seconds,
hg t
1
2
() (^22) ...(ii)
From eq (i) and (ii), we get
Thus, hh−′=−gt gt−=
1
2
1
2
(^22) () 240 m
or 2 g(t – 1) = 40 m
hence, t = 3 s (on putting g = 10 m s–2)
Clearly, h =
1
2
× (10) × (3)^2 m = 45 m- (d) : At the highest point, vx 0 v^0
2
and vy 0 = 0Clearly, v 0 v^0
2cos or q = 60°R v
gv
gv
g ^0 vg2
02
02
0sins 2 in 120 (/ 32 ) () 322 /- (a) : When the balloon descends with acceleration a,
mg – F = ma ...(i)
When the balloon ascends with acceleration a (with
mass mc removed),
F – (m – mc)g = (m – mc)a ...(ii)
From eqns. (i) and (ii), we get
(mg – ma) – mg + mcg = ma – mca
or mc(g + a) = 2ma
or mma
ga