On top of a spiral spring of the force constant 500Nm^{-1} is placed a mass of 5 x 10^{-3} kg. If the spring is compressed downwards by a length of 0.02m and then released, calculate the height to which the mass is projected
A
8 m
B
4 m
C
2 m
D
1 m
CORRECT OPTION:
c
K = 500Nm^{-1}; M = 5 x 10^{-3}kg; e = 0.02m;
g = 10ms^{-2}. The elastic potential energy stored in the compressed spring is given by = ^{1}/_{2}Ke^{2}, and the energy is transformed into gravitational potential energy = mgh, thus from the principle of energy conversion, we have that mgh = ^{1}/_{2}Ke^{2}
5 x 10^{-3} x 10 x h = ^{1}/_{2} x 500 x (0.02)^{2}
∴ h = 1 x 500 x 0.02 x 0.022 x 5 x 10^{-3} x 10
= 5 x 2 x 0.02___10^{-1}
= 10 x 0.02 x 10 = 2
∴ h = 2m
2
A test tube of radius 1.0cm is loaded to 8.8g. If is placed upright in water, find the depth to which it would sink
A
2.8cm
B
5.2cm
C
25.5cm
D
28.0cm
CORRECT OPTION:
a
Pressure = ^{F}/_{A} = pgh, where p = density, h = depth and g = 10ms^{-2}
But F = P x A
∴ Force exerted by the load test tube = cross-sectional Area x Pressure
i.e F = A x pgh
∴ h =
F_
A x pg
=
mg
A x P x g
=
8.8 x 10^{-3} x 10
πr^{2} x p x g
=
_8.8 x 10^{-3} x 10
π x (1.0^{-2})^{2} x 1000 x 10
= 0.028m or 2.8m
3
The length of piece of glass block was measured by means of a vernier calliper as shown above. The length of the glass block is
A
1.63 cm
B
1.64 cm
C
1.65 cm
D
1. 66 cm
CORRECT OPTION:
b
4
A hose of cross-sectional area 0.5m^{2} is used to discharge water from a water tanker at a velocity of 60ms^{-1} in 20s into a container. If the container is filled completely, the volume of the container is
A
240m^{3}
B
600m ^{3}
C
2400m ^{3}
D
6000m^{3}
CORRECT OPTION:
b
Vol = cross-sectional area x length
∴ vol. of water disharged per sec
= 60ms^{-1} x 0.5m
= 30m^{3} 5^{-1}
=> Vol. of water discharged in the said 20s
= 30m^{3} x 20 = 600m^{3}
∴ vol. of the container = 600m^{3}
5
If an object just begins to slide on a surface inclined at 30^{o} to the horizontal,the coefficient of friction is
A
√3
B
^{√3}/_{2}
C
^{1}/_{√2}
D
^{1}/_{√3}
CORRECT OPTION:
d
A body sliding down a plane inclined at 30^{o} to the horizontal, has coefficient of friction
= tan 30^{o} = ^{1}/_{√3}
6
Water does not drop through an open umbrella of silk materials unless the inside of the umbrella is touched. This is due to
A
capilarity
B
osmotic pressure
C
viscosity
D
surface tension
CORRECT OPTION:
d
Surface tension - The tendency of a liquid surface to behave as an elastic skin
7
The stylus of a phonograph record exerts a force of 7.7 x 10^{-2}N on a groove of radius 10^{-5}m. Compute the pressure exerted by the stylus on the groove
A
2.45 x 10^{8} Nm^{-2}
B
3.45 x 10^{8} Nm^{-2}
C
4.90 x 10^{8} Nm^{-2}
D
2.42 x 10^{8} Nm^{-2}
CORRECT OPTION:
a
Pressure P = ^{F}/_{A}
= ^{F}/_{πr2}
=
7.7 x 10^{-2}
π x (10^{-5})^{2}
=
7.7 x 10^{-2}
π x 10^{-10}
= 2.45 x 10^{8} Nm^{-2}
8
A piece of stone attached to one end of a string is whirled round in a horizontal circle and the string suddenly cuts. The stone will fly off in a direction
A
perpendicular to the circular path
B
parallel to the circular path
C
tangential to the circular path
D
towards the center of the circle
CORRECT OPTION:
c
AS the stone is whirled round and te string suddenly cuts, say at A, B, C or D. The stone will fly off in the direction indicated V_{A}, V_{B}, V_{C} or V_{D} respectively which is tangential to the circular path
9
A 90cm uniform lever has a load of 30N suspended at 15cm from one of its ends. If the fulcrum is at the center of gravity, the force that must be applied at its other end to keep it in horizontal equilibrium is
A
15N
B
20N
C
30N
D
60N
CORRECT OPTION:
b
Since the lever is a uniform one, its center of gravity will be acting half way (i.e 45cm)
∴from the principle of moments,
F x 45 = 30 x 30
∴ F = (30 x 30) / 45 = 20N
10
A satellite is in a parking orbit if its period is
A
less than the period of the earth
B
more than the period of the earth
C
equal to the period of the earth
D
the square of the period of the earth
CORRECT OPTION:
c
A satellite is said to be in a parking orbit if its period is exactly equal to the period of the earth as it turns its own orbit;. which is about 24hrs
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