A machine gun fires a bullet of mass 40 g with a velocity 1200 ms^{−1}. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?
one
Four
Two
Three
D.
Three
The force exerted by machine gun on man's hand in firing a bullet = change in momentum per second on a bullet or rate of change of momentum
The force exerted by man on machine gun = 144 N Hence, number of bullets fired =144/48 = 3
A projectile can have the same range R for two angles of projection. If T_{1} and T_{2} be the time of flights in the two cases, then the product of the two time of flights is directly proportional to
1/R^{2}
1/R
R
R^{2}
C.
R
We know in advance that range of projectile is same for complementary angles i.e. for θ and (900 - θ )
Two masses m_{1} = 5 kg and m_{2} = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift free to move?
(g = 9.8 m/s^{2} )
0.2 m/s^{2}
9.8 m/s^{2}
5 m/s^{2}
4.8 m/s^{2}
A.
0.2 m/s^{2}
On release, the motion of the system will be according to the figure.
m_{1}g - T = m_{1}a ...... (i) and
T - m_{2}g = m_{2}a ...... (ii)
On solving;
Here, m_{1} = 5 kg, m_{2} = 4.8 kg, g = 9.8 m/s^{2}
A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?
h/9 metres from the ground
7h/9 metres from the ground
8h/9 metres from the ground
17h/18 metres from the ground.
C.
8h/9 metres from the ground
Which of the following statements is false for a particle moving in a circle with a constant angular speed?
The velocity vector is tangent to the circle.
The acceleration vector is tangent to the circle.
The acceleration vector points to the centre of the circle.
The velocity and acceleration vectors are perpendicular to each other.
B.
The acceleration vector is tangent to the circle.
Which one of the following represents the correct dimensions of the coefficient of viscosity?
ML^{−1} T^{−2}
MLT^{−1}
ML^{−1} T^{−1}
ML^{−2} T^{−2}
C.
ML^{−1} T^{−1}
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to
x^{2}
e^{x}
x
log_{e}x
A.
x^{2}
In this problem acceleration (a) is given in terms of displacement (x) to determine the velocity with respect to position or displacement we have to apply integration method.
From given information a =-kx, where a is acceleration, x is displacement and k is proportionality constant.
Let for any displacement from 0 to x , the velocity changes from v_{o} to v
An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be
20 m
40 m
60 m
80 m
D.
80 m
Third equation of motion gives
v^{2} = u^{2} + 2as ⇒ 2
s ∝ u (since v = 0)
where a = retardation of body in both the cases
therefore, .... (i)
Here, s_{1} = 20 m, u1 = 60 km/h, u_{2} = 120 km/h. Putting the given values in eq. (i), we get
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?
7.2 J
3.6 J
120 J
1200 J
B.
3.6 J
Mass per length
= M/L
= 4/2 = 2 kg/m
The mass of 0.6 m of chain = 0.6 x 2 = 1.2 kg
The centre of mass of hanging part = 0.6 +0 /2 = 0.3 m
Hence, work done in pulling the chain on the table
W =mgh
= 1.2 x 10 x 0.3
= 3.6 J
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