A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 x 10^{-4} J by the end of the second revolution after the beginning of the motion?
0.15 m/s^{2}
0.18 m/s^{2}
0.2 m/s^{2}
0.1 m/s^{2}
D.
0.1 m/s^{2}
Given, mass of particle. m = 0.01 kg
Radius of circle along which particle is moving , r = 6.4 cm
Kinetic energy of particle, K.E. = 8 x 10^{-4} J
Given that, KE of particle is equal to 8 x 10^{-4} J by the end of second revolution after the beginning of the motion of particle.
It means, initial velocity (u) is 0 m/s at this moment.
Now, using the Newton's 3rd equation of motion,
v^{2} = u^{2} + 2as
A particle has initial velocity (2i+3j) and acceleration (0.31i +0.2j). The magnitude of velocity after 10 s will be
5 units
9 Units
B.
unitsFrom the equation of motion, we knows,
v=u+ at
v= (2i+3j)
a=(0.3i +0.2j)
v=(2i+3j) +(0.3i +0.2j) x 10
v=5i+5j
A particle of unit mass undergoes one-dimensional motion such that its velocity according to
V(x) = βx^{-2n} where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by
-2nβ^{2} x^{-2n-1}
-2nβ^{2 }x^{-4n-1}
-2β x^{-2n+1}^{}
-2nβ^{2}e^{-4n+1}
B.
-2nβ^{2 }x^{-4n-1}
Given, v = βx^{-2n}
If the velocity of a particle is v = At + Bt^{2}, where A and B are constants, then the distance travelled by it between 1 s and 2 s is,
3A + 7B
B.
Velocity of the particles is given as,
v = At + Bt^{2, }where a and B are constants.
Integrating both sides, we get
Therefore, distance travelled between 1 s and 2 s is,
The velocity of a projectile at the initial point A is (2 i + 3 j) m/s. Its velocity (in m/s) at point B is,
-2 i - 3 j
-2 i + 3 j
2 i - 3 j
2 i + 3 j
C.
2 i - 3 j
From the fig,, the X component remain unchanged, while the Y-component is reverse. Then, the velocity at point B is (2 i - 3 j) m/s.
A balloon with mass m is descending down with an acceleration a (where a <g). How much mass should be removed from it so that it starts moving up with an acceleration a?
A.
When the balloon is descending down with acceleration a,
So, mg - B = mx A ... (i)
where B is the buoyant force.
We assume here that while removing same mass, the volume of balloon and hence buoyant force will not change.
Let, us assume the new mass of the balloon is m'.
So, mass removed is (m-m')
Therefore,
B-m'g = m'x a ... (ii)
On solving equations (i) and (ii), we have
mg - B = m x a
B - m'g = m' x a
mg - m'g = ma + m'a
(mg - ma) = m' (g+a) = m (g-a) = m' (g+a)
That is,
That is, mass removed is m-m'
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
C.
The question is illustrated in the figure below,
A particle moves so that its position vector is given by r = cos where is a cosntant. which of the following is true?
velocity and acceleration both are parallel to r
velocity is perpendicular to r and acceleration is directed towards the origin
velocity is perpendicular to r and acceleration is directed away from the origin
velocity and acceleration both are perpendicular to r
B.
velocity is perpendicular to r and acceleration is directed towards the origin
Position vector of the article is given by,
where is a constant.
Velocity of the particle is,
The particle is at a point P. That is, its position vector is directed as shown below:
A body of mass 1 kg begins to move under the action of a time dependent force F = N, where are units vectors along X and Y axis. What power will be developed by the force at the time (t)?
(2t^{2} + 4t^{4}) W
(2t^{3} + 3 t^{4}) W
(2t^{3} + 3t^{5}) W
(2t + 3t^{3})W
C.
(2t^{3} + 3t^{5}) W
A body of mass 1 kg begins to move under the action of time dependent force,
F = (2t +3t^{2} ) N,
where are unit vectors along X and Y axis.
F = ma
Acceleration, a =
Integrating both sides, we get
Power developed by the force at the time t will be given as,
P = F.v = (2t + 3t^{2} ).()
= (2t. t^{2} + 3t^{2}.t^{3})
P = (2t^{3} + 3t^{5}) W
A projectile is fired from the surface of the earth with a velocity of 5 m/s and angle with the horizontal. Another projectile fired from another planet with a velocity of 3 m/s at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is,
3.5
5.9
16.3
110.8
A.
3.5
If the trajectory is same for both the projectiles, their maximum height will be the same.
That is,
(H_{max})_{1} = (H_{max})_{2}
i.e.,