A particle moves in a straight line and its position is defined by the equation x = 6 t² - t3 where t is expressed in seconds and x in meters. The maximum velocity during the motion is
6 m/sec
12 m/sec
24 m/sec
48 m/sec
Correct Answer :
B. 12 m/sec
Related Questions
The C.G. of a thin hollow cone of height h, above its base lies on the axis, at a height of
h/3
h/4
2h/3
3h/4
The velocity of a moving body, is
A vector quantity
A scalar quantity
A constant quantity
None of these
A weight of 100 kg is supported by a string whose ends are attached to pegs A and B at the same level shown in below figure. The tension in the string is 
50 kg
75 kg
100 kg
120 kg
The angle of projection at which the horizontal range and maximum height of a projectile are equal to
36°
45°
56°
76°
Periodic time of a particle moving with simple harmonic motion is the time taken by the particle for
Half oscillation
Quarter oscillation
Complete oscillation
None of these
If the gravitational acceleration at any place is doubled, the weight of a body, will
Be reduced to half
Be doubled
Not be affected
None of these
The C.G. of the shaded area of the below figure from the x-axis is 
a/4
3a/4
3b/10
3a/10
In a lifting machine a weight of 5 kN is lifted through 200 mm by an effort of 0.1 kN moving through 15 m. The mechanical advantage and velocity ratio of the machine are respectively
50 and 75
75 and 50
75 and 75
50 and 50
Energy may be defined as
Power of doing work
Capacity of doing work
Rate of doing work
All the above
The velocity ratio of the differential wheel and axle is
R/r1 - r2
2R/r1
3R/r1 - r2
2R/r1 + r2
A load of 500 kg was lifted through a distance of 13 cm. by an effort of 25 kg which moved through a distance of 650 cm. The velocity ratio of the lifting machine is
50
55
60
65
In SI units, the units of force and energy are respectively
Newton and watt
Dyne and erg
Newton and joule
kg wt and joule
The maximum value of the horizontal range for a projectile projected with a velocity of 98 m/sec is
98 m
490 m
980 m
1960 m
From a circular plate of a diameter 6 cm is cut out a circle whose diameter is equal to the radius of the plate. The C.G. of the remainder from the centre of circular plate is at a distance of
2.0 cm
1.5 cm
1.0 cm
0.5 cm
The centre of gravity of a homogenous body is the point at which the whole
Volume of the body is assumed to be concentrated
Area of the surface of the body is assumed to be concentrated
Weight of the body is assumed to be concentrated
All the above
The unit of Moment of Inertia of a body, is
m
m2
m3
m4
A particle is dropped from a height of 3 m on a horizontal floor, which has a coefficient of restitution with the ball of 1/2. The height to which the ball will rebound after striking the floor is
0.5 m
0.75 m
1.0 m
1.5 m
μ is coefficient of friction. A wheeled vehicle travelling on a circular level track will slip and overturn simultaneously if the ratio of its wheel distance to the height of its centroid, is
μ
2μ
3μ
½μ
If g1 and g2 are the gravitational accelerations on two mountains A and B respectively, the weight of a body when transported from A to B will be multiplied by
g1
g2
g1/g2
g2/g1
If a particle moves with a uniform angular velocity ω radians/sec along the circumference of a circle of radius r, the equation for the velocity of the particle, is
v = ω √(y² - r²)
y = ω √(y - r)
v = ω √(r² + y²)
v = ω √(r² - y²)
The intrinsic equation of catenary is
S = c tan ψ
y = c cosh x/c
y = c cosh ψ
y = c sinh ψ
The resolved part of the resultant of two forces inclined at an angle θ in a given direction is
Algebraic sum of the resolved parts of the forces in the direction
Arithmetical sum of the resolved parts of the forces in the direction
Difference of the forces multiplied by cosine θ°
Sum of the forces multiplied by the tangent θ°
In case of S.H.M. the period of oscillation (T), is given by
T = 2ω/π²
T = 2π/ω
T = 2/ω
T = π/2ω
The Law of Polygon of Forces states that
If a polygon representing the forces acting at point in a body is closed, the forces are in equilibrium
If forces acting on a point can be represented in magnitude and direction by the sides of a polygon taken in order, then the resultant of the forces will be represented in magnitude and direction by the closing side of the polygon
If forces acting on a point can be represented of a polygon taken in order, their sides of a polygon taken in order, their resultant will be represented in magnitude and direction by the closing side of the polygon, taken in opposite order
If forces acting on a point can be represented in magnitude and direction by the sides of a polygon in order, the forces are in equilibrium
The moment of inertia of the shaded portion of the area shown in below figure about the X-axis, is
229.34 cm4
329.34 cm4
429.34 cm4
529.34 cm4
On a mass m describing a circular path of radius r, the centrifugal force
Acts tangentially to the circular path
Acts towards the centre of rotation
Acts away from the centre of rotation
Is mw2r/g kgf
One end of an elastic string of natural length / and modulus X is kept fixed while to the other end is attached a particle of mass m which is hanging freely under gravity. The particle is pulled down vertically through a distance x, held at rest and then released. The motion is
A simple harmonic motion
A rectilinear motion with constant speed
A damped oscillatory motion
None of the above
According to Law of Triangle of Forces
Three forces acting at a point, can be rep-resented by the sides of a triangle, each side being in proportion to the force
Three forces acting along the sides of a triangle are always in equilibrium
If three forces acting on a, point can be represented in magnitude and direction, by the sides of a triangle taken in order, these will be in equilibrium
If the forces acting on a particle be represented in magnitude and direction by the two sides of a triangle taken in order, their resultant will be represented in magnitude and direction by the third side of the triangle, taken in opposite order
For a simple pendulum, the period of one oscillation is
2π √(l/2g)
2π √(2g/l)
2π √(l/g)
2π √(g/2l)
Rate of change of angular momentum is equal to
Force
Torque
Linear momentum
Impulse