A man is at a distance of 6 m from a bus. The bus begins to move with a constant acceleration of 3ms.In order to catch the bus, the minimum speed with which the man should run towards the bus is
2 m s-1
4 m s-1
6 m s-1
8 m s-1
C.
6 m s-1
If the man did not run, the bus would be at a distance s1 at time t given by
s1 = 6 + t2
= 6 + × 3 × t2
s1 = 6 + t2
If v is the speed of man, he would cover a distance s2 = vt in time t.
To catch the bus,
s1 = s2
6 + t2 = v t
⇒ t2 t + 4 = 0
which gives
t =
Now t will be real if is positive or zero.
Minimum v corresponds to = 0 which gives v = 6 m s-1
Two identical charged spheres suspended froma common point by two massless strings of lengths l, are initially at a distance d (d<< l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velcoity v. Then, v varies as a function of the distance x between the sphere, as
vx
vx-1
vx-1
B.
Two identical charged spheres suspended from a common point by two massless strings of length L.
In ,
The charge begins to leak from both the sphere at a constant rate. As a result, the spheres approach each other with velocity v.
Therefore, equation (i) can be rewritten as,
A body is at rest at x = 0 and t = 0, it starts moving in the positive X-direction with a constant acceleration. At the same instant, another body passes through x = O moving in the positive X-direction with a constant speed. The position of the first body is given by x1 (t) after time t and that of second body by x2 (t) after the same time interval. Which of the following graphs correctly describes (x1 − x2) as a function of time t ?
C.
The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is
0
5 m/s2
-4 m/s2
-4 m/s2
C.
-4 m/s2
x = 5t – 2t2 y = 10t
Acceleration of particle at t = 2 s is = –4 m/s2
The displacement x of a particle varies with time t as x = ae-αt + beβt where a, b, and β are positive constants. The velocity of the particle will
decrease with time
be independent of
drop to zero when
increase with time
D.
increase with time
Given:-
x = a e-αt + beβt
We know that
velocity v =
v =
= -a e-αt + bβ eβt
v =
v = A + B
where A = - a , B = bβ e-βt
The value ot term A = -a increases and of term B = bβ e-βt increases with time. As a result velocity goes on increasing with time.
A monkey of mass 30 kg climbs on a massless rope whose breaking strength is 450 N. The rope will break if the monkey (Take g = 10 m/s2 )
climbs up with a uniform speed of 5 m/s
climbs down with an acceleration 4 m/s2
climbs up with an acceleration 6 m/s2
climbs down with a uniform speed of 5 m/s2
C.
climbs up with an acceleration 6 m/s2
To move up with an acceleration 'a', the monkey will push the rope downward with a force
F = ma = 30 a
Now, Tmax = mg + ma = 30 × 10 + 30 a
⇒ 450 = 300 + 30 a
⇒ a = 5 m/s2
Hence, the rope will break if the monkey climbs up with an acceleration more than 5 m/s2.
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is,
90o
45o
180o
180o
A.
90o
There are two vectors P and Q.
It is given that,
Let, angle between P and Q is ,
A body is whirled in a horizontal circle of radius 25 cm. It has an angular velocity of 13 rad/s. What is its linear velocity at any point on circular path?
2 m/s
3 m/s
3.25 m/s
4.25 m/s
C.
3.25 m/s
Given:- r = 25 cm
r = 0.25 m
ω = 13 rad/s
Linear speed = Radius × angular speed
V = rω
= 0.25 × 13
3.25 m/s
If A + B = C and that C is perpendicular to A. What is the angle between A and B, if
rad
rad
rad
rad
C.
rad
Given:-
A + B = C
Since, B = C A
Also
B2 = 2 A2
⇒ B = A
Now
A2 + B2 + 2A B cosθ = C2 = A2
∴
This gives
rad
A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of 40 m/s from the bottom of the building. The two balls will meet after
3 s
2 s
2.5 s
5 s
C.
2.5 s
Let balls meet after t s. The distance travelled by the ball coming down is
s1 = gt2
Distance travelled by the other ball
s2 = 40 t gt2
s1 + s2 = 100 m
∴ gt2 + 40 t gt2 = 100m
t =
∴ t = 2.5 s