Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B?
4.9 ms^{-2} in horizontal direction
9.8 ms^{-2} in vertical direction
zero
4.9 ms^{-2} in vertical direction
D.
4.9 ms^{-2} in vertical direction
For the motion of block along inclined plane
mg sin θ =ma
a = g sin θ
where a is along the inclined plane.
The vertical component of acceleration is g sin^{2}θ
Therefore, the relative vertical acceleration of A with respect to B is
g (sin^{2}60 - sin^{2}30) = g/2 = 4.9 ms^{-2} (in vertical direction)
Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is
100N
80 N
120 N
150 N
C.
120 N
In the vertical direction, weight are balanced by frictional forces.
As the blocks are in equilibrium balance forces are in horizontal and vertical direction.
For the system of blocks (A+B)
F = N
For block A, f_{A} = 20 N and for block B.
f_{B} = f_{A} +100 = 120 N
A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to
44%
50%
56%
62%
C.
56%
A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is
g
2/3g
g/3
3/2g
B.
2/3g
A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
2g/3
g/2
5g/6
g
B.
g/2
For the mass m,
mg-T = ma
As we know, a = Rα ... (i)
So, mg-T = mRα
Torque about centre of pully
T x R = mR^{2}α ...... (ii)
From Eqs. (i) and (ii), we get
a = g/2
Hence, the acceleration of the mass of a body fall is g/2.
STATEMENT – 1
Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
STATEMENT – 2
Principle of conservation of momentum holds true for all kinds of collisions.
The statement I is True, Statement II is False.
The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.
The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.
Statement I is False, Statement II is
B.
The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.
A liquid in a beaker has temperature θ(t) at time t and θ_{0} is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ_{0}) and t is
A.
According to Newton's law of cooling.
Hence the plot of ln(θ – θ_{0}) vs t should be a straight line with negative slope.
A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10^{–2N} (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is
0.0125 Nm^{-1}
0.1 Nm^{-1}
0.05 Nm^{-1}
0.025 Nm^{-1}
D.
0.025 Nm^{-1}
⇒F = 2Tl = w
⇒2T(0.3) = 1.5 x 10^{–2}
⇒T = 2.5 x 10^{–2} N/m
A block of mass m is placed on a surface with a vertical cross-section given by y = x^{3}/6. If the coefficient of friction is 0.5, the maximum height above ground at which the block can be placed without slipping is
A.
A block of mass m is placed on a surface with vertical cross section, then
At limiting equilibrium, we get
μ = tan θ
0.5 = x^{2}/2
⇒ x^{2} =1
⇒ x = ±1
Now, putting the value of x in y = x3/6, we get
When x =1
When x =-1
So, the maximum height above the ground at which the block can be placed without slipping is 1/6m.
A pulley of radius 2m is rotated about its axis by a force F = (20t - 5t^{2}) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m^{2}, the number of rotations made by the pulley before its direction of motion if reversed is
more than 3 but less than 6
more than 6 but less than 9
more than 9
less than
A.
more than 3 but less than 6
To reverse the direction