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.
For a particle in uniform circular motion, the acceleration an at point P (R, θ) on the circle of radius R is (Here θ is measured from the x–axis)
C.
A point p moves in counter -clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out the length s = t3 + 5, where s is metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2 s is nearly
13 ms^{-2}
12 ms^{-2}
7.2 ms^{-2}
14 ms^{-2}
D.
14 ms^{-2}
Given that, s =t^{3} +5
therefore speed v, = ds/st = 3t^{2}
and rate of change of speed, at = dv/dt = 6t
∴ Tangential acceleration at t =2s
at = 6 x 2 = 12 ms^{-1}
and at t = 2s, v = 3 (2)2 = 12 ms^{-1}
∴ Centripetal acceleration, ac = v^{2}/R = 144/20 ms^{-2}
∴ Net acceleration =
The figure shows the position –time (x – t) graph of one–dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is
0.4 (Ns)
0.8 Ns
1.6 Ns
0.2 Ns
B.
0.8 Ns
From the graph it is a straight line, so uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
(∴ impulse (l) = f x t
⇒ I = mat
Initial velocity, v_{1} = 2/2 = 1 ms^{-1}
Final velocity v_{2} = 2/2 = - 1 ms^{-1}
p_{i} = mv1 = 0.4 N-s
p_{f} = mv2 = -0.4 N-s
J = p_{f}-p_{i} = -0.4-0.4
= - 0.8 N-s
|J| = 0.8 N-s
A particle is moving with velocity, where K is a constant. The general equation for its path is
y = x^{2} + constant
y^{2} = x + constant
xy = constant
y^{2} = x^{2} + constant
D.
y^{2} = x^{2} + constant
The velocity of the particle,
A ball is made of a material of density ρ where ρ_{oil} < ρ < ρ_{water} with ρ_{oil} and ρ_{water} representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?
C.
ρ_{oil} < ρ < ρ_{water}
Oil is the least dense of them, so it should settle at the top with water at the base. Now, the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So, it will stay at the oil -water interface.
The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10^{–3} are
5,1,2
5,1,5
5,5,2
4,4,2
A.
5,1,2
A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32 V,the efficiency of the engine is
0.5
0.75
0.99
0.25
B.
0.75
A small particle of mass m is projected at an angle θ with the x–axis with an initial velocity v_{0} in the x–y plane as shown in the figure. At a time t<v_{o}sin/g, the angular momentum of the particle is
C.
The angular momentum of the projectile is given by,
L = m (r x v)
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)
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