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’ 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.
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 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.