CBSE
One kg of a diatomic gas is at a pressure of 8 × 10^{4} N/m^{2} . The density of the gas is 4 kg/m^{-3}. What is the energy of the gas due to its thermal motion?
3 × 10^{4} J
5× 10^{4} J
6× 10^{4} J
7× 10^{4} J
B.
5× 10^{4} J
Thermal energy corresponds to internal energy
Mass = 1 kg
density = 8 kg/m^{3}
Let P(r) = Qr/πR^{4} be the charge density distribution for a solid sphere of radius R and total charge Q. for a point ‘p’ inside the sphere at distance r_{1} from the centre of the sphere, the magnitude of electric field is
0
C.
A thin uniform rod of length
D.
If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?
a^{2}T^{2}+ 4π^{2}v^{2}
aT/x
aT + 2πv
aT/v
B.
aT/x
Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate.Assume that the duration of collision is negligible and the collision with the plate is totally elastic.Then the velocity as a function of time and the height as a function of time will be
C.
When ball strikes the surface its velocity will be reversed therefore correct answer is 3.
A particle has an initial velocity of and an acceleration of It speed after 10 s
10 units
7 units
8.5 units
B.
In an experiment, the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half–a–degree (= 0.5o° ), then the least count of the instrument is
one minute
half minute
One degree
half degree
A.
one minute
1VSD = 29/30 MSD
L.C. = 1 MSD – 1 VSD
= 1/30 MSD
A motorcycle starts from rest and accelerates along a straight path at 2 m/s^{2}. At the straight point of the motorcycle, there is a stationary electric siren. How far has the motorcycle gone when the driver hears the frequency of the siren at 94% of its value when the motorcycle was at rest?
49 m
98 m
147 m
196 m
B.
98 m
The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is
2R
R/2
A.
2R
Two points P and Q are maintained at the potential of 10V and -4V respectively. The work done in moving 100 electrons from P to Q is
9.60 × 10^{–17} J
9.60 × 10^{–17} J
– 2.24 × 10^{–16} J
2.24 × 10^{–16} J
D.
2.24 × 10^{–16} J
W = QdV = Q(Vq - VP)
= -100 × (1.6 × 10^{-19}) × (– 4 – 10)
= + 100 × 1.6 × 10^{-19} × 14 = +2.24 10-16 J