CBSE
The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)
B.
The moment of inertia of a uniform cylinder of length
1
C.
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
A.
Velocity at any time t is given by
v = u + at
v = v0 + (–g)t
v = v0 – gt
Straight line negative slope.
A time-dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be
9J
18 J
4.5 J
22 J
C.
4.5 J
F = 6t = ma
⇒ a = 6t
A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λ_{A} to λ_{B} after the collision is
D.
The temperature of an open room of volume 30 m^{3} increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and n_{f} are the numbers of molecules in the room before and after heating, then n_{f} – n_{i} will be
2.5 x 10^{25}
-2.5 x 10^{25}
-1.61 x 10^{23}
1.38 x 10^{23}
B.
-2.5 x 10^{25}
Using ideal gas equation
PV = nRT
(N is number of moles)
P_{0}V_{0} = n_{i}R × 290 ...... (1)
[T_{i} = 273 + 17 = 290 K]
After heating
P_{0}V_{0} = N_{f}R × 300 ...... (2)
[T_{f} = 273 + 27 = 300 K]
from equation (1) and (2)
putting P0 = 105 PA and V0 = 30 m3
Number of molecules nf – ni = – 2.5 × 1025
The following observations were taken for determining surface tension T of water by the capillary method :
Diameter of capillary, D = 1.25 × 10^{–2} m rise of water, h = 1.45 × 10^{–2} m
Using g = 9.80 m/s2 and the simplified relation T = (rhg/2) x103 n/m, the possible error in surface tension is closest to:
2.4%
10%
0.15%
1.5%
D.
1.5%
A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like
B.
A body of mass m = 10^{–2} kg is moving in a medium and experiences a frictional force F = –kv^{2}. Its initial speed is v_{0} = 10 ms^{–1}. If, after 10 s, its energy is 1/8 mv_{0}^{2},the value of k will be
10^{-4} kg m^{-1}
10^{–1} kg m^{–1} s^{–1}
10^{-3} kg m^{-1}
10^{-3} kg s^{-1}
A.
10^{-4} kg m^{-1}