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The dimensional formula for angular momentum is
Concept:
∴ L = I x ω = r x p
Where, L = angular momentum, I = moment of inertia, ω = angular momentum,
r = distance and p = linear momentum = (m×v) .
Explanation:
Now,
Dimensional formula of (r) = [L],
Dimensional formula of (p) = [MLT-1],
Therefore, the dimensional formula of L is
L = [L] × [MLT-1]
∴ L = [ML2T-1]
∴ Dimensional formula of angular momentum L is [ML2T-1].
This are some of the useful dimensions of basic physical quantity, Here M, L, T & I represents mass, Length, Time and Current respectively.
Physical quantity
Formula
Dimension
Velocity(v)
v = dx/dt
[L1T-1]
Angular velocity (ω)
ω = dθ/dt
[T-1]
Acceleration (a)
a =dv/dt
[L1T-2]
Momentum (p)
p = mv
[M1L1T-1]
Angular momentum (L)
L = rp = mvr
[M1L2T-1]
Force(F)
F = ma
[M1L1T-2]
Torque(τ)
τ = r × F
[M1L2T-2]
Work (W)/Energy(E)
W = Fx =1/2 mv2
Pressure (P)/stress
Pressure = Stress = F/Area
[M1L-1T-2]
Electric current (I)
I = dq/dt
[I-1]
Power
P = Work/time = VI
[M1L2T-3]
Area
A = x2
[L2]
The chance of a vehicle to overturn while negotiating a circular path depends on
The maximum velocity of the vehicle by which it can be driven before overturning is given by v = (gra/h)1/2
g = acceleration due to gravity
r = radius of the curved path
2a = length of axle
h = height of the centre of mass of vehicle above ground
Thus, bigger the vehicle, easily it will overturn due to the inverse relation
Larger the radius of the path, easily it can move without overturn
A body is thrown vertically upward with velocity u, the greatest height ‘h’ to which it will rise is:
3 equations of motion govern the position of any particle:
v = u + at
s = ut + 1/2at2
v2 = u2 + 2as Application of Concept:
When the body is thrown upwards, at the top most point its final velocity becomes zero
Using third equation of motion
v2 =u2 + 2as
v = 0
a = -g
− u2 = 2(−g)s
s = u2/2g
The moment of inertia of a rectangle about its XX-axis is given by
The moment of inertia of a rectangle of width b and height d, about each of its centroidal axis, is:
A 10-gm arrow is fired from 150 gm bow with a velocity of 300 m/s. The recoil velocity of the bow is:
From the conservation of momentum
maua + mbub = mava + mbvb
Considering the bow and arrow as one mass system
ma = 10-gm
ua = 300 m/s
mb = 150-gm
ub = ?
10 (300) + 150 (ub) = 0
Recoil Velocity = 20 m/s
The negative sign shows that the bow moves in the opposite direction to the arrow.
Common Mistake:
A common mistake that a student can make is assuming the answer as - 20 m/s
It should be noted that the term "recoil velocity" has been used in the question that will take care of the direction and hence the negative sign.
Droplets of a liquid are usually more spherical in shape than the large drop of same liquid because of the force of
In case of small drops of a liquid, the gravitational potential energy is negligible in comparison to the potential energy due to surface tension. Consequently, to keep the drop-in equilibrium, the liquid drop’s surface tends to contract so that its surface area will be the least for a sphere and the drops will be spherical.
So, small drops of liquid are spherical because surface tension predominates over that of gravity and bigger drops are oval.
________ is the smallest interval of time after which a periodic motion is repeated.
Heat energy of an object is
Heat energy of an object is the overall energy of the molecules that arise due to the motion of the molecules.
Consider 4 molecules of a gas having speed of 6, 10, 8, and 20 km/sec respectively. What is the ratio of their RMS speed to the Average speed?
A heater is marked 620 W. What will be the value of the current drawn when it is connected to 310 V source?
According to Ohm’s Law, Power P = VI
⇒ I = P/V
⇒ I = 620/310
⇒ I = 2A
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