The dimensional formula of gravitational constant is given by,

M^-1 L³ T^-2

Where,

• M = Mass
• L = Length
• T = Time

Given, mass, m = 72.2 kg
Gravity acceleration, g = 9.8 m/${{s}^{2}}$ Scale reading = apparent weight = R = ?
(i) While descending with constant velocity, a =0
R = mg
R = 72.2 x9.8
=>R = 707.56 N
(ii) While ascending with a = 3.2 m/${{s}^{2}}$
R = m(g + a)
R = 72.2 (9.8 + 3.2) = 938.6 N

N = mg

Static friction provides the centripetal acceleration

fs =< μsN

mv2/r =< μsN

v2 =< μsNR/m = μsmgR/m = μsRg

v =< √ μsRg

This is the maximum speed of a car in circular motion on a level road

The moment of inertia of a sphere expression is obtained in two ways.

• First, we take the solid sphere and slice it up into infinitesimally thin solid cylinders.
• Then we have to sum the moments of exceedingly small thin disks in a given axis from left to right.

We will look and understand the derivation below.

First, we take the moment of inertia of a disc that is thin. It is given as;

I = ½ MR²

In this case, we write it as;

dI (infinitesimally moment of inertia element) = ½ r²dm

Find the dm and dv using;

dm = p dV

p = moment of a thin disk of mass dm

dv = expressing mass dm in terms of density and volume

dV = π r² dx

Now we replace dV into dm. We get;

dm = p π r² dx

And finally, we replace dm with dI.

dI = ½ p π r⁴ dx

The next step involves adding x to the equation. If we look at the diagram we see that r, R and x forms a triangle. Now we will use Pythagoras theorem which gives us;

r² = R² – x²

Now if we substitute the values we get;

dI = ½ p π (R² – x²)² dx

This leads to:

I = ½ p π _-R∫^R (R² – x²)² dx

After integration and expanding we get;

I = ½ pπ 8/15 R⁵

Additionally, we have to find the density as well. For that we use;

p = m / V

p = m / m/v πR³

If we substitute all the values;

I = 8/15 [m / m/v πR³ ] R⁵

I = ⅖ MR²

 Let u & v be the velocity of the two trains named A & B.

While overtaking the relative velocity of train A with respect to B = u - v.

While crossing, the relative velocity of train A with respect to B = u + v.

Total distance to be travelled by Train A while crossing = 100 + 100 = 200.

So,

20 = 200 ÷ u-v

=> u - v = 10....1)

Also,

10 = 200 ÷ u + v

=> u + v = 20.......2)

On solving the equation 1 & 2, we get;

u = 15 M/S of the train A / First Train.

v= 5 M/S of the train B / Second Train.

Let 'u' be the initial velocity and 'a' the acceleration.
So we have the distance formula
s = ut + 1/2 at^2
In first 2 seconds,
s = 200
Put the values in the formula.
200 = u x 2 + 1/2 x a x (2)^2
200 = 2u + 1/2 x 4 x a
200 = 2u + 2a
100 = u + a -----(I)
In next 4 seconds
Let the distance traveled be 'y'
Time = 2+4 = 6 sec
So according to the formula ,
y = u x 6 + 1/2a x (6)^2
y = 6u + 1/2 x 36 x a
y = 6u + 18a ------(ii)
Now we know that 220 cm was traveled in between 2 sec - 6 sec
y - 200 = 220
y = 420
We know y = 6u + 18a (from [ii])
So, 6u + 18a = 420
u + 3a = 70 ------(iii)
Equating (I) and (iii)
-2a = 30
a = -15 cm/s^2
u = 100 - (-15)
u = 100 + 15 = 115 cm/sec
Now we know v = u + at
We have to find the "v" after 7th second
So v = 115 + (-15) x 7
v = 115 - 105
v = 10 cm/sec

Here, Mass of machine gun , M = 20 kg
mass of each bullet , m = 35 kg = 0.035 kg
velocity of bullet υ=400m/sυ=400m/s
Number of bullets/sec (n) =400/60=20/3 
Force required = rate of change of linear momentum of bullets =n(mυ)/t = 20/3  × 0.035×400/1 =93.3N 

Current density = Current​/area = Q/ area×t​ = [M⁰L⁻²T⁻¹Q]

The dimensional formula of current density is given by,

[M⁰ L^-2 T⁰ I¹]

Where,

• M = Mass
• I = Current
• L = Length
• T = Time

Statement: For the streamline flow of non-viscous and incompressible liquid, the sum of potential energy, kinetic energy and pressure energy is constant.



Proof: Let us consider the ideal liquid of density ρ flowing through the pipe LM of varying cross-section. Let P₁ and P₂ be the pressures at ends L and M and A₁ and A₂ be the areas of cross-sections at ends L and M respectively. Let the liquid enter with velocity V₁ and leave with velocity v₂. Let A₁ > A₂. By equation of continuity,

Bernoulli’s principle states that The total mechanical energy of the moving fluid comprising the gravitational potential energy of elevation, the energy associated with the fluid pressure and the kinetic energy of the fluid motion, remains constant.

Bernoulli’s equation formula is a relation between pressure, kinetic energy, and gravitational potential energy of a fluid in a container.

The formula for Bernoulli’s principle is given as:

Where,

• p is the pressure exerted by the fluid
• v is the velocity of the fluid
• ρ is the density of the fluid
• h is the height of the container

Moment of inertia of sphere is normally expressed as;

Here, r and m are the radius and mass of the sphere respectively. Students have to keep in mind that we are talking about the moment of inertia of a solid sphere about its central axis above. Additionally, if we talk about the moment of inertia of the sphere about its axis on the surface it is expressed as;

The magnitude of three forces 3N,4N  and 5N  will be zero, if these vectors from a close polygon will all the sides in the same order as shown in figure. Hence,the acceleration of body may be zero

No, the radioactive nucleus does not emit alpha and beta radiations simultaneously. Some nucleus emit alpha particles, others beta particles and changes into a new nucleus and to balance the energy, gamma radiations are emitted.

SI unit of density is Kg/m³
1 m = 100 cm => 1 cm = 10-² m
1 Kg = 1000g => 1g = 10-³ Kg

now,
13.6 g/cm³ = 13.6 × ( 10-³)/(10-²)³ Kg/m³

= 13.6 × 10-³/10^-6 Kg/m³

= 13.6 × 10^(-3 + 6) Kg/m³

= 13.6 × 10³ Kg/m³

Hooke’s Law, F = -kx, where F is the force and x is the elongation.
The work done = energy stored in stretched string = F.dx
The energy stored can be found from integrating by substituting for force,
and we find,
The energy stored = kx²/2, where x is the final elongation.
The energy density = energy/volume
= (kx²/2)/(AL)
=1/2(kx/A)(x/L)
= 1/2(F/A)(x/L)
= 1/2(stress)(strain)