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Let's delve into the physics problem at hand. We have an air chamber with a volume VV and a neck area of cross-section into which a ball of mass mm just fits and can move up and down without any friction. When the ball is pressed down a little and released, it executes Simple Harmonic Motion (SHM). We need to derive an expression for the time period of oscillations, assuming pressure-volume variations of air to be isothermal.

To begin, let's analyze the forces acting on the ball. When the ball is pushed down, it experiences an upward force due to the buoyant force and a downward force due to gravity. At equilibrium, these forces balance out, and the ball remains stationary. When the ball is displaced slightly downwards and released, it experiences an upward force due to the compressed air in the chamber, leading it to oscillate.

The restoring force acting on the ball is due to the pressure difference between the compressed air below the ball and the less compressed air above it. According to Boyle's Law for isothermal processes, PV=constantPV=constant, where PP is pressure and VV is volume. Thus, when the ball is displaced downwards by a distance xx, the volume of air below it decreases, causing an increase in pressure, which provides the restoring force.

Using Hooke's Law, which states that the restoring force is directly proportional to the displacement, we can write:

F=−kxF=−kx

Where kk is the spring constant.

The pressure difference ΔPΔP across the ball can be expressed as:

ΔP=P0−P1ΔP=P0−P1

Where P0P0 is the pressure when the ball is at its equilibrium position, and P1P1 is the pressure when the ball is displaced by xx.

We can express P0P0 and P1P1 using Boyle's Law:

P0=kVandP1=kV−xP0=VkandP1=V−xk

Thus, the pressure difference ΔPΔP is:

ΔP=kV−kV−xΔP=Vk−V−xk

The force exerted by this pressure difference on the ball is:

F=AΔPF=AΔP

Where AA is the cross-sectional area of the neck.

So, we have:

F=A(kV−kV−x)F=A(Vk−V−xk)

F=kA(1V−1V−x)F=kA(V1−V−x1)

Now, equating this to −kx−kx (according to Hooke's Law), we get:

kA(1V−1V−x)=−kxkA(V1−V−x1)=−kx

A(1V−1V−x)=−xA(V1−V−x1)=−x

AV−AxV(V−x)=−xVA−V(V−x)Ax=−x

AV−Ax=−x(V−x)AV−Ax=−x(V−x)

AV−Ax=−xV+x2AV−Ax=−xV+x2

AV=−xV+Ax+x2AV=−xV+Ax+x2

x2−Ax+AV=0x2−Ax+AV=0

This is a quadratic equation in xx. Solving this equation will give us the value of xx, which is the amplitude of the oscillation. Then, we can use the formula for the time period of SHM:

T=2πmkT=2πkm

Here, Volume of water =30 m3
t=15 min=15×60 s=900 s;Height,h=40 m
Efficiency, η=30%
Density of water =103kg m−3
∴ Mass of water pumped =Volume×Density
=(30 m3)(103 kg m−3)=3×104kg
Poutput=Wt=mght=(3×104kg)(10 ms−2)(40 m)900 s
=43×104W
Efficiency, η=PoutputPinput
Pinput=Poutputη=4×1043×30100=49×105
=44.4×103W=44.4 kW.

Linear momentum of particle, →p=px^i+py^j+pz^k

Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y=2×1011Pa

Total force exerted, F=Mg=50000×9.8N

Stress = Force exerted on a single column =50000×9.84=122500N
Young’s modulus, Y=StressStrain

Strain =(F/A)Y
Where,
Area, A=π(R2−r2)=π((0.6)2−(0.3)2)

Strain =122500/[π((0.6)2−(0.3)2)×2×1011]=7.22×10−7

Hence, the compressional strain of each column is 7.22×10−7.

Water is flowing at a rate of 3.0 litre/min.
The geyser heats the water, raising the temperature from 27oC to 77oC.
Initial temperature, T1=27oC
Final temperature, T2=77oC
Rise in temperature,T=T1−T2=77−27=50oC
Heat of combustion =4×104J/g
Specific heat of water, C=4.2J/goC
Mass of flowing water, m= 3.0 litre/min = 3000 g/min
Total heat used, Q=mcT
=3000×4.2×50
=6.3×105J/min
Rate of consumption =6.3×105/(4×104) = 15.75 g/min.

u=126 km/h=126×518m/s=35m/s
v=0
s=200m
Newton's Equation of motion
v2−u2=2as
02−352=2a(200)
a=−3.0625m/s2
Also
v=u+at
0=35−3.06t
t=11.4s