Test: Physics Minor Mock Test- 2 (March 20) - NEET MCQ

We know that,
C_p=c_v+R
E=3Rt/2
also,
E=C_v
c_p=R+3RT/2
=5/2 RT
=2.5R

5 × 4 × 10 + (h₁ - 5) × 10 = h₂ × 1 × 10
200 + 10h₁ – 50 = 10h₂
10h₂ - 10h₁ = 150
h₂ - h₁ = 15
h₂ + h₁ = 20 + 20 + 5 = 45
2h₂ = 60
h₂ = 30
h₁ = 15
h₂/h₁ = 30/15
h₂/h₁ = 2/1

If M is the mass of the Earth and m is the mass of a sphere of radius r and you want the force of gravity to be the same at both the surface of the sphere and the surface of the earth then mg = Mg and m = M. This means that the mass of the sphere would have to be the same as the mass of the Earth no matter what r is. This is impossible since a sub-sphere cannot have the same mass as its parent.
For a sphere of radius r to have force g at its surface, d = (De *Re)/r where De is the density of the entire Earth and Re is the radius of the Earth, d is the density of the sphere of radius r. So there is a radius and a density which will produce a gravitational force of g for any d, r >0 but one cannot be a subset of another.

= d μ 1/r

n is the number of moles.

The force is normal to the surface of the cylinder and hence will pass through the centre. And since the forces pass through the centre it means that the net torque will be zero.

Y=F x l/A x Δ l
Δ l=0.5cm=0.5x10^-2m, l=10M, F=940N
Y=20x10¹⁰pa
20x10¹⁰=940x10/πr²x0.5x10^-10
πr²=94x100/5x10^-3x2x10¹¹=94x10²/10x10⁸
r²=94/π x 10^-7 =2.99 x 10^-6
r2 ≅3x10^-6
r=1.13x10^-10 m
diameter=2r=3.6mm

σ=Stress and ε=strain
σ=F/A= (550kg) × (9.81m/s²)3×10-5m²/=0.18GPA
ε=Δl/l0=σ/Υ​=0.18×10⁹/200×109​=9×10^-4
Δl=εl₀= (9×10^-4) (2m) = 0.0018m=1.8mm

On heating, the copper strip will suffer greater elongation and hence the bimetal strip will bend with the steel strip on the concave side. 

[Bimetal strips are widely used in thermal switching applications such as automatic electric iron].

We know that PV=nRT also PM=dRT
So in the equation The value of R depends on P , V , n , T , d , M
except atomicity 
 so the ans is A

Given Data,
Length of the piece of copper = l = 19.1 mm = 19.1 × 10^-3m
Breadth of the piece of copper = b = 15.2 mm = 15.2× 10^-3m
Tension force applied on the piece of cooper, F = 44500N
Area of rectangular cross section of copper piece,
Area = l× b
⇒ Area = (19.1 × 10-3m) × (15.2× 10-3m)
⇒ Area = 2.9 × 10^-4 m²
Modulus of elasticity of copper from standard list, η = 42× 10⁹ N/m2
By definition, Modulus of elasticity, η = stress/strain

⇒ Strain = F/Aη

⇒ Strain = 3.65 × 10^-3
Hence, the resulting strain is 3.65 × 10^-3

A communications satellite is in a GEO (geostationary equatorial orbit) with a period of 24 hours. The spin rate ω_s about its axis of symmetry is 1 revolution per minute, and the moment of inertia about the spin axis is 550 kg · m².

The body will sink to the bottom as it gains a downward velocity and has no force to bring it back up. The weight becomes greater than upwards thrust.
 

As body is just floating, its density is same as that of the liquid.
If pressed below, it will gain momentum downwards, and continue to sink till bottom.
When the body is slightly pressed, the contraction in volume decreases upthrust, so weight becomes greater than upthrust, body moves down. The upthrust further decreases, since more and more contraction occurs as the body moves down. The body thus, sinks to the bottom.

According to kinetic theory of gases, 0K is that temperature at which for an ideal gas the internal energy is zero because at 0K nearly all molecular motion stops.

When heated the length between adjacent markings increases. Hence the actual length will be less than
30 cm.

Explanation:When external force is applied on the solid bodies, the solid bodies get deformed. The atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic ( or intermolicular ) distances. When the deforming foce is removed, the interatomic forces tend to drive them back to their original postions. Thus the body regains its original shape and size.

The Celsius and Fahrenheit are two important temperature scales. The Fahrenheit scale is used primarily in the United States, while Celsius is used throughout the world. The two scales have different zero points and the Celsius degree is bigger than the Fahrenheit one. There is one point on the Fahrenheit and Celsius scales where the temperatures in degrees are equal. This is -40 degree C and -40 degree F.

h × ρ × 10 = 120/800 × 10^-4 
h × ρ = 12/8 × 10² 
h × ρ = 150m
h × 1000 = 150m
h = 15cm

Force on the body placed on Earth's surface is
F = GMm/R²​
But, F=mg hence,
mg = GMm​/R²
where, variables have their usual meanings.
gR² = GM
Now, force on the body at geo-potential height say h (altitude) where the acceleration due to gravity is 25% of that at the earth's surface i.e.
25//100g ​= g/4​
Hence, we can write
g/4​ = GM/(R+h)²​

g/4​ = gR²/(R+h)²​

(R+h)² = 4R²

Taking roots for both sides we get
R+h = 2R
h = R

When water is heated from 0°C, its volume decreases because its density increases and you can see this effect upto 4°C. Since the density of ice is maximum at 4°C, afterwards as the density decreases the volume increases. The main reason for this is hydrogen bond in ice gets cleaved due to the melting of ice.

Explanation:A perfectly rigid body is hypothetical in nature but for some phenomena ( in rotational bodies ) we assume bodies are perfectly rigid i.e. the intermolecular forces are always in equilibrium irrespective of the external forces due to which their shape and size are constant.

Explanation:More the elastic a material is , more it has the property to regain its original position which is required in construction works.

For the composite system, energy conservation yields no net energy flow in or out of the system.
Let final temperature be T
Then, heat absorbed by A+heat absorbed by B+heat absorbed by C=0

Here in the question m₁=m₂=m₃

The acceleration due to gravity on surface of earth is
g=GM​/ R²
 
g=GρV​/R²
where, G is the gravitational constant, M  is mass of earth, R  is radius of earth, ρ is the density and V is the volume of earth.
∴g=Gρ(4/3)​πR^3​/ R²
 
∴g=(4/3)​GρπR
⇒g∝ρR
since, all other quantities are constant.
Hence, if the radius of the earth be increased by a factor of 5, density will decrease by 51​  to keep the value of g the same.

Consider the mass of the block be m, thus we get a mass of bucket + water is 2m. Now if we consider it as a two mass pulley, we get  the net acceleration a by which the bucket is going down can be found by - 
2mg - mg = 3ma
Thus we get a = g/3
Now we know that the bucket is accelerating with acceleration g/3 downwards hence to treat it as an inertial system we need to apply pseudo force upon it in upward direction. Thus we get in reference to the bucket net acceleration is 2g/3 downwards. Thus the pressure at the bottom is P + x 0.15 x 2g/3 = P + 1 kPa