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CALCULATION:

1. The resistance may be written as

$⇒ R =\frac{V}{I} =\frac{W}{I^2t^{}}$

Where W is the work done and I is the current and t is the time

The dimension of R will be given by

$⇒ \frac{W}{I^2t^{}}= \frac{ML^{2}T^{-2}}{A^{2}T} = ML^{2}T^{-3}A^{-2}$

2. The resistivity is given by

$⇒ \rho = \frac{RA}{l}$

Where A =Area and l = length

The dimension of resistivity is 

$⇒ \rho = \frac{RA}{l} = \frac{ML^{2}T^{-3}A^{2}\times L^{2}}{L} = ML^{3}T^{-3}A^{-2}$

3. As we know, the conductivity is the inverse of resistivity taking the inverse of the dimension of resistivity, then the dimension of electrical conductivity is 

⇒ σ = M-1L-3T3A2

When a body falls freely under gravity, the distance covered in the nth second is given by the formula:

Distance covered in nth second = u + (2gs) (where u is the initial velocity, g is the acceleration due to gravity, and s is the time interval)

For a body falling from rest, u = 0. So, the distance covered in nth second is:

s = (1/2) × g × (2n - 1)

The distances covered in the first, second, and third seconds are in the ratio:

1^st second: (1/2) × g × (2 × 1 - 1) = g/2

2^nd second: (1/2) × g × (2 × 2 - 1) = 3g/2

3^rd second: (1/2) × g × (2 × 3 - 1) = 5g/2

So, the ratio of distances covered in the first, second, and third seconds is:

1 : 3 : 5

Thus, the correct option is Option 1: 1 : 3 : 5

Calculation:

​Given: The projectile is projected at an angle of 30° with initial velocity 40 m/s.

We need to find the velocity of the projectile at t = 2 seconds, given g = 10 m/s².

We can resolve the initial velocity into horizontal and vertical components:

Initial horizontal velocity, v_x0 = 40 × cos(30°) = 40 × √3/2 = 20√3 m/s

Initial vertical velocity, v_y0 = 40 × sin(30°) = 40 × 1/2 = 20 m/s

The horizontal velocity remains constant throughout the motion as no horizontal acceleration exists. Thus, at t = 2 seconds, the horizontal component of velocity is:

v_x = 20√3 m/s

The vertical velocity at any time t is given by:

v_y = v_y0 - g × t

v_y = 20 - 10 × 2 = 20 - 20 = 0 m/s

Therefore, at t = 2 seconds, the vertical velocity is zero.

The total velocity of the projectile at t = 2 seconds is the horizontal velocity alone:

v = v_x = 20√3 m/s ≈ 34.64 m/s

So, the velocity of the projectile at t = 2 seconds is approximately 34.64 m/s, which matches Option 1.

Thus, the correct option is Option 1: 20√3 m/s

Calculation:

Given: Block A has a mass of 4 kg, and block B has a mass of 5 kg. Block B rests on a smooth horizontal table, and the minimum force required to move both blocks together is 12 N.

Let the coefficient of static friction between block A and block B be μ. For the blocks to move together, the frictional force between the two blocks must provide the necessary force to move both blocks as a single unit.

The frictional force, f, between block A and block B is given by:

f = μ × N, where N is the normal force between the two blocks. Since block A is placed on block B, the normal force is equal to the weight of block A, which is:

N = m_A × g = 4 × 10 = 40 N

Therefore, the maximum frictional force is:

f = μ × 40 N

Now, the force required to move both blocks together is:

F = (m_A + m_B) × a, where a is the acceleration of the blocks.

We know that the minimum force required to move the blocks together is 12 N, and this force must overcome the friction between the blocks. Thus, the frictional force must be equal to the applied force for both blocks to move together:

12 N = μ × 40 N

Solving for μ:

μ = 12 / 40 = 0.3

Now, the maximum force that can be applied to block B without sliding is:

F_max = (m_A + m_B) × g × μ = (4 + 5) × 10 × 0.3

F_max = 9 × 10 × 0.3 = 27 N

Thus, the correct option is Option 3: 27 N

Calculation:

Given that a bullet of mass 'a' and velocity 'b' is fired into a large block of wood of mass 'c'. The bullet gets embedded into the block of wood. To find the final velocity of the system, we will use the principle of conservation of momentum.

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.

The initial momentum of the system is:

Momentum before = mass of bullet × velocity of bullet = a × b

The final momentum of the system after the bullet gets embedded in the block is:

Momentum after = (mass of bullet + mass of block) × final velocity = (a + c) × v

By conservation of momentum, we equate the initial momentum to the final momentum:

a × b = (a + c) × v

Solving for the final velocity v:

v = (a × b) / (a + c)

Calculation:

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I₀. The moment of inertia about an axis passing through one of its ends and perpendicular to its length is derived using the parallel axis theorem.

According to the parallel axis theorem, the moment of inertia about an axis through one end of the rod is:

I = I₀ + M × (L/2)²

Substituting the value for I₀ (the moment of inertia through the center), we get:

I = I₀ + M × L²/4

Hence, the correct option is Option 2: I₀ + M × L²/4

Calculation:

A geo-stationary satellite is one that revolves around the Earth at a suitable height, with the same angular velocity as the Earth. This allows it to remain above the same point on the Earth's surface at all times, making it appear stationary from the Earth's surface.

Hence, the correct option is Option 3: It revolves round the Earth at a suitable height with the same angular velocity and in the same direction as the Earth does about its own axis.

Calculation:

Given,

R₁ = 3 cm = 3 × 10⁻² m

R₂ = 5 cm = 5 × 10⁻² m

T = 0.03 N/m (Surface tension)

Original surface area = 2 × 4πR₁

For second bubble = 2 × 4πR₂

Work done = Surface tension × extension in area

Work done = T × ΔA

Substitute the values:

Work done = 0.03 × 2[4πR₂² - 4πR₁²]

Work done = 0.03 × 8π[(5)² - (3)²] × 10⁻⁴

Work done = 0.03 × 8π × 16 × 10⁻⁴

Work done = 0.384π × 10⁻³ J

Hence, the work done is approximately 0.4π mJ.

Calculation:

When two spheres made of the same metal are heated to the same temperature, the coefficient of linear expansion is the same for both spheres. 

The formula for volume expansion is given by:

ΔV = γ · V₀ · ΔT

For the solid sphere:

ΔV_solid = γ · V_solid · ΔT

For the hollow sphere:

ΔV_hollow = γ · V_hollow · ΔT

ΔV_solid = ΔV_hollow

Hence, the correct answer is Option 1: both spheres will expand equally

Calculation:

The change in internal energy (ΔU) of a thermodynamical system can be calculated using the first law of thermodynamics:

ΔU = Q - W

Where:

ΔU is the change in internal energy

Q is the heat absorbed by the system

W is the work done by the system

Given:

Heat absorbed, Q = 2 kcal = 2 × 4.2 × 10³ J = 8400 J

Work done, W = 400 J

Now, applying the values in the equation:

ΔU = 8400 J - 400 J = 8000 J = 8 kJ

Thus, the change in internal energy is 8 kJ.