Physics Test 209

When a magnet is dropped through the hollow metal pipe, the induced eddy currents create a force that opposes the motion of the magnet.

This results in a reduced net force and consequently, the acceleration of the magnet is less than the gravitational acceleration g.

∴ The acceleration of the falling magnet is less than g. Option 2) is correct.

In a simple pendulum, when the bob is displaced by an angle θ and released, the mechanical energy is conserved. The total energy (E) remains constant, and the potential energy lost equals the kinetic energy gained.

The potential energy (PE) at an angle θ is given by:

PE = m × g × l × (1 - cosθ),

where m is the mass, g is the acceleration due to gravity, l is the length of the pendulum, and θ is the angle of displacement.

At the equilibrium position, all the potential energy is converted into kinetic energy (KE), which is given by:

KE = (1/2) × m × v²,

Where v is the speed of the bob at the equilibrium position.

By energy conservation, PE = KE, so:

m × g × l × (1 - cosθ) = (1/2) × m × v²

Canceling m from both sides and solving for v:

v² = 2 × g × l × (1 - cosθ)

v = √(2gl(1 - cosθ))

The speed of the bob at the equilibrium position is v = √(2gl(1 - cosθ)). Option 4 is correct.

(a) Velocity–displacement (v–x):

For SHM, v = ω√(A² − x²) ⇒ v² = ω²(A² − x²).

Rearrange: (v/ω)² + x² = A² ⇒ the v–x curve is a circle in the scaled axes (v/ω, x).

(b) Acceleration–displacement (a–x):

x = A sin(ωt) ⇒ v = dx/dt = ωA cos(ωt) ⇒ a = dv/dt = −ω²A sin(ωt) = −ω²x.

Hence a vs x is a straight line with slope −ω².

(c) Acceleration–time (a–t):

From above, a = −ω²A sin(ωt), i.e., a sinusoid in time.

(d) Acceleration–velocity (a–v):

From v² = ω²(A² − x²) ⇒ x² = A² − v²/ω².

a = −ω²x ⇒ a² = ω⁴x² = ω⁴(A² − v²/ω²).

⇒ (a/ω²)² + (v/ω)² = A² ⇒ an ellipse (reduces to a circle in the scaled axes when ω = 1).

Thus correct matching is a - iii, b - i, c - ii, d - iv.

(i) Coefficient of viscosity
Dimension: [η] = Force × time / area = (M L T^-2) × T / L² = M L^-1 T^-1
⇒ Matches with (d)

(ii) Strain
Strain = ΔL / L (dimensionless quantity)
⇒ Dimension: M⁰ L⁰ T⁰
⇒ Matches with (a)

(iii) Compressibility
Compressibility = 1 / Bulk modulus
Bulk modulus = Pressure = Force / Area = M L^-1 T^-2
⇒ Compressibility = L T² M^-1
⇒ Matches with (b)

(iv) Pressure gradient
Pressure gradient = Pressure / Length
Pressure = M L^-1 T^-2 ⇒ Divide by L
⇒ Dimension: M L^-2 T^-2
⇒ Matches with (c)

Correct Matching:

• (i) → (d)
• (ii) → (a)
• (iii) → (b)
• (iv) → (c)

The Correct answer is Statement I is incorrect but Statement II is true.

Statement I : Area under velocity- time graph gives the distance traveled by the body in a given time. 

Therefore area under the velocity time graph gives displacement.

Hence statement I is false.

Statement II : Area under acceleration- time graph is equal to the change in velocity- in the given time.

We are asked to identify which quantity has the dimensional formula [M L⁻¹ T⁻²].

Option 1: Pressure × Area

• Pressure has the dimensional formula [M L⁻¹ T⁻²].
• Area has the dimensional formula [L²].
• Multiplying them gives [M L¹ T⁻²], which is not the required formula.

Option 2: Force × Pressure

• Force has the dimensional formula [M L T⁻²].
• Pressure has the dimensional formula [M L⁻¹ T⁻²].
• Multiplying them gives [M² L⁰ T⁻⁴], which is not the required formula.

Option 3: Power × Time

• Power has the dimensional formula [M L² T⁻³].
• Time has the dimensional formula [T].
• Multiplying them gives [M L² T⁻²], which is not the required formula.

Option 4: Energy Density

• Energy density has the dimensional formula [M L⁻¹ T⁻²], which matches the required formula.

∴ The correct answer is option 4) Energy Density.

Let’s plug in the dimensions:

[M L^-1 T^-2] = [ (M L^-3)ⁿ ] [ (L T^-1)² ]

Simplify the dimensions on the right-hand side:

[M L^-1 T^-2] = [Mⁿ L^-3n] [L² T^-2]

Combine the terms on the right-hand side:

[M L^-1 T^-2] = [Mⁿ L^-3n+2 T^-2]

Now, equate the exponents of corresponding dimensions from both sides:

For mass (M): 1 = n

For length (L): -1 = -3n + 2

For time (T): -2 = -2 (This is already satisfied)

From the mass dimension equation, we get:

n = 1

To verify, substitute n = 1 in the length dimension equation:

-1 = -3(1) + 2

-1 = -3 + 2

-1 = -1

This is correct, so the value of n is confirmed as:

n = 1

Therefore, the correct option is:

Option 2: n = 1

For jar A, the number of molecules N₁ is:

N₁ = P * V / (k * T)

For jar B, the number of molecules N₂ is:

N₂ = (2P) * (V/4) / (k * (2T))

Simplifying for N₂:

N₂ = (2P * V) / (4k * 2T) = P * V / (4k * T)

The ratio of the number of molecules in jar A to jar B is:

N₁ / N₂ = (P * V / k * T) / (P * V / (4k * T))

N₁ / N₂ = 4

∴ The ratio of the number of molecules in jar A to those in jar B is 4:1. Hence, option 4) is correct.

For the given problem, we are given that:

• The angle of banking, θ = 45°
• The maximum speed is twice the optimum speed, i.e., v_max = 2v_opt

Using the equation for the maximum speed:

v_max = √(rg(tanθ + μ) / (1 - μ tanθ))

Substituting the value of θ = 45° and v_opt = √(rg tan 45°), we get:

v_opt = √(rg)

From the equation for the maximum speed:

(v_max)² = (v_opt)² (tanθ + μ) / (1 - μ tanθ)

Substituting 2v_opt for v_max, we get:

(2v_opt)² = (v_opt)² (tanθ + μ) / (1 - μ tanθ)

This simplifies to:

4(1 - μ) = 1 + μ

4 - 4μ = 1 + μ

3 = 5μ

μ = 0.6

Answer:

The coefficient of static friction between the wheels of the car and the road is 0.6.