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CONCEPT:

Physical Quantities and Dimensions

• Physical quantities are properties or characteristics of a system that can be measured or calculated from other measurements.
• Each physical quantity has a dimension that can be represented by a combination of the basic dimensions (mass, length, time, etc.).
• In order for a mathematical operation involving physical quantities to be physically meaningful, the quantities must have compatible dimensions.

EXPLANATION:

• Let's examine the given options:
• Option 1: (P-Q)/R
• For this operation to be meaningful, P and Q must have the same dimensions because only quantities with the same dimensions can be subtracted. If P and Q have the same dimensions, (P-Q) will have the same dimension as P (or Q), but it is given P and Q have a different dimension thus it cannot be meaningful quantity.
• Option 2: PQ-R
• For this operation to be meaningful, PQ and R must have the same dimensions. PQ represents the product of P and Q, so the dimensions of PQ are the product of the dimensions of P and Q. Unless R has the same dimensions as PQ.
• Option 3: PQ/R
• This operation is meaningful if the dimensions of PQ are compatible with the dimensions of R. Specifically, the dimensions of PQ should be equal to the dimensions of R. This means that the dimensions of P multiplied by the dimensions of Q should result in the dimensions of R.
• Option 4: (PR - Q²)/R
• For this operation is meaningful, PR and Q² must have the same dimensions. Since Q² represents Q multiplied by itself, its dimensions are the square of the dimensions of Q. PR represents the product of P and R, so its dimensions are the product of the dimensions of P and R. Additionally, (PR - Q²)/R must be dimensionally consistent.

Therefore, the correct answer is option 1: (P-Q)/R.

CONCEPT:

Position Vector and Acceleration

• The position vector of a particle provides information about its location in space as a function of time.
• Acceleration is the rate of change of velocity with respect to time.
• To find acceleration, we need to differentiate the position vector twice with respect to time.

EXPLANATION:

• Given position vector: r(t) = 15t² i + (4 - 20t²) j
• First, we find the velocity by differentiating the position vector with respect to time:
• *v(*t) = dr(t)/dt = 30t i - 40t j
• Next, we find the acceleration by differentiating the velocity with respect to time:
• a(t) = dv(t)/dt = 30 i - 40 j
• At t = 1, the acceleration vector is:
• a(1) = 30 i - 40 j
• To find the magnitude of the acceleration:
• |a(1)| = √(30² + (-40)²)
• |a(1)| = √(900 + 1600)
• |a(1)| = √2500
• |a(1)| = 50

Therefore, the correct answer is option 4: 50.

CONCEPT:

Projectile Motion

• When two projectiles are launched with the same initial velocity but at different angles, their ranges and maximum heights can be compared.
• The range (R) of a projectile is given by the formula: R = (v² × sin(2θ)) / g where:
• v is the initial velocity
• θ is the angle of projection
• g is the acceleration due to gravity
• The maximum height (H) of a projectile is given by the formula: H = (v² × sin²(θ)) / (2g)

EXPLANATION:

• Let's examine the given options:
• Option 1: R₁ > R₂ and H₁ = H₂
• This option suggests the range of the first projectile is greater than that of the second, and their maximum heights are equal. This is incorrect because if the angles of projection are different, their maximum heights will also differ.
• Option 2: R₁ = R₂ and H₁ < H₂
• This option suggests the ranges of both projectiles are equal, but the height of the first is less than that of the second. This is correct. When two angles are complementary (e.g., 42° and 48°), the projectiles have equal ranges but different maximum heights.
• Option 3: R₁ < R₂ and H₁ < H₂
• This is incorrect because complementary angles result in equal ranges, so R₁ cannot be less than R₂ in this context.
• Option 4: R₁ = R₂ and H₁ = H₂
• This option assumes both range and height are equal, which is incorrect as maximum height depends on the square of sine of the angle, and hence varies with angle.

Therefore, the correct answer is: Option 2 — R₁ = R₂ and H₁ < H₂.

CONCEPT:

Chance of Overturn for a Cyclist

• When a cyclist takes a turn, the chance of overturn depends on the speed of the cyclist and the centripetal force required to keep the cyclist on the curved path.
• The centripetal force is proportional to the square of the speed of the cyclist.

EXPLANATION:

• Let's examine the given options:
• Option 1: is doubled
• This option suggests that the chance of overturn is directly proportional to the speed. However, the chance of overturn actually depends on the square of the speed.
• Option 2: is halved
• This option suggests that the chance of overturn decreases as the speed increases, which is incorrect because the chance increases with speed.
• Option 3: becomes four times
• This option is correct. Since the chance of overturn is proportional to the square of the speed, doubling the speed will increase the chance of overturn by a factor of four.
• Option 4: becomes 1/4 times
• This option suggests that the chance of overturn decreases by a factor of four, which is incorrect because the chance increases with speed.

Therefore, the correct answer is option 3: becomes four times.

Calculation

From the above diagram, we have

N = mg cosθ + (mv² / R) sinθ                        (1)

f_max = μN ⇒ f_max = μ_s mg cosθ + (μ_s mv² / R) sinθ

mg sinθ + f_max = (mv² / R) cosθ                     (2)

Putting the value

mg sinθ + μ_s mg cosθ + (μ_s mv² / R) sinθ = (mv² / R) cosθ

g sinθ + μ_s g cosθ = (v² / R)(cosθ − μ_s sinθ)

gR [ (tanθ + μ_s) / (1 − μ_s tanθ) ] = v²

v = √[ gR (tanθ + μ_s) / (1 − μ_s tanθ) ]

Calcultion:

Initial momentum of Ball A = 0.2 × 0.3 = 0.06 kg·m/s

Initial momentum of Ball B = 0.4 × v_B

Total initial momentum = 0.06 + 0.4 × v_B

According to conservation of momentum:

0.06 + 0.4 × v_B = 0

0.4 × v_B = -0.06

v_B = -0.06 / 0.4

v_B = -0.15 m/sec

Calculation:

 A sphere rolling without slipping has both translational and rotational kinetic energy.

Rotational Kinetic Energy (K_R) = (1/2) × I × (v² / r²)

Moment of Inertia of a solid sphere, I = (2/5)mr²

So, K_R = (1/2) × (2/5)mr² × (v² / r²) = (1/5)mv²

Translational Kinetic Energy (K_Tr) = (1/2)mv²

Total Energy (K_Total) = K_R + K_Tr = (1/5)mv² + (1/2)mv²

Convert to common denominator:

K_Total = (2/10 + 5/10)mv² = (7/10)mv²

Fraction of Rotational Energy = K_R / K_Total = (1/5)mv² / (7/10)mv²

Cancel mv²:

Fraction = (1/5) / (7/10) = (2/7)

Answer: Option 1) 2/7

Concept:

To escape Earth's gravity, a satellite in low Earth orbit must increase its velocity from orbital velocity (v_o) to escape velocity (v_e).

Formulae:

• Orbital velocity (v_o) = √(GM / R)
• Escape velocity (v_e) = √(2GM / R)

So, v_e = √2 × v_o

Percentage increase in velocity = [(v_e - v_o) / v_o] × 100%

= [(√2 × v_o - v_o) / v_o] × 100%

= (√2 - 1) × 100%

≈ (1.414 - 1) × 100% = 0.414 × 100% = 41.4%

Answer: Option 1) 41%

Calculation:

Orbital velocity (v) of a satellite at distance r from the center of the Earth is given by:

v = √(GM / r)

Let v₀ be the orbital velocity just above the Earth's surface. Then:

v₀ = √(GM / R)

Now consider a satellite at an altitude of half the Earth's radius. So, total distance from Earth's center = R + R/2 = (3/2)R

New orbital velocity (v') = √(GM / (3R/2)) = √((2/3) × GM / R)

= √(2/3) × √(GM / R)

= √(2/3) × v₀

Answer: Option 1) √(2/3) × v₀

CONCEPT:

Capillary Action and Contact Angle

• When a liquid rises or falls in a narrow tube due to the adhesive forces between the liquid and the tube wall and the cohesive forces within the liquid, it is known as capillary action.
• The angle of contact (θ) is the angle formed between the tangent to the liquid surface and the solid surface inside the tube.
• Water typically has a low contact angle with materials it can wet (hydrophilic materials), leading to capillary rise.
• If the tube is made hydrophobic (e.g., by coating with wax), the contact angle increases, leading to a decrease in the height of the liquid rise.

EXPLANATION:

• Let's examine the given options:
• Option 1: θ increases and h also increases
• This option is incorrect because if the contact angle θ increases (due to wax coating), the height h up to which water rises decreases.
• Option 2: θ decreases and h also decreases
• This option is incorrect because if the contact angle θ decreases, the height h up to which water rises increases.
• Option 3: θ increases and h decreases
• This option is correct because if the contact angle θ increases (due to wax coating), the height h up to which water rises decreases.
• Option 4: θ decreases and h increases
• This option is incorrect because if the contact angle θ decreases, the height h up to which water rises increases.

Therefore, the correct answer is option 3: θ increases and h decreases.