Answer: d. 4.All non-zero digits are significant.

Answer: c. $12\text{ m/s}^2$.Velocity ($v$) is the first derivative of position ($X$): $$v = \frac{dX}{dt} = -3 + 12t$$ Acceleration ($a$) is the second derivative: $$a = \frac{dv}{dt} = 12\text{ m/s}^2$$

Answer: c. displacement.The integral of velocity with respect to time yields displacement.

Answer: c. momentum.According to the impulse-momentum theorem, Impulse = $\Delta p$.

Answer: c. zero.Conservative forces are path-independent; thus, a closed loop yields zero net work.

Answer: c. I.It is conventionally represented by K, k, or sometimes I depending on the specific text notation.

Answer: c. Kepler’s law.Specifically, Kepler’s Laws of Planetary Motion.

Answer: a. Modulus of elasticity.Hooke’s law states that stress is directly proportional to strain, and the ratio defines the modulus of elasticity.

Answer: b. the change in internal energy is zero.Because internal energy is a state function and the system returns to its initial state.

Answer: b. temperature.From the formula $$\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$$ at constant pressure, it is directly proportional to temperature.

When a vector is resolved along two mutually perpendicular directions, the resulting components are called rectangular components.If vector $A$ makes an angle $\theta$ with the x-axis, its horizontal component is $A_x = A \cos(\theta)$ and vertical component is $A_y = A \sin(\theta)$.

According to Newton’s Law of Universal Gravitation, $$F = G\frac{m_1 m_2}{r^2}$$ If masses are doubled ($2m_1$, $2m_2$) and distance is doubled ($2r$), the new force is: $$F’ = G\frac{(2m_1)(2m_2)}{(2r)^2} = \frac{4G(m_1 m_2)}{4r^2} = F$$ The force remains unchanged.

The First Law of Thermodynamics is the law of conservation of energy applied to thermodynamic systems.It states that the heat supplied to a system ($\Delta Q$) is equal to the sum of the change in its internal energy ($\Delta U$) and the work done by the system ($\Delta W$). Equation: $$\Delta Q = \Delta U + \Delta W$$
Applications:

• Isothermal Process: Temperature is constant, so $\Delta U = 0$. Thus, $\Delta Q = \Delta W$.
• Isochoric Process: Volume is constant, so $\Delta W = 0$. Thus, $\Delta Q = \Delta U$.

Pascal’s Law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.

Hydraulic Brakes: When the brake pedal is pressed, it exerts force on a small piston in the master cylinder, creating pressure in the brake fluid. According to Pascal’s Law, this exact pressure is transmitted through the fluid lines to larger pistons at the wheels. Because the wheel pistons have a larger area, the resulting force is amplified ($F_2 = P \times A_2$), enabling the vehicle to stop efficiently.

Comparing the given equation $y = 0.40 \sin(440t + 0.61)$ with the standard SHM equation $y = A \sin(\omega t + \phi)$:

• Amplitude ($A$): 0.40 units
• Angular Frequency ($\omega$): $440\text{ rad/s}$
• Frequency ($f$): $f = \frac{\omega}{2\pi} = \frac{440}{2 \times 3.1416} \approx 70.03\text{ Hz}$
• Time Period ($T$): $T = \frac{1}{f} = \frac{1}{70.03} \approx 0.014\text{ s}$

Let initial velocity be $v$ and angle of projection be $\theta$.Components are $v_x = v \cos\theta$ and $v_y = v \sin\theta$.

1. Time of Flight ($T$): Total time the projectile remains in the air. Net vertical displacement is zero. Using $s = ut + \frac{1}{2}at^2$, we get $0 = (v \sin\theta)T – \frac{1}{2}gT^2$. Solving for $T$ yields: $$T = \frac{2v \sin\theta}{g}$$
2. Maximum Height ($H$): At maximum height, final vertical velocity is zero. Using $v^2 – u^2 = 2as$, we get $0 – (v \sin\theta)^2 = -2gH$. Solving for $H$ yields: $$H = \frac{v^2 \sin^2\theta}{2g}$$
3. Horizontal Range ($R$): Horizontal distance covered during the time of flight. $R = v_x \times T = (v \cos\theta) \times \left(\frac{2v \sin\theta}{g}\right)$. Applying the trigonometric identity $2 \sin\theta \cos\theta = \sin(2\theta)$ yields: $$R = \frac{v^2 \sin(2\theta)}{g}$$