JEE Physics - MCQ Practice Questions

Step 1: Differentiate the velocity function

The acceleration is the time derivative of velocity:

**Step 2: Evaluate at $t=2\text{s}$

Result:

Step 1: Recall the kinematic equation

At maximum height, the final velocity is $v=0$. We use the equation

where $u=20\text{m/s}$ is the initial velocity and $g=10{\text{m/s}}^2$.

**Step 2: Substitute and solve for $h$

Result

Step 1: Express momentum in terms of base quantities

Momentum is defined as the product of mass and velocity:

Step 2: Substitute SI base units

The SI unit of mass is $\text{kg}$ and the SI unit of velocity is $\text{m/s}$, so:

Step 3: Note on equivalent units

$\text{N⋅s}$ (newton-second) is dimensionally equivalent to $\text{kg⋅m/s}$, since $1\text{N}=1{\text{kg⋅m/s}}^2$, giving:

However, $\text{N⋅s}$ is a derived combination and is not the fundamental SI expression for momentum.

Result:

The SI unit of momentum is:

Step 1: Write Newton's Second Law

From Newton's second law, $F=ma$, so the acceleration is

Step 2: Set up the ratio of accelerations

Since the force $F$ is the same for both bodies, the acceleration is inversely proportional to the mass:

Step 3: Substitute the mass ratio

Given $m_A:m_B=1:2$, we have

Result

The ratio of their accelerations is

Step 1: Find the acceleration

Using the formula $a=\frac{v-u}{t}$:

Step 2: Find the distance using kinematics

Using $s=ut+\frac{1}{2}a{t}^2$:

Result:

Step 1: Recall Newton's Second Law

Newton's second law states:

Step 2: Solve for mass

Rearranging for $m$:

Step 3: Substitute the known values

With $F=10\text{N}$ and $a=2{\text{m/s}}^2$:

Newton's First Law

A body moving with uniform velocity has a constant velocity, which means its acceleration is zero:

By Newton's Second Law:

Substituting $a=0$:

Therefore, the net force on the body is:

Step 1: Define the velocity direction

For a particle in circular motion, the velocity $\mathbf{v}$ is always tangent to the circle, pointing along the circumference at any given instant.

Step 2: Define centripetal acceleration

The centripetal acceleration ${\mathbf{a}}_c$ is defined as:

where $v$ is the speed, $r$ is the radius of the circle, and $\hat{r}$ is the unit vector pointing radially outward from the center. The negative sign indicates the acceleration points radially inward.

**Step 3: Relationship between ${\mathbf{a}}_c$ and $\mathbf{v}$

Since $\mathbf{v}$ is tangential and ${\mathbf{a}}_c$ is radial, they satisfy:

confirming that ${\mathbf{a}}_c$ is perpendicular to $\mathbf{v}$ at every point on the path.

Conclusion

The centripetal acceleration is directed:

Centripetal acceleration ac = v²/r = (20)²/50 = $\frac{400}{50}$ = 8 m/s²

Range R = u²sin(2θ)/g = (40)²sin(90°)/10 = $\frac{1600}{10}$ = 160 m

Friction is f = μN where μ depends on surface nature and N is normal force. Static friction can be equal to or less than maximum static friction.

Efficiency = (Work output/Work input) × 100. 80 = (8000/W) × 100 → W = 10000 J

v = ωr = 10 × 0.2 = 2 m/s

Total mass = m + 2m + 3m = 6m. Using F = Ma: a = F/(6m)

Impulse = Δ(momentum) = m(v₂ - v₁) = 5(10 - 4) = 30 N·s

Using v² = u² - 2as, where v = 0, u = 20 m/s, a = 5 m/s². Therefore, s = u²/(2a) = $\frac{400}{10}$ = 40 m

At maximum angle, tan(θ) = μs = 0.5, so θ = tan⁻¹(0.5) ≈ 26.6°

In elastic collision between equal masses, if one is at rest, they exchange velocities completely. This satisfies both momentum and energy conservation.

At h/2, PE = mg(h/2), Total energy = mgh. KE = mgh - mg(h/2) = mg(h/2). Fraction = (h/2)/h = $\frac{1}{2}$

Range = u²sin(2θ)/g = (400 × sin(90°))/10 = $\frac{400}{10}$ = 40 m