JEE Physics - MCQ Practice Questions

Step 1: Identify the formula for maximum static friction.

Step 2: Substitute the normal force $N=mg$ for a block on a horizontal surface.

Step 3: Substitute the known values $\mu =0.2$, $m=5\text{kg}$, and $g=10{\text{m/s}}^2$.

Step 1: Recall the range formula

The horizontal range of a projectile is given by

Step 2: Substitute the known values

Here $u=40\text{m/s}$, $\theta =45°$, and $g=10{\text{m/s}}^2$.

Step 3: Final Answer

Step 1: Recall the elastic collision formula

For a perfectly elastic collision, the post-collision velocity of mass $m_1$ is given by

Step 2: Identify the known values

Step 3: Substitute and simplify

Result

The velocity of the $2\text{kg}$ mass after the collision is

Step 1: Write the tension formula

The tension in the rope is related to the weight and horizontal force by

Step 2: Substitute known values

The weight of the mass is $mg=40\times 10=400\text{N}$, and the maximum tension is $T=500\text{N}$. Substituting:

**Step 3: Solve for $F^2$

Squaring both sides:

**Step 4: Solve for $F$

Step 1: Find the system acceleration

The total mass of the system is $m_1+m_2=3+2=5\text{kg}$. Applying Newton's second law to the entire system:

Step 2: Find the tension in the string

The tension $T$ accelerates block $m_2$ alone:

Result:

Step 1: Find time to maximum height

Since the ball returns to the same height in $4\text{s}$, by symmetry the time to reach maximum height is

Step 2: Find initial velocity

At maximum height the velocity is zero, so using $v=u-gt$ with $v=0$:

N - mg = ma. N = m(g+a) = 50(10+2) = 600 N

Centripetal force is perpendicular to velocity (displacement), so W = F·s·cosθ = F·s·cos90° = 0

v = dx/dt = 8t - 2. At t = 1s, v = 8(1) - 2 = 6 m/s

v = u + ∫a dt = 2 + ∫₀² 3t² dt = 2 + [t³]₀² = 2 + 8 = 10 m/s

Both objects have constant acceleration -g. If thrown with same initial velocity but at different times, they will have the same velocity at different times. For same velocity at same instant: vA = u - gt₁ and vB = u - gt₂. Since t₁ ≠ t₂, velocities won't be equal at the same instant.

Deceleration a = g(sinθ + μcosθ) = 10(0.5 + 0.2×0.866) = 10(0.5 + 0.173) = 6.73 m/s². Using v² = u² - 2as: 0 = 100 - 2(6.73)s, s ≈ 3.2 m

For SHM, ωmax = ω × A where ω = √(k/m) = √($\frac{50}{2}$) = 5 rad/s. vmax = 5 × 0.1 = 0.5 m/s

e = √(h₂/h₁) = √(h/2 ÷ h) = √($\frac{1}{2}$) ≈ 0.707

By momentum conservation: 4(5) - 6(3) = (4+6)v → 20 - 18 = 10v → v = 0.2 m/s

Tangential acceleration at = r(dω/dt) = 5 × 2 = 10 m/s² (constant as dω/dt = 2)

Orbital velocity v = √(GM/r). If r → 2r, then v' = √(GM/2r) = v/√2

At maximum height, vertical component vᵧ = 0. Horizontal component vₓ remains 20 m/s. Total velocity = 20 m/s

a = g(sinθ - μcosθ) = 10(sin30° - 0.1×cos30°) = 10(0.5 - 0.1×0.866) = 10(0.5 - 0.0866) = 4.134 ≈ 4.3 m/s²

Total acceleration = F/(m₁+m₂) = $\frac{10}{5}$ = 2 m/s². Tension = F - m₁a = 10 - 2×2 = 6 N (or T = m₂a = 3×2 = 6 N)