JEE Physics — MCQ Practice Questions & Answers

Question 1

A particle moves in a straight line with velocity v = 3t² + 2t m/s. What is its acceleration at t = 2s?

Accepted Answer

Correct answer: A. 14 m/s² — **Step 1: Differentiate the velocity function** The acceleration is the time derivative of velocity: a = \frac{dv}{dt} = \frac{d}{dt}\!\left(3t^2 + 2tight) = 6t + 2 **Step 2: Evaluate at t = 2\,	ext{s} ** \begin{aligned} a\big|_{t=2} &= 6(2) + 2 \ &= 12 + 2 \ &= 14 \ 	ext{m/s}^2 \end{aligned} **Result:** \boxed{a = 14 \ 	ext{m/s}^2}

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B. 16 m/s²

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C. 18 m/s²

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D. 20 m/s²

Question 2

An object is thrown vertically upward with initial velocity 20 m/s. What is the maximum height reached? (g = 10 m/s²)

Accepted Answer

Correct answer: C. 20 m — **Step 1: Recall the kinematic equation** At maximum height, the final velocity is v = 0 . We use the equation v^2 = u^2 - 2gh where u = 20\ 	ext{m/s} is the initial velocity and g = 10\ 	ext{m/s}^2 . **Step 2: Substitute and solve for h ** \begin{aligned} 0 &= u^2 - 2gh \ 0 &= (20)^2 - 2(10)h \ 0 &= 400 - 20h \ 20h &= 400 \ h &= \frac{400}{20} \end{aligned} **Result** \boxed{h = 20\ 	ext{m}}

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A. 10 m

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B. 15 m

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D. 25 m

Question 3

The SI unit of momentum is:

Accepted Answer

Correct answer: A. kg·m/s — **Step 1: Express momentum in terms of base quantities** Momentum is defined as the product of mass and velocity: 	ext{Momentum} = 	ext{mass} 	imes 	ext{velocity} **Step 2: Substitute SI base units** The SI unit of mass is 	ext{kg} and the SI unit of velocity is 	ext{m/s} , so: \begin{aligned} 	ext{Unit of Momentum} &= 	ext{kg} 	imes \frac{	ext{m}}{	ext{s}} \ 6pt] &= 	ext{kg·m/s} \end{aligned} **Step 3: Note on equivalent units** 	ext{N·s} (newton-second) is dimensionally equivalent to 	ext{kg·m/s} , since 1\,	ext{N} = 1\,	ext{kg·m/s}^2 , giving: 1\,	ext{N·s} = 1\,	ext{kg·m/s}^2 	imes 	ext{s} = 1\,	ext{kg·m/s} However, 	ext{N·s} is a derived combination and is **not** the fundamental SI expression for momentum. **Result:** The SI unit of momentum is: \boxed{	ext{kg·m/s}}

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B. kg·m/s²

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C. kg/m·s

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D. N·s

Question 4

Two bodies A and B have masses in ratio 1:2. They experience equal forces. The ratio of their accelerations is:

Accepted Answer

Correct answer: B. 2:1 — **Step 1: Write Newton's Second Law** From Newton's second law, F = ma , so the acceleration is a = \frac{F}{m}. **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: \frac{a_A}{a_B} = \frac{F/m_A}{F/m_B} = \frac{m_B}{m_A}. **Step 3: Substitute the mass ratio** Given m_A : m_B = 1 : 2 , we have \frac{a_A}{a_B} = \frac{m_B}{m_A} = \frac{2}{1}. **Result** The ratio of their accelerations is \boxed{a_A : a_B = 2 : 1.}

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A. 1:2

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C. 1:4

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D. 4:1

Question 5

A car moving at 20 m/s is brought to rest in 4 seconds with uniform deceleration. The distance covered is:

Accepted Answer

Correct answer: C. 40 m — **Step 1: Find the acceleration** Using the formula a = \dfrac{v - u}{t} : a = \frac{0 - 20}{4} = -5 \ 	ext{m/s}^2 **Step 2: Find the distance using kinematics** Using s = ut + \dfrac{1}{2}at^2 : \begin{aligned} s &= (20)(4) + \frac{1}{2}(-5)(4)^2 \ &= 80 + \frac{1}{2}(-5)(16) \ &= 80 - 40 \ &= 40 \ 	ext{m} \end{aligned} **Result:** \boxed{s = 40 \ 	ext{m}}

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A. 20 m

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B. 30 m

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D. 50 m

Question 6

A block of mass 5 kg is placed on a rough surface (μ = 0.2). The maximum static friction is:

Accepted Answer

Correct answer: C. 10 N — **Step 1:** Identify the formula for maximum static friction. f_{\max} = \mu 	imes N **Step 2:** Substitute the normal force N = mg for a block on a horizontal surface. f_{\max} = \mu 	imes mg **Step 3:** Substitute the known values \mu = 0.2 , m = 5\ 	ext{kg} , and g = 10\ 	ext{m/s}^2 . \begin{aligned} f_{\max} &= \mu 	imes mg \ &= 0.2 	imes 5 	imes 10 \ &= \boxed{10 \ 	ext{N}} \end{aligned}

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A. 5 N

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B. 8 N

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D. 12 N

Question 7

A projectile is fired at 45° angle with initial velocity 40 m/s. The range is: (g = 10 m/s²)

Accepted Answer

Correct answer: C. 160 m — **Step 1: Recall the range formula** The horizontal range of a projectile is given by R = \frac{u^2 \sin(2	heta)}{g} **Step 2: Substitute the known values** Here u = 40\ 	ext{m/s} , 	heta = 45° , and g = 10\ 	ext{m/s}^2 . \begin{aligned} R &= \frac{(40)^2 \sin(2 	imes 45°)}{10} \ &= \frac{1600 	imes \sin(90°)}{10} \ &= \frac{1600 	imes 1}{10} \ &= \frac{1600}{10} \end{aligned} **Step 3: Final Answer** \boxed{R = 160\ 	ext{m}}

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A. 80 m

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B. 120 m

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D. 200 m

Question 8

A 2 kg mass moving at 3 m/s collides elastically with a 1 kg mass at rest. The velocity of 2 kg mass after collision is:

Accepted Answer

Correct answer: A. 1 m/s — **Step 1: Recall the elastic collision formula** For a perfectly elastic collision, the post-collision velocity of mass m_1 is given by v_1 = \frac{(m_1 - m_2)\,u_1 + 2m_2\,u_2}{m_1 + m_2} **Step 2: Identify the known values** m_1 = 2\ 	ext{kg}, \quad m_2 = 1\ 	ext{kg}, \quad u_1 = 3\ 	ext{m/s}, \quad u_2 = 0\ 	ext{m/s} **Step 3: Substitute and simplify** \begin{aligned} v_1 &= \frac{(m_1 - m_2)\,u_1 + 2m_2\,u_2}{m_1 + m_2} \ 6pt] &= \frac{(2 - 1)(3) + 2(1)(0)}{2 + 1} \ 6pt] &= \frac{(1)(3) + 0}{3} \ 6pt] &= \frac{3}{3} \ 6pt] &= 1\ 	ext{m/s} \end{aligned} **Result** The velocity of the 2\ 	ext{kg} mass after the collision is \boxed{v_1 = 1\ 	ext{m/s}}

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B. 2 m/s

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C. 3 m/s

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D. 4 m/s

JEE Physics - MCQ Practice Questions