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

An object experiences a net force of 10 N and accelerates at 2 m/s². Its mass is:

Accepted Answer

Correct answer: B. 5 kg — **Step 1: Recall Newton's Second Law** Newton's second law states: F = ma **Step 2: Solve for mass** Rearranging for m : m = \frac{F}{a} **Step 3: Substitute the known values** With F = 10\ 	ext{N} and a = 2\ 	ext{m/s}^2 : \begin{aligned} m &= \frac{10\ 	ext{N}}{2\ 	ext{m/s}^2} \ 6pt] &= \boxed{5\ 	ext{kg}} \end{aligned}

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A. 2 kg

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C. 8 kg

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D. 20 kg

Question 7

A body moving with uniform velocity has a net force of:

Accepted Answer

Correct answer: B. F = 0 — **Newton's First Law** A body moving with uniform velocity has a constant velocity, which means its acceleration is zero: a = 0 By Newton's Second Law: F_{	ext{net}} = ma Substituting a = 0 : \begin{aligned} F_{	ext{net}} &= m \cdot 0 \ &= 0 \end{aligned} Therefore, the net force on the body is: \boxed{F_{	ext{net}} = 0}

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A. F = ma

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C. F = mv

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D. F = mv²/r

Question 8

A particle undergoing circular motion has centripetal acceleration. Its direction is:

Accepted Answer

Correct answer: C. Radially inward — **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: \mathbf{a}_c = -\frac{v^2}{r}\,\hat{r} 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: \mathbf{a}_c \cdot \mathbf{v} = 0 confirming that \mathbf{a}_c is perpendicular to \mathbf{v} at every point on the path. **Conclusion** The centripetal acceleration is directed: \boxed{	ext{toward the center of the circle (radially inward), perpendicular to the velocity}}

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A. Tangential to circle

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B. Radially outward

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D. Perpendicular to velocity

JEE Physics - MCQ Practice Questions