JEE Physics — MCQ Practice Questions & Answers

Question 1

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 2

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 3

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}}

Answer

B. 2 m/s

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C. 3 m/s

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D. 4 m/s

Question 4

A rope can withstand maximum tension of 500 N. A mass of 40 kg is suspended and pulled horizontally. The maximum horizontal force is:

Accepted Answer

Correct answer: A. 300 N — **Step 1: Write the tension formula** The tension in the rope is related to the weight and horizontal force by T = \sqrt{(mg)^2 + F^2} **Step 2: Substitute known values** The weight of the mass is mg = 40 	imes 10 = 400\ 	ext{N} , and the maximum tension is T = 500\ 	ext{N} . Substituting: 500 = \sqrt{(400)^2 + F^2} **Step 3: Solve for F^2 ** Squaring both sides: \begin{aligned} (500)^2 &= (400)^2 + F^2 \ 250000 &= 160000 + F^2 \ F^2 &= 250000 - 160000 \ F^2 &= 90000 \end{aligned} **Step 4: Solve for F ** F = \sqrt{90000} = \boxed{300\ 	ext{N}}

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B. 400 N

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C. 450 N

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D. 500 N

Question 5

Two blocks (m₁ = 3 kg, m₂ = 2 kg) are connected by a string on a frictionless surface. Applied force F = 20 N. Tension in string is:

Accepted Answer

Correct answer: A. 8 N — **Step 1: Find the system acceleration** The total mass of the system is m_1 + m_2 = 3 + 2 = 5\ 	ext{kg} . Applying Newton's second law to the entire system: a = \frac{F}{m_1 + m_2} = \frac{20}{5} = 4\ 	ext{m/s}^2 **Step 2: Find the tension in the string** The tension T accelerates block m_2 alone: \begin{aligned} T &= m_2 \, a \ &= 2 	imes 4 \ &= 8\ 	ext{N} \end{aligned} **Result:** \boxed{T = 8\ 	ext{N}}

Answer

B. 10 N

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C. 12 N

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D. 15 N

Question 6

A ball thrown upward returns to same height in 4s. Initial velocity is: (g = 10 m/s²)

Accepted Answer

Correct answer: C. 20 m/s — **Step 1: Find time to maximum height** Since the ball returns to the same height in 4\ 	ext{s} , by symmetry the time to reach maximum height is t = \frac{4}{2} = 2\ 	ext{s} **Step 2: Find initial velocity** At maximum height the velocity is zero, so using v = u - gt with v = 0 : \begin{aligned} 0 &= u - g t \ u &= g t \ u &= 10 	imes 2 \end{aligned} \boxed{u = 20\ 	ext{m/s}}

Answer

A. 10 m/s

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B. 15 m/s

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D. 25 m/s

Question 7

A 50 kg person stands in an elevator accelerating upward at 2 m/s². Normal force is: (g = 10 m/s²)

Accepted Answer

Correct answer: C. 600 N — N - mg = ma. N = m(g+a) = 50(10+2) = 600 N

Answer

A. 400 N

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B. 500 N

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D. 700 N

Question 8

Work done by centripetal force in circular motion is:

Accepted Answer

Correct answer: C. Zero — Centripetal force is perpendicular to velocity (displacement), so W = F·s·cosθ = F·s·cos90° = 0

Answer

A. Positive

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B. Negative

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D. Depends on radius

JEE Physics - MCQ Practice Questions