Class 11 Physics Half Yearly Question Paper 2025 Answer Key (NCERT Based)
Instructions :
• All questions are compulsory.
• Questions (1-9) carry 1 mark each.
• Questions (10-12) carry 2 marks each.
• Questions (13-14) carry 3 marks each.
• Question (15) case study, carry 4 marks.
• Questions (16-17) carry 5 marks each.
1. The number of significant figures in 0.007 m2 is :
(a) 4
(b) 3
(c) 2
(d) 1
Answer – (d) 1
2. A car is moving with a uniform velocity of 60 km/h. What is its acceleration?
(a) 0 km/h2
(b) 10 km/h2
(c) 20 km/h2
(d) 30 km/h2
Answer – (a) 0 km/h2
3. Which of the following is the essential characteristic of a projectile?
(a) Zero velocity at the highest point
(b) Initial velocity inclined to the horizontal
(c) Constant acceleration perpendicular to the velocity
(d) None of these
Answer – (b) Initial velocity inclined to the horizontal
4. The recoil of cannon after firing is due to :
(a) Newton’s first law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) None of the above
Answer – (c) Newton’s third law of motion
5. SI Unit of Power is …………
Answer – Watt
6. The Physical quantity having same unit as that of Torque is ………….
Answer – Work
7. Relate torque and Angular Momentum.
Answer – Torque (τ) = dL/dt
8. How much is the magnitude of the unit vector?
Answer – The magnitude of the unit vector is 1.
9. Assertion (A) : The speed of a planet increases when it comes closer to sun and vice-versa.
Reason (R) : The angular momentum of planet is conserved.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Answer – (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
10. Check the dimensional validity of S = ut + ½at2 where symbols have their usual meaning in reference to uniformly accelerated motion.
Answer – LHS = [S] = L
RHS = [ut] + [½at2]
[ut] = (LT–1)(T) = L
[½at2] = (LT–2)(T2) = L
Since the dimensions of all terms are the same [L], the equation is dimensionally valid.
11. A car weighs 1800 Kg. The distance between its front and back axle is 1.8 m. Its Centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level found on each front wheel and each back wheel.
Answer – Mass (M) = 1800 kg,
Weight (W) = 1800 × 9.8 = 17640 N
Wheelbase (L) = 1.8 m, CG is 1.05 m from front axle
Front wheel force = Ff (total, so per wheel: Ff/2)
Back wheel force = Fb (per wheel: [Fb/2)
Ff + Fb = W
Back Axle Force × Wheelbase = Weight × Distance to CG
Fb × 1.8 = W × 1.05
Fb × 1.8 = 17640 × 1.05
Fb = 10290 N
Ff = 17640 – 10290 = 7350 N
Each front wheel = 7350/2 = 3675 N
Each back wheel = 10290/2 = 5145 N
12. Why is it difficult to push a lawn mower than to pull it?
Answer – It is more difficult to push a lawn mower than to pull it because of the direction of the applied force. When pushing, the force has a downward component that increases the effective weight and friction on the ground. When pulling, the force has an upward component that reduces the effective weight and friction, making the mower easier to move.
13. State and prove Work Energy Theorem.
Answer – The work done by the net force on a body is equal to the change in its kinetic energy.
Wnet = ∆KE = Kf – Ki = ½mv2 – ½mu2
Proof :
By definition, the work done by a constant net force is :
Wnet = Fnet × S
According to Newton’s Second Law of Motion :
Fnet = ma
Substitute this into the work equation :
Wnet = (ma) × S ……..(i)
For uniformly accelerated motion, the third equation of motion is:
v2 = u2 + 2aS
aS = (v2 – u2)/2 ……..(ii)
Substitute the expression for aS from equation (ii) into the work done equation (i) :
Wnet = m(v2 – u2)/2
Wnet = ½mv2 – ½mu2
Wnet = KEf – KEi = ∆KE
Work done = Change in Kinetic Energy
Hence, the Work-Energy Theorem is proved.
14. Define projectile. An aircraft executes a horizontal loop of radius 1000 m with a steady speed of 900 km/h. Compare its centripetal acceleration with acceleration due to gravity.
Answer – Projectile is a body thrown with initial velocity under gravity, moving in a curved path.
Radius of loop (r) = 1000 m
Speed of aircraft (v) = 900 km/h = 250 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Centripetal acceleration (ac) = v2/r = (250)2/1000 = 62.5 m/s2
The comparison is done by finding the ratio, ac/g = 62.5/9.8 = 6.38
The centripetal acceleration of the aircraft is approximately 6.38 times the acceleration due to gravity (g).
15. According to Newton’s second law of motion, F = ma where F is force required to produce an acceleration in a body of mass m, if a = 0, then F = 0 i.e. no external force is required to move a body uniformly along a straight line. If a force acts on a body for t seconds, the effect of force is given by impulse = F × t = change in linear momentum of body.
Questions :
(i) a cricket ball of mass 150 g is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. If duration on contact between the ball and bat is 0.01 s the impulse of force is :
(a) 7.4 Ns
(b) 4.8 Ns
(c) 1.2 Ns
(d) 4.7 Ns
Answer – (b) 4.8 Ns
Mass = 150 g = 0.15 kg
Initial velocity (u) = +12 m/s
Final velocity (v) = –20 m/s (opposite direction)
Impulse = Change in momentum = mv – mu = m(v – u) = 0.15 (– 20 – 12) = – 4.8 Ns
The magnitude of the impulse is 4.8 Ns.
(ii) Average force exerted by the bat is :
(a) 480 N
(b) 120 N
(c) 1200 N
(d) 840 N
Answer – (a) 480 N
Average force (Favg) = Impulse/t = 4.8/0.01 = 480 N
(iii) In a uniform motion the force acting is :
(a) 0
(b) 1 N
(c) 2 N
(d) can’t be said
Answer – (a) 0
(iv) The unit of impulse is …………..
Answer – Newton-second (Ns)
16. A projectile is thrown at an angle θ from the horizontal with velocity u under the gravitational field of Earth. Define and find expression for Time of flight, Horizontal Range and Maximum Height.
Answer : Time of Flight – The time of flight is the total time taken by the projectile to complete its motion through the air, that is, the time interval between its projection and the moment it returns to the same vertical level from which it was projected.
T = 2usinθ / g
• Horizontal Range – The horizontal range of a projectile is the total horizontal distance covered by it between the point of projection and the point where it again reaches the same vertical level.
R = u2sin2θ / g
• Maximum Height – The maximum height is the greatest vertical distance attained by the projectile above the horizontal plane during its motion.
Hmax = u2sin2θ / 2g
17. Derive all three equations of motion for a uniformly accelerated body.
Answer –
• First Equation (v = u + at) :
Acceleration is defined as the rate of change of velocity with respect to time: a = (v – u)/t
so, v = u + at
• Second Equation (s = ut + ½at2) :
Average velocity (vavg) is (u + v)/2.
Displacement (s) is the product of average velocity and time: s = vavg × t
Substituting v = u + at into the average velocity equation: vavg = (u + u + at)/2 = u + ½at
Therefore, s = (u + ½at) × t = ut + ½at2
• Third Equation (v2 = u2 + 2as):
From the first equation, t = (v – u)/a
Substitute this value of t into the second equation: s = u[(v – u)/a] + ½a[(v – u)/a]2
s = (uv – u2)/a + ½(v2 – 2uv + u2)/a
2as = 2uv – 2u2 + v2 – 2uv + u2
v2 = u2 + 2as