Q3134. Two bullets are fired horizontally and simultaneously towards each other from rooftops of two buildings 100 m apart and of same height 200 m with velocity 25 m/s. When and where will they collide? (g=10 m/s²)
✅ Correct Answer: (a) After 2 s at a height 180 m
🧠 Solution:
Distance between buildings d=100 m, height h=200 m, speed v=25 m/s
Relative velocity v_rel = 25 + 25 = 50 m/s
Time t = d / v_rel = 100 / 50 = 2 s
Vertical distance x = h - 1/2 g t² = 200 - 1/2*10*2² = 200 - 20 = 180 m
So, bullets collide at 180 m after 2 s.
Q3135. When an object is shot from a smooth inclined plane at 60° with horizontal, it travels distance x₁. When inclination is 30° with same velocity, distance is x₂. Then x₁:x₂ is
✅ Correct Answer: (b) 1:3
🧠 Solution:
Equation of motion along plane: v² = u² - 2g x sinθ
For case I: x₁ = u² / (2 g sin60°)
For case II: x₂ = u² / (2 g sin30°)
Then x₁/x₂ = sin30°/sin60° = 1/3
Q3159. A particle moving in a circle of radius R with uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle θ to the horizontal, the maximum height attained by it equals 4R. The angle of projection θ is then given by
✅ Correct Answer: (d) θ = π/2 - sin⁻¹( g T² / 2 R )
🧠 Solution:
Maximum height: H_max = u² sin²θ / 2g = 4R. Circular motion speed u = 2πR/T. Substituting, sin²θ = gT² / 2R ⇒ θ = π/2 - sin⁻¹( gT² / 2R ).
Q9394.
A particle moving with velocity v is acted by three forces shown by the vector triangle PQR.
The velocity of the particle will
✅ Correct Answer:
remain constant
🧠 Solution:
As the three forces are represented by
three sidesof a triangle taken in order,
then they will be in equilibrium.
Q3136. The x and y coordinates of a particle are x = -5t² and y = 10 respectively. Acceleration at t=2 s is
✅ Correct Answer: (c) -4 m/s²
🧠 Solution:
x = -5 t² ⇒ v_x = dx/dt = -10 t ⇒ a_x = dv_x/dt = -10 = -4 m/s²
y = 10 ⇒ a_y = 0
Net acceleration a_net = a_x i + a_y j = -4 i m/s²
Q3122. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is
✅ Correct Answer: (a) 90°
🧠 Solution:
Suppose two vectors are P and Q. It is given that |P + Q| = |P - Q|. Let angle between P and Q is φ.
P² + Q² + 2PQ cos φ = P² + Q² - 2PQ cos φ ⇒ 4PQ cos φ = 0 ⇒ cos φ = 0 ⇒ φ = 90°
Q3153. The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is
✅ Correct Answer: (d) 30º west
🧠 Solution:
Given: vR = 10 m/s, vSN = 20 m/s. For shortest path, sinθ = vR / vSN = 10/20 = 1/2 ⇒ θ = 30°. As river flows east, swimmer must swim towards west, so angle = 30º west.
Q3154. Two particles A and B move with constant velocities v1 and v2. At the initial moment, their position vectors are r1 and r2 respectively. The condition for particles A and B for their collision is
✅ Correct Answer: (a) (r1 - r2)/|r1 - r2| = (v1 - v2)/|v1 - v2|
🧠 Solution:
For collision, the direction of the relative velocity of one particle w.r.t the other should be along the relative position vector. That is, (r1 - r2)/|r1 - r2| is along (v1 - v2)/|v1 - v2|. Using resultant displacement R = (r1 - r2) - (v1 - v2)t = 0, we get the same condition.
Q3138. Position vector of a particle R = 4√2 i + 4√2 j sin(πt) cos(πt). Which statement is wrong?
✅ Correct Answer: (c) Magnitude of velocity is 8 m/s
🧠 Solution:
x = 4√2 sin πt, y = 4√2 cos πt ⇒ x² + y² = 16 ⇒ radius 4 m
Acceleration a_R = -v²/R, magnitude of v = √(v_x² + v_y²) = 8π
So option (c) is incorrect (should be 8π)
Q3137. A ship A is moving westwards at 10 km/h and ship B is 100 km south of A, moving northwards at 10 km/h. Time after which distance between them is shortest is
✅ Correct Answer: (b) 5 h
🧠 Solution:
Shortest distance PQ between A and B is perpendicular. PQ = 100/√2 = 50√2 km
Relative speed v_AB = √(10² + 10²) = 10√2 km/h
Time t = PQ / v_AB = 50√2 / 10√2 = 5 h
Q3123. If vectors A = i cos ωt + j sin ωt and B = i cos ωt + j sin ωt are functions of time, the value of t at which they are orthogonal to each other is
✅ Correct Answer: (c) t = π/ω
🧠 Solution:
For perpendicular vectors, A · B = 0
[(cos ωt i + sin ωt j) · (cos ωt i + sin ωt j)] = 0
cos²ωt + sin²ωt = 0 ⇒ cos 2ωt = 0 ⇒ 2ωt = π ⇒ t = π/ω
Q3140. Particle moves (2,3) at t=0, (6,7) at t=2 s, (13,14) at t=5 s. Average velocity vector v_av from t=0 to t=5 s is
✅ Correct Answer: (d) (11/5)(i + j)
🧠 Solution:
Displacement = (13-2)i + (14-3)j = 11 i + 11 j, Time = 5 s
v_av = (11/5)(i + j)
Q3139. Projectile fired from Earth with velocity 5 m/s at angle θ. Another from another planet at 3 m/s same angle follows identical trajectory. Acceleration due to gravity on planet is
✅ Correct Answer: (a) 3.5
🧠 Solution:
For identical trajectories u²/g = u'²/g' ⇒ g' = u'²/u² * g = 3²/5² * 9.8 = 35 m/s²
Q3142. Horizontal range = maximum height for a projectile. Angle of projection is
✅ Correct Answer: (b) θ = -tan⁻¹(1/4)
🧠 Solution:
R = H ⇒ u² sin2θ/g = u² sin²θ / (2g) ⇒ tan θ = 4 ⇒ θ = -tan⁻¹(1/4)
Q3160.
A particle moves in a circle of radius 5 cm with constant speed and time period 0.2 s. The acceleration of the particle is ?
✅ Correct Answer:
(c) 5 m/s²
🧠 Solution:
Given, r = 5 × 10⁻² m, T = 0.2 s. Acceleration a_r = ω² r = 4 π² r / T² = 4 × π² × 5 × 10⁻² / (0.2)² = 5 m/s².
Q3143. Missile fired for maximum range with u=20 m/s, g=10 m/s². Range is
✅ Correct Answer: (d) 40 m
🧠 Solution:
Maximum range θ=45°, R_max = u²/g = 20²/10 = 40 m
Q3144. Particle has initial velocity 3 i + 4 j and acceleration 0.4 i + 0.3 j. Speed after 10 s is
✅ Correct Answer: (b) 7√2 units
🧠 Solution:
v = u + at = (3 + 0.4*10) i + (4 + 0.3*10) j = 7 i + 7 j ⇒ |v| = 7√2
Q3145. A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground, the magnitude of the change in its momentum will be
✅ Correct Answer: (c) mv√2
🧠 Solution:
∆p = m(v - u) = | -mv[cos 45° i + sin 45° j] + mv[cos 45° i + sin 45° j]| = mv√2
Q3146. A particle starting from the origin (0,0) moves in a straight line in the xy-plane. Its coordinates at a later time are given. The path of the particle makes an angle with the x-axis
✅ Correct Answer: (c) 60°
🧠 Solution:
tanθ = slope of line OA = Δy / Δx = 3/0 ⇒ θ = 60°
Q3124. A and B are two vectors and θ is the angle between them. If |A × B| = 3 A · B, then the value of θ is
✅ Correct Answer: (a) 60°
🧠 Solution:
|A × B| = AB sin θ = 3 AB cos θ ⇒ tan θ = 3 ⇒ θ = 60°
Q3161. A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is?
✅ Correct Answer: (b) 0, 10 m/s
🧠 Solution:
Average velocity = Displacement / Time-interval = 0 / t = 0.
Average speed = Distance / Time-interval = 2 π r / t = 2 × π × 100 / 62.8 ≈ 10 m/s.
Q3147. For angles of projection of a projectile at (45°−θ) and (45°+θ), the horizontal ranges described by the projectile are in the ratio
✅ Correct Answer: (a) 1:1
🧠 Solution:
Horizontal ranges for complementary angles of projection are same. (45°−θ) and (45°+θ) are complementary. Hence R1:R2 = 1:1
Q3163. The circular motion of a particle with constant speed is
✅ Correct Answer: (d) periodic but not simple harmonic
🧠 Solution:
Particle repeats motion after equal intervals of time, so motion is periodic. It is not SHM as it does not move to-and-fro about a fixed point.
Q3162. A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with constant speed. If the stone makes 22 revolutions in 44 s, what is the magnitude and direction of acceleration of the stone?
✅ Correct Answer: (c) π²/2 m/s² along radius towards centre
🧠 Solution:
Radial acceleration a_r = ω² r = (2 π n / t)² × r = (2 π × 22 / 44)² × 1 = π²/2 m/s². Direction along radius towards centre.
Q3125. If a vector 2i + 3j + 8k is perpendicular to the vector 4j + 4i - k α, then the value of α is
✅ Correct Answer: (c) -1/2
🧠 Solution:
Dot product of perpendicular vectors = 0
(2i + 3j + 8k) · (4i + 4j - α k) = 0 ⇒ 8 + 12 - 8 α = 0 ⇒ α = 1/2 ⇒ but negative sign ⇒ α = -1/2
Q3126. If |A × B| = 3 AB, then the value of |A + B| is
✅ Correct Answer: (a) (A² + B² + 2AB cos 60°)/2
🧠 Solution:
Given |A × B| = AB sin θ = 3, and A · B = AB cos θ. Using |A + B|² = A² + B² + 2AB cos θ = A² + B² + 2AB cos 60° ⇒ |A + B| = √(A² + B² + 2AB cos 60°)
Q3164. A particle moves along a circle of radius 20π m with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is
✅ Correct Answer: (b) 40 2 m/s²
🧠 Solution:
Angular acceleration α = (ω² - ω₀²) / 2θ = 4π² / (2 × 2 × π) = 2 rad/s². Tangential acceleration a_T = α × r = 2 × 20 = 40 2 m/s².
Q3127. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
✅ Correct Answer: (d) are equal to each other in magnitude
🧠 Solution:
Let F1 = A + B, F2 = A - B. Given F1 · F2 = 0 ⇒ (A + B) · (A - B) = 0 ⇒ |A|² - |B|² = 0 ⇒ |A| = |B|. Thus, the forces are equal in magnitude.
Q3165. P is the point of contact of a wheel and the ground. The radius of wheel is 1 m. The wheel rolls on the ground without slipping. The displacement of point P when wheel completes half rotation is
✅ Correct Answer: (b) π²/4 + m
🧠 Solution:
Half rotation: horizontal displacement x_R = πR, vertical displacement y_R = 2R. Total displacement s = √(x_R² + y_R²) = √(π² + 4) = π²/4 + m (for R = 1 m).
Q3166. A particle of mass M is revolving along a circle of radius R and another particle of mass m is revolving in a circle of radius r. If time periods of both particles are same, then the ratio of their angular velocities is
✅ Correct Answer: (a) 1
🧠 Solution:
ω = 2π/T. Angular velocity depends on T only, not mass or radius. Since T is same, ω₁/ω₂ = 1.
Q3148. Two particles are projected with same initial velocities at angles 30° and 60° with the horizontal. Then
✅ Correct Answer: (b) their ranges will be equal
🧠 Solution:
Maximum height H = u² sin²θ / (2g) ⇒ H1/H2 = sin²30° / sin²60° = 1/3 (different)
Range R = u² sin2θ / g ⇒ R1/R2 = sin60° / sin120° = 1 ⇒ Ranges equal
Time of flight T = 2u sinθ / g ⇒ T1/T2 = sin30° / sin60° = 1/√3 (different)
Hence horizontal ranges are equal.
Q3167. What is the linear velocity if angular velocity vector ω = -3 i + 4 j + k and position vector r = 5 i + 6 j + 6 k?
✅ Correct Answer: (b) -18/13/2 i j k
🧠 Solution:
Linear velocity v = r × ω = (-3 i + 4 j + k) × (5 i + 6 j + 6 k) = -18/13/2 i j k.
Q3155. A person swims in a river aiming to reach exactly opposite point on the bank of a river. His speed of swimming is 0.5 m/s at an angle 120° with the direction of flow of water. The speed of water in stream is
✅ Correct Answer: (c) 0.25 m/s
🧠 Solution:
Let u be the speed of stream and v = 0.5 m/s be the speed of swimmer. Since he wants to reach directly opposite, u = v sin 30° = 0.5 × 1/2 = 0.25 m/s.
Q3128. If a unit vector is represented by 0.5 i + 0.8 j + c k, then the value of c is
✅ Correct Answer: (b) 0.11
🧠 Solution:
Magnitude of unit vector = 1 ⇒ √(0.5² + 0.8² + c²) = 1 ⇒ √(0.25 + 0.64 + c²) = 1 ⇒ √(0.89 + c²) = 1 ⇒ c² = 0.11 ⇒ c = 0.11
Q3156. The speed of a boat is 5 km/h in still water. It crosses a river of width 1.0 km along the shortest possible path in 15 min. The velocity of the river water is (in km/h)
✅ Correct Answer: (c) 3
🧠 Solution:
Time t = 15/60 h = 1/4 h. Shortest path: v_b² = v_r² + v_br² ⇒ 1/4² = 5² - v_r² ⇒ v_r² = 25 -16 = 9 ⇒ v_r = 3 km/h.
Q3168. A body is whirled in a horizontal circle of radius 20 cm. It has an angular velocity of 10 rad/s. What is its linear velocity at any point on the circular path?
✅ Correct Answer: (b) 2 m/s
🧠 Solution:
Linear speed v = r ω = 0.20 × 10 = 2 m/s.
Q3150. The position vector of a particle is r = a t (cos ω t) i + a t (sin ω t) j. The velocity of the particle is
✅ Correct Answer: (d) perpendicular to the position vector
🧠 Solution:
Velocity v = dr/dt = d/dt [a t (cos ω t) i + a t (sin ω t) j] = -a t ω sin ω t i + a t ω cos ω t j.
Slope of position vector = (a t sin ω t)/(a t cos ω t) = tan ω t, slope of velocity vector = -a t cos ω t / a t sin ω t = -1/tan ω t.
∴ Velocity is perpendicular to displacement. Alternative check: v⋅r = 0, confirming perpendicularity.
Q3149. A bullet is fired from a gun with a speed of 1000 m/s in order to hit a target 100 m away. At what height above the target should the gun be aimed? (The resistance of air is negligible and g = 10 m/s²)
✅ Correct Answer: (a) 5 cm
🧠 Solution:
Time taken by bullet to cover horizontal distance, t = 100/1000 = 1/10 s.
Vertical displacement due to gravity, h = 1/2 g t² = 1/2 × 10 × (1/10)² = 0.05 m = 5 cm.
Q3129. Which of the following is not a vector quantity?
✅ Correct Answer: (a) Speed
🧠 Solution:
Speed is a scalar quantity. It gives no idea about direction. Velocity, displacement, and torque are vector quantities.
Q3130. The angle between the two vectors A = 3i + 4j + 5k and B = 3i + 4j - 5k will be
✅ Correct Answer: (c) 90°
🧠 Solution:
cos θ = (A · B) / (|A||B|) = (9 + 16 - 25)/(50) = 0 ⇒ θ = 90°
Q3157. A boat is sent across a river with a velocity of 8 km/h. If the resultant velocity of boat is 10 km/h, then the velocity of river is
✅ Correct Answer: (b) 6 km/h
🧠 Solution:
From relative velocity: v_rb² = v_r² + v_b² ⇒ v_r = √(v_rb² - v_b²) = √(10² - 8²) = √36 = 6 km/h.
Q3131. The resultant of A × 0 will be equal to
✅ Correct Answer: (c) zero vector
🧠 Solution:
Cross product of any vector with zero vector is a null vector or zero vector.
Q3169. When milk is churned, cream gets separated due to
✅ Correct Answer: (b) centrifugal force
🧠 Solution:
Centrifugal force moves lighter cream particles outward. Net force on cream = m r ω² - m' r ω² causes cream to move to top.
Q3170. A electric fan has blades of length 30 cm measured from the axis of rotation. If the fan is rotating at 120 rev/min, the acceleration of a point on the tip of the blade is
✅ Correct Answer: (b) 47.4 m/s²
🧠 Solution:
Centripetal acceleration a_c = v²/r = ω² r = (2 π ν)² × 0.30 = 47.4 m/s², where ν = 2 rev/s.
Q3152. The maximum range of a gun on horizontal terrain is 16 km. If g = 10 m/s², the muzzle velocity of a shell must be
✅ Correct Answer: (c) 400 m/s
🧠 Solution:
Maximum range Rmax = u²/g (for θ=45°). Given Rmax = 16 km = 16000 m, g = 10 m/s². So, u² = Rmax × g = 16000 × 10 = 160000 ⇒ u = √160000 = 400 m/s
Q3151. Two bodies of same mass are projected with the same velocity at angles 30° and 60° respectively. The ratio of their horizontal ranges will be
✅ Correct Answer: (a) 1:1
🧠 Solution:
Horizontal range R = u² sin 2θ / g. For θ and (90°-θ), R1 = R2 = u² sin 60°/g = u² sin 60°/g. ∴ R1:R2 = 1:1
Q3158. A bus is moving on a straight road towards North with a uniform speed of 50 km/h. If the speed remains unchanged after turning through 90°, the increase in the velocity of bus in the turning process is
✅ Correct Answer: (a) 70.7 km/h along South-West direction
🧠 Solution:
v1 = 50 km/h North, v2 = 50 km/h West. Magnitude of change in velocity = |v2 - v1| = √(50² + 50²) = 70.7 km/h along South-West.
Q3132. The angle between A and B is θ. The value of the triple product A · (B × A) is
✅ Correct Answer: (b) zero
🧠 Solution:
Scalar triple product A · (B × A) = 0 because cross product A × A = 0. Therefore, A · (B × A) = 0
Q3133. The magnitudes of vectors A, B and C are 3, 4 and 5 units respectively. If A + B = C, the angle between A and B is
✅ Correct Answer: (a) π/2
🧠 Solution:
Given |A|=3, |B|=4, |C|=5 and A + B = C
So, |C|² = |A|² + |B|² + 2AB cosθ ⇒ 25 = 9 + 16 + 2*3*4*cosθ ⇒ cosθ=0 ⇒ θ=π/2
Thus, A is perpendicular to B.