Question 1

A person travels 30 m due north, then 20 m due east, and finally 30√2 m in the south-west direction. What is the person's net displacement from the starting point?

Accepted Answer

10 m toward the west. North 30 -> (0,30); East 20 -> (20,30); 30sqrt2 SW = (-30,-30) -> (-10,0). Net displacement is 10 m due west.

Question 2

Two particles are set in motion at the same instant from a common point along two different straight paths. One particle moves with constant speed v, while the other starts with uniform acceleration a. If the angle between their directions of motion is α, then the time at which the relativ

Accepted Answer

(v)/(a)cosα. The correct option is derived from analyzing the relative velocity between the two particles. The minimum relative velocity occurs when the rate of change of the distance between them is zero, which can be mathematically expressed using the cosine of the angle between their paths, leading to the formula (v)/(a) ext{cos} alpha.

Question 3

Two time-dependent vectors are given by A = cos(ω t) î + sin(ω t) ĵ and B = cos ((ω t)/(2)) î + sin ((ω t)/(2)) ĵ. At what value of t are these vectors perpendicular to each other?

Accepted Answer

t = (π)/(ω). A.B = cos(wt)cos(wt/2) + sin(wt)sin(wt/2) = cos(wt - wt/2) = cos(wt/2). Setting this to zero gives wt/2 = pi/2, hence t = pi/w.

Question 4

A bus travels north along a straight road at a constant speed of 50 km/h and then takes a 90° turn. If its speed does not change after the turn, the change in the bus’s velocity during the turn is

Accepted Answer

70.7 km/h toward the south-west. Initial velocity 50 km/h north, final 50 km/h (after 90 deg turn, e.g. west). Change = vf - vi = (west) + (-north), a vector of magnitude sqrt(50^2 + 50^2) = 70.7 km/h directed toward the south-west (not north-west, since we subtract the original northward velocity).

Question 5

Two vectors A and B combine to give a resultant that is perpendicular to A, and the magnitude of this resultant is one-half the magnitude of B. What is the angle between A and B?

Accepted Answer

150°. R=A+B perp to A: A^2 + A.B = 0. |R|^2 = A^2+2A.B+B^2 = B^2-A^2 = B^2/4 -> A^2 = 3B^2/4. cos(theta)=A.B/(AB) = -A^2/(AB) = -A/B = -sqrt3/2, so theta = 150 deg.

Question 6

A person moves on a straight path with constant velocity u î. He observes that rain appears to fall straight downward along -ĵ. When he increases his speed to twice the original value, the rain is seen to make an angle θ with the vertical. The velocity of the rain relative to the ground

Accepted Answer

u î - (u)/(tanθ) ĵ. When the man moves at u and rain looks vertical, the rain's horizontal velocity equals u. At speed 2u the apparent horizontal component is u (backward), and tan(theta) = u/|v_y| gives |v_y| = u/tan(theta) downward. So the rain's ground velocity is u i - (u/tan(theta)) j.

Question 7

Two projectiles A and B are launched with initial speeds in the ratio 1: √2 and reach the same maximum height. If projectile A is projected at 45° to the horizontal, what is the launch angle of projectile B?

Accepted Answer

30°. Maximum height H = u^2 sin^2(theta)/(2g). Equal heights give uA^2 sin^2(45) = uB^2 sin^2(thetaB). With uB = sqrt(2) uA: uA^2 (1/2) = 2 uA^2 sin^2(thetaB) -> sin^2(thetaB) = 1/4 -> sin(thetaB) = 1/2 -> thetaB = 30 degrees.

Question 8

A stone attached to the end of a 1 m long string is moved in a horizontal circle at constant speed. If it completes 22 revolutions in 44 s, what is the magnitude of its acceleration?

Accepted Answer

π² m s⁻² and directed along the radius toward the centre. 22 rev in 44 s gives T = 2 s, so w = 2*pi/T = pi rad/s. Centripetal acceleration = w^2 r = pi^2 * 1 = pi^2 m/s^2, directed along the radius toward the centre.

Question 9

A particle has position vector r = cos(ωt) x̂ + sin(ωt) ŷ, where ω is constant. Which statement is correct?

Accepted Answer

The velocity is perpendicular to r and the acceleration points toward the origin. r=(cos wt, sin wt). v=(-w sin wt, w cos wt) gives v.r=0, so velocity is perpendicular to r. a=(-w^2 cos wt, -w^2 sin wt)=-w^2 r, i.e. centripetal, pointing toward the origin (antiparallel to r), not perpendicular.

Question 10

Ship A is sailing toward the west at 10 km h⁻¹. Ship B is located 100 km due south of A and is moving northward at 10 km h⁻¹. After how much time will the separation between the two ships be minimum?

Accepted Answer

5 h. Take A at origin moving west (-10,0) and B at (0,-100) moving north (0,10). Separation squared = (10t)^2 + (10t-100)^2; minimizing gives 200t + 200t - 2000 = 0, so t = 5 h.

Question 11

A projectile is launched at an angle of 45° to the horizontal. The angle of elevation of the projectile from the point of launch when it is at its maximum height is

Accepted Answer

tan⁻¹(1/2). For launch angle 45 deg, range R = u^2/g and max height H = u^2/(4g) at horizontal distance R/2 = u^2/(2g). The elevation angle from launch point satisfies tan(theta) = H/(R/2) = (u^2/4g)/(u^2/2g) = 1/2, so theta = tan^-1(1/2).

Question 12

A particle has position vector as a function of time given by R = 4 sin(2πt) î + 4 cos(2πt) ĵ, where R is measured in metres, t in seconds, and î and ĵ are unit vectors along the x- and y-axes. Which of the following statements is incorrect for this motion?

Accepted Answer

The magnitude of the particle’s velocity is 8 m/s. R = 4 sin(2*pi*t) i + 4 cos(2*pi*t) j describes uniform circular motion of radius 4 m with omega = 2*pi rad/s, so speed v = omega*R = 8*pi ~ 25.1 m/s. The statement 'velocity is 8 m/s' is therefore the incorrect one; the centripetal results (a = v^2/R, directed along -R) are all correct.

Question 13

Two vectors A and B satisfy the condition that the magnitude of their sum is equal to the magnitude of their difference. What is the angle between A and B?

Accepted Answer

90°. Squaring |A+B|=|A-B| gives 4 A.B = 0, so A.B = 0 and the angle between A and B is 90 deg.

Question 14

A stone is attached to a string of length l and is rotated in a vertical circle with the other end of the string fixed. At one instant, the stone is at the lowest point and its speed is u. What is the magnitude of the change in its velocity when it reaches the point where the string become

Accepted Answer

√(2(u² − gl)). The correct option is derived from the conservation of mechanical energy. At the lowest point, the stone has kinetic energy due to its speed u, and as it rises to the horizontal position, some of this energy is converted into potential energy, resulting in a change in velocity that can be expressed as √(2(u² - gl)).

Question 15

A rigid rod of fixed length ℓ has its ends A and B constrained to slide along the x-axis and y-axis, respectively. The angular speed of the rod is constant, with dθ/dt = 2 rad s⁻¹ at all times. A point P is fixed on the rod, and M is the foot of the perpendicular from P onto the x-axis. Du

Accepted Answer

M performs simple harmonic motion. The foot M tracks the x-coordinate of P, which equals a*cos(theta). Since dtheta/dt is constant, theta = w*t and x_M = a*cos(w*t), so M executes simple harmonic motion.

Question 16

A ring carrying uniform charge has radius 10 cm and spins about its own axis with a frequency of 10⁴ revolutions per second. At a point on the axis, 20 cm from the centre, the ratio of the electric-field energy density to the magnetic-field energy density is 9.1 × 10¹⁰. Determine the value

Accepted Answer

9. On axis, E/B = x*c^2/(2*pi*f*R^2), so E/(cB) = x*c/(2*pi*f*R^2) = (0.2*3e8)/(2*pi*1e4*0.01) = 9.55e4. Then u_E/u_B = (E/(cB))^2 = 9.1e9, giving a = 9.

Question 17

A cyclotron operates with an oscillator frequency of 10 MHz. If each dee has a radius of 60 cm, what kinetic energy does the accelerated proton beam acquire?

Accepted Answer

7.421 MeV. Using B = 2*pi*m*f/q, KE = (B^2 q^2 r^2)/(2m) = 2*pi^2*m*r^2*f^2 = 2*pi^2*(1.67e-27)*(0.6)^2*(1e7)^2 = 1.19e-12 J = about 7.42 MeV.

Question 18

Consider the following two statements:

Statement 1: If two stones are thrown from the same point on the ground at the same instant but with different initial velocities, they cannot meet each other in mid-air. (Ignore air resistance.)

Statement 2: When two particles start from the same p

Accepted Answer

(b) Statement 1 is true, Statement 2 is true, and Statement 2 correctly explains Statement 1.. Statement 1 is true because two stones with different velocities will follow different trajectories and cannot meet in mid-air. Statement 2 is also true as it describes that if two particles have zero relative acceleration, their distance will either remain constant or increase, which aligns with the idea that differing velocities prevent them from converging.

Question 19

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30° with the horizontal. How far from the throwing point will the ball be at the height of 10 m from ground?
[g = 10 m/s², sin 30° = 1/2, cos 30° = √3/2]

Accepted Answer

8.66 m. The correct option is right because the horizontal distance traveled by the ball can be calculated using the horizontal component of the initial velocity and the time it takes to reach the height of 10 m. By applying the kinematic equations for projectile motion, we find that the ball lands 8.66 m away from the throwing point.

Question 20

The co-ordinates of a moving particle at any time 't' are given by x = αt³ and y = βt³. The speed of the particle at time 't' is given by

Accepted Answer

3t²√(α² + β²). The speed of the particle is derived from the derivatives of its position coordinates with respect to time. By differentiating x and y with respect to t, we find the velocity components, and using the Pythagorean theorem to combine them gives the speed as 3t²√(α² + β²).

Question 21

A projectile can have the same range 'R' for two angles of projection. If 'T₁' and 'T₂' be the time of flights in the two cases, then the product of the two times of flights is directly proportional to

Accepted Answer

R. The time of flight for a projectile is influenced by the range and the angle of projection. When two angles yield the same range, the product of their respective times of flight is directly proportional to the range itself, as derived from the equations of motion for projectile motion.

Question 22

Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed ?

Accepted Answer

The acceleration vector is tangent to the circle. The statement is false because, for a particle moving in a circle at constant angular speed, the acceleration vector, known as centripetal acceleration, always points towards the center of the circle, not tangentially along the path.

Question 23

A ball is thrown from a point with a speed 'v₀' at an elevation angle of θ. From the same point and at the same instant, a person starts running with a constant speed 'v₀' to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?

Accepted Answer

Yes, 60°. The person can catch the ball if they run at the same speed as the ball's horizontal component of motion. At an angle of 60°, the vertical and horizontal components of the ball's trajectory allow the runner to reach the same horizontal distance as the ball when it lands, making it possible for the catch.

Question 24

A particle has initial velocity of 3î + 4ĵ and an acceleration of 0.4î + 0.3ĵ. Its speed after 10 s is:

Accepted Answer

7√2 units. The final velocity can be calculated by adding the initial velocity to the product of acceleration and time. After 10 seconds, the resultant velocity vector's magnitude is found to be 7√2 units, confirming that option A is correct.

Question 25

A particle is moving with velocity v = k(yî + xĵ), where k is a constant. The general equation for its path is

Accepted Answer

y² = x² + constant. vx = dx/dt = k y and vy = dy/dt = k x, so dy/dx = x/y -> y dy = x dx -> y^2/2 = x^2/2 + C -> y^2 = x^2 + constant.

Question 26

For a particle in uniform circular motion, the acceleration a at a point P(R,θ) on the circle of radius R is (Here θ is measured with the x-axis)

Accepted Answer

-(v²)/(R)cosθ î - (v²)/(R)sinθ ĵ. In uniform circular motion the acceleration is purely centripetal, directed from P(R,theta) toward the center, i.e. opposite the radial unit vector (cos theta i + sin theta j). Hence a = -(v^2/R)cos theta i - (v^2/R)sin theta j.

Question 27

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is:

Accepted Answer

π (v⁴)/(g²). Water reaches farthest at 45 deg, giving max range R = v^2/g. The wet region is a circle of that radius, so area = pi*R^2 = pi*v^4/g^2.

Question 28

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be

Accepted Answer

20 m. Thrown straight up to H = 10 m means u^2/(2g) = 10, so u^2 = 20g. The maximum horizontal range (at 45 degrees) is R = u^2/g = 20 m.

Question 29

A projectile is given an initial velocity of (î + 2ĵ) m/s, where î is along the ground and ĵ is along the vertical. If g = 10 m/s², the equation of trajectory is:

Accepted Answer

y = 2x - 5x². Here ux = 1, uy = 2, so tan(theta) = 2 and the trajectory is y = 2x - (g/(2 ux^2)) x^2 = 2x - (10/2) x^2 = 2x - 5x^2.

Question 30

A particle is moving with a velocity v = K(yî + xĵ), where K is a constant. The general equation for its path is:

Accepted Answer

y² = x² + constant. The correct option describes a relationship between x and y that arises from the particle's velocity components, which are proportional to each other. By integrating the velocity components, we find that the path of the particle follows a quadratic relationship, specifically in the form of a circle or parabola, leading to the equation y² = x² + constant.

Question 31

A thin circular disk of radius R carries a uniform positive surface charge density σ. If the disk is set into steady rotation about its symmetry axis with angular speed ω, its magnetic dipole moment is

Accepted Answer

(c) πR⁴σω / 4. For a ring radius r: dq = sigma*2pi r dr, current di = dq*omega/2pi = sigma*omega*r dr, dm = di*pi r^2 = pi*sigma*omega*r^3 dr. Integrating 0 to R gives m = pi*sigma*omega*R^4/4.

Question 32

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity u and the other from rest with uniform acceleration f. Let α be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is

Accepted Answer

u cos α / f. Relative speed squared = u^2 + (ft)^2 - 2u(ft)cos(alpha). Differentiating with respect to t and setting to zero: 2f^2 t - 2uf cos(alpha)=0, so t = u cos(alpha)/f.

Question 33

The upper 3/4 th portion of a vertical pole subtends an angle tan⁻¹(3/5) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is

Accepted Answer

40 m. The height of the pole can be determined using the tangent of the angle subtended, which relates the height of the pole to the distance from the foot. Given that the angle is tan⁻¹(3/5), we can set up the relationship using the tangent function, leading to a height of 40 m for the upper 3/4 portion of the pole.

Question 34

A particle moves from the point (2.0i + 4.0j) m. at t = 0, with an initial velocity (5.0i + 4.0j) m s⁻¹. It is acted upon by a constant force which produces a constant acceleration (4.0i + 4.0j) ms⁻². What is the distance of the particle from the origin at time 2 s?

Accepted Answer

20√2 m. The correct option is right because, using the kinematic equations, the position of the particle after 2 seconds can be calculated by adding the initial position to the displacement caused by the initial velocity and the acceleration. This results in a final position that, when calculated, gives a distance of 20√2 m from the origin.

Question 35

The position co-ordinates of a particle moving in a 3-D coordinate system is given by x = a cosωt, y = a sinωt and z = aωt. The speed of the particle is:

Accepted Answer

√2 aω. vx=-a*w*sin(wt), vy=a*w*cos(wt), vz=a*w. Speed = sqrt(a^2 w^2 (sin^2+cos^2) + a^2 w^2) = sqrt(2) * a * w.

Question 36

A particle of mass m is moving along a trajectory given by
x = x0 + a cos ωt

y = y0 + b sin ωt
The torque, acting on the particle about the origin, at t = 0 is:

Accepted Answer

m y0 aω2 k. At t = 0, the particle's position is (x0, y0), and the torque is calculated using the formula τ = r × F, where r is the position vector and F is the force. The force acting on the particle is its mass times acceleration, which is derived from the second derivatives of the position equations, leading to the correct torque expression of m y0 aω² k.

Question 37

The position vector of a particle changes with time according to the relation r(t) = 15t² î + (4 − 20t²) ĵ. What is the magnitude of the acceleration at t = 1 ?

Accepted Answer

50. r=15t^2 i +(4-20t^2) j gives a = r'' = 30 i - 40 j (constant). |a| = sqrt(30^2+40^2) = sqrt(2500) = 50.

Question 38

Starting from the origin at time t = 0, with initial velocity 5j m s⁻¹, a particle moves in the x-y plane with a constant acceleration of (10i + 4j) ms⁻². At time t, its coordinates are (20 m, y0 m). The values of t and y0 are, respectively.
(1) 4s and 52 m
(2) 5s and 25m
(3) 2s and 18m
(4

Accepted Answer

2s and 18m. The particle's x-coordinate can be calculated using the equation of motion, which shows that with a constant acceleration of 10 m/s², it reaches 20 m in 2 seconds. The y-coordinate can then be determined using the initial velocity and acceleration, resulting in a position of 18 m at that same time.

Question 39

A particle moves such that its position vector r(t) = cos ωt î + sin ωt ĵ where ω is a constant and t is time. Then which of the following statements is true for the velocity v(t) and acceleration a(t) of the particle:

Accepted Answer

v is perpendicular to r and a is directed towards the origin. The velocity vector v(t) is derived from the position vector r(t) and represents the tangent to the circular path, making it perpendicular to r. The acceleration vector a(t), which is centripetal in this case, points towards the center of the circular motion, thus directed towards the origin.

Question 40

A particle starts from the origin at t = 0 with an initial velocity of 3.0 î m/s and moves in the x-y plane with a constant acceleration (6.0 î + 4.0 ĵ) m/s². The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D metres. The value of D is

Accepted Answer

60. From y = (1/2)(4)t^2 = 2t^2 = 32, t = 4 s. Then x = 3t + (1/2)(6)t^2 = 12 + 48 = 60 m.

Question 41

A particle of mass m is projected with a speed u from the ground at an angle θ = π/3 w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity u i. The horizontal distance covered by the comb

Accepted Answer

3√3 u² / 8g. At apex vx=u/2, vy=0, height H=u^2 sin^2(60)/(2g)=3u^2/(8g). Inelastic collision with mass m moving at u gives combined vx=(u/2+u)/2=3u/4. Fall time t=sqrt(2H/g)=u*sqrt(3)/(2g), so horizontal distance = (3u/4)*t = 3*sqrt(3)*u^2/(8g).

Question 42

Two identical antennas mounted on identical towers are separated from each other by a distance of 45 km. What should be the nearby minimum height of receiving antenna to receive the signals in line of sight ? (Assume radius of earth is 6400 km)

Accepted Answer

39.55 m. The minimum height of the receiving antenna is determined by the line of sight distance, which depends on the curvature of the Earth. Using the formula for line of sight, the height required for a 45 km distance results in approximately 39.55 m, ensuring that the antenna can receive signals without obstruction.

Question 43

Let a = î + 2ĵ − 3k̂ and b = 2î − 3ĵ + 5k̂. If r × a = b × r, r.(αî + 2ĵ + k̂) = 3 and r.(2î + 5ĵ − αk̂) = −1, α ∈ R, then the value of α + |r|² is equal to:

Accepted Answer

15. r x a = b x r means r x a = -(r x b), so r x (a+b) = 0 and r is parallel to a+b = (3,-1,2); write r = t(3,-1,2). The conditions r.(alpha,2,1)=3 and r.(2,5,-alpha)=-1 give t=1, alpha=1, so |r|^2 = 9+1+4 = 14 and alpha + |r|^2 = 15.

Question 44

Two vectors P and Q have equal magnitudes. If the magnitude of P + Q is n times the magnitude of P − Q, then angle between P and Q is -

Accepted Answer

cos⁻¹((n²−1)/(n²+1)). The correct option is derived from the relationship between the magnitudes of the sum and difference of two vectors, which can be expressed in terms of the cosine of the angle between them. By applying the law of cosines and manipulating the resulting equations, we find that the angle can be expressed as cos⁻¹((n²−1)/(n²+1)), which matches the given condition.

Question 45

A butterfly flying with a velocity 4√2 m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is:

Accepted Answer

15m. The butterfly's velocity can be broken down into its north and east components, and when combined with the wind's southward velocity, the resultant velocity leads to a net displacement. Over 3 seconds, this results in a total displacement of 15 meters.

Question 46

Choose the correct option:

Accepted Answer

True dip is less than apparent dip.. Apparent dip is measured in a vertical plane not containing the magnetic meridian: tan(delta') = tan(delta)/cos(theta) >= tan(delta). Hence the true dip is always less than (or equal to) the apparent dip.

Question 47

What should be the height of transmitting antenna and population covered if television telecast is to cover a radius of 150 km? The average population density in tower is 2000/km and the value of Re = 6.5 × 10⁶ m.

Accepted Answer

Height = 1731 m
Population Covered = 1413 × 10⁵. Coverage radius d = sqrt(2*Re*h) -> h = d^2/(2*Re) = (150000)^2/(2*6.5e6) = 1731 m. Population = 2000 * pi * d^2 = 2000 * pi * (150)^2 = 1413 x 10^5.

Question 48

Two vectors X and Y have equal magnitude. The magnitude of (X − Y) is n times the magnitude of (X + Y). The angle between X and Y is

Accepted Answer

cos⁻¹((n²−1)/(−n²+1)). The correct option is derived from the relationship between the magnitudes of the vectors and the cosine of the angle between them. By applying the law of cosines to the expressions for the magnitudes of (X - Y) and (X + Y), we can establish a relationship that simplifies to the given expression for the angle.

Question 49

A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a:

Accepted Answer

straight line vertically down the plane. To the observer in the plane, the bomb has zero horizontal velocity relative to the plane, so it appears to drop vertically straight down.

Question 50

A particle is moving with uniform speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to R³. Its time period of revolution will be given by:

Accepted Answer

T ∝ R². The time period of revolution for a particle moving in a circular path is related to the radius and the centripetal force acting on it. Since the fictitious force is inversely proportional to R³, the centripetal acceleration is proportional to F/R, leading to a relationship where the time period T is proportional to R².

JEE Main Physics: Motion in a Plane questions with solutions

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