Question 1

An inclined plane makes an angle θ with the horizontal. Its top half is frictionless, while its bottom half has friction. A body released from rest at the top just reaches the bottom and comes to rest there again. The coefficient of friction on the lower half must be

Accepted Answer

2 tan θ. The correct option is 2 tan θ because the body gains kinetic energy while sliding down the frictionless top half, which is then dissipated by friction on the lower half. The frictional force must be sufficient to counteract the kinetic energy gained, leading to the relationship that the coefficient of friction equals twice the tangent of the angle of inclination.

Question 2

A block slides down a rough inclined plane of length l and angle of inclination θ. If the coefficient of friction between the block and the plane is μ, what is the speed of the block when it reaches the lower end of the plane?

Accepted Answer

[2g(l sin θ − μ cos θ)]¹/2. The correct option accounts for the net force acting on the block as it slides down the incline, which includes the gravitational component along the incline and the opposing frictional force. By applying energy conservation principles and considering the work done against friction, the derived expression accurately reflects the block's speed at the bottom of the incline.

Question 3

A machine supplies a particle of mass m with constant power k watts. If the particle is initially at rest, what is the magnitude of the force acting on it at time t?

Accepted Answer

√(mk/2) t^(-1/2). The correct option is derived from the relationship between power, force, and velocity. Since power is the product of force and velocity, and knowing that the particle starts from rest and accelerates under constant power, we can express the force in terms of the power and the time, leading to the result of √(mk/2) t^(-1/2).

Question 4

A body of mass m1 moving with speed v1 collides head-on with another body of mass m2 initially at rest. If the collision is perfectly elastic and the first body’s speed after impact becomes 2v1/3, what must be the ratio m1/m2?

Accepted Answer

5. For an elastic head-on collision with target at rest, v1' = (m1-m2)/(m1+m2) v1. Since the speed stays 2v1/3 in the same direction, (m1-m2)/(m1+m2) = 2/3, giving 3m1-3m2 = 2m1+2m2, so m1 = 5m2, i.e. m1/m2 = 5.

Question 5

Two springs P and Q are alike in all respects except that their spring constants satisfy K_P > K_Q. They are extended first through the same extension in case (a), and then by applying the same force in case (b). If the work done by the springs is W_P and W_Q, then the correct relation in

Accepted Answer

W_P > W_Q; W_Q > W_P. For equal extension x, W = (1/2)k x^2, so larger k gives more work: W_P > W_Q. For equal force F, W = F^2/(2k), so larger k gives less work: W_Q > W_P. Thus W_P > W_Q in case (a) and W_Q > W_P in case (b).

Question 6

A body is dropped from rest from a height of 10 m. If 25% of its energy is lost during collision with the ground, the greatest height it will reach after the first bounce is

Accepted Answer

7.5 m. Losing 25% of the energy leaves 75% of the kinetic energy at impact, which converts to potential energy on the rebound. Maximum height = 0.75 x 10 = 7.5 m.

Question 7

Water descends through a vertical drop of 60 m at a mass flow rate of 15 kg/s to drive a turbine. If 10% of the available energy is lost because of friction, what power is delivered by the turbine? (Take g = 10 m/s²)

Accepted Answer

8.1 kW. The power delivered by the turbine is calculated using the potential energy lost due to the height drop, adjusted for the mass flow rate and accounting for the 10% energy loss. The potential energy per second is given by the formula P = mgh, where m is the mass flow rate, g is the acceleration due to gravity, and h is the height. After calculating the total power and applying the energy loss, the result is 8.1 kW.

Question 8

A glass marble is released from a height above a level floor and takes time t to hit the floor for the first time. After that, it keeps rebounding vertically. If the coefficient of restitution is e, the total time after release for which the marble keeps moving before it finally comes to r

Accepted Answer

[(1+e)/(1-e)]t. The first fall takes time t. Each rebound's up-down flight time forms a geometric series: T = t + 2*t*(e + e^2 + ...) = t + 2*t*e/(1-e) = t*(1+e)/(1-e).

Question 9

A 1 kg particle starts from rest and is acted upon by a time-varying force F=(2t î+3t² ĵ) N, where î and ĵ are unit vectors along the x- and y-axes. What is the instantaneous power delivered by the force at time t?

Accepted Answer

(2t³+3t⁵) W. The instantaneous power delivered by a force is calculated as the dot product of the force vector and the velocity vector. By first determining the velocity from the force using Newton's second law and then substituting it into the power formula, we find that the correct expression for power at time t is ((2t³+3t⁵) ext{ W}.

Question 10

A 20 g bullet traveling at 600 m/s strikes a 4 kg block suspended by a string. If, after the impact, the block swings up to a height of 0.2 m, what is the speed of the bullet as it emerges from the block?

Accepted Answer

200 m/s. Block speed after impact V = sqrt(2*10*0.2) = 2 m/s. Momentum: 0.02*600 = 0.02*v + 4*2, so 12 = 0.02v + 8, giving v = 200 m/s for the emerging bullet.

Question 11

A particle of mass 1 kg can move only along the x-axis, and its potential energy is V(x) = (x⁴/4 - x²/2) J. If the particle’s total mechanical energy is 2 J, what is its greatest possible speed (in m/s)?

Accepted Answer

3/√2. V'(x)=x^3-x=0 gives minima at x=+-1 with V=-1/4. Max KE = E - V_min = 2 - (-1/4) = 9/4, so (1/2)(1)v^2 = 9/4 and v = sqrt(9/2) = 3/sqrt(2) m/s.

Question 12

A vehicle of mass m begins from rest and moves with such acceleration that the power supplied to it remains constant at P0. Its speed at time t is

Accepted Answer

t^(1/2). Power P0 = F v = m v (dv/dt). So m v dv = P0 dt, giving (1/2) m v^2 = P0 t, hence v = sqrt(2 P0 t / m), which varies as t^(1/2).

Question 13

A steel sphere of mass 5 g is projected vertically downward at 10 m/s from a height of 19.5 m. It sinks into sand by 50 cm. Taking g = 10 m/s², the loss in mechanical energy is:

Accepted Answer

1.25 J. Taking the final rest point (0.5 m inside sand) as reference, total initial mechanical energy = m*g*(19.5+0.5) + (1/2)*m*u^2 = 0.005*10*20 + 0.5*0.005*100 = 1.0 + 0.25 = 1.25 J, all lost when the sphere stops.

Question 14

A 10 H.P. motor is used to lift water from a well that is 20 m deep and deliver it into a tank of capacity 22380 litres placed 10 m above the ground. Taking g = 10 m/s², how long will the motor take to completely fill the empty tank?

Accepted Answer

15 minutes. The motor's power allows it to lift water against gravity, and by calculating the total work done to lift the water to the tank and dividing it by the motor's power, we find that it takes 15 minutes to fill the tank.

Question 15

A block of mass 2 kg moves along a level surface with speed 4 m/s and then comes into contact with an initially uncompressed spring. The block continues until it momentarily stops. If the kinetic friction force is 15 N and the spring constant is 10,000 N/m, what is the maximum compression

Accepted Answer

5.5 cm. Energy: (1/2)(2)(4^2) = (1/2)(10000)x^2 + 15x, i.e. 5000x^2 + 15x - 16 = 0. Solving gives x ~= 0.055 m = 5.5 cm.

Question 16

An engine drives water through a hose. The water emerges from the hose at a speed of 2 m/s. If the mass of water contained per unit length of the hose is 100 kg/m, what power does the engine deliver?

Accepted Answer

400 W. Mass flow rate = (mass per length)(speed) = 100*2 = 200 kg/s. Power delivered = rate of kinetic energy = (1/2)(dm/dt)v^2 = (1/2)(200)(2^2) = 400 W.

Question 17

A body of mass m starts from rest and moves through a fixed distance under a constant force. The kinetic energy gained by it is directly proportional to

Accepted Answer

independent of m. The kinetic energy gained by the body is determined by the work done on it, which is the product of the force and the distance moved. Since the force is constant and the distance is fixed, the kinetic energy is independent of the mass of the body.

Question 18

A spring has force constant K. What is the work required to increase its extension from l1 to l2?

Accepted Answer

(K/2)(l2² − l1²). Spring potential energy is U = (1/2)K l^2, so the work needed to change the extension from l1 to l2 is U2 - U1 = (K/2)(l2^2 - l1^2).

Question 19

A 4 kg block is suspended from a light spring that follows Hooke’s law, and the spring extends by 2 cm. If an external agent stretches the same spring by 5 cm, the work done is (take g = 9.8 m/s²):

Accepted Answer

2.450 joule. The work done on a spring is calculated using the formula W = 0.5 * k * x², where k is the spring constant and x is the extension. Given that the spring extends by 2 cm under a 4 kg weight, we can find the spring constant and then use it to calculate the work done when the spring is stretched by 5 cm, resulting in 2.450 joules.

Question 20

A particle having mass m and charge q is initially at rest in a constant electric field E. If it is allowed to move freely, what will be its kinetic energy after it has travelled a distance x?

Accepted Answer

qE x. Constant force qE acts over distance x, so KE gained = work done = qE*x.

Question 21

An electron gun has a potential difference of 3000 V between its filament and plate. What speed will an electron acquire as it leaves the gun?

Accepted Answer

3.26 × 10⁷ m/s. v = sqrt(2eV/m) = sqrt(2 * 1.6e-19 * 3000 / 9.11e-31) = 3.25 x 10^7 m/s, which matches 3.26 x 10^7 m/s.

Question 22

An α-particle of mass m approaches a very heavy nucleus of charge Ze with speed v. The minimum distance it can get to the nucleus is proportional to which power of m?

Accepted Answer

1/m. At closest approach all KE converts to potential energy: (1/2)m v^2 = k(2e)(Ze)/rmin, so rmin = 4kZe^2/(m v^2), i.e. rmin is proportional to 1/m.

Question 23

A neutron moving with speed v and kinetic energy E undergoes a perfectly elastic, head-on collision with a stationary atomic nucleus of mass number A. The approximate fraction of the neutron’s total energy that remains after the collision is

Accepted Answer

[(A − 1)/(A + 1)]². For a neutron (mass ~1) in a head-on elastic collision with a nucleus of mass A, the fraction of energy retained is [(m-M)/(m+M)]^2 = [(1-A)/(1+A)]^2 = [(A-1)/(A+1)]^2.

Question 24

Identify the pair whose dimensions are equal

Accepted Answer

torque and work. Torque and work both have dimensions of energy, as they can be expressed in terms of force multiplied by distance. This similarity in dimensional analysis makes them equal in terms of their fundamental units.

Question 25

A spherical ball of mass 20 kg starts from rest at the top of a hill 100 m high. It moves down a frictionless track to the ground, then goes up a second hill of height 30 m, and finally descends to a horizontal platform that is 20 m above the ground. The speed of the ball at that point is

Accepted Answer

40 m/s. Frictionless track conserves energy; speed depends only on net descent to the 20 m platform: v = sqrt(2g(100-20)) = sqrt(2*10*80) = sqrt(1600) = 40 m/s.

Question 26

A mass m starts from rest and is brought to speed v with constant acceleration over a time interval T. The instantaneous power supplied to the body at time t is

Accepted Answer

(mv²)/(T²) t. Constant acceleration a = v/T, so velocity at time t is v(t)=vt/T and force F=ma=mv/T. Instantaneous power P = F*v(t) = (mv/T)(vt/T) = m v^2 t / T^2.

Question 27

A particle of mass m travels with speed v and undergoes an inelastic collision with an identical particle initially at rest. After the collision, the first particle moves at speed v/√(3) in a direction at right angles to its original path. What is the speed of the second particle after the

Accepted Answer

2v/√(3). In an inelastic collision, momentum is conserved. The initial momentum of the system is entirely from the first particle, and after the collision, the momentum must be distributed between both particles. By applying conservation of momentum in both the x and y directions, we find that the second particle must move with a speed of 2v/√3 to balance the momentum equations.

Question 28

A 16 kg bomb initially at rest bursts into two fragments of masses 4 kg and 12 kg. If the 12 kg fragment moves with a speed of 4 m s⁻¹, what is the kinetic energy of the 4 kg fragment?

Accepted Answer

288 J. By momentum conservation 4*v = 12*4 -> v = 12 m/s. KE = 0.5*4*12^2 = 288 J.

Question 29

A body of mass 100 g is projected straight upward with an initial speed of 5 m/s. The work done by gravity while the body is moving upward is

Accepted Answer

-1.25 J. The stone rises to h = u^2/(2g) = 25/20 = 1.25 m. While moving upward, gravity acts opposite to motion, so W = -mgh = -(0.1)(10)(1.25) = -1.25 J.

Question 30

A particle of mass 1 kg can move only along the x-axis, and its potential energy is V(x) = (x⁴/4 - x²/2) J. If the particle’s total mechanical energy is 2 J, what is its greatest possible speed (in m/s)?

Accepted Answer

3/√(2). V(x) = x^4/4 - x^2/2 has minima at x = +/-1 where V = -1/4 J. Max KE = E - V_min = 2 - (-1/4) = 9/4 J, so (1/2)(1)v^2 = 9/4 gives v = 3/sqrt(2) m/s.

Question 31

A block of mass 2 kg moves along a rough horizontal surface with a speed of 4 m/s. It then hits a compressed spring and comes to rest after compressing it. If the kinetic friction force is 15 N and the spring constant is 10,000 N/m, what is the compression of the spring?

Accepted Answer

5.5 cm. Initial KE = friction work + spring PE: 0.5*2*4^2 = 15*x + 0.5*10000*x^2. So 5000 x^2 + 15 x - 16 = 0, giving x = 0.0551 m ≈ 5.5 cm.

Question 32

A 0.50 kg block travels at 2.00 m s⁻¹ on a frictionless horizontal surface. It collides with a 1.00 kg block, and after impact the two stick together and move as one. The amount of kinetic energy dissipated in the collision is

Accepted Answer

0.67 J. The kinetic energy before the collision is calculated using the formula KE = 0.5 * m * v² for the moving block, and after the collision, the combined mass moves with a new velocity. The difference in kinetic energy before and after the collision gives the amount of energy dissipated, which in this case is 0.67 J.

Question 33

For the interaction between two atoms in a diatomic molecule, the potential energy is approximately U(x) = a/x¹² - b/x⁶, where a and b are constants and x is the separation between the atoms. If the dissociation energy is defined as D = U(x = ∞) - U at the equilibrium separation, then D eq

Accepted Answer

b²/4a. Set dU/dx=0: -12a/x^13 + 6b/x^7 = 0 -> x^6 = 2a/b. Then U_eq = a/x^12 - b/x^6 = a(b/2a)^2 - b(b/2a) = b^2/4a - b^2/2a = -b^2/4a. Since U(infinity)=0, D = 0 - (-b^2/4a) = b^2/4a.

Question 34

A person trying to lose weight by burning fat lifts a mass of 10 kg up to a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8

Accepted Answer

12.89 × 10⁻³ kg. The correct option is right because the total work done in lifting the mass is calculated using the formula for gravitational potential energy, which is then adjusted for the efficiency of energy conversion from fat. After determining the total energy required and considering the efficiency, the amount of fat needed to provide that energy is accurately calculated, resulting in 12.89 × 10⁻³ kg.

Question 35

A body of mass m = 10⁻² kg is moving in a medium and experiences a frictional force F = -kv². Its initial speed is v0 = 10 ms⁻¹. If, after 10 s, its energy is 1/8 mv0², the value of k will be:

Accepted Answer

10⁻⁴ kg m⁻¹. Energy (1/8)mv0^2 means v = v0/2 = 5 m/s. From m dv/dt = -k v^2: 1/v - 1/v0 = (k/m)t -> 1/5 - 1/10 = (k/0.01)*10 -> 0.1 = 1000k -> k = 1e-4 kg/m.

Question 36

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 second will be

Accepted Answer

4.5 J. With F = 6t and m = 1 kg, a = 6t so v = 3t^2; at t = 1 s, v = 3 m/s. By work-energy theorem, work = final KE = (1/2)(1)(3^2) = 4.5 J.

Question 37

A particle is moving in a circular path of radius a under the action of an attractive potential U = -k/(2r²). Its total energy is:

Accepted Answer

zero. F = -dU/dr = -k/r^3 (attractive). Circular motion: mv^2/r = k/r^3, so KE = (1/2)mv^2 = k/(2a^2). Total E = KE + U = k/(2a^2) - k/(2a^2) = 0.

Question 38

In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is:

Accepted Answer

√2 v0. Momentum: v1 + v2 = v0. KE up 50%: v1^2 + v2^2 = 1.5 v0^2. Then (v1 - v2)^2 = 2(v1^2 + v2^2) - (v1 + v2)^2 = 3 v0^2 - v0^2 = 2 v0^2, so |v1 - v2| = sqrt(2) v0.

Question 39

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is Pd, while for its similar collision with carbon nucleus at rest, fractional loss of energy is Pc. The values of Pd and Pc are respectively:

Accepted Answer

(89, -28). For a head-on elastic collision the fractional energy loss is 4*m1*m2/(m1+m2)^2. Neutron-deuteron: 4*1*2/9 = 8/9 ~ 89%. Neutron-carbon: 4*1*12/169 = 48/169 ~ 28%. So (Pd, Pc) = (89, 28).

Question 40

A uniform cable of mass ‘M’ and length ‘L’ is placed on a horizontal surface such that its (1/n)th part is hanging below the edge of the surface. To lift the hanging part of the cable to the surface, the work done should be:

Accepted Answer

MgL/(2n²). The work done to lift the hanging part of the cable is calculated by considering the weight of the hanging segment and the distance it needs to be raised. Since only a fraction (1/n) of the cable is hanging, the average height of this segment is (L/(2n)), leading to the total work done being MgL/(2n²).

Question 41

A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body?

Accepted Answer

1.2 kg. For an elastic collision with a stationary target, v1' = (m1-m2)/(m1+m2)*v = v/4. So (2-m2)/(2+m2)=1/4 -> 8-4 m2 = 2 + m2 -> m2 = 1.2 kg.

Question 42

A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is

Accepted Answer

8 J. The work done on a spring is calculated using the formula W = (1/2)k(x2² - x1²), where k is the spring constant and x is the extension. By substituting the values for the initial and final extensions (0.05 m and 0.15 m) into the formula, the work done is found to be 8 J.

Question 43

A wire is fixed at one end and stretches by length ℓ by applying a force F. The work done in stretching is

Accepted Answer

Fℓ/2. The work done in stretching the wire is calculated as the average force applied over the distance stretched. Since the force increases linearly from 0 to F as the wire stretches by length ℓ, the average force is F/2, resulting in the work done being Fℓ/2.

Question 44

A beam carrying energy E is incident perpendicular to a perfectly reflecting surface. The momentum delivered to the surface is

Accepted Answer

2E/c. When a beam of energy E strikes a perfectly reflecting surface, it imparts momentum to the surface upon reflection. The momentum change is twice the momentum of the incident beam, which is given by E/c, leading to a total momentum of 2E/c.

Question 45

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λ_A to λ_B after the collision is

Accepted Answer

λ_A/λ_B = 2. Elastic head-on collision, A (mass m, speed v) hits B (mass m/2) at rest: v_A' = (m - m/2)/(m + m/2) v = v/3, v_B' = 2m/(3m/2) v = 4v/3. de-Broglie lambda is inversely proportional to momentum, so lambda_A/lambda_B = p_B/p_A = ((m/2)(4v/3))/(m(v/3)) = (2v/3)/(v/3) = 2.

Question 46

A particle is acted upon by constant forces 4î + ĵ - 3k̂ and 3î + ĵ - k̂ which displace it from a point î + 2ĵ + 3k̂ to the point 5î + 4ĵ + k̂. The work done in standard units by the forces is given by

Accepted Answer

40. Total force = 7i+2j-4k, displacement = 4i+2j-2k. Work = 28+4+8 = 40.

Question 47

This question has Statement-I and Statement-II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-I: A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss p

Accepted Answer

Statement-I is false but statement-II is true.. Statement-I incorrectly states the formula for maximum energy loss in a collision, while Statement-II accurately describes that maximum energy loss occurs when the particles stick together, making it true.

Question 48

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to -

Accepted Answer

56 %. In a perfectly inelastic collision, the two particles stick together, resulting in a loss of kinetic energy due to the conversion of some kinetic energy into other forms of energy. By calculating the initial and final kinetic energies, we find that the percentage loss in energy is approximately 56%, which reflects the significant energy transformation during the collision.

Question 49

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 ×

Accepted Answer

12.89 × 10⁻³ kg. The correct option is right because the total work done in lifting the mass is calculated, and considering the efficiency of energy conversion, the amount of fat used is derived from the energy required to perform that work. The calculations show that the energy expended corresponds to approximately 12.89 grams of fat.

Question 50

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelength of A to B after the collision is -

Accepted Answer

λA/λB = 2. In an elastic collision, momentum and kinetic energy are conserved. After the collision, particle A, having a mass m, moves with a reduced velocity, while particle B, with mass m/2, gains velocity. The de-Broglie wavelength is inversely proportional to momentum, and since particle B has a greater velocity after the collision, its wavelength becomes shorter, leading to the ratio of the wavelengths being 2.

JEE Main Physics: Work, Energy and Power questions with solutions

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