Exams › JEE Main › Physics

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:

1. 10⁻⁴ kg m⁻¹
2. 10⁻¹ kg m⁻¹
3. 10⁻³ kg m⁻¹
4. 10⁻³ kg s⁻¹

Correct answer: 10⁻⁴ kg m⁻¹

Solution

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.

Related JEE Main Physics questions

• 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
• 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?
• 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?
• 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?
• 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 cases (a) and (b), respectively, is
• 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

⚔️ Practice JEE Main Physics free + battle 1v1 →