IIT JEE Main 2017 Physics Question 4 SetA

IIT JEE Main 2017 Physics Question 4 SetA: 

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:

(1) 4.5 J
(2) 22 J
(3) 9 J
(4) 18 J

Solution

Given in problem that:
the mass of the particle is m = 1 kg,
initially the particle is at rest i.e. u =0
and let the final velocity achieved by particle be v after a time t.
then as we know that, Force = mass x acceleration, we have
=> ∫v0 dv = ∫ 6t dt
=> v = 6[t2/2]10
=> v = 3 ms-1
Thus Work done, W = change in Kinetic Energy = (1/2)* m * v * v
=> W = 0.5 * 1 * 3 * 3
=> W = 4.5 J

Answer to IIT JEE Main 2017 Physics Question 4 SetA

Correct answer to given problem No 4 of IIT JEE Main Physics paper of 2017 SetA is option number (1) that is the work done by the force will be 4.5 J.

IIT JEE Main 2017 Physics Question 3 SetA

IIT JEE Main 2017 Physics Question 3 SetA: 

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 1/8mv20, the value of k will be:

(1) 10–3 kg m–1
(2) 10–3 kg s–1
(3) 10–4 kg m–1
(4) 10–1 kg m–1 s–1

Solution

Given in problem that the body of mass m is experiencing a frictional force of F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is 1/8mv20, the value of k can be obtained by following method:
kf/ki = 1/8mv20/1/2mv20,
=> kf/ki = 1/4
=> vf/vi = 1/2
=> vf = v0/2
=> -kv2 = m dv/dt
=> ∫v0v0/2 dv/v2 = ∫0t0-k dt/m
=> 1/v0 – 2//v0 = -kt0/m
=> 1/v0 = -kt0/m
=> k = m/(v0t0)
=> k = 10-2/(10 x 10)
=> k = 10-4 kg m-1

Answer to IIT JEE Main 2017 Physics Question 3 SetA

Correct answer to given problem No 3 of IIT JEE Main Physics paper of 2017 SetA is option number (3) that is the value of k is 10–4 kg m–1.