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 = –kv^{2}. Its initial speed is v_{0} = 10 ms^{–1} . If, after 10 s, its energy is 1/8mv^{2}_{0}, 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 = –kv^{2}. Its initial speed is v_{0} = 10 ms^{–1}. If, after 10 s, its energy is 1/8mv^{2}_{0}, the value of k can be obtained by following method:

k_{f}/k_{i} = 1/8mv^{2}_{0}/1/2mv^{2}_{0},

=> k_{f}/k_{i} = 1/4

=> v_{f}/v_{i} = 1/2

=> v_{f} = v_{0}/2

=> -kv^{2} = m dv/dt

=> ∫_{v0}^{v0/2} dv/v^{2} = ∫_{0}^{t0}-k dt/m

=> 1/v_{0} – 2//v_{0} = -kt_{0}/m

=> 1/v_{0} = -kt_{0}/m

=> k = m/(v_{0}t_{0})

=> 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}.