Laws Of Motion Question 47

Question: A bullet is fired from a gun. The force on the bullet is given by $ F=600-2\times 10^{5}t $ , where F is in newtons and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet[CBSE PMT 1998]

Options:

A) 9 Ns

B) Zero

C) 0.9 Ns

D) 1.8 Ns

Show Answer

Answer:

Correct Answer: C

Solution:

$ F=600-2\times 10^{5}t=0 $

therefore $ t=3\times {{10}^{-3}}\sec $

Impulse $ I=\int _{0}^{t}{Fdt}=\int _{0}^{3\times {{10}^{-3}}}{(600-2\times 10^{3}t)dt} $

$ =[600t-10^{5}t^{2}] _{0}^{3\times {{10}^{-3}}}=0.9N\times \sec $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक