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JEE Main 2012Offline
Exam
JEE Main
Year
2012
Mode
Offline
Subject
Physics
Chapter
Laws of Motion

JEE Main 2012 · Offline · Physics · Laws of Motion

A particle of mass m is at rest at the origin at time t=0 . It is...

A particle of mass m\mathrm{m} is at rest at the origin at time t=0\mathrm{t}=0. It is subjected to a force F(t)=F0e−bt\mathrm{F}(\mathrm{t})=\mathrm{F}_0 \mathrm{e}^{-\mathrm{bt}} in the xx direction. Its speed v(t)v(t) is depicted by which of the following curves?A particle of mass m is at rest at the origin at time t=0 . It is subjected to a force F(t)=F_0 e^ -bt in the xA particle of mass m is at rest at the origin at time t=0 . It is subjected to a force F(t)=F_0 e^ -bt in the xA particle of mass m is at rest at the origin at time t=0 . It is subjected to a force F(t)=F_0 e^ -bt in the xA particle of mass m is at rest at the origin at time t=0 . It is subjected to a force F(t)=F_0 e^ -bt in the x

Options

  1. AHow many litres of water must be added to 1 litre of aqueous solution of HCl with a pH of 1 to create an aqueous
  2. BHow many litres of water must be added to 1 litre of aqueous solution of HCl with a pH of 1 to create an aqueous
  3. CHow many litres of water must be added to 1 litre of aqueous solution of HCl with a pH of 1 to create an aqueous
  4. DHow many litres of water must be added to 1 litre of aqueous solution of HCl with a pH of 1 to create an aqueous

Correct answer

C. How many litres of water must be added to 1 litre of aqueous solution of HCl with a pH of 1 to create an aqueous

Step-by-step solution

F=F0e−bt⇒a=Fm=F0 me−bt⇒dvdt=F0 me−bt∫dv=∫0tFme−btdt⇒v=Fm[−1b][e−bt]0t⇒v=Fmb[e−bt]v=0 at t=0\mathrm{F}=\mathrm{F}_0 \mathrm{e}^{-\mathrm{bt}} \Rightarrow \quad \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{F}_0}{\mathrm{~m}} \mathrm{e}^{-\mathrm{bt}} \Rightarrow \quad \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{F}_0}{\mathrm{~m}} \mathrm{e}^{-\mathrm{bt}} \quad\int \mathrm{dv}=\int_0^{\mathrm{t}} \frac{\mathrm{F}}{\mathrm{m}} \mathrm{e}^{-\mathrm{bt}} \mathrm{dt} \Rightarrow v=\frac{F}{m}\left[\frac{-1}{b}\right]\left[e^{-b t}\right]_0^t \Rightarrow \quad v=\frac{F}{m b}\left[e^{-b t}\right] v=0 \text{ at } t=0 and v→Fmb\quad v \rightarrow \frac{F}{m b} as t→∞t \rightarrow \infty So, velocity increases continuously and attains a maximum value of v=Fmbv=\frac{F}{m b} as t→∞t \rightarrow \infty.
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