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JEE Main 201212 MayOnline
Exam
JEE Main
Year
2012
Date
12 May
Mode
Online
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2012 · 12 May · Online · Physics · Motion in One Dimension

The distance travelled by a body moving along a line in time t is...

The distance travelled by a body moving along a line in time tt is proportional to t3t^3. The acceleration-time (a,t)(a, t) graph for the motion of the body will beThe distance travelled by a body moving along a line in time t is proportional to t^3 . The acceleration-time (a, t)The distance travelled by a body moving along a line in time t is proportional to t^3 . The acceleration-time (a, t)The distance travelled by a body moving along a line in time t is proportional to t^3 . The acceleration-time (a, t)The distance travelled by a body moving along a line in time t is proportional to t^3 . The acceleration-time (a, t)

Options

  1. AOne mole of an ideal gas at 300K in thermal contact with its surroundings expands isothermally from 1 . 0L to 2 . 0L
  2. BOne mole of an ideal gas at 300K in thermal contact with its surroundings expands isothermally from 1 . 0L to 2 . 0L
  3. COne mole of an ideal gas at 300K in thermal contact with its surroundings expands isothermally from 1 . 0L to 2 . 0L
  4. DOne mole of an ideal gas at 300K in thermal contact with its surroundings expands isothermally from 1 . 0L to 2 . 0L

Correct answer

B. One mole of an ideal gas at 300K in thermal contact with its surroundings expands isothermally from 1 . 0L to 2 . 0L

Step-by-step solution

Distance along a line i.e., displacement ( ss ) =t3(∵s∝t3=t^3\left(\because s \propto t^3\right. given )) By double differentiation of displacement, we get acceleration. V=dsdt=dt3dt=3t2andV=\frac{d s}{d t}=\frac{d t^3}{d t}=3 t^2 anda=dvdt=d3t2dt=6ta=6t or  a∝t\begin{aligned}& a=\frac{d v}{d t}=\frac{d 3 t^2}{d t}=6 t \\ & a=6 t \text{\text{ or } }a \propto t \end{aligned}$Hence graph (b) is correct.
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