JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 7)
A particle of mass m is acted upon by a force F given by the empirical law
F =$${R \over {{t^2}}}\,v\left( t \right).$$ If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot :
$$\upsilon $$(t) against t2
log $$\upsilon $$(t) against $${1 \over {{t^2}}}$$
log $$\upsilon $$(t) against t
log $$\upsilon $$(t) against $${1 \over {{t}}}$$
Explanation
Given,
F = $${R \over {{t^2}}}$$ v(t)
$$ \Rightarrow $$ m $${{dv} \over {dt}}$$ = $${R \over {{t^2}}}$$ (v)
$$ \Rightarrow $$ $${{dv} \over v}$$ = $${R \over m}$$ $${{dt} \over {{t^2}}}$$
Intergrating both sides,
$$\int {{{dv} \over v} = {R \over m}\int {{{dt} \over {{t^2}}}} } $$
$$ \Rightarrow $$ lnv = $${{R \over m}}$$ $$ \times $$ $$\left( { - {1 \over t}} \right)$$ + C
$$ \Rightarrow $$ lnv = $$-$$ $${{R \over m}}$$ $$\left( {{1 \over t}} \right)$$ + C
Graph between lnv and $${{1 \over t}}$$ will be straight line curve.
F = $${R \over {{t^2}}}$$ v(t)
$$ \Rightarrow $$ m $${{dv} \over {dt}}$$ = $${R \over {{t^2}}}$$ (v)
$$ \Rightarrow $$ $${{dv} \over v}$$ = $${R \over m}$$ $${{dt} \over {{t^2}}}$$
Intergrating both sides,
$$\int {{{dv} \over v} = {R \over m}\int {{{dt} \over {{t^2}}}} } $$
$$ \Rightarrow $$ lnv = $${{R \over m}}$$ $$ \times $$ $$\left( { - {1 \over t}} \right)$$ + C
$$ \Rightarrow $$ lnv = $$-$$ $${{R \over m}}$$ $$\left( {{1 \over t}} \right)$$ + C
Graph between lnv and $${{1 \over t}}$$ will be straight line curve.
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