JEE MAIN - Physics (2004 - No. 44)
A body of mass $$' m ',$$ acceleration uniformly from rest to $$'{v_1}'$$ in time $${T}$$. The instantaneous power delivered to the body as a function of time is given by
$${{m{v_1}{t^2}} \over {{T}}}$$
$${{mv_1^2t} \over {T^2}}$$
$${{m{v_1}t} \over {{T}}}$$
$${{mv_1^2t} \over {{T}}}$$
Explanation
Assume acceleration of body be $$a$$
$$\therefore$$ $${v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}$$
$$\therefore$$ $$v = at \Rightarrow v = {{{v_1}t} \over {{T}}}$$
$${P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v $$
$$= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)$$
$$ = m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$$
$$\therefore$$ $${v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}$$
$$\therefore$$ $$v = at \Rightarrow v = {{{v_1}t} \over {{T}}}$$
$${P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v $$
$$= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)$$
$$ = m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$$
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