The linear momentum `P’ of a moving body in one dimensional frame varies with time as $P = x - y t^{2}$ , where x and y are position constants. The rate of change of momentum is the force that acts on the body.
In such a case the force is

Proportional to

t^{2}

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Proportional to t

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Inversely proportional to

t^{2}

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Inversely proportional to t

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Solution

The correct option is B Proportional to t
$P = x + y t^{2}$
$f = \frac{d p}{d t}$
$d p - 2$
$\frac{d p}{d t} = 2 y t$
as `2y’ is constant
$F \propto t$

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The linear momentum `P’ of a moving body in one dimensional frame varies with time as P=x−yt2, where x and y are position constants. The rate of change of momentum is the force that acts on the body. In such a case the force is

The correct option is B Proportional to tP=x+yt2 f=dpdt dp−2 dpdt=2yt as `2y’ is constant F∝t

The linear momentum `P’ of a moving body in one dimensional frame varies with time as $P = x - y t^{2}$ , where x and y are position constants. The rate of change of momentum is the force that acts on the body.
In such a case the force is

The correct option is B Proportional to t
$P = x + y t^{2}$
$f = \frac{d p}{d t}$
$d p - 2$
$\frac{d p}{d t} = 2 y t$
as `2y’ is constant
$F \propto t$