If the kinetic energy of a body increases by $0.1 %$ , the percentage increase of its momentum will be

10 %

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1.0 %

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0.1 %

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0.05 %

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Solution

The correct option is A $0.05 %$
momentum, $p = m v$

$v = \sqrt{\frac{2 E}{m}}$

$p = m \sqrt{\frac{2 E}{m}} = \sqrt{2 m E}$

taking log and then differentiating on both the sides, we get,

$\frac{Δ p}{p} = \frac{1}{2} (\frac{Δ m}{m}) + \frac{1}{2} (\frac{Δ E}{E})$

$\frac{Δ p}{p} = \frac{1}{2} \times 0.1 = 0.05$

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If the kinetic energy of a body increases by 0.1%, the percentage increase of its momentum will be

The correct option is A 0.05%momentum, p=mvv=√2Emp=m√2Em=√2mEtaking log and then differentiating on both the sides, we get,Δpp=12(Δmm)+12(ΔEE)Δpp=12×0.1=0.05

If the kinetic energy of a body increases by $0.1 %$ , the percentage increase of its momentum will be