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Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A function is $y = x^{3}$ . The approximate value of the function at $x = 2.0001$ is possibly (choose the closest answer possible)

Options:

Answer:

Solution:

$consider y = x^{3}$ at $x = 2$

$y + ∆ y = {(2.0001)}^{3} = {(2 + 0.0001)}^{3} = {(x + ∆ x)}^{3}$
$and ∆ y = 3 x^{2} ∆ x$
$⟹ ∆ y = 3 {(2)}^{2} \times (0.0001) = 0.0012$
$∴ y = 2^{3} + △ y = 8.0012$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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