If E= $a t + b t^{2} =$ , What is the temperature of inversion

- \frac{a}{2 b}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ \frac{a}{2 b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{a}{b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ \frac{a}{b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- \frac{a}{2 b}$
$\begin{matrix} \frac{d E}{d t} = 0 a + 2 b t = 0 t = \frac{- a}{2 b} H e n c e, o p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

For a single step reaction to the type A + 2B → E + 2F

The horizontal distance x and the vertical height y of a projectile at a time t are given by
$x = a t a n d y = b t^{2} + c t$
where a, b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired?

Q. Angular displacement

θ

of a flywheel varies with time as

θ = a t + b t^{2} + c t^{3}

, then angular acceleration of the flywheel is given by

Q. If

x = \frac{\sqrt{a + 2 b} + \sqrt{a - 2 b}}{\sqrt{a + 2 b} - \sqrt{a - 2 b}}

, prove that

b x^{2} - a x + b = 0

[\vec{a} \vec{b} \vec{a} \times \vec{b}] + {(\vec{a} . \vec{b})}^{2} =

(a)

{|\vec{a}|}^{2} {|\vec{b}|}^{2}

(b)

{|\vec{a} + \vec{b}|}^{2}

(c)

{|\vec{a}|}^{2} + {|\vec{b}|}^{2}

(d)

2 {|\vec{a}|}^{2} {|\vec{b}|}^{2}

Join BYJU'S Learning Program

If E=at+bt2=, What is the temperature of inversion

The correct option is A −a2bdEdt=0a+2bt=0t=−a2bHence,optionAiscorrectanswer.

If E= $a t + b t^{2} =$ , What is the temperature of inversion

The correct option is A $- \frac{a}{2 b}$
$\begin{matrix} \frac{d E}{d t} = 0 a + 2 b t = 0 t = \frac{- a}{2 b} H e n c e, o p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$