Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Let $I$ be the purchase value of an equipment and $V(t)$ be the value after it has been used for $t$ years. The value $V(t)$ depreciates at a rate given by differential equation $\frac{dV(t)}{dt}=-k(T-t);$ where $k> 0$ is a constant and $T$ is the total life in years of the equipment. Then the scrap value $V(T)$ of the equipment is

Option 1)
$I-\frac{k(T-t)^{2}}{2}\;$

Option 2)
$\; \; e^{-kT}\;$

Option 3)
$\; T^{2}-\frac{I}{k}\;$

Option 4)
$\; I-\frac{kT^{2}}{2}$

Answers (1)

best_answer

As we learnt in

Temperature Problems -

$\frac{dT}{dt}= -k\left ( T-T_{m} \right )$

- wherein

K is Proportionality constant

T = Temperature of body

$T_{m}=$ Temperature of Surrounding

$\int_{0}^{T}V(t)=\int_{0}^{T} -K(T-t)dt$

$V(t) -V(0)= -K(Tt-\frac{t^2}{2})|^T_0$

$V(t) -V(0)= -K(T^2-\frac{T^2}{2})$

$= \frac{-K.T^2}{2}$

$V(T )= V_0 -\frac{-K.T^2}{2}$

$V(T)= I-\frac{KT^2}{2}$

Option 1)

$I-\frac{k(T-t)^{2}}{2}\;$

Incorrect

Option 2)

$\; \; e^{-kT}\;$

Incorrect

Option 3)

$\; T^{2}-\frac{I}{k}\;$

Incorrect

Option 4)

$\; I-\frac{kT^{2}}{2}$

Correct

Posted by

Plabita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question