The current density in a circular wire is given by J(r)= (1 × 10^5 ~A / m^3 ) r , where r is the radial distance and the wire's radius is 2 ~mm . If the potential applied across the wire is 70 V , then the energy consumed by the wire in 1000 seconds is

Correct answer: (B) 37 π kJ. I=∫ JdA =∫_0^2 ~mm 10^5 r × 2 π r r =2 π × 10^5 × [(r^3)/3 ]_0^2 × 10^-3 =2/3 π × 10^5 × [8 × 10^-9 ] =16/3 π × 10^-4 ~A So, energy consumed = VIt =16/3 π × 10^-4 × 70 × 1000=37 π Joule

Q147 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Physics

The current density in a circular wire is given by $J(r)=$ $\left(1 \times 10^5 \mathrm{~A} / \mathrm{m}^3\right) r$, where $r$ is the radial distance and the wire's radius is $2 \mathrm{~mm}$. If the potential applied across the wire is 70 $\mathrm{V}$, then the energy consumed by the wire in 1000 seconds is

A
$25 \mathrm{~kJ}$
Correct answer
B
$37 \pi \mathrm{kJ}$
C
$18 \pi \mathrm{kJ}$
D
$88 \mathrm{~kJ}$

Solution

$\mathrm{I}=\int \mathrm{JdA}$ $$ \begin{aligned} & =\int_0^{2 \mathrm{~mm}} 10^5 \mathrm{r} \times 2 \pi \mathrm{r} \mathrm{r} \\ & =2 \pi \times 10^5 \times\left[\frac{\mathrm{r}^3}{3}\right]_0^{2 \times 10^{-3}} \\ & =\frac{2}{3} \pi \times 10^5 \times\left[8 \times 10^{-9}\right] \\ & =\frac{16}{3} \pi \times 10^{-4} \mathrm{~A} \end{aligned} $$ So, energy consumed $=$ VIt $$ =\frac{16}{3} \pi \times 10^{-4} \times 70 \times 1000=37 \pi \text { Joule } $$

Hence, the correct answer is option: 'B'.

Solution

Relevant Questions