The temperature gradient in a text0.5 m long rod is 80 ° C m- 1 . If the temperature of the hotter end of the rod is 30 ° C , then the temperature of the cooler end is

Question

The temperature gradient in a text0.5 m long rod is 80 ° C m- 1 . If the temperature of the hotter end of the rod is 30 ° C , then the temperature of the cooler end is (A) 40° C (B) -10 ° C (C) 10° C (D) 0° C

Tardigrade Admin · Accepted Answer

⇒ (d T/d x)=80 ⇒ (T2 - T1/l)=80 ⇒ (30 - T1/0.5)=80 ⇒ 30-T1=40 ⇒ T1=-10 ° C

Q. The temperature gradient in a $0.5 m$ long rod is $80^{\circ} C m^{- 1}$ . If the temperature of the hotter end of the rod is $30^{\circ} C$ , then the temperature of the cooler end is

$4 0^{\circ} C$

$- 10^{\circ} C$

$1 0^{\circ} C$

$0^{\circ} C$

$\Rightarrow \frac{d T}{d x} = 80$
$\Rightarrow \frac{T _{2} - T _{1}}{l} = 80$
$\Rightarrow \frac{30 - T _{1}}{0.5} = 80$
$\Rightarrow 30 - T_{1} = 40$
$\Rightarrow T_{1} = - 10^{\circ} C$

Q. The temperature gradient in a 0.5m long rod is 80∘Cm−1 . If the temperature of the hotter end of the rod is 30∘C , then the temperature of the cooler end is

40∘C

−10∘C

10∘C

0∘C