Please solve q 12?

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Please solve q 12?

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Answer 1 · 2018-03-30T17:59:46

The answer is option $(B)$ $(B)$

Explanation:

The heat produced is

$W = U i t = R i^{2} t = \frac{U^{2}}{R} t$ $W=Uit=Ri^2t=U^2/Rt$

But, the resistance is

$R = ρ (\frac{l}{s})$ $R=rho (l/s)$

The heat is doubled :

$W_{1} = 2 W = 2 \frac{U^{2}}{R} t = \frac{U^{2}}{\frac{R}{2}} \cdot t$ $W_1=2W=2U^2/Rt=U^2/(R/2)*t$

when the resistance is halved.

$R_{1} = \frac{R}{2} = \frac{1}{2} ρ (\frac{l}{π r^{2}}) = ρ (\frac{l}{2 π r^{2}})$ $R_1=R/2=1/2rho (l/(pir^2))=rho(l/(2pir^2))$

-Both the length and the radius are halved

$\frac{l}{2}$ $l/2$ and $\frac{r}{2}$ $r/2$

$R_{\frac{l}{2}, \frac{r}{2}} = ρ ⎛ ⎝ \frac{\frac{l}{2}}{π {(\frac{r}{2})}^{2}} ⎞ ⎠ = ρ (\frac{l}{2} \cdot (\frac{1}{π \frac{r^{2}}{4}})) = 2 ρ (\frac{l}{π r^{2}}) = 2 R$ $R_(l/2, r/2)=rho((l/2)/(pi(r/2)^2))=rho(l/2*(1/(pir^2/4)))=2rho(l/(pir^2))=2R$

-Both the length and the radius are doubled

$2 l$ $2l$ and $2 r$ $2r$

$R_{2 l, 2 r} = ρ (\frac{2 l}{π {(2 r)}^{2}}) = ρ (2 l \cdot (\frac{1}{4 π r^{2}})) = \frac{1}{2} ρ (\frac{l}{π r^{2}}) = \frac{R}{2} = R_{1}$ $R_(2l,2r)=rho((2l)/(pi(2r)^2))=rho(2l*(1/(4pir^2)))=1/2rho(l/(pir^2))=R/2=R_1$

When the length is doubled

$R_{2 l} = ρ \frac{2 l}{s} = \frac{2 ρ l}{π r^{2}} = 2 R$ $R_(2l)=rho(2l)/s=(2rhol)/(pir^2)=2R$

When the radius is doubled

$R_{2 r} = \frac{ρ l}{π {(2 r)}^{2}} = \frac{1}{4} R$ $R_(2r)=(rhol)/(pi(2r)^2)=1/4R$

The answer is option $B$ $B$ , both the length and the radius are doubled.

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Please solve q 12?

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