Please solve q 12?

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Answer 1 · 2018-03-30T17:59:46

The answer is option #(B)#

The heat produced is

#W=Uit=Ri^2t=U^2/Rt#

But, the resistance is

#R=rho (l/s)#

The heat is doubled :

#W_1=2W=2U^2/Rt=U^2/(R/2)*t#

when the resistance is halved.

#R_1=R/2=1/2rho (l/(pir^2))=rho(l/(2pir^2))#

-Both the length and the radius are halved

#l/2# and #r/2#

#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

#2l# and #2r#

#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_(2l)=rho(2l)/s=(2rhol)/(pir^2)=2R#

When the radius is doubled

#R_(2r)=(rhol)/(pi(2r)^2)=1/4R#

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