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A cylindrical rod with one end in steam and other end in ice results in melting of 0.1 g of ice per second. If the rod is replaced by another with half the length and double the radius and 1//4th the conductivity than that of the first, the rate at which the ice melts in g/sec will be : |
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Answer» 3.2 This HEAT is used in melting ice `:.Q=mL`. `RARR""(mL)/(t)=(kA(T_(1)-T_(2)))/(d)` or `m/t=(kA(T_(1)-T_(2)))/(Ld)` `:.""(m_(1))/(t_(1))=(k_(1)pir_(1)^(2)(T_(1)-T_(2)))/(Ld_(1))` `(m_(2))/(t_(2))=(k_(2)pir_(2)^(2)(T_(1)-T_(2)))/(Ld_(2))` `:.(m_(1))/(t_(1))xx(t_(2))/(m_(2))=(k_(1))/(k_(2))((r_(1))/(r_(2)))^(2)*(d_(2))/(d_(1))=0.1xx(t_(2))/(m_(2))=(1)/(1//4)xx(1/2)^(2)xx(1//2)/(1)=1/2` `(m_(2))/(t_(2))=0.2" g/sec"` Thus correct CHOICE is (C ). |
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