Water boils in an electric kettle in 20 minut | vedclass

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(C) The power consumed by the heating element is given by $P = V^2/R$,where $V$ is the supply voltage and $R$ is the resistance.Since $R = \rho \ell/A

Vedclass · Accepted Answer

decreased,$3/4$

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(C) The power consumed by the heating element is given by $P = V^2/R$,where $V$ is the supply voltage and $R$ is the resistance.Since $R = ho \ell/A$,where $ho$ is resistivity,$\ell$ is length,and $A$ is the cross-sectional area,we have $P \propto 1/\ell$.The heat required to boil the water is $H = P 	imes t$. Since $H$ is constant,$P_1 t_1 = P_2 t_2$.Therefore,$P_1/P_2 = t_2/t_1 = 15/20 = 3/4$.Since $P \propto 1/\ell$,we have $P_1/P_2 = \ell_2/\ell_1$.Thus,$\ell_2/\ell_1 = 3/4$.Since the new length $\ell_2$ is $3/4$ of the initial length $\ell_1$,the length must be decreased to $3/4$ of its initial length.

Water boils in an electric kettle in $20$ minutes after being switched on. Using the same main supply,the length of the heating element should be . . . . . . to . . . . . . times of its initial length if the water is to be boiled in $15$ minutes.

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