Two heater coils $A$ and $B$ made of the same material are connected in parallel across the same mains. The length and diameter of the wire used in $A$ are doubled as that of the wire used in $B$ . If $H_{1}$ and $H_{2}$ are the quantities of heat liberated per second in $A$ and $B$ , respectively, then $H_{1} : H_{2} =$

1 : 4

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2 : 1

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1 : 3

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1 : 1

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Solution

The correct option is B $2 : 1$
Since the coils are in parallel across the same mains, $V$ is the same for both
$\frac{H_{1}}{H_{2}} = \frac{R_{2}}{R_{1}} (∵ H = \frac{V^{2}}{R})$

$\frac{H_{1}}{H_{2}} = \frac{2}{1}$

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Two heater coils A and B made of the same material are connected in parallel across the same mains. The length and diameter of the wire used in A are doubled as that of the wire used in B. If H1 and H2 are the quantities of heat liberated per second in A and B, respectively, then H1:H2=

The correct option is B 2:1Since the coils are in parallel across the same mains, V is the same for bothH1H2=R2R1(∵H=V2R)H1H2=21

The correct option is B $2 : 1$
Since the coils are in parallel across the same mains, $V$ is the same for both
$\frac{H_{1}}{H_{2}} = \frac{R_{2}}{R_{1}} (∵ H = \frac{V^{2}}{R})$

$\frac{H_{1}}{H_{2}} = \frac{2}{1}$