(B) The heat energy $H$ required to boil a given quantity of water is constant.Using the formula for heat energy $H = \frac{V^2}{R} t$,where $V$ is th

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(B) The heat energy $H$ required to boil a given quantity of water is constant.Using the formula for heat energy $H = \frac{V^2}{R} t$,where $V$ is the voltage,$R$ is the resistance,and $t$ is the time.Since $H$ and $R$ are constant,we have $H_1 = H_2$.$\frac{V_1^2}{R} t_1 = \frac{V_2^2}{R} t_2$Given $V_1 = V$,$t_1 = 5 \text{ min}$,and $V_2 = \frac{V}{2}$.Substituting these values: $\frac{V^2}{R} (5) = \frac{(\frac{V}{2})^2}{R} t_2$$5 = \frac{V^2}{4R} \cdot \frac{R}{V^2} \cdot t_2$$5 = \frac{t_2}{4}$$t_2 = 20 \text{ min}$.

Class 12 Physics Current Electricity Electrical Energy and Power

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$A$ given quantity of water is boiled by an electric heater in $5 \text{ min}$. If the supply voltage of the heater is reduced to half,then the time taken to boil the same quantity of water will be . . . . . . $\text{min}$. (Assume the resistance of the heater remains constant.)

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A
$40$
B
$20$
C
$10$
D
$2.5$

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The $H$ amount of thermal energy is developed by a resistor in $10\,s$ when a current of $4\,A$ is passed through it. If the current is increased to $16\,A$,the thermal energy developed by the resistor in $10\,s$ will be $........\,H$

MediumJEE MAIN 2023

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How many calories of heat will be produced approximately in a $210\, W$ electric bulb in $5$ minutes?

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In an electrical circuit,a battery is connected to pass $20\, C$ of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at $15\, V$. The work done by the battery is ........... $J$.

EasyJEE MAIN 2021

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The resistance of a bulb filament changes with temperature. If a bulb rated at $220 \ V$ and $100 \ W$ is connected to a supply of $(220 \times 0.8) \ V$,the actual power consumed will be:

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If a wire of resistance $20\,\Omega$ is covered with ice and a voltage of $210\,V$ is applied across the wire,then the rate of melting of ice is .................. $g/s$. (Take latent heat of fusion of ice $L = 80\,cal/g$ and $1\,cal = 4.2\,J$)

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$A$ given quantity of water is boiled by an electric heater in $5 \text{ min}$. If the supply voltage of the heater is reduced to half,then the time taken to boil the same quantity of water will be . . . . . . $\text{min}$. (Assume the resistance of the heater remains constant.)

Similar Questions

The $H$ amount of thermal energy is developed by a resistor in $10\,s$ when a current of $4\,A$ is passed through it. If the current is increased to $16\,A$,the thermal energy developed by the resistor in $10\,s$ will be $........\,H$

How many calories of heat will be produced approximately in a $210\, W$ electric bulb in $5$ minutes?

In an electrical circuit,a battery is connected to pass $20\, C$ of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at $15\, V$. The work done by the battery is ........... $J$.

The resistance of a bulb filament changes with temperature. If a bulb rated at $220 \ V$ and $100 \ W$ is connected to a supply of $(220 \times 0.8) \ V$,the actual power consumed will be:

If a wire of resistance $20\,\Omega$ is covered with ice and a voltage of $210\,V$ is applied across the wire,then the rate of melting of ice is .................. $g/s$. (Take latent heat of fusion of ice $L = 80\,cal/g$ and $1\,cal = 4.2\,J$)

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