Tag: physics

Questions Related to physics

An electric heater operating at $220\ V$ boils $5\ l$ of water in $5\ \text{minutes}$. If it is used on a $110\ V$ line, it will boil the same amount of water in:

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $10\ \text{minutes}$

  2. $20\ \text{minutes}$

  3. $5\ \text{minutes}$

  4. $1\ \text{minute}$

Show answer
Correct Option: B
Explanation:
Heat required to boil the water will be same in the two cases.
Heat produced by heater is:
$Q = \cfrac{V^2}{R} t$
$V^2 = \cfrac{QR}{t}$
$V^2 \propto \dfrac{1}{t}$

$\cfrac{t _2}{t _1} = \cfrac{V _1^2}{V _2^2}$
$t _2 = \dfrac{220^2 \times 5}{110^2}$
$t _2 = 20\ min$

The maximum current $I$, which can be passed through a fuse without melting varies with its radius $r$ as:

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $I \propto r$

  2. $I \propto r^{3/2}$

  3. $I \propto r^2$

  4. $I \propto (1/r^2)$

Show answer
Correct Option: B
Explanation:

Heat lost per second per unit surface area of fuse wire is
$H = \frac{I^2 \rho }{2 \pi ^2 r^3}$
$\Rightarrow  I^2 \propto r^3$
$\Rightarrow  I \propto r^{\frac{3}{2}}$

If current is flowing through a 10 $\Omega$ resistor, then indicate in which case the maximum heat will be generated?

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. Current of $5\ A$ flows for $2\ \text{minutes}$.

  2. Current of $4\ A$ flows for $3\ \text{minutes}$.

  3. Current of $3\ A$ flows for $6\ \text{minutes}$.

  4. Current of $2\ A$ flows for $5\ \text{minutes}$.

Show answer
Correct Option: C
Explanation:

Heat produced is given by
$Q = I^2 R t$

Option A:
$5\ A$ flows for $2\ \text{minutes}$
$Q = 5^2 \times 10 \times 2 = 500\ J$

Option B:

$4\ A$ flows for $3\ \text{minutes}$

$Q = 4^2 \times 10 \times 3 = 480\ J$


Option C:
$3\ A$ flows for $6\ \text{minutes}$
$Q = 3^2 \times 10 \times 6 = 540\ J$

Option D:
$2\ A$ flows for $5\ \text{minutes}$
 $Q = 2^2 \times 10 \times 5 = 200\ J$


Hence, option C is correct.

The energy expended in $1\  kW$ electric heater in $30\ \text{seconds}$ will be:

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $\displaystyle 3\times 10^4\ J$

  2. $\displaystyle 3\times 10^4\ erg$

  3. $\displaystyle 3\times 10^4\ eV$

  4. $0$

Show answer
Correct Option: A
Explanation:

$\text{Energy = Power} \times \text{time}$
$\text{Energy} = 1000 \times 30 = 30000\ J = 3 \times {10}^{4}\ J$

A resistor has resistance R. When the potential difference across the resistor is V, the current in
the resistor is I. The power dissipated in the resistor is P. Work W is done when charge Q flows
through the resistor.
What is not a valid relationship between these variables? 

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $I =\frac {P}{V}$

  2. $Q =\frac {W}{V}$

  3. $R =\frac {P}{I^2}$

  4. $R =\frac {V}{P}$

Show answer
Correct Option: D
Explanation:

We know that 

$P= VI$
$\implies P= V\times \dfrac{V}{R}$
$\implies P = \dfrac{V^2}{R}$
$\implies R= \dfrac{V^2}{P}$..............(1)
Therefore the option D is wrong .

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. The energy supplied in kWh to the three heaters in 5 hours is :

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $3.75 kWh$

  2. $4 kWh$

  3. $0.6 kWh$

  4. $5.74 kWh$

Show answer
Correct Option: A
Explanation:

Power is given as: $P=VI$


Substituting, $I=\dfrac{V}{R}$ in the above formula, we get, $P=\dfrac { { V }^{ 2 } }{ R } $

Given that the voltage is $250\ V$ and the power is $60\ W$, the resistance of the bulb is calculated as follows.

$R=\dfrac { { V }^{ 2 } }{ P } =\dfrac { { 100 }^{ 2 } }{ 250 } =40\Omega$

Hence, the resistance of each resistor is 40 ohms.

The energy consumed by the appliance in kWh is given by the formula: $P=\dfrac { { V }^{ 2 } }{ R } \times t=\dfrac { { 100 }^{ 2 } }{ 40 } \times t= 1.25\ kWh$.

When three heater are connected in parallel, then the total energy consumed is given as $1.25\ kWh\times 3=3.75\ kWh.$.

Hence, the total power consumed is 3.75 kWh.

A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. The cost of energy consumed at Rs. 4.20 per kWh for 5 hours will be :

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $Rs. 250$

  2. $Rs. 300$

  3. $Rs. 310$

  4. $Rs. 315$

Show answer
Correct Option: D
Explanation:

The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. 


The power is given by the product of applied voltage and the electric current.

 That is, $P=VI. The\ power\ of\ the\ geyser\ is\ given\ as\ 1500 W$

The energy consumed by the geyser in kWh is given by the formula $Q=P\times t\quad =\quad 1500W\times 50hours\quad =\quad 75\quad kWh$. 

Hence, the energy consumed by the geyser in 5 hours is 75 kWh. 

The cost of energy consumed per kWh is given as Rs. 4.20.

That is, $4.20\times 75\quad kWh=\quad Rs.315$

Hence, the total cost is given as Rs. 315.

A lamp of $100\ W$ and a heater of $1\ kW$ are in simultaneous use for $10\ hrs$. The units of electricity consumed according to the meter in the house is:

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $10$

  2. $9$

  3. $11$

  4. $1$

Show answer
Correct Option: C
Explanation:

For lamp, electrical energy consumed is $E _L = P _Lt =100\times 10 = 1\ kWh$

For heater, electrical energy consumed is $E _H = P _Ht =1000\times 10 = 10\ kWh$
Hence, total electrical energy consumed is: $E = E _L+E _H=11\ kWh$

A motor of $50\ W$ runs for $20\ hrs$. How many 'units' ($kWh$) of electrical energy are consumed?

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $5\ kWh$

  2. $2\ kWh$

  3. $1\ kWh$

  4. $2.5\ kWh$

Show answer
Correct Option: C
Explanation:

Given :       $P = 0.050\ kW$ 

                   $t = 20\ h$

Energy consumed,      
$E = Pt =0.050\times 20 = 1\ kWh$

A heater is marked 1000 W. The energy consumed by it in ten hours is

physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. 10 J

  2. 10 KJ

  3. 10 kWh

  4. 100 KJ

Show answer
Correct Option: C
Explanation:

Given :    $P = 1000$ W $=1$ kW              $t = 10$ h

$\therefore$ Energy consumed in 10 h       $E = Pt = 1\times 10 =10$ kWh