Tag: physics

Questions Related to physics

Name the effect of current responsible for the glow of the bulb in an electric circuit.

  1. Chemical

  2. Heating

  3. Magnetic

  4. Heating and Magnetic


Correct Option: B
Explanation:

The electric bulb has a filament called tungsten when electricity passes through this filament, it heats up and glows. 

This heat is generated due to the passage of electric current, the drift of electrons due to the current and the resistance it has.
Hence, the heating effect of electric current is responsible for the glow of the bulb in an electric circuit.

A house is fitted with 10 tubes of 40 W. If all tubes are lighted for 10 hours and if the cost of one unit of electricity energy is Rs. 2.50 the total cost of electricity consumption is ...

  1. Rs. $100$

  2. Rs. $20$

  3. Rs. $25$

  4. Rs. $10$


Correct Option: D

The resistance of a hot tungsten filament is about $10$ times the cold resistance. What will be the resistance of $100\ W,\ $$200\ V$ lamp when not in use?

  1. $20\Omega$

  2. $40\Omega$

  3. $200 \Omega$

  4. $400 \Omega$


Correct Option: B
Explanation:
When the lamp is in use, it is hot.

Using Joule's Law of heating,
$P = \cfrac{V^2}{R}$
$R= \cfrac{200^2}{100}=400\Omega$ when operating 

It is given resistance of hot lamp is 10 times the cold lamp.
Hence, $R _{cold}=\cfrac{400}{10}=40\Omega$

Calculate the energy consumed by $2\ kW$ heater used for $1\ hr$ every day in a period of $30\ \text{days}$.

  1. $60\ \text{units}$

  2. $120\ \text{units}$

  3. $15\ \text{units}$

  4. $30\ \text{units}$


Correct Option: A
Explanation:

$E= Pt = 2 kW\times (30 \times 1 hr)=60 kW h=60\ \text{units}$

You have the following appliances each of $500\ W$ running on  $220\ V$ a.c.: 

(1) Electric iron.
(2) Electric lamp. 
(3) Electric room heater. 
The electric resistance is:

  1. maximum for the heater.

  2. maximum for the electric lamp.

  3. maximum for the electric iron.

  4. same in all the three cases.


Correct Option: D
Explanation:

Resistance$,\ R= \dfrac{V^2}{P} =\dfrac{(220)^2}{P}$. Since $P$ of each appliance is the same, hence $R$ is same for all the three appliances.

An electric lamp is marked $60\ W,\ 220\ V$. The cost of kilo watt hour of electricity is $Rs.\ 1.25$. The cost of using thing lamp on $220\ V$ for $8\ hrs$ is:

  1. $Rs.\ 0.25$

  2. $Rs.\ 0.60$

  3. $Rs.\ 1.2$

  4. $Rs.\ 4.00$


Correct Option: B
Explanation:

Given:

Power, $P=60\ W$
Time, $t=8\ hrs$

Energy consumed per day, $E=Pt=60\times 8=480\ Whr=\dfrac{480}{1000}=0.48\ kW hr$
Hence total cost, $C=0.48\times 1.25=Rs. 0.60 $

An electric bulb of $60\ W$ is used for $6\ hrs/day$. Calculate the units of energy consumed in one day by the bulb.

  1. $0.18\ \text{units}$

  2. $0.36\ \text{units}$

  3. $0.54\ \text{units}$

  4. $0.72\ \text{units}$


Correct Option: B
Explanation:

Given :    $t = 6\ hrs$             

                $P = 0.06\ kW$
Thus energy consumed per day is:

$E=  Pt = 0.06\times 6  = 0.36\ kWh$$ = 0.36\ \text{units}$

An electric bulb of $30\ W$ consumes $0.72$ units of energy in a day. Find the number of hours it is working in a day?

  1. $6\ hrs$

  2. $12\ hrs$

  3. $18\ hrs$

  4. $24\ hrs$


Correct Option: D
Explanation:
Given :   $P =0.030$ kW
Energy consumed per day       $E= 0.72\ \text{units}$$ = 0.72\ kWh$
Using   $E = P\times t$
$\therefore$   $0.72 = 0.03\times t$               
$\implies t = 24\ hrs$

For a heater rated at 4.4 kW; 220 V  . Calculate  the energy consumed by the heater in 4 hours? 

  1. 12 kWh

  2. 17.6 kWh

  3. 0 kWh

  4. 4.4 kWh


Correct Option: B
Explanation:

Given :   $P = 4.4$ kW             $t = 4$ h

$\therefore$ Energy consumed in 4 h       $E= Pt =4.4\times 4 = 17.6$ kWh

How much units of energy is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

  1. 100 kwh

  2. 150 kwh

  3. 160 kwh

  4. 500 kwh


Correct Option: B
Explanation:
Given :    $P = 0.05$ kW               $t = 10$ hours per day 
Number of bulbs     $n = 10$
Energy consumed per day       $E= nPt = 10 \times 0.05\times 10 = 5$ kwh  per day
Total energy consumed in 30 days    $E _t = 5\times 30 = 150$ kWh