Motion Class 9 Numerical Part - 1 |...

Motion Class 9 Numerical Part - 1 | Physicscatalyst

**Class 10 Science**Solved Numerical Questions for electricity

a. Concepts questions

b. Calculation/Numerical problems

c. Multiple choice questions

d. Long answer questions

e. Fill in the blanks

**Question 1**

A wire of length 3 m and area of cross-section 1.7 × 10^{-6} m^{2} has a resistance 3 × 10^{-2} ohm.

a. What is the formula for resistivity of the wire and what is the unit of it

b. Calculate the resistivity of the wire**Solution**

a. Resistivity of the wire is given by

$\rho =R\frac{A}{L}$

And It unit is Ohm-m

b. In this case

L=3 m

$A=1.7 \times 10^{-6}$ m^{2}

$R=3 \times 10^{-2}$ ohm

So

$\rho= 1.7 \times 10^{-8}$ Ohm-m**Question 2**

The table given below shows the resistivity of three Material X, Y and Z?

Samples | X | Y | Z |

Resistivity | 3 X 10^{-9} | 11.1 X 0^{-6} | 18X10^{-17} |

a. Arrange the samples in increasing order of conductivity

b. Which of these is best conductor?

c. Which are these is best insulator?**Solution**

a.Conductivity is inversely proportional to resistivity

So

Y< X< Z

b. Z is the best conductor as it has least resistivity

c. Y is the best insulator as it has highest resistivity

**Question 3.**

There are m resistor each of resistance R. First they all are connected in series and equivalent resistance is X. Now they are connected in parallel and equivalent resistance is Y. What is the ratio of X and Y?**Solution**

Series combination

$X=R+R+R+......m \; times = mR$

Parallel combination

$\frac{1}{Y}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+...=\frac{m}{R}$

Or $Y=\frac {R}{m}$

So X : Y=m^{2} : 1

**Question 4**

We have four resistors A ,B ,C and D of resistance 4 ohm,8 ohm ,12 ohm and 24 ohm respectively?

1 | Lowest resistance which can be obtained by combining these four resistors | |

2 | highest resistance which can be obtained by combining these four resistors |

**Solution**

Lowest resistance is obtained in parallel combination

$\frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}$

Or R=2Ω

Highest resistance is obtained in series combination

$R=4+8+12+24=48\Omega$**Question 5**

Three resistors 5Ω, 10Ω and 30Ω are connected in parallel with the battery of Voltage 6V?

S.no | Questions | |

1 | The value of current across each resistor | |

2 | The value of Potential difference across each resistor | |

3 | Total current in the circuit | |

4 | Effective resistance of the circuit |

**Solution**

Potential difference remains same across parallel combination

So current in each resistor is calculated as

I_{1}=V/R_{1}=6/5=1.2 A

I_{2}=V/R_{2}=6/10=.6 A

I_{3}=V/R_{3}=6/30=.2 A

Total current in the circuit

I=I_{1}+I_{2}+I_{3}=1.2+.6+.2=2 A

Effective resistance

$\frac{1}{R}=\frac{1}{5}+\frac{1}{10}+\frac{1}{30}$

R= 3Ω

**Question 6**

An electric bulb draws a current of .8 A and works on 250 V on the average 8 hours a day.

a. Find the power consumed by the bulb

b. If the electric distribution company changes Rs 5 for 6 KWH, what is the monthly bill for 60 days**Solution** Power of the electrical bulb is given by

$P=V \times I=.8 \times 250 = 200 W =.2 KW$

Total energy consumption by the bulb in 60 days

$E=P \times t= .2 \times 8 \times 60=96 KWH$

So, cost will be = $\frac {5 \times 96}{6}=Rs 80$

**Question 7**

A_{1} ,A_{2 },A_{3} and A are ammeters connected in the circuit

B_{1 },B_{2} and B_{3} are three identical bulbs

They all are connected to Voltage source as shown in Figure

When the three bulb are working good and glowing ,the current recorded in Ammeter A is 6 A__Answer Following questions__

a. Same amount of current will go through each Bulb. And the value is 2 A .True or False

b. If the Bulb B_{3}is blown away, the bulb B_{1} and B_{2} will start glowing more. True or False

c. What will happen to all the ammeter reading if Bulb B_{1} is blown away

d. The current shown in Ammeter A remains even any bulb goes down. True or False**Solution**

a. Since Bulb are identical and connected in parallel with Voltage. Same current will flow through each bulb. Since the total current is 6 A.Individual current will be 2 A

b. If the Bulb B_{3} is blown away, The potential difference across other bulb still remains same, So same current will flow and they will glow as it is. No change

c. when Bulb B_{1} goes down, the current in that part become zero.

So reading of Ammeter A_{1} becomes zero

Reading of Ammeter A_{2} will remain same i.e. 2 A

Reading of Ammeter A_{3} will remain same i.e. 2 A

Reading of Ammeter A will be = 2+2=4 A

d. As shown above, the reading of Ammeter A will change

**Question 8**

Give the formula for each

1 | Ohm’s Law | |

2 | Resistance in terms of Length,Area,resistivity | |

3 | Current in terms of Resistance and Voltage | |

4 | Equivalent Resistance for Resistors in Series | |

5 | Equivalent Resistance for Resistors in Parallel | |

6 | Power produced in the resistance |

**Solution**

1 | Ohm’s Law | $V=IR$ |

2 | Resistance in terms of Length,Area,resistivity | $R=\rho \frac{L}{A}$ |

3 | Current in terms of Resistance and Voltage | $I=\frac{V}{R}$ |

4 | Equivalent Resistance for Resistors in Series | R=R_{1}+R_{2}+R_{3}….. |

5 | Equivalent Resistance for Resistors in Parallel | $\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ |

6 | Power produced in the resistance | $P=I^2R$ |

**Question 9**

Two students (P) and (Q) connect their two given resistors $R_1$ and $R_2$ in the manners shown below :

Student (P) connects the terminals marked ($b_1$) and ($c_1$) while student (Q) connects the terminals marked ($d_2$) and ($c_2$) in their respective circuits at the points marked X and Y.

Which one of the following is correct in relation to the above arrangements?

a. both the students will determine the equivalent resistance of the series combination of the two resistors.

b. both the students will determine the equivalent resistance of the parallel combination of the two resistors

c. student (P) will determine the equivalent resistance of the series combination while student (Q) will determine the equivalent resistance of the parallel combination of the two resistors.

d. student (P) will determine the equivalent resistance of the parallel combination while student (Q) will determine the equivalent resistance of the series combination of the two resistors.**Solution**

In combination made by student A, only one end of the resistors are in contact so they are connected in series. But, in the combination made by student B, both the ends of both the resistances are connected together, so they are connected in parallel.

So, Student P will measure the equivalent resistance for series combination and Student Q will measure for parallel combination

Hence, the correct answer is option c.

**Question 10**

which one of the below circuit is properly connected with the electrical components

a. P

b. Q

c. R

d. S**Solution**

Ammeter should be in series and Voltmeter in parallel. Also current flows from + to -. So (b) is the correct option

**Question 11**

Judge the equivalent resistance when the following are connected in parallel

(a) 1 Ω and $10^6$ Ω

(b) 1 Ω, $10^3$ Ω and $10^6$ Ω.**Solution**

(a) $\frac{1}{R}=\ \frac{1}{1}+\ \frac{1}{10^6\ }=\ \frac{10^6+1}{10^6}$

$= \frac{1000000+1}{1000000}=\ \frac{1000001}{1000000}$

$= \frac{1000000}{1000001}=0.9 \Omega$

(b) $\frac{1}{R}=\ \frac{1}{1}+\ \frac{1}{10^3}+\ \frac{1}{10^6}=\ \frac{10^6+10^3+1}{10^6}$

$= \frac{1001001}{1000000}$

$R = \frac{1000000}{1001001}=\ 0.9 \Omega$

**Question 12**

How can three resistors of resistances 2 Ω, 3 Ω and 6Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?**Solution**

(a) for 4 Ω reistance:

3Ω and 6 Ω resistances are connected in parallel and this combination is connected in series to 2 Ω resistor.

(b) For 1 Ω resistance:

All these three resistances are connected in parallel to get equivalent resistance of 1 Ω

**Question 13**

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?**Solution**

(a) For the highest resistance, resistors are connected in series:

$R = R_1 + R_2 + R_3 + R_4$

= 4 + 8 + 12 + 24

R = 48 Ω

(b) The lowest resistance will be obtained when these are connected in parallel:

$\frac{1}{R}=\ \frac{1}{R_1}+\ \frac{1}{R_2}+\ \frac{1}{R_3}+\ \frac{1}{R_4}$

$\frac{1}{R}=\ \frac{1}{4}+\ \frac{1}{8}+\ \frac{1}{12}+\ \frac{1}{24} $

$\frac{1}{R}=\ \frac{6+3+2+1}{24}=\ \frac{12}{24}$

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

R = 2 Ω

**Question 14**

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.**Solution**

Power=VI

$P= 220 \times 5$

P= 1100 Watts

$\text {Energy} = \text {Power} \times \text{Time}$

$= 1100 \times 2 \times\ 3600 $ sec

$= 3960000 = 3.96 \times\ 10^6$ joules

$= 3.96\times\ 10^3$ kilojoules

**Question 15**

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as must current as all three appliances and what is the current through it?**Solution**

The electric iron will have equivalent resistance to the same as 100 Ω, 50 Ω and 500 Ω resistors are in parallel.

Let it be R Ω.

Therefore

$\frac{1}{R}=\ \frac{1}{100}+\ \frac{1}{50}+\ \frac{1}{500}$

$= \frac{5+10+1}{500}=\ \frac{16}{500}$

$R = \frac{500}{16}\ =31.25 $ Ω

Also Current is given as

$I = \frac{V}{R}=\ \frac {220}{31.25}$

I = 7.04 A

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**link to this page by copying the following text****Also Read**

**Notes**- Chapter 12 electricity notes

**Assignments**- Electricity NCERT Solutions(intext)
- Electricity NCERT Solutions(Exercise)
- Electricity Formative Assignment
- Electricity Conceptual questions
- Electricity Numerical questions
- Electricity Short questions
- Electricity Long questions
- Electricity Numerical Worksheet

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