10th Science Important Numerical Problems with Solutions

10th Science Problem Questions Full PDF

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1. LAWS OF MOTION

Problem from Two Marks

1 If a 5 N and a 15 N forces are acting opposite to one another. Find the resultant force and the direction of action of the resultant force. Given:

\(F_{1} = 5N\)
\(F_{2} = 15N\)

Solution:

Two forces acting opposite to one another.
Therefore, Resultant force is,
\(F_{net} = F_{2} - F_{1}\)
\(F_{net} = 15 - 5 = 10N\)

The magnitude of resultant force is 10 N and its direction is along 15 N force.

Numerical Problems

1 Two bodies have a mass ratio of 3:4. The force applied on the bigger mass produces an acceleration of \(12~m~s^{-2}\). What could be the acceleration of the other body, if the same force acts on it? Given:

\(m_{1}:m_{2} = 3:4\) ; \(F_{1} = F_{2}\)
Let \(m_{2}\) be the bigger mass, acceleration due to bigger body, \(a_{2} = 12~ms^{-2}\)

Solution:

We know, \(F = ma \Rightarrow F_{1} = m_{1}a_{1} ; F_{2} = m_{2}a_{2}\)
\(\frac{F_{1}}{F_{2}} = \frac{m_{1}}{m_{2}} \times \frac{a_{1}}{a_{2}}\)
\(1 = \frac{3}{4} \times \frac{a_{1}}{12} \Rightarrow a_{1} = 16~ms^{-2}\)

Acceleration of the other body, \(a_{1}\) is \(16~ms^{-2}\)

2 A ball of mass 1 kg moving with a speed of \(10~m~s^{-1}\) rebounds after a perfect elastic collision with the floor. Calculate the change in linear momentum of the ball. Given:

Mass of ball (m) = 1 kg
Initial velocity \((u) = 10~m~s^{-1}\)
It is perfect elastic collision, ball rebounds with the same speed but in opposite direction.
Final velocity \((v) = -10~ms^{-1}\)

Solution:

\(\Delta p = mv - mu\)
\(= 1 \times (-10) - 1 \times (10) = -10 - 10 = -20~kg~m~s^{-1}\)
(Negative sign just indicates the direction of momentum)

Change in linear momentum of the ball is \(20~kg~ms^{-1}\)

3 A mechanic unscrew a nut by applying a force of 140 N with a spanner of length 40 cm. What should be the length of the spanner if a force of 40 N is applied to unscrew the same nut? Given:

Force \(F_{1} = 140~N\), Length \(d_{1} = 40~cm\)
Force \(F_{2} = 40~N\), Length \(d_{2} = ?\)

Solution:

Moment of couple is same for both the spanner, and so \(F_{1}d_{1} = F_{2}d_{2}\)
\(d_{2} = \frac{F_{1}d_{1}}{F_{2}} = \frac{40 \times 140}{40} = 140~cm\)

If a force of 40 N is applied, the length of the spanner should be 140 cm / 1.4 m

4 The ratio of masses of two planets is 2:3 and the ratio of their radii is 4:7. Find the ratio of their accelerations due to gravity. Given:

Ratio of radii, \(R_{1}:R_{2} = 4:7\)
Ratio of masses, \(m_{1}:m_{2} = 2:3\)
Ratio of acceleration due to the gravity, \(g_{1}:g_{2} = ?\)

Solution:

\(g_{1} = \frac{GM_{1}}{R_{1}^{2}}\) ...(1)
\(g_{2} = \frac{GM_{2}}{R_{2}^{2}}\) ...(2)
Eqn (1) ÷ (2) \(\Rightarrow \frac{g_{1}}{g_{2}} = \frac{M_{1}}{M_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}}\)
\(\frac{g_{1}}{g_{2}} = \frac{2}{3} \times \frac{7^{2}}{4^{2}} \Rightarrow \frac{2}{3} \times \frac{49}{16} = \frac{49}{24}\)

The ratio of acceleration due to gravity, \(g_{1}:g_{2} = 49:24\)

Example Problems

5 Calculate the velocity of moving body of mass 5 kg whose linear momentum is \(2.5~kg~m~s^{-1}\). Solution:

Mass = 5 kg
Linear momentum = \(2.5~kg~m~s^{-1}\)
Linear momentum = mass × velocity
Velocity = linear momentum / mass
\(V = \frac{2.5}{5} = 0.5~m~s^{-1}\)

Velocity of moving body is \(0.5~ms^{-1}\).

6 A door is pushed, at a point whose distance from the hinges is 90 cm, with a force of 40N. Calculate the moment of the force about the hinges. Given:

\(F = 40~N\); \(d = 90~cm = 0.9~m\)

Solution:

The moment of a force \(M = F \times d\)
\(M = 40 \times 0.9 = 36~Nm\)

7 At what height from the centre of the Earth surface, the acceleration due to gravity will be \(1/4^{th}\) of its value as at the Earth. Given:

Height from the centre of the earth, \(R' = R + h\)
Acceleration due to gravity at that height, \(g' = \frac{g}{4}\)

Solution:

\(g = \frac{GM}{R^{2}}, g' = \frac{GM}{R'^{2}} \Rightarrow \frac{g}{g'} = (\frac{R'}{R})^{2}\)
\(\frac{g}{g/4} = (\frac{R+h}{R})^{2} = (1 + \frac{h}{R})^{2}\)
\(4 = (1 + \frac{h}{R})^{2}\) (take square root on both sides)
\(2 = 1 + \frac{h}{R} \Rightarrow h = R\)
\(R' = R + R = 2R\)

The acceleration due to gravity will be \(1/4^{th}\) of its value as at the Earth, when the object is placed at twice the radius of the earth from its centre.

Additional Problems

8 A lift is moving downwards with an acceleration of \(1.8~m~s^{-2}\). What is apparent weight realised by a man of mass 50 kg? Given:

Acceleration \((a) = 1.8~m~s^{-2}\), Mass \(m = 50~kg\)

Solution:

If Lift is moving downward with an acceleration 'a' then,
The Apparent weight is, \(R = m(g - a)\)
\(= 50(9.8 - 1.8)\)
\(R = 50 \times 8 = 400~N\)

9 A weight of a man is 686 N on the surface of the earth. Calculate the weight of the same person on moon. ('g' value of a moon is \(1.625~m~s^{-2}\)) Given:

\(W = mg = 686~N\)

Solution:

\(m = \frac{W}{g} = \frac{686}{9.8} = 70~kg\)
Weight on moon \(W = mg = 70 \times 1.625 = 113.75~N\)

10 A force of 5 N applied on a body produces an acceleration \(5~cm~s^{-2}\). Calculate the mass of the body. Given:

\(F = 5~N\); \(a = 5~cms^{-2} = 0.05~ms^{-2}\)

Solution:

\(F = ma\)
\(m = \frac{F}{a} = \frac{5}{0.05} = 100~kg\)

Hot Questions

1 Two blocks of masses 8 kg and 2 kg respectively lie on a smooth horizontal surface in contact with one other. They are pushed by a horizontally applied force of 15 N. Calculate the force exerted on the 2 kg mass. Given:

\(m_{1} = 8~kg, m_{2} = 2~kg\), Force \(F = 15~N\)

Solution:

\(F = ma = (m_{1} + m_{2})a\)
\(a = \frac{F}{m_{1}+m_{2}} = \frac{15}{8+2} = \frac{15}{10} = 1.5~ms^{-2}\)
Force on 2 kg mass (\(m=2~kg\), \(a=1.5~ms^{-2}\)):
\(F = ma = 2 \times 1.5 = 3N\)

2 A heavy truck and bike are moving with the same kinetic energy. If the mass of the truck is four times that of the bike, then calculate the ratio of their momenta. Given:

Mass of bike = \(m_{B}\), Mass of truck = \(m_{T}\)
\(\frac{m_{T}}{m_{B}} = 4\)

Solution:

Kinetic Energy = \(\frac{1}{2}mv^{2}\)
K.E of truck = K.E of bike
\(\frac{1}{2}m_{T}{v_{T}}^{2} = \frac{1}{2}m_{B}{v_{B}}^{2}\)
\((\frac{V_{B}}{V_{T}})^{2} = \frac{m_{T}}{m_{B}} = 4\)
\(\frac{V_{B}}{V_{T}} = 2 \Rightarrow \frac{V_{T}}{V_{B}} = \frac{1}{2}\)
Ratio of their momentum is: \(\frac{p_{T}}{p_{B}} = \frac{m_{T}V_{T}}{m_{B}V_{B}} = 4 \times \frac{1}{2} = 2\)

Ratio of their momentum is 2 : 1.

2. OPTICS

Numerical Problems

1 An object is placed at a distance 20 cm from a convex lens of focal length 10 cm. Find the image distance and nature of the image. Given:

Focal length of the convex lens \(f = 10~cm\)
Distance between object and lens \(u = -20~cm\)

Solution:

Lens formula, \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)
\(= \frac{1}{10} + \frac{1}{-20} = \frac{2-1}{20} = \frac{1}{20}\)
\(v = 20~cm\)

The image distance is 20 cm. Nature: Real and inverted image (The image is at 2F).

2 An object of height 3 cm is placed at 10 cm from a concave lens of focal length 15 cm. Find the size of the image. Given:

Focal length of concave lens, \(f = -15~cm\)
Distance between object and lens \(u = -10~cm\)
Height of the object, \(h = 3~cm\)

Solution:

Lens formula, \(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)
\(= \frac{1}{-15} + \frac{1}{-10} = \frac{-2-3}{30} = -\frac{5}{30} = -\frac{1}{6}\)
\(v = -6~cm\)
Magnification \(m = \frac{v}{u} = \frac{-6}{-10} = 0.6\)
Magnification \(m = \frac{h'}{h} \Rightarrow 0.6 = \frac{h'}{3}\)
\(h' = 0.6 \times 3 = 1.8~cm\)

Height of the image h' is 1.8 cm.

Example Problems

3 Light rays travel from vacuum into a glass whose refractive index is 1.5. If the angle of incidence is \(30^{\circ}\). Calculate the angle of refraction inside the glass. Solution:

\(\mu_{1} = 1\), \(\mu_{2} = 1.5\), \(i = 30^{\circ}\)
Snell's law, \(\frac{\sin i}{\sin r} = \frac{\mu_{2}}{\mu_{1}}\)
\(\sin r = \frac{\mu_{1}}{\mu_{2}} \times \sin i = \frac{1}{1.5} \times \sin 30^{\circ}\)
\(\sin r = \frac{1}{1.5} \times 0.5 = 0.333\)
\(r = \sin^{-1}(0.333) \Rightarrow r = 19.45^{\circ}\)

4 A beam of light passing through a diverging lens of focal length 0.3 m appear to be focused at a distance 0.2 m behind the lens. Find the position of the object. Solution:

Diverging (concave) lens: \(f = -0.3~m\), \(v = -0.2~m\)
\(\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-0.2} - \frac{1}{-0.3}\)
\(\frac{1}{u} = \frac{-0.3+0.2}{0.06} = \frac{-0.1}{0.06} = -\frac{10}{6}\)
\(u = -0.6~m\)

5 A person with myopia can see objects placed at a distance of 4 m. If he wants to see objects at a distance of 20 m. What should be the focal length and Power of the concave lens he must wear? Given:

\(x = 4~m\), \(y = 20~m\)

Solution:

\(f = \frac{xy}{x-y} = \frac{4 \times 20}{4-20} = \frac{80}{-16} = -5~m\)
Power \(P = \frac{1}{f} = \frac{1}{-5} = -0.2~D\)

6 For a person with Hypermetropia, the near point has moved to 1.5 m. Calculate the focal length of the correction lens in order to make his eyes normal. Given:

\(d = 1.5~m\), \(D\) (for normal vision) = 0.25 m

Solution:

\(f = \frac{dD}{d-D} = \frac{1.5 \times 0.25}{1.5 - 0.25} = \frac{0.375}{1.25} = 0.3~m\)

3. THERMAL PHYSICS

1 Find the final temperature of a copper rod, whose area of cross section changes from \(10~m^{2}\) to 11 \(m^{2}\) due to heating. The copper rod is initially kept at 90 K. (Coefficient of superficial expansion is \(0.0021~K^{-1}\)) Solution:

\(A_{o} = 10~m^{2}, A = 11~m^{2} \Rightarrow \Delta A = 1~m^{2}\)
\(T_{o} = 90~K\), \(\alpha_{A} = 0.0021~K^{-1}\)
\(\Delta A = A_{o} \alpha_{A} \Delta T \Rightarrow \Delta T = \frac{\Delta A}{A_{o}\alpha_{A}}\)
\(\Delta T = \frac{1}{10 \times 0.0021} = \frac{1}{0.021} = 47.61~K\)
\(T = \Delta T + T_{o} = 47.61 + 90 = 137.6~K\)

2 Calculate the coefficient of cubical expansion of a zinc bar, whose volume is increased from \(0.25~m^{3}\) to \(0.3~m^{3}\) due to the change in its temperature of 50 K. Solution:

\(\alpha_{v} = \frac{\Delta v}{v_{o}\Delta T} = \frac{0.3 - 0.25}{0.25 \times 50} = \frac{0.05}{12.5} = 0.004~K^{-1}\)

3 A container whose capacity is 70 ml is filled with a liquid up to 50 ml. Then, the liquid in the container is heated. Initially, the level of the liquid falls from 50 ml to 48.5 ml. Then we heat more, the level of the liquid rises to 51.2 ml. Find the apparent and real expansion. Solution:

\(L_{1} = 50~ml, L_{2} = 48.5~ml, L_{3} = 51.2~ml\)
Apparent expansion = \(L_{3} - L_{1} = 51.2 - 50 = 1.2~ml\)
Real expansion = \(L_{3} - L_{2} = 51.2 - 48.5 = 2.7~ml\)

4 Keeping the temperature as constant, a gas is compressed four times of its initial pressure. The volume of gas in the container changing from 20 cc to \(V_{2}\) cc. Find \(V_{2}\). Solution:

\(P_{1} = P, P_{2} = 4P, V_{1} = 20~cc\)
Boyle's Law: \(P_{1}V_{1} = P_{2}V_{2}\)
\(V_{2} = \frac{P_{1}V_{1}}{P_{2}} = \frac{P \times 20}{4P} = 5~cc\)

4. ELECTRICITY

1 An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case? Solution:

Max rate: \(I = \frac{P}{V} = \frac{420}{220} = 1.909~A\)
Min rate: \(I = \frac{P}{V} = \frac{180}{220} = 0.818~A\)

2 A 100 watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January. Solution:

January = 31 days.
Energy (100W bulb) = \(100W \times 5h \times 31 \times 1 = 15500~Wh = 15.5~kWh\)
Energy (60W bulbs) = \(60W \times 5h \times 31 \times 4 = 37200~Wh = 37.2~kWh\)
Total Energy = \(15.5 + 37.2 = 52.7~kWh\)

3 A torch bulb is rated at 3 V and 600 mA. Calculate its a) Power b) Resistance c) Energy consumed if it is used for 4 hour. Solution:

\(V = 3V, I = 600mA = 0.6A\)
a) \(P = VI = 3 \times 0.6 = 1.8~Watt\)
b) \(R = \frac{V}{I} = \frac{3}{0.6} = 5~\Omega\)
c) \(E = P \times t = 1.8 \times 4 = 7.2~Wh\)

4 A piece of wire having a resistance R is cut into five equal parts. Questions:

• a) How will the resistance of each part change?
• b) If placed in parallel, how will the combination change?
• c) Ratio of effective resistance in series to parallel?

Solution:

a) \(L' = L/5\), so \(R' = R/5\). Resistance of each part is reduced to 1/5th.
b) In Parallel: \(\frac{1}{R_{P}} = \frac{5}{R/5} = \frac{25}{R} \Rightarrow R_{P} = \frac{R}{25}\)
c) Series \(R_{S} = R\). Ratio \(R_{S}:R_{P} = R : \frac{R}{25} = 25:1\)

14 (Circuit Problem) Three resistors \(R_{1}\) (5 \(\Omega\)), \(R_{2}\) (10 \(\Omega\)) and \(R_{3}\) (20 \(\Omega\)) are connected in parallel with a 10V battery. Calculate:

a) Current through each: \(I_{1} = 10/5 = 2A\), \(I_{2} = 10/10 = 1A\), \(I_{3} = 10/20 = 0.5A\)
b) Total current: \(I = 2 + 1 + 0.5 = 3.5A\)
c) Total Resistance: \(\frac{1}{R_{P}} = \frac{1}{5} + \frac{1}{10} + \frac{1}{20} = \frac{7}{20} \Rightarrow R_{P} = 2.857~\Omega\)

5. ACOUSTICS

1 A sound wave has a frequency of 200 Hz and a speed of \(400~m~s^{-1}\). Find the wavelength.

\(\lambda = \frac{v}{n} = \frac{400}{200} = 2~m\)

2 The thunder of cloud is heard 9.8 seconds later than the flash of lightning. If the speed of sound in air is \(330~m~s^{-1}\) what will be the height of the cloud?

\(d = v \times t = 330 \times 9.8 = 3234~m\)

5 A man is standing between two vertical walls 680 m apart. He claps his hands and hears two distinct echoes after 0.9 seconds and 1.1 second respectively. What is the speed of sound in the air? Solution:

\(d_{1} + d_{2} = 680~m\)
\(d_{1} = \frac{v \times 0.9}{2}\), \(d_{2} = \frac{v \times 1.1}{2}\)
\(\frac{v}{2}(0.9 + 1.1) = 680 \Rightarrow v = 680~m~s^{-1}\)

6. NUCLEAR PHYSICS

1 \({}_{88}Ra^{226}\) experiences three \(\alpha\)- decay. Find the number of neutrons in the daughter element. Solution:

Reaction: \({}_{88}Ra^{226} \rightarrow {}_{Z}Y^{A} + 3({}_{2}He^{4})\)
Mass number: \(226 = A + 3(4) \Rightarrow A = 226 - 12 = 214\)
Atomic number: \(88 = Z + 3(2) \Rightarrow Z = 88 - 6 = 82\)
Neutrons = \(A - Z = 214 - 82 = 132\)

6 Calculate the amount of energy released when a radioactive substance undergoes fusion and results in a mass defect of 2 kg. Solution:

\(E = mc^{2} = 2 \times (3 \times 10^{8})^{2} = 1.8 \times 10^{17} J\)

7. ATOMS AND MOLECULES

1 Find the percentage of nitrogen in ammonia (\(NH_{3}\)). Solution:

Molecular mass of \(NH_{3} = 14 + 3 = 17~g\)
% of N = \(\frac{14}{17} \times 100 = 82.35\%\)

2 Calculate the number of water molecule present in one drop of water, which weighs 0.18 g. Solution:

Moles = \(0.18 / 18\)
Molecules = Moles \(\times\) Avogadro's No
\(= \frac{0.18}{18} \times 6.023 \times 10^{23} = 6.023 \times 10^{21}\)

9. SOLUTIONS

1 A solution is prepared by dissolving 45 g of sugar in 180 g of water. Calculate the mass percentage of solute. Solution:

Mass % = \(\frac{\text{Mass solute}}{\text{Mass solute} + \text{Mass solvent}} \times 100\)
\(= \frac{45}{45+180} \times 100 = \frac{45}{225} \times 100 = 20\%\)

10. TYPES OF CHEMICAL REACTIONS

1 Lemon juice has a pH 2, what is the concentration of \(H^{+}\) ions? Solution:

\(pH = -\log_{10}[H^{+}] = 2\)
\([H^{+}] = 10^{-2} = 0.01~M\)

4 The hydroxide ion concentration of a solution is \(1 \times 10^{-11} M\). What is the pH of the solution? Solution:

\(pOH = -\log[OH^{-}] = -\log(10^{-11}) = 11\)
\(pH = 14 - pOH = 14 - 11 = 3\)

Title: 10th Science Important Numerical Problems with Solutions Labels: 10th Science, Numerical Problems, Physics Problems, Chemistry Problems, Board Exam Prep Permanent Link: 10th-science-numerical-problems-solutions Search Description: Comprehensive collection of 10th Standard Science numerical problems and solutions for Physics and Chemistry chapters including Motion, Optics, Electricity, and more.

50 Numerical Questions On Electricity Class 10

50 Numerical Questions

On Electricity Class 10 with Solutions

Source: omtexclasses.com

Numerical Problems based on Electric Current

This provides a comprehensive guide on solving numerical problems related to electricity, including electric current, potential difference, Ohm's law, resistance, and resistivity. It presents various formulas and examples to calculate charge, current, voltage, and resistance in different scenarios. Additionally, it covers the combination of resistances in series and parallel configurations.

Formulas to Remember

To solve the numericals on electricity, we will use the following formulas:

$$ I = \frac{Q}{t} $$

$$ I = \frac{ne}{t} $$

Where:
I = Electric current – Ampere
Q = charge – Coulomb
T = time - second
N = number of electrons
E = charge on an electron = \( 1.6 \times 10^{-19} \)

1. The filament of bulb draws a current of 0.5 ampere. Calculate the amount of charge if bulb glows for 2.5 hrs.

Solution

Given values Electric current (I) = 0.5 A
Time (t) = 2.5 Hrs = \( 2.5 \times 60 \times 60 = 9000 \) sec
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = It $$

$$ Q = 0.5 \times 9000 $$

Q = 4500 Coulomb

2. 10 Coulombs charge passing through a point in a circuit in 5 seconds. Find the electric current flowing in the circuit.

Solution

Given values Time (t) = 5 sec
Charge (Q) = 10 C
Electric current (I) = ?

$$ I = \frac{Q}{t} $$

$$ I = \frac{10}{5} $$

I = 2 Amp.

3. A current of 0.5 Ampere is drawn by a filament of an electric bulb for 15 minutes. Find the amount of electric charge.

Solution

Given values Electric current (I) = 0.5 A
Time (t) = 15 min = \( 15 \times 60 = 900 \) sec
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = I \times t $$

$$ Q = 0.5 \times 900 $$

Q = 450 Coulomb

4. A current of 1 Ampere is drawn by a filament of an electric bulb. Find the Number of electrons passing through a cross section of the filament in 10 seconds.

Solution

Given values Electric current (I) = 1 A
Time (t) = 15 s
Number of electrons (n) = ?
Charge on electrons (e) = \( 1.6 \times 10^{-19} \) C

$$ I = \frac{ne}{t} $$

After cross multiplication, we find:

$$ n = \frac{I \times t}{e} $$

Now put the given values in the formula:

$$ n = \frac{1 \times 10}{1.6 \times 10^{-19}} $$

$$ n = \frac{100}{16 \times 10^{-19}} $$

$$ n = \frac{6.25}{10^{-19}} $$

\( n = 6.25 \times 10^{19} \)

Number of electrons (n) has no unit required.

5. 150000 coulomb charge is required to deposit one mole of copper from the CuSO4 solution. Find the time to deposit 0.2 mole copper from the copper solution if a current of 5 Ampere is flowing the solution.

Solution

Given values Electric current (I) = 5 Amp
Charge (Q) = 150000 C

Charge required for 1 mole copper = 150000 C

Charge required for 0.2 mole copper = \( 150000 \times 0.2 = 30000 \) C

$$ I = \frac{Q}{t} $$

$$ t = \frac{Q}{I} = \frac{30000}{5} $$

t = 6000 S

6. An heater draws a current 10 A for 5 minutes. Calculate the charge flowing in heater.

Solutions

Given values Electric current (I) = 10 A
Time (t) = 5 min = \( 5 \times 60 = 300 \) s
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = I \times t $$

$$ Q = 10 \times 300 $$

Q = 3000 C

7. \( 1.25 \times 10^{18} \) electrons are passed from one end to another end of a conductor in 10 seconds. Find the current flowing through the conductor.

Solution

Given values Number of electrons (n) = \( 1.25 \times 10^{18} \)
Time (t) = 10 s
Electric current (I) = ?

$$ I = \frac{ne}{t} $$

[e = \( 1.6 \times 10^{-19} \)]

$$ I = \frac{1.25 \times 10^{18} \times 1.6 \times 10^{-19}}{10} $$

I = 0.2 A

8. Calculate the number of electrons present in 0.1 coulomb of charge.

Solution

Given values Charge (Q) = 0.1
Number of electrons (n) = ?

Number of elections in 1 Coulombs is = \( 6.25 \times 10^{18} \)

Number of electrons in 0.1 Coulomb charge is = \( 6.25 \times 10^{18} \times 0.1 \)

= \( 6.25 \times 10^{17} \)

Numerical Problems based on Potential difference and work

Formulas to Remember

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

Where:
V = Potential difference – volt
W = work done - Joule
Q = charge – Coulomb

9. Calculate the amount of work done in carrying 8C charge from a terminal of 150 volt to 200 volt.

Solution

Given values Charge (Q) = 8 C
Potential (V1) = 150 volt
Potential (V2) = 200 volt

Potential difference (V) = V2 - V1

200 – 150 = 50 volt

Work (W) = ?

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

$$ W = VQ $$

$$ W = 50 \times 8 = 400 J $$

10. How much energy is given to each coulomb of charge passing through a 10 Volt battery?

Solution

Given values Potential difference/voltage (V) = 10 volt
Charge (Q) = 1 C
Energy (E) = ?

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

Work is defined as energy transferred by force, so We can put E in the place of W

$$ E = VQ $$

= 10 × 1 = 10 J

11. Work done in moving 5 coulomb charge from point A to B is 80 Joule, if potential at point A is 25 volt then find the potential at B.

Solution

Given values Work (W) = 80 J
Charge (Q) = 5 C
Potential (VA) = 25 volt
Potential (VB) = ?

$$ V = V_B - V_A = \frac{W}{Q} $$

$$ V_B - 25 = \frac{80}{5} $$

$$ V_B - 25 = 16 $$

$$ V_B = 16 + 25 = 41 \text{ volt} $$

Numerical Problems based on Ohm’s law

Formulas to Remember

$$ V = IR $$

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

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

Where:
V = Voltage (Potential difference) = volt
R = Resistance – Ohm (Ω)
I = Electric current – Ampere

12. When a cell of 1.5 volt is applied in a circuit, a current of 0.5 ampere flows through it. Calculate the resistance of the circuit.

Solution

Given values Voltage (V) = 1.5 volt
Current (I) = 0.5 A
Resistance (R) = ?

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

Put the values in the formula of resistance:

$$ R = \frac{1.5}{0.5} = 3 \text{ Ohm} $$

13. How much current will an electric bulb draw from a 220-volt source. If the resistance of the filament is 1200 ohm?

Solution

Given values Voltage (V) = 220 volt
Resistance (R) = 1200 Ohm
Current (I) = ?

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

$$ I = \frac{220}{1200} = 0.183 A $$

14. What is the potential difference between the ends of a conductor of 15 Ohm resistance when a current of 2 ampere flows through it.

Solution

Given values Resistance (R) = 15 Ohm
Electric current (I) = 2 A
Potential difference (V) = ?

$$ V = IR $$

$$ V = 2 \times 15 = 30 \text{ volt} $$

15. A heater draws a current of 5 A when it is connected to 110 volts. What current will the heater draw when it is connected to 220 volts.

Solution

Given values Current (I1) = 5 A
Voltage (V1) = 110 volts
Voltage (V2) = 220 volts
Current (I2) = ?

According to Ohm’s law, V=IR

$$ R = \frac{110}{5} = 22 \text{ Ohm} $$

When the voltage is 220 volt, The current will be:

$$ I_2 = \frac{220}{22} = 10 A $$

16. Calculate the potential difference required across a conductor of resistance 6 ohm to make a current of 2.5 A flow through it.

Solution

Given values Resistance (R) = 6 ohm
Current (I) = 2.5 A
Potential difference (V) = ?

$$ V = IR $$

$$ V = 2.5 \times 6 = 15 \text{ volt} $$

Numerical Problems based on Resistance and Resistivity

Formulas to Remember

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

Where:
R = Resistance - Ohm
\( \rho \) = Resisitivity - Ohm m
l = length – m
A = area of cross section - m²

17. Calculate the resistance of a copper wire of length 30 cm and area of cross section \( 3 \times 10^{-4} \text{ m}^2 \). The resistivity of copper is \( 1.7 \times 10^{-8} \text{ ohm m} \).

Solution

Given values Length of wire (l) = 30 cm = 0.3m
Area of cross section (A) = \( 3 \times 10^{-4} \text{ m}^2 \)
Resistivity of copper (\( \rho \)) = \( 1.7 \times 10^{-8} \text{ ohm m} \)
Resistance (R) = ?

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

$$ R = 1.7 \times 10^{-8} \times \frac{0.3}{3 \times 10^{-4}} $$

$$ R = \frac{0.51 \times 10^{-8}}{3 \times 10^{-4}} $$

$$ R = \frac{0.17 \times 10^{-8}}{10^{-4}} $$

$$ R = 0.17 \times 10^{-8} \times 10^4 $$

R = \( 0.17 \times 10^{-4} \text{ Ohm} \)

18. Calculate the resistivity of wire having length 1 m and area of cross section \( 1.20 \times 10^{-6} \text{ m}^2 \), if its resistance is 0.013 ohm.

Solution

Given values Area of cross section (A) = \( 1.20 \times 10^{-6} \text{ m}^2 \)
Length (l) = 1m
Resistance (R) = 0.013 ohm
Resistivity (\( \rho \)) = ?

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

Now put the values in the formula:

$$ 0.013 = \rho \frac{1}{1.20 \times 10^{-6}} $$

$$ \rho = 0.013 \times \frac{1.20 \times 10^{-6}}{1} $$

\( \rho = 1.56 \times 10^{-8} \text{ ohm m} \)

19. Calculate the area of cross section of a wire of 1 m and resistance 25 ohm, if the resistivity of material of the wire is \( 1.84 \times 10^{-6} \text{ ohm m} \).

Solution

Given values Length (l) = 1m
Resistance (R) = 25 ohm
Resistivity (\( \rho \)) = \( 1.84 \times 10^{-6} \text{ ohm m} \)
Area of cross section (A) = ?

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

$$ 25 = 1.84 \times 10^{-6} \times \frac{1}{A} $$

$$ A = 1.84 \times 10^{-6} \times \frac{1}{25} $$

A = \( 7.36 \times 10^{-8} \text{ m}^2 \)

20. 1 m long wire with resistance 0.85 ohm and diameter 0.2 mm, what will be the resistivity of the metal at 20°C.

Solution

Given values Length (l) = 1m
Resistance (R) = 0.85 ohm
Diameter (d) = 0.2 mm = \( 2 \times 10^{-4} \text{m} \)
Radius (r) = d/2 = \( 1 \times 10^{-4} \text{m} \)
Area of cross section (A) = ? (\( \pi r^2 \))

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

$$ R = \rho \frac{l}{\pi r^2} $$

$$ 0.85 = \rho \frac{7}{22 \times (1 \times 10^{-4})^2} \times 1 $$

$$ \rho = \frac{0.85 \times 22}{7 \times 10^{-8}} $$

\( \rho = 2.67 \times 10^{-8} \text{ ohm m} \)

Numerical Problems based on the Combination of Resistances

Formulas to Remember

$$ R = R_1 + R_2 + R_3 + \dots R_n $$

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

Where:
\( R_1, R_2, R_3 \dots R_n \) are different resistances
R = total resistance of the combination

21. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in series, find the total resistance.

Solution

Given value R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
R = ?

$$ R = R_1 + R_2 + R_3 $$

$$ R = 5 + 10 + 15 = 30 \text{ ohm} $$

22. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in parallel, find the total resistance.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
R = ?

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

$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{15} $$

$$ \frac{1}{R} = \frac{6+3+2}{30} = \frac{11}{30} $$

$$ \frac{1}{R} = \frac{11}{30} $$

$$ R = \frac{30}{11} = 2.72 \text{ ohm} $$

23. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in series, and system is connected with 90 volt battery. What will the current flowing in the circuit.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 90 volt
I = ?

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

We don’t have the value of ‘R’ so first we will find the value:

$$ R = 5 + 10 + 15 = 30 \text{ ohm} $$

$$ I = \frac{V}{R} = \frac{90}{30} = 3 A $$

24. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in parallel combination and the circuit is connected with 100 volt battery. Find electric current flowing in each of the resistance.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 100 volt
I1 = ?
I2 = ?
I3 = ?

$$ I_1 = \frac{V}{R_1} $$

$$ I_1 = \frac{100}{5} = 20 A $$

$$ I_2 = \frac{V}{R_2} $$

$$ I_2 = \frac{100}{10} = 10 A $$

$$ I_3 = \frac{V}{R_3} $$

$$ I_3 = \frac{100}{15} = 6.66 A $$

(Note: Current in all resistors is different in the parallel combination.)

25. Three resistors of 5ohm, 10 ohm and 15 ohm are connected in series combination and 10 ampere current is flowing in the system when connected with 110 Volt. Find the potential difference between two ends of each resistor.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 110 volt
I = 10 A
V1 = ?
V2 = ?
V3 = ?

$$ V_1 = I R_1 $$

$$ V_1 = 10 \times 5 = 50 \text{ volt} $$

$$ V_2 = I R_2 $$

$$ V_2 = 10 \times 10 = 100 \text{ volt} $$

$$ V_3 = I R_3 $$

$$ V_3 = 10 \times 15 = 150 \text{ volt} $$

26. Three resistances of 2 ohm, 3 ohm and 6 ohm are connected in series and then in parallel. Find the total resistance in both the arrangements.

Solution

Given values R1 = 2 ohm
R2 = 3 ohm
R3 = 6 ohm
R = Total resistance = ?

For series combination:

$$ R = R_1 + R_2 + R_3 $$

$$ R = 2 + 3 + 6 = 11 \text{ ohm} $$

For parallel combination:

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

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

$$ \frac{1}{R} = \frac{6+4+2}{12} = \frac{12}{12} $$

R = 1 ohm

27. A 18 Volt battery is connected across a lamp whose resistance is 50 ohm, through a variable resistor. If the current flowing through the circuit is 0.3 A. Calculate the value of resistance used from the variable resistor.

Solution

Given values Voltage (V) = 18 volt
Current (I) = 0.3 A
Resistance of bulb (r) = 50 ohm
R1 = resistance form variable resistor = ?

Let R1 is the resistance from variable resistor that is used.

Both R1 and r are is series combination so the resultant resistance is:

$$ R = R_1 + r $$

According to Ohm’s law V=IR

$$ 15 = 0.3 \times (R_1 + r) $$

$$ R_1 + 50 = \frac{18}{0.3} = 60 $$

$$ R_1 = 60 - 50 = 10 \text{ ohm} $$

Numerical Problems based on the Circuit diagrams

28. Calculate the total resistance of the following circuit.

[Insert Image from Page 25 here: Series circuit with 5Ω, 10Ω, 20Ω]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 20 ohm
R = ?

All resistances in the given circuit are connected in the series so total resistance:

$$ R = R_1 + R_2 + R_3 $$

$$ R = 5 + 10 + 20 = 35 \text{ ohm} $$

29. Calculate the total resistance in the given circuit

[Insert Image from Page 26 here: Parallel circuit with 5Ω, 10Ω, 20Ω]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 20 ohm
R = ?

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

$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{20} $$

$$ \frac{1}{R} = \frac{4+2+1}{20} = \frac{7}{20} $$

$$ R = \frac{20}{7} = 2.85 \text{ ohm} $$

30. You have been given the following circuit diagram. Find the following:
(i) Electric current through each resistor
(ii) Total resistance
(iii) Total current

[Insert Image from Page 27 here: Parallel circuit with 5Ω, 10Ω, 30Ω connected to 6V]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 30 ohm
V = 6 volt
R = ?
I1 = ?, I2 = ? and I3 = ?
I = ?

All the resistors are connected in parallel combination.

(i) Current through each resistor

$$ V = I_1 R_1 $$

$$ 6 = I_1 \times 5 $$

$$ I_1 = \frac{6}{5} = 1.2 A $$

$$ V = I_2 R_2 $$

$$ 6 = I_2 \times 10 $$

$$ I_2 = \frac{6}{10} = 0.6 A $$

$$ V = I_3 R_3 $$

$$ 6 = I_3 \times 30 $$

$$ I_3 = \frac{6}{30} = 0.2 A $$

(ii) Total resistance (R)

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

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

$$ \frac{1}{R} = \frac{6 + 3 + 1}{30} = \frac{10}{30} $$

$$ R = \frac{30}{10} = 3 \text{ Ohm} $$

(iii) Total current (I)

$$ V = IR $$

$$ 6 = I \times 3 $$

$$ I = \frac{6}{3} = 2 A $$

31. Find the equivalent resistance in the given circuit and current flowing through the circuit.

[Insert Image from Page 29 here: 3Ω and 6Ω in parallel, connected to 4.5V]

Solution

Given values R1 = 3 ohm
R2 = 6 ohm
V = 4.5 volt
R = ?

We can see that the resistors of 3 ohm and 6 ohm are connected in parallel combination so the equivalent resistance R

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

$$ \frac{1}{R} = \frac{1}{3} + \frac{1}{6} $$

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

$$ R = \frac{6}{3} = 2 \text{ ohm} $$

According to Ohm’s law, V=IR

$$ 4.5 = I \times 2 $$

$$ I = \frac{4.5}{2} = 2.25 A $$

32. You have been given the following circuit of resistors connected with a battery.
Calculate:
(i) total resistance of the circuit
(ii) total current
(iii) voltage across 5 ohm resistor

[Insert Image from Page 31 here: 10Ω and 10Ω in parallel, connected in series with 5Ω, V=6V]

Solution

Given values R1 = 10 ohm
R2 = 10 ohm
R3 = 5 ohm
V = 6 volt

(i) total resistance

We can see R1 and R2 are in parallel so the total resistance RA

$$ \frac{1}{R_A} = \frac{1}{R_1} + \frac{1}{R_2} $$

$$ \frac{1}{R_A} = \frac{1}{10} + \frac{1}{10} $$

$$ \frac{1}{R_A} = \frac{1+1}{10} = \frac{2}{10} $$

$$ R_A = \frac{10}{2} = 5 \text{ Ohm} $$

Now RA and R3 in series so the total resistance R

R = RA + R3

R = 5 + 5 = 10 ohm

(ii) total current

$$ I = \frac{V}{R} = \frac{6}{10} = 0.6 A $$

(iii) Voltage across 5 ohm resistor

V1 = IR1

V1 = 0.6 × 5 = 3 volt

33. Find the equivalent resistance of the following circuit of resistors.

[Insert Image from Page 33 here: Triangle configuration. 4Ω on one side, 8Ω and 8Ω on others]

Solution

Given values R1 = 4 Ohm
R2 = 8 Ohm
R3 = 8 Ohm
R = ?

In the given circuit diagram we can see R2 and R3 are series so:

RA = R2 + R3

RA = 8 + 8 = 16 Ohm

Now the RA is parallel to R1 so the equivalent resistance R

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_A} $$

$$ \frac{1}{R} = \frac{1}{4} + \frac{1}{16} $$

$$ \frac{1}{R} = \frac{4+1}{16} = \frac{5}{16} $$

$$ R = \frac{16}{5} = 3.2 \text{ Ohm} $$

Numerical Problems based on Work, Power and Energy

Formulas to Remember

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

$$ W = V \times Q $$

$$ W = I^2Rt $$

$$ W = \frac{V^2t}{R} $$

Note = We can put E in place of W

$$ P = \frac{W}{t} $$

$$ P = \frac{V^2}{R} $$

$$ P = I^2R $$

Where:
P = power – Watt
W = work – Joule
V = voltage – Volt
I = current – Ampere
Q = charge = Coulomb

34. An electric bulb is connected to 110 volt electric source. The current is 0.5 A, then what is the power of the bulb?

Solution

Given values Voltage (V) = 110 volt
Current (I) = 0.5 A
Power (P) = ?

P = VI

P = 110 × 0.5

P = 55 Watt

35. A bulb rated 5 volt – 100mA, calculate its (i) power (ii) resistance

Solution

Given values Voltage (V) = 5 volt
Current (I) = 100mA = \( 100 \times 10^{-3} \) A

P = VI

P = \( 5 \times 100 \times 10^{-3} \)

P = 0.5 Watt

36. Two electric bulbs rated 60W, 220V and 100W, 220 V. which one of them has higher resistance?

Solution

Given values (i) P1 = 60 W, V = 220 volt
(ii) P2 = 110 W , 220 volt
R1 = ?
R2 = ?

$$ R_1 = \frac{V^2}{P_1} $$

$$ R_1 = \frac{(220)^2}{60} $$

$$ R_1 = 806.67 \text{ Ohm} $$

$$ R_2 = \frac{V^2}{P_2} $$

$$ R_2 = \frac{(220)^2}{100} $$

$$ R_2 = 484 \text{ Ohm} $$

60 watt bulb has higher resistance than 100 watt bulb.

37. Two lamps rated 100 W, 220 Volt and 60W, 220 V connected in parallel. Calculate the current drawn by the circuit.

Solution

Given values P1 = 100W
P2 = 60 W
V = 220volt
I = ?

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

In the above formula we don’t have value of R so we will have to find the value of R

$$ R_1 = \frac{V^2}{P_1} = \frac{(220)^2}{100} \text{ Ohm} $$

$$ R_2 = \frac{V^2}{P_2} = \frac{(220)^2}{60} \text{ Ohm} $$

Both the lamps are in parallel, so the resultant resistance is

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

$$ \frac{1}{R} = \frac{100}{220 \times 220} + \frac{60}{220 \times 220} $$

$$ \frac{1}{R} = \frac{100+60}{220 \times 220} $$

$$ \frac{1}{R} = \frac{160}{220 \times 220} \text{ Ohm} $$

Now we can find the value of current (I)

$$ I = \frac{220}{\frac{220 \times 220}{160}} $$

$$ I = \frac{220 \times 160}{220 \times 220} $$

$$ I = \frac{160}{220} = 0.72 A $$

38. An electric lamp is rated 100W, 220V. it is used for 10 hours daily then calculate energy consumed in KWh per day.

Solution

Given values P = 100 W
V = 220 Volt

$$ \text{Energy consumed per day} = \frac{P \times t}{1000} $$

$$ = \frac{100 \times 10}{1000} $$

Energy consumed per day = 1KWh or 1 Unit

39. A bulb rated 2.5 V and 650mA. Calculate (i) power (ii) resistance (iii) energy consumed for 5 hours.

Solution

Given values V = 2.5 Volt
I = 650 mA = 0.65A
T = 5 hours

(i) P=VI

P = 2.5 × 0.65 = 1.625 W

(ii) Resistance

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

$$ R = \frac{2.5}{0.65} = 3.84 \text{ Ohm} $$

(iii) Energy consumed in 5 hours

$$ = \frac{P \times t}{1000} $$

$$ = \frac{1.625 \times 5}{1000} = 0.008125 \text{ KWh} $$

40. An immersion rod of 750W is used for one hour. Find the energy consumed in (i) KWh (ii) Joule

Solution

Given values P = 750 W
T = 1 hour

(i) Energy consumed in KWh

$$ \text{Energy consumed for 1 hour} = \frac{P \times t}{1000} $$

$$ = \frac{750 \times 1}{1000} = 0.75 \text{ KWh} $$

(iii) Energy consumed in Joule

1 KWh = \( 3.6 \times 10^6 \) Joule

So \( 0.75 \times 3.6 \times 10^6 \) Joule

= \( 2.7 \times 10^6 \) Joule

41. One heater is rated 220 V- 5 A. calculate the energy consumed if it is used for 5 hours.

Solution

Given values V = 220 volt
I = 5 A

$$ \text{Energy consumed for 5 hour} = \frac{P \times t}{1000} $$

But in this formula we need value of Power (P)

P = VI

P = 220 × 5

P = 1100W

Now we can put the value of P in the formula

$$ \text{Energy consumed for 5 hour} = \frac{1100 \times 5}{1000} $$

= 5.5 KWh

42. Which uses more energy a 250W computer in 1 hour or a 1200W heater in 10 minutes.

Solution

Given values P1 = 250W
T1 = 1hour
P2 = 1200W
T2 = 10 min = 10/60 hour

(i) A 250 W computer is used for 1 hour

= P1 × t1 = 250 × 1

= 250Wh

(ii) A 1200W heater is used for 10 min

= P2 × t2 = 1200 × 10/60

= 200Wh

So, computer uses more energy.

43. An electric heater is rated 1kW, 220V. calculate its resistance.

Solution

Given values P = 1kW = 1000W
V = 220 volt
R = ?

$$ R = \frac{V^2}{P} $$

$$ R = \frac{220^2}{1000} $$

R = 48.4 Ohm

44. An electric bulb rated 220 V- 100 W. calculate its resistance.

Solution

Given values Power (P) = 100W
Voltage (V) = 220volt
Resistance (R) = ?

$$ P = \frac{V^2}{R} $$

$$ 100 = \frac{220^2}{R} $$

$$ R = \frac{220^2}{100} $$

R = 484 Ohm

Numerical Problems based on Heating effect of current

45. How much heat will a device of 15 W produces in one minute if it is connected to a battery of 15 Volt.

Solution

Given values P = 15W
V = 15 volt
T = 1 min = 60 s

H = P × t

H = 15 × 60 = 900 J

46. 1 kJ hat is produced each second in 50 Ohm resistor. Calculate the potential difference across the resistor.

Solution

Given values H = 1kJ = 1000J
Time (t) = 1 s
Resistance (R) = 50 Ohm
Voltage (V) = ?

$$ H = I^2Rt $$

$$ 1000 = I^2 \times 50 \times 1 $$

$$ I^2 = \frac{1000}{50} = 20 $$

$$ I = \sqrt{20} = 4.47A $$

So, potential difference across the resistor

V = IR

V = 4.47 × 50 = 223.5 volt

47. A heater of resistance 10 Ohm draws 10 A current from electric source for 2 hours. Calculate the heat produced in the heater.

Solution

Given values Resistance (R) = 10 Ohm
Current (I) = 10 A
Time (t) = 2 hours = 2 × 60 × 60 = 7200 s
Heat (H) = ?

$$ H = I^2Rt $$

$$ H = (10)^2 \times 10 \times 7200 $$

H = \( 72 \times 10^5 \) J

48. 100 Joule of heat is produced each second in 10 Ohm resistance. find the potential difference across the resistor.

Solution

Given values Heat (H) = 100J
Time (t) = 1s
Resistance (R) = 10 Ohm
Potential difference (V) = ?

V = IR

In the above formula we need the value of current (I) so at first we must find the value of I form the given values

$$ H = I^2Rt $$

$$ I = \sqrt{\frac{H}{Rt}} $$

$$ I = \sqrt{\frac{100}{10 \times 1}} $$

$$ I = \sqrt{10} $$

$$ I = 3.16 A $$

Now we can find the value of potential difference

V = IR

V = 3.16 × 10

V = 31.6 volt

49. Find the heat produced in the following combination of resistors if current drawn for 10 seconds.

[Insert Image from Page 47 here: Two resistors in series (2Ω and 2Ω) connected to 10V]

Solution

Given values R1 = 2 Ohm
R2 = 2 Ohm
Potential difference (V) = 10 volt
Time (t) = 10 s
Heat (H) = ?

H = V I t

We will have to find the value of current (I)

R = R1 + R2

R = 2 + 2 = 4 Ohm

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

$$ I = \frac{10}{4} = 2.5 A $$

H = V I t

H = 10 × 2.5 × 10

H = 250 J

50. How much energy is given to 10 coulomb charge passing through 15 volt battery?

Solution

Given values Charge (Q) = 10 Coulomb
Voltage (V) = 15 volt
Energy (E) = ?

E = VQ

E = 15 × 10 = 150 J

TRUTH TABLE | LOGIC HSC | 12TH STANDARD MATHS | Mathematical Logic.

Mathematical Logic: Truth Tables & Solutions (HSC 12th)

Solutions and Explanations

Below are the detailed explanations and solutions for Mathematical Logic, specifically focusing on logical connectives and the construction of truth tables, which is a core part of the HSC 12th Standard Mathematics syllabus.

1. Introduction to Logical Connectives

A simple statement is a declarative sentence which is either true or false, but not both simultaneously. We use logical connectives to join simple statements to form compound statements.

Key Symbols:

• Conjunction (AND): \( \wedge \)
• Disjunction (OR): \( \vee \)
• Negation (NOT): \( \sim \)
• Conditional (Implication): \( \rightarrow \)
• Biconditional (Double Implication): \( \leftrightarrow \)

2. Fundamental Truth Tables

Before solving complex problems, we must understand the standard truth values for each connective.

A. Conjunction (\( p \wedge q \)) and Disjunction (\( p \vee q \))

p	q	p \( \wedge \) q (AND)	p \( \vee \) q (OR)
T	T	T	T
T	F	F	T
F	T	F	T
F	F	F	F

Note: Conjunction is True only if both are True. Disjunction is False only if both are False.

B. Conditional (\( p \rightarrow q \)) and Biconditional (\( p \leftrightarrow q \))

p	q	p \( \rightarrow \) q	p \( \leftrightarrow \) q
T	T	T	T
T	F	F	F
F	T	T	F
F	F	T	T

Note: Implication is False only when Hypothesis (p) is True and Conclusion (q) is False. Biconditional is True when both have the same truth value.

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Solved Examples: Constructing Truth Tables

Question 1: Construct the truth table for \( (p \wedge q) \vee \sim p \)

Solution:

We need columns for \( p \), \( q \), \( \sim p \), \( p \wedge q \), and finally the whole statement.

p	q	\( \sim p \)	\( p \wedge q \)	\( (p \wedge q) \vee \sim p \)
T	T	F	T	T
T	F	F	F	F
F	T	T	F	T
F	F	T	F	T

Question 2: Examine whether the statement pattern is a Tautology, Contradiction, or Contingency.

Statement: \( (p \rightarrow q) \leftrightarrow (\sim p \vee q) \)

Solution:

A Tautology is true in all cases. A Contradiction is false in all cases. A Contingency is a mix.

p	q	\( \sim p \)	\( p \rightarrow q \) (I)	\( \sim p \vee q \) (II)	(I) \( \leftrightarrow \) (II)
T	T	F	T	T	T
T	F	F	F	F	T
F	T	T	T	T	T
F	F	T	T	T	T

Conclusion: Since all the entries in the last column are 'T', the given statement pattern is a Tautology. This also proves that \( p \rightarrow q \) is logically equivalent to \( \sim p \vee q \).

Question 3: Three variable Truth Table

Construct the truth table for: \( (p \vee q) \rightarrow r \)

Solution:

Since there are 3 statements (p, q, r), there will be \( 2^3 = 8 \) rows.

p	q	r	p \( \vee \) q	\( (p \vee q) \rightarrow r \)
T	T	T	T	T
T	T	F	T	F
T	F	T	T	T
T	F	F	T	F
F	T	T	T	T
F	T	F	T	F
F	F	T	F	T
F	F	F	F	T

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Important Rules to Remember (Summary)

• Negation: \( \sim T = F \) and \( \sim F = T \)
• Conjunction: T only if T \( \wedge \) T
• Disjunction: F only if F \( \vee \) F
• Conditional: F only if T \( \rightarrow \) F
• Biconditional: T if values match (T \( \leftrightarrow \) T or F \( \leftrightarrow \) F)

29 Coordinate Geometry. Complete Coordinate Geometry: 10 Solved Examples & 50 Practice Questions (Class 10)

Coordinate Geometry: Complete Guide for Class 10

Master Coordinate Geometry with these 10 fully solved examples covering Distance Formula, Section Formula, and Area of Triangles. Afterward, test your skills with 50 practice questions provided with an answer key.

Part 1: 10 Fully Solved Important Questions

Q1 - Distance Formula

Find the distance between the points $A(2, 3)$ and $B(4, 1)$.

Solution: Let $A(x_1, y_1) = (2, 3)$ and $B(x_2, y_2) = (4, 1)$.
Using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d = \sqrt{(4 - 2)^2 + (1 - 3)^2}$$ $$d = \sqrt{(2)^2 + (-2)^2}$$ $$d = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units.}$$

Q2 - Midpoint Formula

Find the coordinates of the midpoint of the line segment joining $P(-5, 7)$ and $Q(-1, 3)$.

Solution: Using the midpoint formula $M(x, y) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$: $$x = \frac{-5 + (-1)}{2} = \frac{-6}{2} = -3$$ $$y = \frac{7 + 3}{2} = \frac{10}{2} = 5$$ The midpoint is $(-3, 5)$.

Q3 - Section Formula

Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2:3$.

Solution: Let the points be $A(-1, 7)$ and $B(4, -3)$. Ratio $m_1:m_2 = 2:3$.
Using Section Formula: $$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1+m_2}$$ $$x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$$ $$y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$$ The required point is $(1, 3)$.

Q4 - Value of k

Find the value of $k$ if the points $A(2, 3)$, $B(4, k)$, and $C(6, -3)$ are collinear.

Solution: For collinear points, the area of the triangle formed by them is 0. $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$$ $$2(k - (-3)) + 4(-3 - 3) + 6(3 - k) = 0$$ $$2(k + 3) + 4(-6) + 18 - 6k = 0$$ $$2k + 6 - 24 + 18 - 6k = 0$$ $$-4k = 0 \Rightarrow k = 0$$

Q5 - Equidistant Points

Find a point on the y-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$.

Solution: Let the point on y-axis be $P(0, y)$.
Given $PA = PB$, so $PA^2 = PB^2$. $$(6 - 0)^2 + (5 - y)^2 = (-4 - 0)^2 + (3 - y)^2$$ $$36 + 25 + y^2 - 10y = 16 + 9 + y^2 - 6y$$ $$61 - 10y = 25 - 6y$$ $$36 = 4y \Rightarrow y = 9$$ The point is $(0, 9)$.

Q6 - Ratio Finding

In what ratio does the point $(-4, 6)$ divide the line segment joining the points $A(-6, 10)$ and $B(3, -8)$?

Solution: Let the ratio be $k:1$. Using the Section Formula for the x-coordinate: $$-4 = \frac{k(3) + 1(-6)}{k+1}$$ $$-4(k+1) = 3k - 6$$ $$-4k - 4 = 3k - 6$$ $$2 = 7k \Rightarrow k = \frac{2}{7}$$ Therefore, the ratio is $2:7$.

Q7 - Parallelogram Vertex

If $(1, 2)$, $(4, y)$, $(x, 6)$, and $(3, 5)$ are vertices of a parallelogram taken in order, find $x$ and $y$.

Solution: Diagonals of a parallelogram bisect each other. Midpoint of AC = Midpoint of BD. $$\left(\frac{1+x}{2}, \frac{2+6}{2}\right) = \left(\frac{4+3}{2}, \frac{y+5}{2}\right)$$ $$\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6$$ $$\frac{8}{2} = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y = 3$$ So, $x=6, y=3$.

Q8 - Centroid of Triangle

Find the centroid of the triangle formed by the vertices $(3, -5)$, $(-7, 4)$, and $(10, -2)$.

Solution: Centroid $G(x,y) = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$. $$x = \frac{3 + (-7) + 10}{3} = \frac{6}{3} = 2$$ $$y = \frac{-5 + 4 + (-2)}{3} = \frac{-3}{3} = -1$$ Centroid is $(2, -1)$.

Q9 - Circle Diameter

Find the coordinates of a point A, where AB is the diameter of a circle whose center is $(2, -3)$ and B is $(1, 4)$.

Solution: Let $A = (x, y)$. The center $C(2, -3)$ is the midpoint of $AB$. $$\frac{x+1}{2} = 2 \Rightarrow x+1=4 \Rightarrow x=3$$ $$\frac{y+4}{2} = -3 \Rightarrow y+4=-6 \Rightarrow y=-10$$ Point A is $(3, -10)$.

Q10 - Area of Triangle

Find the area of a triangle whose vertices are $(1, -1)$, $(-4, 6)$, and $(-3, -5)$.

Solution: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ $$= \frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$$ $$= \frac{1}{2} |1(11) - 4(-4) - 3(-7)|$$ $$= \frac{1}{2} |11 + 16 + 21| = \frac{1}{2} |48| = 24 \text{ sq units.}$$

Part 2: 50 Practice Questions

• Find the distance between points (0, 0) and (36, 15).
• Find the distance between (a, b) and (-a, -b).
• Calculate the distance of point P(6, -6) from the origin.
• Find x if the distance between (x, 7) and (1, 15) is 10 units.
• Find the midpoint of the line segment joining (3, 4) and (5, 2).
• The midpoint of line segment joining (2a, 4) and (-2, 2b) is (1, 2a+1). Find a and b.
• Find the centroid of the triangle with vertices (1, 4), (-1, -1), and (3, -2).
• Find the coordinates of the point dividing the line joining (-1, 3) and (4, -7) in ratio 3:4.
• Determine the ratio in which the line segment joining (1, -5) and (-4, 5) is divided by the x-axis.
• Find the coordinates of the point of trisection of the line segment joining (2, -2) and (-7, 4).
• Find the area of the triangle formed by vertices (2, 3), (-1, 0), (2, -4).
• Find the value of k if points (7, -2), (5, 1), and (3, k) are collinear.
• If the distance between (4, p) and (1, 0) is 5, find p.
• Check if the points (5, -2), (6, 4), and (7, -2) form an isosceles triangle.
• Find a point on the x-axis which is equidistant from (2, -5) and (-2, 9).
• Find the perimeter of the triangle with vertices (0, 0), (3, 0), and (0, 4).
• Find the fourth vertex of the rectangle with three vertices (0,0), (2,0), and (0,3).
• The coordinates of one end of a diameter of a circle are (2, 3) and the center is (-2, 5). Find the other end.
• Find the value of y for which the distance between P(2, -3) and Q(10, y) is 10 units.
• Show that points (1, 7), (4, 2), (-1, -1) are vertices of a square (Distance check).
• Find the ratio in which the y-axis divides the line segment joining (5, -6) and (-1, -4).
• If (1, 2), (4, y), (x, 6), and (3, 5) are vertices of a parallelogram, find x+y.
• Find the distance of point (2, 3) from the x-axis.
• Find the distance of point (-5, 4) from the y-axis.
• What is the distance between the points ($a \cos \theta, 0$) and ($0, a \sin \theta$)?
• Find the centroid of a triangle with vertices (3, -7), (-8, 6), and (5, 10).
• If the origin is the centroid of the triangle with vertices (x, 1), (y, -2), (2, 3), find x and y.
• The line segment joining (2, -3) and (5, 6) is divided by the x-axis in what ratio?
• Find the coordinates of a point on the x-axis which is equidistant from (5, 4) and (-2, 3).
• The area of a triangle with vertices (a, 0), (0, b), and (1, 1) is collinear. Find the relation between a and b.
• Name the type of triangle formed by (3, 2), (-2, -3), (2, 3).
• If P(9a - 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3:1, find a and b.
• Find the distance between A(2a, 6a) and B(2a + \sqrt{3}a, 5a).
• If A(-2, 1), B(a, 0), C(4, b), and D(1, 2) are vertices of a parallelogram, find a and b.
• Find the coordinates of the point which is equidistant from the three vertices of $\Delta$ AOB where A=(0,2y), O=(0,0), B=(2x,0).
• Find the area of the triangle formed by (0, 0), (4, 0), and (0, 3).
• Points A(4, 3), B(6, 4), C(5, -6) and D(-3, 5) are vertices of a parallelogram? (True/False).
• Find the perpendicular distance of A(5, 12) from the origin.
• Find the value of k for which A(-5, 1), B(1, k), and C(4, -2) are collinear.
• The midpoint of (3p, 4) and (-2, 2q) is (2, 6). Find p + q.
• Find the coordinates of the circumcenter of the triangle formed by (0, 0), (4, 0), and (0, 4).
• In what ratio does the point P(2, -5) divide the line joining A(-3, 5) and B(4, -9)?
• The distance between points (5, 3) and (x, -1) is 5. Find x.
• Find the area of the quadrilateral ABCD with vertices A(-5, 7), B(-4, -5), C(-1, -6), and D(4, 5).
• Determine if the points (1, 5), (2, 3), and (-2, -11) are collinear.
• Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP = 2/5 AB.
• Find the relation between x and y such that the point (x, y) is equidistant from (7, 1) and (3, 5).
• Find the coordinates of the point of intersection of the medians of a triangle with vertices (-1, 0), (5, -2), and (8, 2).
• If the area of a triangle formed by (x, 2x), (-2, 6), and (3, 1) is 5 sq units, find x.
• Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).

Answer Key

Q.No	Answer	Q.No	Answer	Q.No	Answer	Q.No	Answer	Q.No	Answer
1	39	11	10.5 sq units	21	5:1	31	scalene	41	p=2, q=4, sum=6
2	$2\sqrt{a^2+b^2}$	12	4	22	9	32	a=1, b=-3	42	(2, 2)
3	$6\sqrt{2}$	13	$\pm 4$	23	3 units	33	2a	43	2:5 approx (check coords)
4	7 or -5	14	Yes	24	5 units	34	a=1, b=1	44	2 or 8
5	(4, 3)	15	(-7, 0)	25	a	35	(x, y)	45	72 sq units
6	a=2, b=2	16	12 units	26	(0, 3)	36	6 sq units	46	Yes
7	(1, 0)	17	(2, 3)	27	x=-2, y=-1	37	False	47	(3, 4)
8	(1.14, -1.28)	18	(-6, 7)	28	1:2	38	13	48	x - y = 2
9	1:1	19	3 or -9	29	(2, 0) approx	39	-1	49	(4, 0)
10	(-1, 0) & (-4, 2)	20	Proof	30	1/a + 1/b = 1	40	6	50	(-7, 0)