10th Science Problem Questions Full PDF
Download the complete collection of numerical problems and solutions.
Download PDF1. 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:
\(F_{2} = 15N\)
Two forces acting opposite to one another.
Therefore, Resultant force is,
\(F_{net} = F_{2} - F_{1}\)
\(F_{net} = 15 - 5 = 10N\)
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:
Let \(m_{2}\) be the bigger mass, acceleration due to bigger body, \(a_{2} = 12~ms^{-2}\)
We know, \(F = ma \Rightarrow F_{1} = m_{1}a_{1} ; F_{2} = m_{2}a_{2}\)
\(\frac{F_{1}}{F_{2}} = \frac{m_{1}}{m_{2}} \times \frac{a_{1}}{a_{2}}\)
\(1 = \frac{3}{4} \times \frac{a_{1}}{12} \Rightarrow a_{1} = 16~ms^{-2}\)
\(\frac{F_{1}}{F_{2}} = \frac{m_{1}}{m_{2}} \times \frac{a_{1}}{a_{2}}\)
\(1 = \frac{3}{4} \times \frac{a_{1}}{12} \Rightarrow a_{1} = 16~ms^{-2}\)
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:
Initial velocity \((u) = 10~m~s^{-1}\)
It is perfect elastic collision, ball rebounds with the same speed but in opposite direction.
Final velocity \((v) = -10~ms^{-1}\)
\(\Delta p = mv - mu\)
\(= 1 \times (-10) - 1 \times (10) = -10 - 10 = -20~kg~m~s^{-1}\)
(Negative sign just indicates the direction of momentum)
\(= 1 \times (-10) - 1 \times (10) = -10 - 10 = -20~kg~m~s^{-1}\)
(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:
Force \(F_{2} = 40~N\), Length \(d_{2} = ?\)
Moment of couple is same for both the spanner, and so \(F_{1}d_{1} = F_{2}d_{2}\)
\(d_{2} = \frac{F_{1}d_{1}}{F_{2}} = \frac{40 \times 140}{40} = 140~cm\)
\(d_{2} = \frac{F_{1}d_{1}}{F_{2}} = \frac{40 \times 140}{40} = 140~cm\)
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:
Ratio of masses, \(m_{1}:m_{2} = 2:3\)
Ratio of acceleration due to the gravity, \(g_{1}:g_{2} = ?\)
\(g_{1} = \frac{GM_{1}}{R_{1}^{2}}\) ...(1)
\(g_{2} = \frac{GM_{2}}{R_{2}^{2}}\) ...(2)
Eqn (1) ÷ (2) \(\Rightarrow \frac{g_{1}}{g_{2}} = \frac{M_{1}}{M_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}}\)
\(\frac{g_{1}}{g_{2}} = \frac{2}{3} \times \frac{7^{2}}{4^{2}} \Rightarrow \frac{2}{3} \times \frac{49}{16} = \frac{49}{24}\)
\(g_{2} = \frac{GM_{2}}{R_{2}^{2}}\) ...(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
\(V = \frac{2.5}{5} = 0.5~m~s^{-1}\)
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 \(M = F \times d\)
\(M = 40 \times 0.9 = 36~Nm\)
\(M = 40 \times 0.9 = 36~Nm\)
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:
Acceleration due to gravity at that height, \(g' = \frac{g}{4}\)
\(g = \frac{GM}{R^{2}}, g' = \frac{GM}{R'^{2}} \Rightarrow \frac{g}{g'} = (\frac{R'}{R})^{2}\)
\(\frac{g}{g/4} = (\frac{R+h}{R})^{2} = (1 + \frac{h}{R})^{2}\)
\(4 = (1 + \frac{h}{R})^{2}\) (take square root on both sides)
\(2 = 1 + \frac{h}{R} \Rightarrow h = R\)
\(R' = R + R = 2R\)
\(\frac{g}{g/4} = (\frac{R+h}{R})^{2} = (1 + \frac{h}{R})^{2}\)
\(4 = (1 + \frac{h}{R})^{2}\) (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, \(R = m(g - a)\)
\(= 50(9.8 - 1.8)\)
\(R = 50 \times 8 = 400~N\)
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 \(W = mg = 70 \times 1.625 = 113.75~N\)
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:
\(F = ma\)
\(m = \frac{F}{a} = \frac{5}{0.05} = 100~kg\)
\(m = \frac{F}{a} = \frac{5}{0.05} = 100~kg\)
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 (\(m=2~kg\), \(a=1.5~ms^{-2}\)):
\(F = ma = 2 \times 1.5 = 3N\)
\(a = \frac{F}{m_{1}+m_{2}} = \frac{15}{8+2} = \frac{15}{10} = 1.5~ms^{-2}\)
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:
\(\frac{m_{T}}{m_{B}} = 4\)
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: \(\frac{p_{T}}{p_{B}} = \frac{m_{T}V_{T}}{m_{B}V_{B}} = 4 \times \frac{1}{2} = 2\)
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:
Distance between object and lens \(u = -20~cm\)
Lens formula, \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)
\(= \frac{1}{10} + \frac{1}{-20} = \frac{2-1}{20} = \frac{1}{20}\)
\(v = 20~cm\)
\(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)
\(= \frac{1}{10} + \frac{1}{-20} = \frac{2-1}{20} = \frac{1}{20}\)
\(v = 20~cm\)
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:
Distance between object and lens \(u = -10~cm\)
Height of the object, \(h = 3~cm\)
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 \(m = \frac{h'}{h} \Rightarrow 0.6 = \frac{h'}{3}\)
\(h' = 0.6 \times 3 = 1.8~cm\)
\(= \frac{1}{-15} + \frac{1}{-10} = \frac{-2-3}{30} = -\frac{5}{30} = -\frac{1}{6}\)
\(v = -6~cm\)
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, \(\frac{\sin i}{\sin r} = \frac{\mu_{2}}{\mu_{1}}\)
\(\sin r = \frac{\mu_{1}}{\mu_{2}} \times \sin i = \frac{1}{1.5} \times \sin 30^{\circ}\)
\(\sin r = \frac{1}{1.5} \times 0.5 = 0.333\)
\(r = \sin^{-1}(0.333) \Rightarrow r = 19.45^{\circ}\)
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: \(f = -0.3~m\), \(v = -0.2~m\)
\(\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-0.2} - \frac{1}{-0.3}\)
\(\frac{1}{u} = \frac{-0.3+0.2}{0.06} = \frac{-0.1}{0.06} = -\frac{10}{6}\)
\(u = -0.6~m\)
\(\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-0.2} - \frac{1}{-0.3}\)
\(\frac{1}{u} = \frac{-0.3+0.2}{0.06} = \frac{-0.1}{0.06} = -\frac{10}{6}\)
\(u = -0.6~m\)
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 \(P = \frac{1}{f} = \frac{1}{-5} = -0.2~D\)
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:
\(A_{o} = 10~m^{2}, A = 11~m^{2} \Rightarrow \Delta A = 1~m^{2}\)
\(T_{o} = 90~K\), \(\alpha_{A} = 0.0021~K^{-1}\)
\(\Delta A = A_{o} \alpha_{A} \Delta T \Rightarrow \Delta T = \frac{\Delta A}{A_{o}\alpha_{A}}\)
\(\Delta T = \frac{1}{10 \times 0.0021} = \frac{1}{0.021} = 47.61~K\)
\(T = \Delta T + T_{o} = 47.61 + 90 = 137.6~K\)
\(T_{o} = 90~K\), \(\alpha_{A} = 0.0021~K^{-1}\)
\(\Delta A = A_{o} \alpha_{A} \Delta T \Rightarrow \Delta T = \frac{\Delta A}{A_{o}\alpha_{A}}\)
\(\Delta T = \frac{1}{10 \times 0.0021} = \frac{1}{0.021} = 47.61~K\)
\(T = \Delta T + T_{o} = 47.61 + 90 = 137.6~K\)
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 = \(L_{3} - L_{2} = 51.2 - 48.5 = 2.7~ml\)
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: \(P_{1}V_{1} = P_{2}V_{2}\)
\(V_{2} = \frac{P_{1}V_{1}}{P_{2}} = \frac{P \times 20}{4P} = 5~cc\)
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: \(I = \frac{P}{V} = \frac{180}{220} = 0.818~A\)
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 = \(15.5 + 37.2 = 52.7~kWh\)
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) \(E = P \times t = 1.8 \times 4 = 7.2~Wh\)
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?
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 \(R_{S}:R_{P} = R : \frac{R}{25} = 25:1\)
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: \(\frac{1}{R_{P}} = \frac{1}{5} + \frac{1}{10} + \frac{1}{20} = \frac{7}{20} \Rightarrow R_{P} = 2.857~\Omega\)
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:
\(d_{1} + d_{2} = 680~m\)
\(d_{1} = \frac{v \times 0.9}{2}\), \(d_{2} = \frac{v \times 1.1}{2}\)
\(\frac{v}{2}(0.9 + 1.1) = 680 \Rightarrow v = 680~m~s^{-1}\)
\(d_{1} = \frac{v \times 0.9}{2}\), \(d_{2} = \frac{v \times 1.1}{2}\)
\(\frac{v}{2}(0.9 + 1.1) = 680 \Rightarrow v = 680~m~s^{-1}\)
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 = \(A - Z = 214 - 82 = 132\)
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 = \(\frac{14}{17} \times 100 = 82.35\%\)
% 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
\(= \frac{0.18}{18} \times 6.023 \times 10^{23} = 6.023 \times 10^{21}\)
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 % = \(\frac{\text{Mass solute}}{\text{Mass solute} + \text{Mass solvent}} \times 100\)
\(= \frac{45}{45+180} \times 100 = \frac{45}{225} \times 100 = 20\%\)
\(= \frac{45}{45+180} \times 100 = \frac{45}{225} \times 100 = 20\%\)
10. TYPES OF CHEMICAL REACTIONS
1 Lemon juice has a pH 2, what is the concentration of \(H^{+}\) ions?
Solution:
\(pH = -\log_{10}[H^{+}] = 2\)
\([H^{+}] = 10^{-2} = 0.01~M\)
\([H^{+}] = 10^{-2} = 0.01~M\)
4 The hydroxide ion concentration of a solution is \(1 \times 10^{-11} M\). What is the pH of the solution?
Solution:
\(pOH = -\log[OH^{-}] = -\log(10^{-11}) = 11\)
\(pH = 14 - pOH = 14 - 11 = 3\)
\(pH = 14 - pOH = 14 - 11 = 3\)
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