(a) Dissolving common salt in water.
(b) Burning of Sugarcane.
(c) Burning of candle.
(d) Purification of metal.
(i) In compressed spring, the energy is stored by applying a non zero external force. By vertue of the rest position, this will corresponde to the Potential Energy.
(ii) In flowing of river water, water is flowing by vertue of its motion, so this energy will corresponde to Kinetic Energy.
The speed-time graph for a car is given as follows:
Calculate:
(a) Distance travelled in 3 seconds.
(b) Acceleration.
(a) Distance travelled in 3 s = area of ∆ ABC
= 6 m
(b) Acceleration = change in speed/time taken
a = (6-0)/3
a = 2 m/s2
K shell | L shell | M shell |
2 | 8 | 1 |
Valency of the sodium = 1 (one).
Mass of water = 240 g
Total mass of solution = 240 + 20 = 260 g
Mass percentage of sugar x 100
= (20 x 100)/260
= 7.7 %
Calculate the number of particles in each of the following:
(a) 7g of the nitrogen atoms.
(b) 0.5 mole of carbon dioxide
(a) Atomic mass of nitrogen = 14 u
Thus no. of particles in 14 g = 6.023 x 1023
No. of particles in 1 g =6.023 x 1023/14
No. of particles in 7 g of nitrogen atoms =
No. of particles in 7 g of nitrogen atoms = 3.012 x 1023
(b) 1 mole of carbon dioxide = 6.023 x 1023 particles
0.5 moles of carbon dioxide = particles
= 3.012 x 1023 particles
Prokaryotic cell | Eukaryotic cell |
(a) Small in size. | (a) Generally large in size. |
(b) Nuclear region is not surrounded by a nuclear membrane and is called nucleoid and contains only a single chromosome. | (b) Nuclear region is surrounded by a nuclear membrane and contains more than one chromosome. |
(c) Membrane bound cell organelles are absent. | (c) Membrane bound cell organelles are present. |
Mass of the hammer (m) = 500 g = 0.5 Kg
Initial velocity of the hammer (u) = 50 m/s
Final velocity of the hammer (v) = 0
Time taken to stop the hammer (t) = 0.01 s
Acceleration on the hammer (a) = (v-u)/t
= (0-50)/0.01
= -5000 m/s2
Force exerted by the nail to stop hammer (F) = m x a
F = 0.5 x 5000
F = 2500 N
A car moving with a speed of 12 m/s comes to rest when brakes are applied to produce an acceleration of 3 m/s in opposite direction. Calculate:
(A) the car to come to rest.
(b) Distance travelled before coming to rest.
(a) Here initial velocity (u) = 12 m/s
Final velocity = 0 (as the car has come to rest)
We know, acceleration =
time = 12/3
time = 4 s
(b) Distance travelled before coming to rest =
=
= 48 + 24
= 72 m
Solution | Suspension | Colloid |
(a) It is a homogeneous mixture. | (a) It is heterogeneous mixture. | (a) It is heterogeneous mixture. |
(b) The particles of a solution cannot be seen with the naked eye. | (b) The particles of a suspension can be seen with the naked eye. | (b) The particles of a colloid cannot be seen with the naked eye. |
(c) These do not show tyndall effect. | (c) These do not show tyndall effect. | (c) These show tyndall effect. |
(d) The particles of a solution cannot be separated by filtration. | (d) The particles of a solution can be separated by filtration. | (d) The particles of a colloid cannot be separated by filtration. |
g | G |
(i) It is the acceleration produced due to the gravitational force. | (i) It is the universal gravitation constant. |
(ii) Its value is not constant. Its SI unit is m/s2. | (ii) Its value is constant everywhere in the universe. Its SI unit is Nm2/Kg. |
(c) The mass of the object (m) = 40 Kg
Weight of the object on the earth (W) = m x g
W = 40 x 9.8
W = 392 N
Since the acceleration due to gravity on the moon is 1/6th of the acceleration due to gravity on the earth, weight of the object on the moon = m x g/6
= (40 x 9.8)/6
= 65.33 N
Light waves | Sound waves |
(i) Light waves travel in the form of transverse waves.
| (i) Sound waves travel in the form of longitudinal waves.
|
(ii) These do not require any medium for their propagation. | (ii) These require a medium for their propagation. |
(c) (i) We hear different sound from different sources because of the differences in the properties like pitch, amplitude and wavelength of the sound waves coming from different sources.
(ii) Speed of light is more than the speed of sound due to which light travels faster than sound and hence we are able to hear the sound of the thunder a few seconds after we see the lightning. Striated muscles | Unstriated muscles |
(i) The cells are long, cylindrical, and multinucleate. | (i) The cells are spindle shaped and uninucleate. |
(ii) These move according to our will. | (ii) These cannot move according to our will. |
(iii) These have striations on them e.g. muscles found in our limbs. | (iii) These do not have striations e.g. muscles present in alimentary canal. |
(b)
Xylem | Phloem |
(i) These consist of tracheids, vessels, xylem parenchyma. | (i) These consist of sieve tubes, companion cells, phloem fibres and parenchyma. |
(ii) These help in the transport of water and minerals. | (ii) These help in the transport of food. |
(iii) Materials move in one direction. | (iii) Materials can move in both the directions. |
(ii) Development of disease and draught resistance in crop plants.
(iii) Development of crops having shorter duration of maturity so that mutiple rounds of crops can be grown in one year.
A ball of mass 2 kg is kept on a tower of height 30 m. Find its potential energy at this point. If it is allowed to fall freely, find its kinetic energy when it just touches the ground?
m = 2 kg; h = 30 m; g = 10 m/s2
P.E = m × g × h
= 2 kg × 10 m/s2 × 30 m
= 600 J
When ball just touches the ground its kinetic energy, K.E. = (1/2)mv2
v2 - u2 = 2as [Third equation of kinematics]
v2 = 2 × 10 m/s2 × 30 m [As u = 0 m/s]
= 600 m/s
So, K.E. = (1/2) × 2 kg × 600 m/s
= 600 J
The potential energy is completely converted into kinetic energy. This is in accordance to law of conservation of energy.
(i) A solution has been prepared by dissolving 5 g of urea in 95 g of water. What is the mass percent of urea in the solution ?
(ii)Will blood show Tyndall effect?
(i)
(ii) Blood is a colloidal solution, therefore it will show Tyndall effect.
(i) Ammonium sulphate
(ii) Magnesium nitrate
(iii) Aluminium bromide
(ii)
(iii)
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b)
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Who proposed five-kingdom classification? What are the names of kingdoms?
R.H. Whittaker (1969) proposed five-kingdom classification , they are:
i) Monera
ii) Protista
iii) Fungi
iv) Plantae
v) Animalia
If in an office, 10 tubes of 40 W, 5 fans of 75 W and 2 ACs of 1500 W are used for 8 hours a day. Calculate the energy consumed per day in commercial units of energy.
Power consumed by tubes = p1 = 10 × 40W = 400 W
Power consumed by fans = p2 = 5 × 75W = 375 W
Power consumed by ACs = p3 = 2 × 1500W = 3000 W
Total energy consumption per day, E = p1t + p2t + p3t
or, E = (p1 + p2 + p3)t
= ( 400W + 375W + 3000W ) × 8 h
= 3775 W × 8 h = 30200 Wh
or, E = 30200 Wh/1000 = 30.2 kWh
An iron piece weighs 600 g in air and 400 g in water. Find volume of the solid.
Given:
Weight of solid in water = 400 g
Weight of solid in air = 600 g
Loss of weight of solid in water = 600– 400g
= 200g
= Mass of water displaced.
A - Fusion or melting
B - Vaporization or boiling
C - Condensation
D - Solidification
E - Sublimation
Give a brief account of the observations made by Rutherford in his alpha particle scattering experiment.
The observations made by the Rutherford is shown by the figure below,
He observed that:
1) Most of the particles passed through the gold foil directly without changing their path, which proved that most of the atom is empty.
2) Slight deflection of some of the particles, and large scale deflection of some particles ,proved that central mass of an atom is very high (called as nucleus) and central mass of an atom is positively charged.
3) Very few -particles (1 in 1 lac) bounced back completely. This shows that nucleus is very small compared to the whole atom.
Tom was verifying the law of conservation of mass by reacting barium chloride and sodium sulphate.
(i)The colours of barium chloride and sodium sulphate are
(a) blue and white respectively.
(b) blue and blue respectively.
(c) white and white respectively.
(d) white and blue respectively.
(ii)During the experiment, he reacted 209 g of BaCl2 and 145 g of Na2SO4 . If after reaction he obtained 120 g of NaCl, find the amount of BaSO4
he obtained.
(i)
(c) Both the salts are white in colour.
(ii)
Can law of conservation of mass be verified by heating crystals of ferrous sulphate in a china dish? Give reasons.
When a chemical reaction is carried out in an open container like china dish, some mass will evaporate in the form of vapours or gas. Therefore, the mass taken in the container will no longer remain constant. It means law of conservation cannot be verified by heating crystals of ferrous sulphate in a china dish.
The teacher gave a de-shelled egg to Ritika and an unknown solution in a glass and asked to dip the egg in the solution and observe it after 7 minutes. After 7 minutes the egg was found to be shrunk.
Which of the following inference can be drawn from above observation?
Given solution was hypertonic
To move a wooden block X of weight 2 kg placed on a horizontal table, John hangs a weight of unknown mass on the other side of the attached string.
(i) The block X suddenly moves rapidly means:
(a) Weight used is less than 2 kg.
(b) Weight used is equal to 2 kg.
(c) Weight used is greater than 2 kg.
(d) Other parameters are required to know the weight.
(ii) Name the law on which a spring balance works.
(i) The correct option is (c).
Explanation: From Newton’s third law of motion, action and reaction are equal and opposite and acts on different bodies. Therefore, the block will move only when the applied force is more than its weight.
(ii) A spring balance works on Newton’s third law of motion.
(i) Jerry observed using a spring balance that a weight of 5 kg was just enough to slide a block kept on a horizontal surface. The calculated force comes out to be:
(a) 0.98 N
(b) 98 N
(c) 49 N
(d) 0.49 N
(ii) If the weight of the block is doubled, what will be the magnitude of new force?
(i) The correct option is (c).
Explanation: F = mg
= 5 x 9.8 = 49 N
(ii) If the weight of the block is doubled, the new force will also be doubled, i.e., 98 N.
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