Download ICSE Guess Paper - 2008

ICSE Guess Paper - 2008
Class - X
Subject - Physics
CH. – 9. ELECTRICAL POWER and HOUSEHOLD ELECTRICITY
Question – 1.Name the material commonly used for making (i) filament of bulb, (ii) fuse
wire and (iii) heater element. State two properties of the material which make them suitable.
Question – 2. Why is the filament of an electric iron or kettle put between mica and sheets ?
Question – 3. State the purpose of a fuse in an electric circuit. Explain how a fuse protects an
electric circuit ?
Question – 4. Explain the function of the following in the household wiring (i) a three pin
plug and (ii) main switch. In a three pin plug, why is the earth pin made longer and thicker
than the other two pins ?
Question – 5. What is high tension wires ? Give two characteristics of these wires ?
NUMERICALS
Numerical – 1. An electrical appliance is rated 1500 W, 250 V. This appliance is connected
to 250 mains. Calculate :
(i) the current drawn,
(ii) the electrical energy consumed in 60 hours,
(iii) the cost of electrical energy consumed at Rs.2.50 per kWh.
Solution :- Here, P = 1500 W = 1.5 kW, V = 250 V
(i) Using, P = V I , we have
I = P/V = 1500/250 = 6 A.
(ii) E = P × t = 1.5 × 60 = 90 kWh
(iii) Cost = E × cost per unit = 90 × 2.50 = Rs.225.
Numerical – 2. An electric kettle is rated 2.5 kW, 250 V. Find the cost of running the kettle
for two hours at 60 paisa per unit.
Solution :- Here, P = 2.5 kW, V = 250 V, t = 2 hours, cost per unit = Re.0.60.
Using, E = P × t = 2.5 × 2 = 5 kWh,
Therefore, cost of running the kettle = 5 × 0.60 = Rs.3.0
Numerical – 3. Calculate the electrical energy in SI units consumed by a 100 W bulb and a
60 W fan connected in parallel in 5 minutes.
Solution:- Here, P1 = 100 W, P2 = 60 W, t = 5 minutes = 5 × 60 = 300 s.
Total power in parallel P = P1 + P2 = 100 + 60 = 160 W
E = P × t = 160 × 300 = 48,000 J.
Numerical – 4. A bulb is marked 100 W, 220 V and an electric heater is marked 2000 W,
220 V. (i) What is the ratio of their resistances? (ii) In which of the above a thicker
connecting wire or lead is required?
Solution :- (i) Let RB and RH be the resistances of bulb and heater respectively.
Using, R = V2/P , for the bulb marked 100 W, 220 V we get
RB = (220)2/100 = 484 Ω
And for the heater marked 2000 W, 220 V we get
RH = (220)2/2000 = 24.2 Ω
Therefore, RB/RH = 484/24.2 = 20.
(ii) Current flowing through the heater is 20 times than that through the bulb.
Therefore, a thicker connecting wires (or lesad) will be required for the electric heater.
Numerical – 5. What quantity of heat will be produced in a coil of resistance 80 ohm if
current of 3 A is passed through it for 4 second.
Solution :- Here, R = 80 Ω, I = 3 A, t = 4 s
Using H = I2Rt = (3)2 × 80 × 4 = 2880 J
Numerical – 6. A refrigerator is marked 80 W, 220 V.
(i) How much energy does it consume in one day if on an average it is used for 20 hours a
day ?
(ii) What is likely to happen if the voltage drops to 50 V.
Solution :- Here, P = 80 W, v= 220 V, t= 20 hours, E = ?
(i) Using, E = P × t = 80 × 20 = 1600 Wh = 1.6 kWh
(ii) If voltage drops a large current will flow through it and damage the refrigerator.
ICSE Guess Paper - 2008
Class - X
Subject - Physics
CH. – 8 CURRENT ELECTRICITY
Question – 1. Mention two factors on which the resistance of a wire depends.
Question – 2. Mention two factors on which the internal resistance of a cell depends.
Question – 3. State Ohm’s law. Write an expression for the resistance of a conducting wire
in terms of its length and area of cross section.
Question – 4. Distinguish between potential difference and emf of a cell.
Question – 5. Derive an expression for equivalent resistance when two resistors of
resistance R1 and R2 are connected in (i) parallel and (ii) series.
NUMERICALS
Numerical – 1. A cell supplies a current of 0.6 A through a 2 ohm coil and a current of 0.3 A
through a 8 ohm coil. Calculate the emf and the internal resistance of the cell.
Solution :- Here, I1 = 0.6 A, R1 = 2 Ω, I2 = 0.3 A, R2 = 8 Ω, r = ?
Using, I = E/(R + r), we get
0.6 = E /(2 + r) --- --- --- --- (i)
0.3 = E / (8 + r) --- --- --- --- (ii)
from (i) and (ii) we get 0.6(2 + r) = 0.3(8 + r)
or, 0.3 r = 1.2
or,
r = 4 ohm.
Substituting the value of r in equation (i) we get
E = 0.6(2 + 4) 3.6 V
Hence, r = 4 ohm, E = 3.6 V. [Ans.]
Numerical – 2. Four cells each of emf 2 V and internal resistance 0.1 ohm are connected in
series. The combination is connected in series to an ammeter of negligible resistance, a 1.6
ohm resistor and an unknown resistor R1 . The current in the circuit is 2 A. Draw a labelled
circuit diagram for the above arrangement and calculate :
(i) the total resistance of the circuit
(ii) the total emf
(iii) the value of R1
(iv) the potential difference across R1.
Solution :- The labelled circuit diagram is given below :
Fig
Total internal resistance = 4 × 0.1 Ω = 0.4 Ω
(i) Total resistance = 1.6 + 0.4 + R1 = (2 + R1) Ω
(ii) Total emf = (2 + 2 + 2 + 2) V = 8 V
(iii) Using, I = total emf / total resistance
or, 2 = 8 / (2 + R1)
or, 2R1 + 4 = 8
or,
R1 = 2 Ω
(iv) Potential difference across R1 = I R1 = 2 × 2 = 4 V
from (i) total resistance = 2 + R1 = 2 + 2 = 4 Ω
Numerical – 3. Four cells each of emf 1.5V and internal resistance 2 ohm are connected in
parallel. The battery of cells is connected to an external resistance of 2.5 ohm. Calculate :
(i) the total resistance of the circuit
(ii) the current flowing in the external circuit and
(iii) the drop in potential across the terminal of the cells.
Solution :- The arrangement of the four cells in parallel is as under :
Fig.
(i) As the cells are connected in parallel, the net emf of battery = 1.5 V,
Let the net internal resistance be r
Therefore, 1/r = 1/2 + 1/2 + 1/2 + 1/2 = 2 or, r = 0.5 Ω
(ii) The total resistance R of the circuit is given by
R = 0.5 + 2.5 = 3 Ω
Current flowing in the circuit, I = V/R = 1.5/ 3 = 0.5 A
(iii) Potential difference across the battery, V = IR = 0.5 ×2.5 = 1.25 V
Numerical – 4. Four resistances of 2.0 Ω each are joined end to end to form a square ABCD.
Calculate the equivalent resistance of the combination between any two adjacent corners.
Solution :- In this situation three resistors are connected in series and their combination in
parallel with the fourth resistor as given below :
Fig
.
By series combination we get, 2 + 2 + 2 = 6 ohm
This 6 ohm resistor is connected in parallel with fourth 2 ohm resistor. Let the resistance
of the combination be R
Therefore, 1/R = 1/6 + 1/2 = 2/3
Hence,
R = 3/2 ohm.
Numerical – 5. The following circuit diagram shows three resistors 2 ohm, 4 ohm and R ohm
connected to a battery of emf 2 V and internal resistance 3 ohm. A main current of 0.25 A
flows through the circuit.
Fig
(i) What is the pd across the 4 ohm resistor ?
(ii) Calculate the pd across the internal resistance of the cell.
(iii) What is the pd across R and 2 ohm resistors ?
(iv) calculate the value of R.
Solution :-
.
ICSE Guess Paper - 2008
Class - X
Subject - Physics
CH. – 7. SOUND.
Question – 1. A vibrating tuning fork is placed over the mouth of a burette filled with water.
The tap is opened and the water level gradually falls. It is observed that the sound becomes
the loudest for particular length of air column :
(i) Name the above phenomenon.
(ii) How does it occur ?
(iii) Why does the sound become the loudest ?
(iv) How does the frequency of the loud sound compare with that of the tuning fork ?
(v) What is the name of the phenomenon taking place when sound is produced for another
length of air column and is not the loudest ?
(vi) State the unit for measuring the loudness.
Questing – 2. Under what condition does resonance occur ?
Question – 3. Why is loud sound heard at acoustic resonance ?
Question – 4. Differentiate between forced vibration and resonance.
Question – 5. What is sonar ? State the principle on which it is based .
NUMERICALS
Numerical – 1. A radar is able to detect the reflected waves from an enemy aeroplane, after
a time interval of 0.02 milliseconds. If the velocity of waves is 3 × 10 8 m s-1, calculate the
distance of the plane from the radar.
Solution :- Here, t = 0.02 millisecond
= 0.02 × 10-3 s,
c = 3 × 108 m s-1,
d=?
[here d is the distance of radar from plane]
Or,
2d = c ×t
= 3 ×108 × 0.02 × 10-3
= 6000 m
d = 3000 m = 3 km. [Ans.]
Numerical – 2. The wavelength of waves produced on the surface of water is 20 cm. If the
wave velocity is 24 m s-1, calculate :
(i) the number of waves produced in one second and
(ii) the time required to produce one wave.
Solution :- Here, λ = 20 cm = 0.20 m, c = 24 m s-1, v = ? , T = ?
(i) Using, c = v λ,
v = c/λ = 24/0.20 = 120 Hz
(ii) T = 1/v = 1/120 s.
So, v = 120 Hz, T = 1/120 s [Ans.]
Numerical – 3. The longitudinal wave of wavelength 1 cm travels in air with a speed of 330 m
s-1. Calculate the frequency of the wave. Can this wave be heard by a normal human being ?
Solution :- Here, λ = 1 cm = 0.01 m, c = 330 m s-1
Using c = v λ , we have v = c/λ = 330/0.01 = 33000 Hz [Ans.]
This wave cannot be heard by normal human being.
Numerical – 4. A sound wave of wavelength 0.332 m has a time period of 10-3 s. If the time
period is decreased to 10-4s, calculate the wavelength and frequency of the new wave.
Solution :- Here, λ = 0.332 m, T = 10-3 s
Using , c = λ/T = 0.332/10-3 = 332 m s-1.
Again , T = 10-4s, λ = ?, v = ?
Using , c = λ/T, we have λ = cT
= 332 × 10-4 = 0.0332 m.
Also
v = 1/T = 1/10-4 = 104 Hz.
So, wavelength = 0.0332 m and frequency = 104 Hz [Ans.]
Numerical – 5. A pendulum has a frequency of 5 vibrations per second. An observer starts
the pendulum and fires a gun simultaneously. He hears the echo from the cliff after 8
vibrations of the pendulum. If the velocity of sound in air is 340 m s-1, what is the distance
between the cliff and the observer ?
Solution :- Here, v = 5, therefore, T = 1/v = 1/5 = 0.2 s
Therefore time for 8 interval = 0.2 × 8= 1.6 s,
c = 340 m s-1, d = ?
2 d = c × t = 340 × 1.6 = 544 m
Therefore, d = 272 m. [Ans.]