Previous Year Paper year:2017 : ICSE-X-Physics

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ICSE-X-Physics

Previous Year Paper year:2017

with Solutions - page 3

#4-e [2]
When does the nucleus of an atom tend to be radioactive?
Ans : A nucleus tends to be radioactive when the number of neutrons inside it is
more than the number of protons.

# [40]
Section : II
Attempt any four questions from this Section

#5

#5-a [3]
A uniform half metre rule balances horizontally on a knife edge at ``\pu{29 cm}``
mark when a weight of ``\pu{20 gf}`` is suspended from one end.

#5-a-i
Draw a diagram of the arrangement
Ans : Diagram of the arrangement mentioned in the corresponding question:

#5-a-ii
What is the weight of the half metre rule?
Ans : The centre of gravity(CG) of the uniform half meter rule = 25 cm.
The rule balances horizontally on a knife edge = 29 cm
weight of suspension = 20 kgf
The difference in the CG /point of balance=
29 - 25 = 4 cm(=Y).
The weight is suspended at one of the extremes,
let us assume at the 50 cm end.
Therefore, the distance from the point of knife edge to the weight
= 50-29 = 21 cm(=X).
As formula:
weight of meter rule = ``\frac{X} {Y} *(suspended weight )``
= ``\frac{21}{4} * 20 = 105 gf ``

#5-b

#5-b-i [3]
A boy uses a single fixed pulley to lift a load of ``\pu{50 kgf}`` to some height. Another boy uses a single movable pulley to lift the same load to the same height. Compare the effort applied by them. Give a reason to support your answer.
Ans : The ratio of load to effort is called mechanical advantage.
Mechanical advantage (MA) = ``\frac{Load(L)}{effort(E)}``
Hence Effort is inveresely proportional to MA.
``\frac{MA_1}{MA_2} = \frac{E_2}{E_1}``
MA of single fixed Pulley ``(M_{sfp}) = 1``
``E_{sfp} = 50 kgf``
MA of single Movable Pulley ``(M_{smp}) = 2``
``\frac{MA_{sfp}}{MA_{smp}} = \frac{E_{smp}}{E_{sfp}}``
so ``E_{smp} = \frac12 * 50 = 25kgf``
Therefore, the boy using a single movable pulley requires lesser effort by magnitude of half to lift an equal load of 50 kgf when compared with the effort required by
the boy using a single fixed pulley.

#5-b-ii
How does uniform circular motion differ from uniform linear motion?
Ans : When a force acts on a body at rest which is free to move, the body
starts moving at a constant speed in a straight path in the direction of
force. This type of motion is called uniform linear or uniform
translational motion.
Whereas when a body or a particle moves with a constant speed in a
circular path, its motion is said to be uniform circular motion.

#5-b-iii
Name the process used for producing electricity using nuclear energy.
Ans : Nuclear fission is the process used for producing electricity using nuclear
energy.

#5-c [4]
A pulley system with VR = 4 is used to lift a load of ``\pu{175 kgf}`` through a vertical height of ``\pu{15 m}``. The effort required is ``\pu{50 kgf}`` in the downward direction. (g = ``\pu{10 N kg-1}``)
Calculate

#5-c-i
Distance moved by the effort
Ans : Displacement due to the effort,VR = 4
Load (L)= 175 kgf
Effort (E)= 50 kgf
Load displacement (d )=15 m
$$VR= \frac{d_E}{d_L}$$
Effort displacement ``(d_E) = 4 * d_L = 4*15 = 60 m``
Hence the distance moved by the effort is 60 m.

#5-c-ii
Work done by the effort
Ans : Work done by the effort is calculated using the following formula.
Work(w) = Force(F): Dispalcement(d)
w = 50*60 = 3000 J

#5-c-iii
M.A. of the pulley system
Ans : Mechanical advantage of the pulley system is calculated as
``MA = \frac {Load}{Effort}``
$$ = \frac {175}{50} = 3.5 $$

#5-c-iv
Efficiency of the pulley system
Ans : Efficiency of the pulley is the ratio of work output to the work input
required by the system.
$$Efficiency = \frac{Output- Work done by load}{Input-Work done by effort} * 100$$
Work done by load(Output) = Force*d = 175*15= 2625 kgf
Work done by Effort(Input) = 3000 kgf
hence efficiency =
$$\frac{2625}{3000} * 100 = 87.5 \% $$