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WINTER BREAK ASSIGNMENT
CLASS - XI
SUBJECT - PHYSICS
Work Done by a Variable Force
Formulae Used

 
W   F .s
i

s2
 
W   F .d s
s1

W = Area under the force-displacement curve between the initial and final
positions of the body.
 A 2kg particle starts at the origin and moves along the positive x-axis. The net
force acting on it measured at intervals of 1 m is: 27.9, 28.3, 30.9, 34.0, 34.5,
46.9, 48.2, 50.0, 63.5, 13.6, 12.2, 32.7, 46.6 and 27.0 (in Newton’s). What is
the total work done on the particle in this interval?
 A body moves from point A to B under the action of a force, varying in
magnitude as shown in figure. Obtain the work done. Force is expressed in
Newton and displacement in metre.
20
Q
15
10
A
P
5
R
0
-5
1
2
3
4
5
-10
-15
B
s
 A woman pushes a trunk on railway platform which has a rough surface. She
supplies a force of 100 N over a distance of 10m. Thereafter she gets progressively
tired and her applied force reduces linearly with distance to 50N. The total distance
by which the trunk has been moved in 20m. plot the force applied by the woman
and the frictional force, which is 50N.
Calculate the work done by the two forces over 20m.
 A particle moves along the X-axis from x = 0 to x = 5 m under the influence of a
force given by f = 7 – 2x + 3x2. Find the work done in t he process.
 The two thigh bones (femurs), each of cross-sectional area 10 cm support the
upper part of a human body of mass 40 kg. Estimate the average pressure
sustained by the femurs. Take g = 10 ms-2.
[Ans. 2 × 105 Nm-2]
 How much pressure will a man of weight 80 kgf exert on the ground when (i) he
is lying and (ii) he is standing on his feet? Given that the area of the body of the
man is 0.6 m2 and that of a foot is 80 cm2.
[Ans. 4.9 × 104 Nm-2]
 Find the pressure exerted at the tip of a drawing pin if it is pushed against a
board with a force of 20 N. Assume the area of the tip to be 0. [Ans. 2 × 108 Pa]
 Atmospheric pressure is nearly 100 kPa. How large the force does the air in a
room exert on the inside of a window pan that is 40 cm × 80 cm?[Ans. 32 KN]
 The force on a phonograph needle is 1.2 N. The point has a circular crosssection whose radius is 0.1 mm. Find the pressure (in atm) it exerts on the
records. Given 1 atm = 1.013 × 105 Pa.
[Ans. 377 atm]
A cylindrical vessel containing liquid is closed by a smooth piston of mass m. The
area of cross- section of the piston is A. If the atmospheric pressure is P 0, find the
pressure of the liquid just below the piston.
CENTRE OF MASS
 Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with
dimensions as shown in the given figure. The mass of lamina is 3 kg.
y
2m
F(0, 2)
E(1, 2)
C3
C1
D(1, 1)
B(2, 1)
1m
C2
x
O(0, 0)
A(2, 0)
[Ans.:
5
m]
6
 Show that the centre of mass of a uniform rod of mass M and length L lies at
the middle point of the rod.
[Ans.:
L
]
2
 Four particles of masses m, m, 2 m and 2m are placed at the four corners of a
square of side a. Find the centre of mass of the system.

 Ans.:


a 2 
 , a  with first mass m at the origin 
2 3 

 Four particles of masses m, 2m, 3m and 4m respectively are placed at the
corners of a square of side a, as shown in the given figure. Locate the centre of
mass.
Y
4m
D
C
3m
a
m
2m
A
B
a
 The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in
16 seconds.
(i) What is its angular acceleration, assuming the acceleration to be uniform?
(ii) How many revolutions does the wheel make during this time?

[Ans.: 576]
A constant torque is acting on a wheel. If starting from rest, the wheel makes n
rotations in t second, show that the angular acceleration is given by:

4 n
rad s 2
t2
4 n

2 
 Ans.:   t 2 rad s 
 The motor of an engine is rotating about its axis with an angular velocity of 100
rpm. It comes to rest in 15 s after being switched off. Assuming constant
angular deceleration, calculate the number of revolutions made by it before
coming to rest.
MECHANICAL PROPERTIES OF MATTER (SURFACE TENSION)
 A soap film is formed on a rectangular frame of length 7 cm dippling in soap
solution. The frame hangs from the arm of a balance. An extra weight of 0.38 g
is to be placed in the opposite pan to balance the pull on the frame. Calculate
the surface tension of the soap solution. Given g = 980 cms-2
[Ans. 26.6 dyne cm-1]
 A soap film is on a rectangular wire ring of size 5 cm × 4cm. If the size of the
film is changed to 5 cm × 5 cm, then calculate the work done in this process.
The surface tension of soap film is 5 102 Nm1 .
[Ans. 5 105 J ]
 A soap bubble is blown to a diameter of 7 cm. If 36,960 erg of work is done is
blowing it further, find the new radius if the surface tension of soap solution is
40 dyne cm-1.
[Ans. 7 cm]
 A soap bubble of radius 1/  cm is expanded to radius 2 /  cm. Calculate the
work done. Surface tension of soap solution = 30 dyne cm
[Ans. 720 erg]
 What amount of energy will be liberated if 1000 droplets of water, each of
diameter 10-8 cm, coalesce to form a bigger drop? Surface tension of water is
0.072 Nm-1.
[Ans. 2.035 × 10-14J]