305
(ii)
Measure height of I2 and divide by height of object
3.
4.
5.
2004: Place candle at a distance as shown below
u
v
306

Adjust until a sharp image is formed

Measure U and V and record

Repeat this procedure for other values of u

For each set of values find using
1
/
f
=
1
/
u
+
1
/
v

Calculate the
average value
(b)
Draw one ray from top of object straight through the optical centre
of the lens a
 Draw another ray parallel to the principal axis which should pass
through principal axis after refraction
 Where they meet, draw the image.
(c)
Short sightedness
6.
(a)
(i)
307
(ii)
(b)
(i)
ADiaphragm
BFilm
(ii)
The distance between the lens is adjusted so that the image is
formed on the film
(iii)
Shutter Opens for some given time to alloy rays from the object to
fall on the film creating the image impression.
A
(Diaphragm) controls intensity of light entering the camera
B
(Film)
–
coated with light sensitive components which react with
light to create the impression
(c)
(i)
Magnification = V = 3
U
Since v + u = 80
U = 80
–
v
V
= 3
80
–
v
V= 240
–
3v
V= 60 cm
308
(ii)
From above u = 20cm
I = I + I + I + I
F
v
u
60
20
7.
f= 15cm
8.
f= 14.3cm
9.
Given information
10.
u= 16cm
f = 12cm
Applying l = l + l
f
u
v
then I = l + l
v
f
u
= 1
–
1
12
V16
= l = 4 3
= 1
V
48
48
Hence distance = 48
(ii)
Nature: image is real

Inverted

48 cm from the lens

Magnified
(iii)
Magnification, m= v = 48 cm = 3
u
16 cm
309
11.
Solution
Given information
V= 6 cm
U = + 15cm
Applying I = I + I
F
u
v
= I
–
I
15
6
25
30
I = 3 = 1
F
30
10
F= 10
F= 10 cm
12.
Solution
P= I
F
P =
1
/
10
=
1
/
10/100
m
= 1
10
/
100
m
= 1 x
100
/
10
Power = 10 dioptre (D)
310
13.
Similarity
Differences
 Both have converging lens
 In the eye, retina act as the
screen while
in the camera a
light sensitive
film act as one
 Focal length of the eye lens is
variable while that of the camera
is fixed.
311
2.
UNIFORM CIRCULAR MOTION
1.
When tension is max then r= l
T= F = Mv
2
r
2 = 0.1 x V2
1.25
V2 = 2 x 1.25
0.1
V= 5m/s
2.
(a) (i) & (ii)
(b) (i)
S= Ut + ½ at
2
t =
√
(2 x 5)
10
= 1 seconds
(ii)
S = Ut = 30 x 1 = 30m
312
(iii)
V
2
= U
2
+ 2as
V= √(2 x 10 x 5)
= 10m
3.
Principal of conical pendulum Or principle of circular motion
4.
V= wr
But w =
θ
= 2
x 7 = 14 rads
1
T
1
V= 14
x 0.20
= 8.8 m/s
5.
Centripetal acceleration of bucket is equal or higher than gravitational
acceleration of the water.
6.
(a)
Keep angular velocity ω
1
constant; Centripetal force provided by
mg; Fix the mass m and measure r; repeat for different values of m.
Tabulate the values
(b)
Force Mg. Calculate and fill a column for force and another for
radius in m
Plot the graph
313
(ii)
Gradient of the graph
= 0.625
–
0.1 = 1.167 N 1m
0.525
–
0.075
Force F, on the body = m
b
w
2
r
Where m
b
= mass of the body
M
b
w
2
= Gradient of the graph = 1.167
W
2
= 1.167 = 11.67
0.1
W=
√11. 67
= 3.42 rads
1
7.
2
f = 2
x 8
= rw
2
= 0.2 (16
)
2
=
505.3
m/s
8.
(a)
Angular displacement per second.
(b)
In 13s, angle turned = 300170 = 130 rad
w=
θ
= 130 = 10rads1
t
13
So
300
= 10
t+13
10t + 130 = 300
10t = 170
t = 17 sec
314
(c)
(i)
Plot the graph which is a straight line.
(ii)
From T = (mr)w
2
–
C and y = mx + c
Y = T, grad = mr, y = intercept = C
Obtain gradient and equate it to mr.
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 Spring '20
 Karani
 Electron, Magnetic Field