## Physics 2010 | Study Mode

Question 21
A
retina
B
iris
C
cornea
D
pupil
Explanation.
The retina is light sensitive layer living the inner wall of the eye where image is formed. in the camera, image is formed on a light sensitive film.

Question 22
A
2 cm
B
3 cm
C
4 cm
D
5 cm
Explanation.
(1)/v + (1)/u = (1)/f
(1)/v + (1)/15 = (1)/5
(1)/v = (1)/5 - (1)/15 = (3 - 1)/15 = (2)/15
v = (15)/2 = 7.5 cm
magnification = (v)/u = (image size)/object size
(7.5)/15 = (X)/4
image size. X = (7.5 x 4)/15 = 2 cm

Question 23
A
2.50 cm
B
6.67 cm
C
7.50 cm
D
7.63 cm
Explanation.
Refraction index, n = (real depth)/apparent depth
1.5 = (10)/X
X = (10)/1.5 = 6.67 cm

Question 24
A
convex lens
B
concave lens
C
convex mirror
D
concave mirror
Explanation.
Concave lens diverges parallel rays of light. A short sighted person has the image formed before the retina,and the rays have to be diverged so that they intersect to form image onto the retina

Question 25
A
different hidden colours in the glass
B
different speeds of the colour in the glass
C
defects in the glass
D
high density of the glass
Explanation.
Dispersion is the splitting of white light into its spectral component colours and is caused by different speed of the colours in the prisma

Question 26
A
be induced
B
converge
C
diverge
D
remain constant
Explanation.
Since the charged rod is not in contact with the electroscope, the leaves divergence will be induced

Question 27
A
make pass through the trasformers
B
increase the power supply
C
make it travel fast
D
prevent overheating of the coil
Explanation.
H eat generated in a wire carrying current, H = 12 Rt.
so, when voltage is high and current is low heat generated in the wire will be minimal.
Therefore, high voltage and low current prevent overheating of the coil.

Question 28
A
B
C
D
10
Explanation.
R = (Vg)/lg = G, where lg = current through galvanometer
R = resistance of the multiplier
G = resistance of the galvanometer
G = (vg-R)/lg 5(1000/10) - 490 = 500 -490 = 10

Question 29
A
500.0J
B
250.0J
C
62.5J
D
50.0J
E
no correct option
Explanation.
Total inductance,L = L1 + L2 = m
15 + 5 = 20mH = (20)H/1000
1 = 5A
Energy of inductor, (1)/2 L I2
= (1)/2 X (20)/1000 X 5-2
= 0.25 J

Question 30
B
()/2
C
()/3
D
Explanation.

No official Explanation yet!

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