IIT-JEE 2010 - Career Point

2. Show that the temperature of a planet varies inversely
as the square root of its distance from the Sun.
Sol. Let R s be the radius of the Sun. Consider the Sun as a
black body.
Energy emitted per sec by it equals 4 π R σ T
4
This energy falls uniformly on the inner surface of
spheres centred on the Sun. If d is the distance of the
planet from the Sun, then energy falling on unit area
of the sphere of radius d is :
2 4
sσTs
2
4πR
σR sTs
=
2
4πd
d
This energy received by the planet is given by
2
4
2
Q = πr 2 σR sTs
= where r is the radius of planet.
2
d
If T is the temperature of the planet, then energy lost
by it per sec is 4 π r 2 σ T 4
In the steady state the rate of reception of energy is
equal to the loss of energy
Hence, 4 π r 2 σ T 4 πr σR
=
2
d
1
Thus, T ∝
d
4
2
2 4
s Ts
3. A sphere of specific gravity s just fits into a vertical
cylinder with lower end closed. The sphere is allowed
to drop slowly until it is held in equilibrium by the
thrust of the compressed air. There is no leakage of
air. If the diameter of the sphere is d, the length of
the cylinder is L and the height of the water
barometer is h, then what will be the position of
sphere ?
Sol. Initially, the cylinder contained air at atmospheric
pressure. When the sphere comes down into the
cylinder by the action of its own weight, it presses the
air downwards. Suppose the sphere comes to position
C which is at height x above the closed end. Let the
sphere remain in equilibrium in this position.
Volume of sphere = (4/3)πr 3
Weight of the sphere = (4/3)πr 3 sg
Volume of cylinder = πr 2 L
Volume of air inside the cylinder when the sphere is
in position A = πr 2 L – (2/3)πr 3
2
s
s
Volume of compressed air when the sphere is in
position C = [πr 2 x – (2/3)πr 3 ]
Atmospheric pressure = h cm of water
Let the pressure of compressed air be p cm of water.
L
d
According to Boyle's Law (assuming no change in
the temperature of compressed air)
p[πr 2 x – (2/3)πr 3 ] = h[πr 2 L – (2/3)πr 3 ]
or p[x – (2r/3) = h[L – (2r/3)] ...(1)
In the equilibrium position C, the weight of the
sphere is balanced by the difference of vertical thrust
on either side due to atmospheric and compressed air.
Hence, πr 2 (p – h)g = mg = (4/3)πr 3 sg
or p – h = (4/3)rs ...(2)
From equation (1) and (2), we get
h(3L
− 2r)
x =
+
3p
h(3L − 2r)
or x =
+
3h + 4rs
or x =
2
9hL + 2d s
9h + 6ds
2r
3
A
C
x
2 r 9hL + 8r s
=
3 3(3h + 4rs)
4. Given the position of the object O and the image I as
shown in the figure. Find (a) the position of the
convex lens (b) its focal length and the magnification
of the image. Verify graphically.
2
1
0
–1
–2
Y
O
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Sol. A ray of light from the object passes undeviated
through the optic centre of the lens (C) and also the
image I. So join OI.
So it cuts the principal axis XY and C. So AB is the
position of the lens. A ray parallel to the principal
axis from the object after refraction meets the
principal axis at F. F is the focus.
2
I
X
XtraEdge for IIT-JEE 23 DECEMBER 2009