When X-rays of wavelength 0.5 Å would be transmitted by an aluminium tube of thickness 7 mm, its intensity remains one-fourth. The attenuation coefficient of aluminium for these X-rays is

0.188 mm^–1
0.189 mm^–1
0.198 mm^–1
None of these

Solution

The B – H curve (i) and (ii) shown in fig.associated with

(i) diamagnetic and (ii) paramagnetic substance
(i) paramagnetic and (ii) ferromagnetic substance
(i) soft iron and (ii) steel respectively
(i) steel and (ii) soft iron respectively

Solution

The loop (i) is for soft iron and the loop (ii) is for steel in Fig.

A source S₁ is producing, 10¹⁵ photons per second of wavelength 5000 Å. Another source S² is producing 1.02 × 10¹⁵ photons per second of wavelength 5100 Å Then,(power of S²) to the (power of S¹) is equal to :

1.00
1.02
1.04
0.98

Solution

The ratio of principal molar heat capacities of a gas is maximum for

adiatomic gas
a monatomic gas
apolyatomic gas having linear molecules.
apolyatomic gas having non-linear molecules.

A graph of acceleration versus time of a particle starting from rest at t =0 is as shown in Fig. The speed of the particle at t = 14 second is

2ms^–1
34 ms^–1
20 ms^–1
42 ms^–1

Solution

Area under a-t graph is change in velocity.

Area=¹⁄₂ (4 × 4) + 6 × 4 + ¹⁄₂ × 2 × 4-¹⁄₂ × 2 × 2

=36 - 2 = 34 ms^–1

As initial velocity is zero therefore, the velocity at 14 second is 34 ms^–1

Block A is moving with acceleration A along a friction less horizontal surface. When a second block, B is placed on top of Block A the acceleration of the combined blocks drops to 1/5 the original value. What is the ratio of the mass of A to the mass of B?

5 : 1
1 : 4
3 : 1
2 : 1

Solution

Apply Newton’s second law

F_A= F_AB, therefore:

m_A a_A= (m_A+ m_B)a_AB and a_AB= a_A / 5

Therefore : m_A a_A = (m_A+ m_B)a_A/5 which reduces to 4 m_A = m_B or 1 : 4

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

Ec
2E/c
E/c
E/c²

Solution

Momentum of photon = ^E⁄_C

Change in momentum = ^2E⁄_C

momentum transferred to the surface(the photon will reflect with same magnitude of momentum in opposite direction).

Two glass plates are separated by water. If surface tension of water is 75 dyne/cm and area of each plate wetted by water is 8 cm² and the distance between the plates is 0.12mm, then the force applied to,separate the two plates is

10² dyne
10⁴ dyne
10⁵ dyne
10⁶ dyne

Solution

A small photocell is placed at a distance of 4 m from a photosensitive surface. When light falls on the surface the current is 5 mA. If the distance of cell is decreased to 1m,the current will become

1.25 mA
(⁵⁄₁₆)5 mA
20 mA
80 mA

Solution

\(I\alpha \frac{1}{d^{2}}\)

The correct statement of the law of equipartition of energy is

the total energy of a gas is equally divided among all the molecules.
The gas possess equal energies in all the three directions x,y and z-axis.
the total energy of a gas is equally divided between kinetic and potential energies.
the total kinetic energy of a gas molecules is equally divided among translational and rotational kinetic energies.

In a n-type semiconductor, which of the following statement is true?

Electrons are minority carriers and pentavalent atoms are dopants.
Holes are minority carriers and pentavalent atoms are dopants.
Holes are majority carriers and trivalent atoms are dopants.
Electrons are majority carriers and trivalent atoms are dopants.

Solution

In a n-type semiconductor holes are minority carriers and pentavalent atoms are dopants.

It is given that t = px²+ qx, where x is displacement and t is time. The acceleration of particle at origin is

\(-\frac{2p}{q^{3}}\)
\(-\frac{3p}{q^{3h}}\)
\(-\frac{8p}{q^{3}}\)
\(-\frac{8p}{q^{3}}\)

Solution

Differentiate two times and put x = 0.

A thermodynamic system goes from states (i) P₁, V to 2P₁,V (ii) P, V₁ to P, 2V₁. Then work done in the two cases is

zero, zero
zero, PV₁
PV₁, zero
PV₁, P₁V₁

Solution

An achromatic convergent lens of focal length 20 cms is made of two lenses(in contact) of materials having dispersive powers in the ratio of 1: 2 and having focal lengths f₁ and f₂
.Which of the following is true ?

f₁ = 10 cms,f₂= –20 cms
f₁ = 20 cms,f₂= 10cms
f₁ = –10 cms,f₂= –20 cms
f₁ = 20 cms,f₂= –20 cms

Solution

An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram :

If Q₁, Q₂, Q₃ indicate the heat a absorbed by the gas along the three processes and ΔUQ₁, ΔU₂, ΔU₂ indicate the change in internal energy along the three processes respectively, then

Q₁ > Q₂ > Q₃ and ΔU₁ = ΔU₂ = ΔU₃
Q₃ > Q₂ > Q₁ and ΔU₁= ΔU₂ = ΔU₃
Q₁ = Q₂ = Q₃ and ΔU₁ > ΔU₂ > ΔU₃
Q₃ > Q₂ > Q₁ and ΔU₁ > ΔU₂ > ΔU₃

Solution

Initial and final condition is same for all process

ΔU₁ = ΔU₂= ΔU₃

from first law of thermodynamics

ΔQ= ΔU + ΔW

Work done

ΔW₁ > ΔW₂ > ΔW₃(Area of P.V. graph)

So ΔQ₁ > ΔQ₂ > ΔQ₃

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