Physics Test

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Physics Test - 3

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Question 1
1 / -0.25

In amplitude modulation, the amplitude of the carrier wave is A_c,and that of the modulating signal is A_m. In practice, the ratio of A_mto A_cis kept less than or equal to one, to avoid

A
A. distortion

B
B. attenuation

C
C. fading

D
D. noise

Solution

Modulation index, μ= Am/Ac ≤ 1 is carried by the carrier wave, i.e., there is no signal distortion.
Question 2
1 / -0.25

The frequency of a sonometer wire is f but when the weights producing tension are completely immersed in water the frequency becomes f/2 and on immersing the weights in certain liquid the frequency becomes f/3. The specific gravity of the liquid is

A
A. 1.320

B
B. 1.333

C
C. 0.889

D
D. 1.185

Solution

Let ρ the density of weight.
σ be the density of liquid.
and V be the volume of the weights.
From (1) and (2)
From (1) and (3)
Question 3
1 / -0.25

A heat engine has an efficiency η. Temperatures of source and sink are each decreased by 100 K. Then, the efficiency of the engine.

A
A. Increases

B
B. Decreases

C
C. Remains constant

D
D. Becomes 1

Solution

where T₁ and T₂ are the temperatures of a source and sink respectively.
When T₁ and T₂ both are decreased by 100 K each, (T₁ - T₂ ) stays constant. T₁ decreases.
∴ η increases.
Question 4
1 / -0.25

Which of these is correct in regard to a magnet?

A
A. Geometric length = 0.8 times the magnetic length

B
B. Magnetic length = 0.8 times the geometric length

C
C. Magnetic length= Geometric length

D
D. Geometric length = 10/9 of magnetic length

Solution

Magnetic length is the distance between the two equal and opposite poles of the magnet.
Poles are not actually at the end of the magnet they are little inside from the end this makes the magnetic length of the magnet little less than the Geometric length.
Geometric length is the actual length of the magnet.
So when the relationship between these two measured it comes to be,
Magnetic length = 0.8× Geometric length
Question 5
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The mean lives of radioactive substances are 1620 y and 405 y for α- emission and β- emission respectively. Find out the time during which three-fourth of a sample will decay if it is decaying both by α- emission and β- emission simultaneously.

A
A. 449 y

B
B. 399 y

C
C. 549 y

D
D. 579 y

Solution

Let at some instant of time tt, number of atoms of the radioactive substance are N. It may decay either by α - emission on by β- emission. So, we can write,
If the effective decay constant is λ, then
Now,
Question 6
1 / -0.25

When the velocity of an electron increases, its de-Broglie wavelength

A
A. increases.

B
B. decreases.

C
C. remains same.

D
D. may increase or decrease.

Solution

The de Broglie wavelength is given by
So if the velocity of the electron increases, the de Broglie wavelength decreases.
Question 7
1 / -0.25

A charge q is located at the centre of a cube. The electric flux through any face is

A
A.

B
B.

C
C.

D
D.

Solution

The total flux through the cube
∴ The electric flux, through any face
Question 8
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Glass has refractive index μwith respect to air and the critical angle for a ray of light going from glass to air is an if a ray of light is incident from the air on the glass with the angle of incidence, a corresponding angle of refraction is

A
A. sin^-1 (μ)

B
B. sin^-1 (1/μ² )

C
C. sin^-1 (1/√μ)

D
D. sin^-1 (1/μ)

Solution

Refractive index of glass with respect to air = μ Critical angle = θ
∴ μ = (1/sinθ) …(i)
When a ray of light incident from the air on the glass with the angle of incidence θ. If r be the angle of refraction, then by Snell’s law,
μ = sinθ/sinr
⇒ [fromEq. (i)]
⇒ sinr = 1/μ² r = sin^-1 (1/μ² )
Question 9
1 / -0.25

A molecule of water on the surface experiences a net

A
A. downward resultant unbalanced adhesive force

B
B. upward resultant unbalanced cohesive force

C
C. downward resultant unbalanced cohesive force

D
D. upward resultant unbalanced adhesive force

Solution

A molecule of the surface experiences a nat force (Cohesive force) in a downward direction due to the net number of molecules pulling it downward (molecule C as shown in figure underneath). In contrast, a molecule well inside is equally pulled in all directions by molecules around at (molecule A as shown in the figure).
Question 10
1 / -0.25

The S I unit and dimensions of Stefan 's constant σin case of Stefan 's law of radiation is

A
(J/m³ s⁴ )(M ’L⁰ T^-3 K^-4 )

B
(J/m³ s⁴ K)(M ’L⁰ T^-3 K^-3 )

C
(J/m³ s⁴ K⁴ )(M ’L⁰ T^-3 K^-4 )

D
(J/m ² sK⁴ )(M¹ L⁰ T^-3 K^-4 )

Solution

According to Stefan 's law, the energy emitted by a body per unit area per second is proportional to the fourth power of the absolute temperature.
E = σT⁴ where E = energy emitted area/second,
T = absolute temperature in kelvin
and σ= Stefan 's constant ⇒σ= E/T⁴
Unit of Stefan 's constant
⇒J/m² s/K⁴ = J/m² sK⁴ Dimensions of Stefan ’s constant,
Question 11
1 / -0.25

V-I characteristics of LED is shown correctly by graph

A
A.

B
B.

C
C.

D
D.

Solution

Since, an LED is forward biased diode, so its characteristics are those of forward bias, i,e„ Here, option (a) is characteristics of a diode in forward and reverse bias.
Option (c) characteristics of the solar cell.
Option (d) → reverse bias characteristics 1 drawn in the first quadrant can be photodiode or Zener diode.
Question 12
1 / -0.25

The refractive index of the crystal material is 1.68, and that of castor oil is 1.2. When a ray of light passes from oil to glass, its velocity will change by a factor

A
A. 2/3

B
B. 3/4

C
C. 5/6

D
D. 5/7

Solution

We know, the refractive index of crystal w.r.t. Oil, where, μ_c = absolute refractive index of crystal w.r.t. Vacuum and μ_o = absolute refractive index of oil w. r. t, vacuumThe fractionalonal change in velocity, when light travel from oil to crystal = speed in crystal/speed in oil = 5/7.
Question 13
1 / -0.25

Six very long insulated (topper wires are bound together to form a cable. The currents carried by the wires are l₁= +10 A, l₂= - 13A , l₃= +10A, l₄= + 7A, l₅= 12A and l₆= 18A. The magnetic induction at a perpendicular distance of 10 cm from the cable is (μ₀= 4πx 10^-7Wb/A-m)

A
A. 40 μt

B
B. 375 μt

C
C. 30 μt

D
D. 35 μt

Solution

Net current due to all wires, i_net = i1 + i2 + i3 + i4 + i5 + i6
I_net = 10 - 13 + 10 + 7 - 12 + 18 = 20A
We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance r from it is given by
Where, i = current in wire and r = perpendicular distance.
B = 40μT
Question 14
1 / -0.25

In case of dimension of electric field and electric dipole moment the power of mass is respectively,

A
A. 1, 1

B
B. 1, 0

C
C. 0, 1

D
D. 0, 0

Solution

Electric field, E = F/q
Where, F = force and q = electric charge
Dimensions of E or [E] = [MLT^-2 ]/[AT] = [MLT^-0 A^-1 ]
Power of mass = 1
Electric dipole moment, m = l x A
Where, l = current and A = area
Dimensions of m = [M] = [A L² ] = [M⁰ L² T⁰ A]
Hence, power of mass = 0
Question 15
1 / -0.25

For homogeneous isotropic material, which one of the following cannot be the value of Poisson ’s ratio?

A
0.1

B
-1

C
0.5

D
0.8

Solution

The Poisson ’s ratio for homogeneous isotropic material must be between - 10 to + 05 because of the requirement. For Young ’s modulus, the shear modulus and bulk modulus have positive values. Hence, 0.8 cannot be the value of Poisson ’s ratio for homogeneous isotropic material.
Question 16
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A circular coil and a square coil is prepared from two identical metal wires and a current is passed through The ratio of magnetic dipole moment associated with circular coil to that of square coil is

A
A. 2/π

B
B. π/2

C
C. π

D
D. 4/π

Solution

Let l be the length of metal wire. When wire is bent into a circular coil of radius r, then r = l/2π
∴ Area,
∴ Magnetic dipole moment associated with circular coil,
μ_c = il² /4π …(i)
When metal wire is bent into a square coil then side of square a = l/4
∴ Area, A = a² = l² /16
∴ Magnetic dipole moment associated with square coil, μ_c = iA = i(l² /16)
∴
Hnce, the ratio of magnetic dipole moment of circular coil and square coil is 4 : π.
Question 17
1 / -0.25

Figure show the circular coil carrying current l kept very close but not touching at a point A on a straight conductor carrying the same current I. The magnitude of magnetic induction at the centre of the circular coil will be

A
A.

B
B.

C
C.

D
D. zero

Solution

Magnetic induction at the centre of circular current carrying loop,
Magnetic induction at the centre O due to straight wire,
Net magnetic induction at O,