Physics Test 256

Result

Physics Test 256

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

A receiving station on the ground is receiving a signal of frequency 5 MHz from a transmitter at a height of 300 m above the surface of the earth (of radius 6.4 × 10⁶ m) at a distance of 100 km from the receiver. Then the signal is coming via

A
ground wave propagation

B
sky wave propagation

C
both ground wave and sky wave propagation

D
neither ground wave nor sky wave propagation.

Solution

which is less than 100 km. Hence choice (a) is wrong. The maximum frequency which can be propagated via sky waves is v_max≃ 9MHz which is more than 5 MHz. Hence a 5 MHz signal can be propagated via sky waves and not via ground waves. Thus the correct choice is (b).
Question 2
4 / -1

Two gases A and B have equal pressure P, temperature T, and volume V. The two gases are mixed together and the resulting mixture has the same temperature T and volume V as before. The ratio of pressure exerted by the mixture to either of the two gases is:

A
1 : 1

B
2 : 1

C
3 : 1

D
1 : 2

Solution

Let P₁ and P₂ be the partial pressures of gases A and B, respectively. Under equal conditions of temperature and volume, P₁ = P₂

Let P be the total pressure exerted by the mixture of two gases A and B. Using Dalton's law of partial pressures, P = P₁ + P₂

⇒ P = 2P₁ ⇒ P/P₁ = 2/1 ⇒ P : P₁ = 2 : 1
Question 3
4 / -1

A book with many printing errors contains four different formulae for the displacement of a particle undergoing a certain periodic motion x. Which one is the wrong formula on dimensional grounds (A = amplitude, ω = angular velocity, T = time period of motion)?

A

B

C

D

Solution

The argument in the trigonometric function should be a dimensionless quantity.

The dimensions of time (t) angular velocity (ω) and amplitude (A) are [t] = [T], [ω] = [T^-1], [A] = [L]

A. The dimension of the argument in the equation:

x = A√2 (sin [2πt/T] + cos [2πt/T]) is zero.

Thus, option 'A' is not wrong on dimensional grounds.

Option B Analysis:

In the equation:

x = A/T sin (t/A),

the dimension of the argument in the sine function is:

[t/A] = |T|/L

⇒ [t/A] = [L⁻¹ T]

Thus, option 'B' is wrong on dimensional grounds.

Option C Analysis:

The dimension of the argument in the equation:

x = A sin (2πt/T)

⇒ [2πt/T] = |T|/T = dimensionless quantity

Thus, option 'C' is correct on dimensional grounds.

D. The dimension of the argument in equation: x = A sin (ωt) is [ωt] = [T^-1] [T] = dimensionless quantity

Thus, option 'D' is not wrong on dimensional ground. All the equations, except x = A/T sin (t/A) i.e. option 'B' have dimensionless argument.
Question 4
4 / -1

The figure below shows four plates each of area A and separated from one another by a distance d.

What is the capacitance between P and Q?

A

B

C

D

Solution

The given figure is shown below:

The above arrangement is equivalent to two parallel plate capacitors connected in parallel with the same separation d and area A.

Now, the capacitance of each capacitor is:

C₁ = (ε_₀ A) / d , C₂ = (ε_₀ A) / d

The equivalent capacitance of C₁ and C₂ is:

C = C₁ + C₂

⇒ C = (ε_₀ A) / d + (ε_₀ A) / d = (2ε_₀ A) / d
Question 5
4 / -1

Two sources of sound A and B produce progressive waves given by y₁ = 6 cos(100πt) and y₂ = 4 cos(102πt) ear the ears of an observer. It will hear

A
1 beat/s and intensity ratio of maxima to minima = 9/4

B
1 beat/s and intensity ratio of maxima to minima = 25/1

C
2 beat/s and intensity ratio of maxima to minima = 9/4

D
4 beat/s and intensity ratio of maxima to minima = 3/2

Solution

y₁ = 6 cos(100πt) ...... (i)

y₂ = 4 cos(102πt) ...... (ii)

Standard wave equation is given by y = A_o cos(ωt) ....... (iii)

From equation (i) and (iii), we get angular frequency ω₁ = 100π = 2πf₁ ⇒ f₁= 50 Hz

Amplitude, A₀₁ = 6 unit From equation (ii) and (iii), we get ω₂=102π = 2πf₂ ⇒ f₂ =51 Hz

Amplitude, A₀₂ = 4 unit

Beat frequency = difference in frequencies of the two given waves = f₂ - f₁ = 51 Hz - 50 Hz = 1 Hz, i.e 1 beat per sec

Resultant amplitude A of the superposition of two waves of amplitudes A₀₁ and A₀₂ is:

A = √(A₀₁² + A₀₂² + 2A₀₁A₀₂ cos φ)

For maximum resultant amplitude, when phase φ = 0°:

Aₘₐₓ = √(36 + 16 + 48) = √100 = 10 units

For minimum resultant amplitude, when phase φ = π:

Aₘᵢₙ = √(36 + 16 - 48) = √4 = 2 units

[Since cos π = -1]

Also, Intensity is proportional to (Aₘₐₓ)²:

So,

Iₘₐₓ / Iₘᵢₙ = (Aₘₐₓ / Aₘᵢₙ)²

⇒ Iₘₐₓ / Iₘᵢₙ = (10 / 2)² = 25/1
Question 6
4 / -1

Two beams of red and violet colors are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be:

A
30° for both the color

B
Greater for the violet color

C
Greater for the red color

D
Equal but not 30° for both the color

Solution

Two beams of red and violet colors are made to pass separately through a prism (angle of the prism is 60°).

We have to find the angle of refraction at the position of minimum deviation.

In the position of minimum deviation, light passes symmetrically through the prism irrespective of the light wavelength used.
Question 7
4 / -1

A planet of radius R = 1/10 × ( radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R/5 on it and lower a wire of the same length and of linear mass density 10^-3kg m^-1into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is: (take the radius of Earth = 6 × 10⁶ m and the acceleration due to gravity on Earth is 10 ms^-2)

A
96 N

B
108 N

C
120 N

D
150 N

Solution

Radius of the planet R = 1/10 x (radius of Earth)

Depth of the well = R/5 Linear mass density of the wire, μ = 10^-3 kg m^-1

We have to find the force applies at the top of the wire by a person holding it in.

Acceleration due to gravity on the planet will be given by g = GM/R²

Acceleration due to Gravity

⇒ g = (G / R²) × (ρ × (4/3) πR³)

⇒ g = (4/3) π G ρ R

(i) Acceleration due to gravity of Earth

gₑ = (4/3) π G ρ Rₑ

Dividing equation (i) by (ii), we get:

g / gₑ = R / Rₑ

Now, we will consider a mass element dm of width dx at depth x below the planet.

Mass of this element will be dm = μ dx

Also, acceleration due to gravity at depth x below the planet will be:

gₓ = (1 - x/R) gₚ

Force on the small segment of wire:

dF = dm gₓ

dF = (μ dx) (1 - x/R) gₚ

dF = μ g (1 - x/R) dx

Total force on the wire is obtained by integrating the above equation:

∫₀ᴿ dF = ∫₀ᴿ μ g (1 - x/R) dx

After solving, we get:

F = μ g (R/2)

Substituting all the values, we get:

F = 10⁻³ × 1 × R (9/50)

⇒ F = (9/50) × 10⁻³ × 10 × (6 × 10⁶)

⇒ F = 108 N
Question 8
4 / -1

A stone of mass m is attached to one end of a wire of cross-sectional area A and Young's Modulus Y. The stone is revolved in a horizontal circle at speed so that the wire makes an angle θ with the vertical, the strain produced in the wire is

A

B

C

D

Solution

The free-body diagram of the system is shown below:

[ac is centripetal acceleration]

Centripetal acceleration is in the horizontal direction as seen in the diagram. So, no acceleration is there in a vertical direction. So, the net force in a vertical direction is zero.

Thus

T cos θ = mg ⇒ T = mg / cos θ .....(i)

If L is the original length of the string, then the increase in length is given by:

ΔL = (T / A) × (L / Y)

where Y is Young's modulus of the wire, and A is the cross-sectional area of the wire.

⇒ ΔL / L = (mg / cos θ) x 1/(Y A) [using equation (i)]

⇒ Strain = mg / (Y A cos θ)
Question 9
4 / -1

The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0°C to 50°C when no work is done is ______. (R is the universal gas constant.

A
750 R

B
175 R

C
500 R

D
250 R

Solution

ΔQ = ΔU + ΔW

Here, ΔW = 0

ΔQ = ΔU = n Cv ΔT

ΔQ = 4 × (5R / 2) × 50 = 500R
Question 10
4 / -1

If the diameter of the Earth becomes half without changing its mass, the value of acceleration due to gravity on the surface becomes:

A
2g

B
g/2

C
g/4

D
4g

Solution

If the diameter of the Earth becomes half, the radius R also becomes half (R′ = R/2). Since the mass of the Earth remains the same, the new acceleration due to gravity g′ can be calculated as:

g′ = GM / (R′)²

Substitute R′ = R/2 into the formula:

g' = GM / (R/2)²

g' = GM / (R² / 4)

g' = (4GM) / R²

g' = 4g

Thus, the new value of acceleration due to gravity becomes 4 times the original value, i.e., 4g.