Physics Test 210

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Physics Test 210

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Weekly Quiz Competition

Question 1
4 / -1

A cylinder of fixed capacity 44.81 contains hydrogen gas at STP. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 20°C? (R = 8.31 J mol^-1 K^-1)

A
541 J

B
374 J

C
831 J

D
743 J

Solution

We are given the following data:

Volume of the cylinder = 44.81 L

Temperature change, ΔT = 20°C (which is the same in Kelvin)

R = 8.31 J/mol·K

At STP, the volume of one mole of gas is 22.4 L, so the number of moles, n, is:

n = Volume of gas / Volume of one mole at STP = 44.81 L / 22.4 L/mol = 2 mol

C_V = (5/2) × R = (5/2) × 8.31 J/mol·K = 20.775 J/mol·K

The heat required:

Q = n × C_V × ΔT

Q = 2 mol × 20.775 J/mol·K × 20 K = 831 J

∴ The amount of heat needed to raise the temperature of the hydrogen gas in the cylinder by 20°C is 831 J.

Hence, the correct option is 3) 831 J.
Question 2
4 / -1

The total internal energy of 4 moles of a diatomic gas at a temperature of 27^∘C is (Universal gas constant 8.31J mol⁻¹K⁻¹)

A
13.47 kJ

B
4.98 kJ

C
24.93 kJ

D
14.96 kJ

Solution

Given,

Number of moles, n = 4 moles

Temperature, T = 27°C = 27 + 273.15 = 300.15 K

Universal gas constant, R = 8.31 J/mol·K

Now, using the formula for internal energy of a diatomic gas:

U = (5/2) × n × R × T

U = (5/2) × 4 × 8.31 × 300.15

U = 2 × 8.31 × 300.15 = 24.93 × 10³ J

∴ The total internal energy of 4 moles of the diatomic gas is 24.93 kJ.

Hence, correct answer is 3) 24.93 kJ.
Question 3
4 / -1

Assertion: In a vernier caliper, one vernier scale division is smaller than one main scale division.

Reason: The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions.

A
Only Assertion is correct

B
Only reason correct.

C
Both assertion and reason are correct

D
None of the above

Solution

To evaluate the statements:

Statement I: Statement I is true because in a vernier caliper, the vernier scale division is indeed smaller than the main scale division, allowing for more precise measurements.

Statement II: Statement II is false because the vernier constant is actually calculated as the difference between one main scale division and one vernier scale division, not by multiplying one main scale division by the number of vernier scale divisions.

Only Assertion is correct
Question 4
4 / -1

A balloon is rising vertically up with a velocity of 29ms⁻¹. A stone is dropped from it and it reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it, was (g = 9.8 ms⁻²)

A
100 m

B
200 m

C
400 m

D
150 m

Solution

h = 29 × 10 - (1/2) × 9.8 × 10²

h = 290 - 490 = -200 m

∴ The height of the balloon when the stone was dropped is 200 m. Option 2 is correct.
Question 5
4 / -1

If two soap bubbles of different radii are connected by a tube

A
air flows from the smaller bubble to the bigger

B
air flows from bigger bubble to the smaller bubble till the sizes are interchanged

C
air flows from the bigger bubble to the smaller bubble till the sizes become equal

D
there is no flow of air.

Solution

Concept:

The pressure inside a soap bubble is inversely proportional to its radius, given by the formula: P = 4σ/r, where P is the pressure, σ is the surface tension, and r is the radius of the bubble. When two soap bubbles of different radii are connected by a tube, the bubble with the smaller radius will have a higher internal pressure.

Explanation:

- Air always flows from the region of higher pressure to the region of lower pressure.

- Since the smaller bubble has higher pressure due to its smaller radius, the air will flow from the smaller bubble to the bigger bubble, which has lower pressure, until the pressures are balanced.

Conclusion:

The correct option is: air flows from the smaller bubble to the bigger.
Question 6
4 / -1

In a vessel, the ideal gas is at a pressure P. If the mass of all the molecules is halved and their speed is doubled, then resultant pressure of the gas will be

A
4P

B
2P

C
P

D
P/2

Solution

Initial pressure: P = (1/3) * (N/V) * m * v²

After halving the mass and doubling the speed:

New pressure: P_new = (1/3) * (N/V) * (m/2) * (2v)^2

P_new = (1/3) * (N/V) * (m/2) * 4v²

P_new = 2 * (1/3) * (N/V) * m * v² = 2P

∴ The new pressure is: 2P. So, the correct option is 2).
Question 7
4 / -1

A satellite is revolving around the earth in a circular orbit with kinetic energy of 1.69 × 10¹⁰ J. The additional kinetic energy required for just escaping into the outer space is

A
3.38 × 10¹⁰ J

B
1.69 × 10¹⁰ J

C
0.89 × 10¹⁰ J

D
1.35 × 10¹⁰ J

Solution

Given,

The satellite's current kinetic energy is KE = 1.69 × 10¹⁰ J.

The total energy E of the satellite in orbit is:

E = -KE = -1.69 × 10¹⁰ J

For the satellite to just escape from Earth's gravity, its total energy E should become zero. Thus, the additional kinetic energy required ΔKE is the amount of energy required to raise the total energy to zero:

ΔKE = |E| = 1.69 × 10¹⁰ J

∴ The additional kinetic energy required for the satellite to escape into outer space is 1.69 × 10¹⁰ J.

Hence, the correct option is 2) 1.69 × 10¹⁰ J.
Question 8
4 / -1

The displacement y of a particle in a medium can be expressed as, y = 10⁻⁶sin m where t is in second and x in meter. The speed of the wave is

A
20 m ∕ s

B
5 m ∕ s

C
2000 m ∕ s

D
5πm ∕ s

Solution

From the given equation:

ω = 100 rad/s

k = 20 m^-1

Using the formula:

v = ω / k

v = 100 / 20

v = 5 m/s

∴ The speed of the wave is 5 m/s.
Question 9
4 / -1

One mole of a gas having is mixed with one mole of a gas having The value of y for the mixture is (y is the ratio of the specific heats of the gas)

A

B

C

D

Solution

Given:

Gas 1: γ₁ = 7/5

Gas 2: γ₂ = 4/3

⇒ C_p1 / C_v1 = 7/5 ⇒ C_p1 = (7/5) C_v1

⇒ C_p2 / C_v2 = 4/3 ⇒ C_p2 = (4/3) C_v2

Using the formula for a mixture:

⇒ C_p(mix) = [(1 × 7C_v/5) + (1 × 4C_v/3)] / (1+1)

⇒ C_p(mix) = [(21C_v + 20C_v) / 15] / 2

⇒ C_p(mix) = (41C_v / 30)

Similarly,

⇒ C_v(mix) = [(1 × C_v) + (1 × C_v)] / (1+1)

⇒ C_v(mix) = 2C_v / 2

⇒ C_v(mix) = C_v

Now, calculating γ for the mixture:

⇒ γ_mix = C_p(mix) / C_v(mix)

⇒ γ_mix = (41C_v / 30) / (11C_v / 11)

⇒ γ_mix = 15/11

∴ The correct answer is 15/11.
Question 10
4 / -1

The energy required to take a body from the surface of the earth to a height equal to the radius of the earth is 'W'. The energy required to take this body from the surface of the earth to a height equal to twice the radius of the earth is

A

B

C
W

D

Solution

Let the mass of the object be m and the radius of the Earth be R.

The energy required to take the body from the Earth's surface to a height h = R is given by:

Initial potential energy at the surface, U_i = - GMm / R

Final potential energy at height h = R, U_f = - GMm / 2R

The energy required for this displacement is:

E₁ = U_f - U_i = (- GMm / 2R) - (- GMm / R) = GMm / 2R

The energy required to take the body from the Earth's surface to a height h = 2R is:

Initial potential energy at the surface, U_i = - GMm / R

Final potential energy at height h = 2R, U_f = - GMm / 3R

The energy required for this displacement is:

E₂ = U_f - U_i = (- GMm / 3R) - (- GMm / R) = 2GMm / 3R

The energy required to take the body to a height of h = 2R is E₂ = 4/3 E₁

∴ The energy required is 4W / 3.

Hence, the correct option is 4.