Physics Test 271

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Physics Test 271

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Question 1
4 / -1

Kinetic theory explains the behavior

A
of liquids based on the idea that the liquids consist of rapidly moving atoms or molecules

B
of gases based on the idea that the gas consists of rapidly moving atoms or molecules

C
of solids based on the idea that the solid consists of rapidly vibrating atoms or molecules

D
of solids and liquids based on the idea that they gas consist of rapidly vibrating atoms or molecules

Solution

Explanation:

The theory for ideal gases makes the following assumptions

1. Gases consist of particles in constant, random motion. They continue in a straight line until they collide with something—usually each other or the walls of their container.

2. Particles are point masses with no volume. The particles are so small compared to the space between them, that we do not consider their size in ideal gases.

3. No molecular forces are at work. This means that there is no attraction or repulsion between the particles.

4. Gas pressure is due to the molecules colliding with the walls of the container. All of these collisions are perfectly elastic, meaning that there is no change in energy of either the particles or the wall upon collision. No energy is lost or gained from collisions.

5. The time it takes to collide is negligible compared with the time between collisions.

6. The kinetic energy of a gas is a measure of its Kelvin temperature. Individual gas molecules have different speeds, but the temperature and kinetic energy of the gas refer to the average of these speeds.

7. The average kinetic energy of a gas particle is directly proportional to the temperature. An increase in temperature increases the speed in which the gas molecules move.

8. All gases at a given temperature have the same average kinetic energy.

9. Lighter gas molecules move faster than heavier molecules.
Question 2
4 / -1

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.

A
1.339 × 10⁻⁴ m

B
1.439 × 10⁻⁴ m

C
1.879 × 10⁻⁴ m

D
1.239 × 10⁻⁴ m

Solution

Mass, m = 14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s

Cross-sectional area of the wire, a = 0.065 cm² = 0.065 × 10^-4 m²

Let Δl be the elongation of the wire when the mass is at the lowest point of its path.

When the mass is placed at the position of the vertical circle, the total force on the mass is:

F = mg + mlω²

= 14.5 × 9.8 + 14.5 × 1 × (12.56)²

= 2429.53 N

Young’s modulus = Stress / Strain

Y = (F/A) / (∆l/l)

∴ ∆l = Fl / AY

Young’s modulus for steel = 2 × 10¹¹ Pa

∆l = 2429.53 × 1 / (0.065 × 10^-4 × 2 × 10¹¹) = 1.87 × 10^-3 m

Hence, the elongation of the wire is 1.87 × 10^–3 m

Hence 1.87 × 10^–3 m
Question 3
4 / -1

Two bullets A and B are fired horizontally with speed v and 2v respectively.which of the following is true

A
Both will reach the ground in Same time

B
Bullet with speed 2v will cover more horizontal distance on the ground

C
B will reach the ground in less time then A

D
A will reach the ground in less time then B
Solution
In projectile motion, the time taken for an object to reach the ground is independent of its horizontal velocity. The only factor that determines the time it takes for an object to reach the ground is the vertical motion (due to gravity).

Since both bullets A and B are fired horizontally, their initial vertical velocities are zero. The time taken for both bullets to fall will only depend on the height from which they are fired and the acceleration due to gravity.

The time t taken for an object to reach the ground from a certain height h is given by:

Where:
- h is the height from which the bullets are fired,
- g is the acceleration due to gravity.
This equation shows that the time to reach the ground depends only on the height and gravitational acceleration, not the horizontal velocity.

Thus, both bullets A and B will reach the ground in the same time. Therefore, option A is correct.
Question 4
4 / -1

The displacement of a damped harmonic oscillator is given by:

x(t) = e^-0.1t cos (10πt + φ). Hence, t is in seconds.

The time taken for its amplitude of vibration to drop to half of its initial value is close to:

A
4 s

B
13 s

C
27 s

D
7 s

Solution
Question 5
4 / -1

1 mole of a monoatomic gas is mixed with 3 moles of a diatomic gas. What is the molecular specific heat of the mixture at constant volume?

A
12.5 J / mol K

B
18.7 J / mol K

C
15.2 J / mol K

D
22.6 J / mol K

Solution
Question 6
4 / -1

The average distance a molecule can travel without colliding is called the

A
mean free length

B
mean free path

C
mean free distance

D
mean free motion

Solution

Explanation:the mean free path is the average distance traveled by a moving particle (such as an atom, a molecule, a photon) between successive impacts (collisions), which modify its direction or energy or other particle properties
Question 7
4 / -1

Value of gas constant, R for one mole of a gas is independent of the

A
Atomicity of the gas

B
Mass of gas

C
Distance between two molecules of gas at 273 K

D
Volume of gas

Solution

We know that PV=nRT also PM=dRT

So in the equation The value of R depends on P , V , n , T , d , M

except atomicity

so the ans is A
Question 8
4 / -1

According to kinetic theory of gases, 0K is that temperature at which for an ideal gas

A
pressure is not zero

B
internal energy is zero

C
nearly all molecular motion becomes very rapid

D
volume is not zero

Solution

According to kinetic theory of gases, 0K is that temperature at which for an ideal gas the internal energy is zero because at 0K nearly all molecular motion stops.
Question 9
4 / -1

Four moles of an ideal diatomic gas is heated at constant volume from 20° C to 30° C. The molar specific heat of the gas at constant pressure (Cp) is 30.3 Jmol^-1K^-1 and the universal gas constant (R) is 8.3 Jmol^-1K^-1. The increase in internal energy of the gas is

A
332 J

B
80.3 J

C
303 J

D
880 J

Solution

The value of Cp is 30.3

and as Cp-Cv = R(8.3)

hence Cv = 30.3-8.3

Cv is 22

change in internal energy = no of moles × Cv × change in temperature

hence

change in internal energy = 22 × 4 × 10

= 880j

Hence Option D is correct.
Question 10
4 / -1

Three moles of an ideal monoatomic gas is initially in the state A shown in the adjoining pressure-temperature graph. It is taken to state B without changing its pressure. If R is the universal gas constant, the work done by the gas in this process is

A
600 R

B
400 R

C
300 R

D
200 R

Solution

The work done by the gas in taking it from state A to state B = PΔV where ΔW is the increase in volume at constant pressure P.

We have PV = μRT where p is the number of moles in the sample of the gas and R is the universal gas constant.

Therefore we have PΔV = μR ΔT = 3 xR(450 - 250) = 600R

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Physics Test 271

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Question 1

of liquids based on the idea that the liquids consist of rapidly moving atoms or molecules

of gases based on the idea that the gas consists of rapidly moving atoms or molecules

of solids based on the idea that the solid consists of rapidly vibrating atoms or molecules

of solids and liquids based on the idea that they gas consist of rapidly vibrating atoms or molecules

Solution

Question 2

1.339 × 10−4 m

1.439 × 10−4 m

1.879 × 10−4 m

1.239 × 10−4 m

Solution

Question 3

Both will reach the ground in Same time

Bullet with speed 2v will cover more horizontal distance on the ground

B will reach the ground in less time then A

A will reach the ground in less time then B

Solution

Question 4

4 s

13 s

27 s

7 s

Solution

Question 5

12.5 J / mol K

18.7 J / mol K

15.2 J / mol K

22.6 J / mol K

Solution

Question 6

mean free length

mean free path

mean free distance

mean free motion

Solution

Question 7

Atomicity of the gas

Mass of gas

Distance between two molecules of gas at 273 K

Volume of gas

Solution

Question 8

pressure is not zero

internal energy is zero

nearly all molecular motion becomes very rapid​

volume is not zero

Solution

Question 9

332 J

80.3 J

303 J

880 J

Solution

Question 10

600 R

400 R

300 R

200 R

Solution

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1.339 × 10⁻⁴ m

1.439 × 10⁻⁴ m

1.879 × 10⁻⁴ m

1.239 × 10⁻⁴ m

nearly all molecular motion becomes very rapid