Physics Test 202

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Physics Test 202

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Weekly Quiz Competition

Question 1
4 / -1

If the r.m.s speed of oxygen at NTP is x m/s. If the gas is heated at constant pressure till its volume is four fold, what will be its final temperature and r.m.s speed?

A
2xm/s

B
3xm/s

C
8xm/s

D
4xm/s

Solution
Question 2
4 / -1

Which one of the following quantities can be zero on an average for the molecules of an ideal gas in equilibrium?

A
Speed

B
Momentum

C
Kinetic Energy

D
Density

Solution

In case of ideal gases the average velocity is always zero. Hence the average momentum is zero.

Whereas average speed is non- zero so the kinetic energy is also non-zero, as these two are scalar quantities.
Question 3
4 / -1

The path difference between two waves y₁= A₁ sin wt and y₂= A₂ cos (wt + f) will be

A
(λ/2π) f

B
(λ/2π) (f + π/2)

C
(2π/λ) (f - π/2)

D
(2π/λ) f

Solution

y₁ = A₁ sin wt

y₂ = A₂ cos (wt + f)

Now, y₂ can be rewritten as A₂ sin (wt + f + π/2)

So, the phase difference is f + π/2

Path Difference = (λ/2π)(f + π/2)

Hence, the correct option is b.
Question 4
4 / -1

At constant volume, temperature is increased then

A
collision on walls will be less [1989]

B
number of collisions per unit time will increase

C
collisions will be in straight lines

D
collisions will not change.

Solution

As the temperature increases, the average velocity increases. So, the collisions are faster.
Question 5
4 / -1

The inputs of a NAND gate are connected together. The resultant circuit is

A
NOT gate

B
AND gate

C
OR gate

D
None of the above

Solution

The equation of NAND gate is

Y=Aˉ.Bˉ

When both the terminals are connected together.

A=B=A

Y=Aˉ.Aˉ=Aˉ.Aˉ=Aˉ

Which is nothing but the equation of NOT gate.

Y=Aˉ

Option A is the correct answer.
Question 6
4 / -1

The path difference between two wavefronts emitted by coherent sources of wavelength 5460 Å is 2.1 micron . The phase difference between the wavefronts at that point is _

A
7.692

B
7.692 π

C

D

Solution
Question 7
4 / -1

A particle of mass 10 gm lies in a potential field v = (50x²+100) J/kg. The value of frequency of oscillations in cycle/sec is

A
5/π

B
50/π

C
10/π

D
100/π

Solution

Correct option: A

Mass is m = 10 g = 0.01 kg.

The actual potential energy is U = m × (potential per unit mass); the quadratic contribution is U_q= 50 m x².

Compare with U = (1/2) k x². Thus (1/2) k = 50 m, so k = 100 m.

Angular frequency satisfies ω² = k/m. Substituting gives ω² = 100, hence ω = 10 rad s^-1.

Frequency is f = ω / (2π) = 10 / (2π) = 5/π Hz. Therefore option A is correct.
Question 8
4 / -1

Two containers A and B are partly filled with water and closed. The volume of A is twice that of B and it contains half the amount of water in B. If both are at the same temperature, the water vapour in the containers will have pressure in the ratio of [1988]

A
1 : 2

B
1 : 1

C
2 : 1

D
4 : 1

Solution

Vapour pressure does not depend on the amount of substance. It depends on the temperature alone.
Question 9
4 / -1

Energy is supplied to the damped oscillatory system at the same rate at which it is dissipating energy, then the amplitude of such oscillations would become constant. Such oscillations are called

A
Damped oscillations

B
Undamped oscillations

C
Coupled oscillations

D
Maintained oscillations

Solution

Energy is supplied to the damped oscillatory system at the same rate at which it is dissipating energy, and then the amplitude of such oscillations would become constant. Such oscillations are called maintained oscillations. By the definition of maintained oscillations.
Question 10
4 / -1

A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now:

A
f/2

B
f

C
2f

D
3f/2

Solution

As we know, f=v/2l

Now, it will act as one end and one end closed.

So, f₀=v/2l’=v/4½=v/2l=f

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Re-Attempt Weekly Quiz Competition

Physics Test 202

Result

Physics Test 202

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Question 1

2xm/s

3xm/s

8xm/s

4xm/s

Solution

Question 2

Speed

Momentum

Kinetic Energy

Density

Solution

Question 3

(λ/2π) f

(λ/2π) (f + π/2)

(2π/λ) (f - π/2)

(2π/λ) f

Solution

Question 4

collision on walls will be less [1989]

number of collisions per unit time will increase

collisions will be in straight lines

collisions will not change.

Solution

Question 5

NOT gate

AND gate

OR gate

None of the above

Solution

Question 6

7.692

7.692 π

Solution

Question 7

5/π

50/π

10/π

100/π

Solution

Question 8

1 : 2

1 : 1

2 : 1

4 : 1

Solution

Question 9

Damped oscillations

Undamped oscillations

Coupled oscillations

Maintained oscillations

Solution

Question 10

f/2

f

2f

3f/2

Solution

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