Physics Test 297

Result

Physics Test 297

Score
-
out of -
Rank
-
out of -

TIME Taken - -

Weekly Quiz Competition

Question 1
4 / -1

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heat 1.4. The vessel is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by :

(R = universal gas constant)

A

B

C

D

Solution

Given: γ = 1.4

Let us take "n" moles of the gas.

The energy loss is equal to the energy gain in the system and it is written as;

γ = 1.4 = 7/5 (given), from this we see that it is a diatomic gas, and specific heat capacity at constant volume is written as;

C_v = 5R/2

Here equation (2) becomes,
Question 2
4 / -1

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x/2 times its original time period. Then the value of x is:

A
√3

B
√2

C
2√3

D
4

Solution

Here,

T = 2 π √ l /g

T ′ = 2 π √ l / g

T ′ = T /√ 2

T /√ 2 = (x/2)T

⇒ x = √ 2

∴ The correct option is 2
Question 3
4 / -1

A tuning fork arrangement produces 4 beats/second with one fork of frequency 288 Hz. A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is _______ Hz.

A
286

B
292

C
294

D
288

Solution

Initial beat frequency = 4

⇒ |f − 288| = 4

⇒ f = 288 ± 4

⇒ f = 292 Hz or 284 Hz

After applying wax, frequency decreases and beat frequency becomes 2.

Check f = 292 Hz:

New frequency decreases ⇒ moves towards 288

⇒ beat decreases from 4 to 2 ✔

Check f = 284 Hz:

New frequency decreases ⇒ moves away from 288

⇒ beat increases (not possible) ✖

Conclusion: Correct frequency = 292 Hz
Question 4
4 / -1

Time period of a simple pendulum in a stationary lift is ‘T'. If the lift accelerates with g/6 vertically upwards then the time period will be :

(Where g = acceleration due to gravity)

A

B

C

D

Solution
Question 5
4 / -1

An expression for a dimensionless quantity P is given by where α and β are constants, x is distance; k is the Boltzmann constant and t is the temperature. Then the dimensions of α will be :

A
[M⁰L^-1T⁰]

B
[M L⁰ T^-2]

C
[M L T^-2]

D
[M L² T^-2]

Solution
Question 6
4 / -1

Two plates are at potentials -10V and +30V. If the separation between the plates is 2 cm then find the electric field between them.

A
2000 V/m

B
1000 V/m

C
500 V/m

D
3000 V/m

Solution

Given potential of the plates,

⇒ V₁ = −10 V, V₂ = 30 V

⇒ Distance between the plates, d = 2 cm = 0.02 m

The electric field between the plates is

⇒ E = (V₂ − V₁) / d

⇒ E = (30 − (−10)) / 0.02

⇒ E = 2000 V/m
Question 7
4 / -1

A man wearing a bullet-proof vest stands on roller skates. The total mass is 80 kg. A bullet of mass 20 g is fired at 400 m/s. It is stopped by the vest and falls to the ground. Then what is the velocity of the man?

A
1 m/s

B
0.1 m/s

C
0.01 m/s

D
0 m/s

Solution

Given,

The mass of man, M = 80 kg

The mass of the bullet, m = 20g

The velocity of the bullet, v = 400 m/s

From the conservation of momentum, Pi = P_f

Hence, the velocity of the man is 0.1 m/s.
Question 8
4 / -1

A bar magnet is dropped and it passes through a copper ring as shown in the diagram. The acceleration of the falling magnet while passing through the ring is

A
equal to acceleration due to gravity

B
more than the acceleration due to gravity

C
less than the acceleration due to gravity

D
zero

Solution

As the magnet falls the flux through the ring changes.

Regardless of whether the flux increases or decreases, by Lenz's law a current is induced in the ring that opposes the change in flux.

This induced current induced a magnetic field that interacts with the falling magnet and decreases the acceleration of fall below g.

Hence, the acceleration of the falling magnet while passing through the ring is less than the acceleration due to gravity.
Question 9
4 / -1

Two narrow parallel slits illuminated by a coherent monochromatic light produces an interference pattern on a screen placed at a distance D from the slits. The separation between the dark lines of the interference pattern can be increased by

A
decreasing the distance between the screen and the slits

B
increasing the distance between the slits

C
using monochromatic light of a longer wavelength

D
using monochromatic light of higher frequency

Solution

The position of the dark lines (or fringes) on the screen is given by the condition for destructive interference:

y = (m + 1/2) λ D / d

y is the position of the dark fringe on the screen, m is the order of the dark fringe (0, 1, 2,...), λ is the wavelength of the light, D is the distance between the slits and the screen, and d is the separation between the slits.

The separation between the adjacent dark fringes (Δy) is given by:

Δy = λ D / d

Now, considering the given options:

Option 1 (Decreasing the distance between the screen and the slits) would decrease Δy, so this option is incorrect.

Option 2 (Increasing the distance between the slits) would decrease Δy, so this option is incorrect.

Option 3 (Using monochromatic light of a longer wavelength) will increase Δy, as the wavelength λ is directly proportional to the separation between the dark lines. This option is correct.

Option 4 (Using monochromatic light of higher frequency) would decrease Δy, because frequency and wavelength are inversely related, so this option is incorrect.

∴ The correct answer is Option 3) Using monochromatic light of a longer wavelength.
Question 10
4 / -1

A full wave rectifier circuit with diodes (D₁) and (D₂) is shown in the figure. If input supply voltage Vin = 220sin (100πt) volt, then at t = 15 m sec

A
D₁ is forward biased, D₂ is reverse biased

B
D₁ is reverse biased, D₂ is forward biased

C
D₁ and D₂ both are forward biased

D
D₁ and D₂ both are reverse biased

Solution

Correct option is: (2) D₁ is reverse biased, D₂ is forward biased

V_in = 220 × sin(100πt) volt

Given: t = 15 ms = 0.015 s

ω = 100π

2π / T = 100π

⇒ T = 1 / 50 s

⇒ T = 0.02 s

Now, t = (3T / 4)

i.e., the signal is in the negative half cycle.

So now the negative half cycle is fed to the circuit, making D₁ reverse biased and D₂ forward biased.