Physics Test 292

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Physics Test 292

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Weekly Quiz Competition

Question 1
4 / -1

In Young's double-slit experiment, 12 fringes are obtained on a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed on the same segment of the screen is given by

A
18

B
24

C
30

D
36
Solution
Given:
- Number of fringes with wavelength 600 nm is 12.
- New wavelength is 400 nm.
We know that the number of fringes is inversely proportional to the wavelength of light used. Therefore, we can set up the following ratio:

Thus, the number of fringes observed on the same segment of the screen when the wavelength is changed to 400 nm is 18.

Therefore, the answer is 18.
Question 2
4 / -1

Samples A and B are initially kept in the same state. They expand through adiabatic and isothermal processes respectively to the same final volume. If P_A is final pressure of sample A and P_B that of sample B. Then

A
P_A > P_B

B
P_A = P_B

C
P_A < P_B

D
The relation between P_A and P_B cannot be deduced

Solution

Step 1: Both samples start from the same initial state (P₀, V₀, T₀) and expand to the same final volume V_f.
Question 3
4 / -1

In Young's double slit arrangement, water is filled in the space between screen and slits. Then

A
fringe pattern shifts upwards but fringe width remains unchanged.

B
fringe width decreases and central bright fringe shifts upwards.

C
fringe width increases and central bright fringe does not shift.

D
fringe width decreases and central bright fringe does not shift.

Solution

Step 4: The central bright fringe occurs where the optical-path difference is zero. Since both waves traverse the same extra optical path in water, the zero-path-difference point stays at the original centre. Hence central bright fringe does not shift.

Conclusion: fringe width decreases and central bright fringe does not shift.
Question 4
4 / -1

In an interference experiment, the ratio of amplitudes of coherent waves is The ratio of maximum and minimum intensities of fringes will be

A
2

B
18

C
4

D
9

Solution
Question 5
4 / -1

A uniform rod of length L, cross-sectional area A and Young’s modulus Y is suspended vertically from a rigid support. A collar of mass M is attached at the lower end. A block of mass m is dropped from height h above the collar and sticks to it. The maximum elongation produced in the rod is found to be ΔL. Neglecting energy loss, find the value of h in terms of the given parameters and ΔL.

A

B

C

D

Solution

Step 1: Treat the rod as a spring of stiffness

Thus option D is correct.
Question 6
4 / -1

According to Kepler's Second Law of Planetary Motion, which of the following statements is true regarding a planet's motion around the Sun?

A
The planet moves at a constant speed throughout its elliptical orbit.

B
The planet's speed is greatest when it is farthest from the Sun (at aphelion).

C
The line joining the planet and the Sun sweeps equal areas in equal intervals of time.

D
The square of the orbital period is directly proportional to the orbital radius, not area.

Solution

Step 1: Kepler's Second Law fundamentally describes the variation of orbital speed. It states the Law of Equal Areas: a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This physical law means the planet accelerates when closer to the Sun and decelerates when further away.

Step 2: Options (A) and (B) are incorrect because the speed changes and is greatest at perihelion (closest point), not aphelion or constant.

Step 3: Option (D) refers to Kepler's Third Law (Harmonic Law), which relates the period and average radius ( T²∝ r³ ).
Question 7
4 / -1

When a dipole is aligned parallel to a uniform electric field, which of the following is true?

A
Net force is zero and torque is maximum.

B
Net force is non-zero and torque is zero.

C
Both net force and torque are non-zero.

D
Both net force and torque are zero.

Solution

In a uniform electric field, the net force on a dipole is always zero because forces on +q and −q cancel.

When aligned parallel (θ = 0^∘), torque τ = pEsin⁡0^∘= 0.

Hence, both net force and torque are zero.
Question 8
4 / -1

In a parallel combination of capacitors, which of the following is the same for all capacitors?

A
Charge

B
Voltage

C
Capacitance

D
Energy stored

Solution

In parallel combination, the potential difference (voltage) across each capacitor is the same as the applied voltage. Charges differ based on capacitance.
Question 9
4 / -1

What is the critical angle for diamond with respect to air if its refractive index is 2.42?

A
24.41°

B
41.14°

C
37.31°

D
48.75°

Solution

Step 1: Apply Snell's Law at the critical angle.

At the critical angle, the angle of refraction in the rarer medium (air) is 90^∘.
Question 10
4 / -1

A
127.5 MeV

B
931.5 MeV

C
15.99493 MeV

D
16.12744 MeV

Solution