Physics Test 239

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Physics Test 239

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Question 1
4 / -1

Directions For Questions

If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then

A. the charge stored in it, increases.

B. the energy stored in it, decreases.

C. its capacitance increases.

D. the ratio of charge to its potential remains the same.

E. the product of charge and voltage increases.

...view full instructions

Choose the most appropriate answer from the options given below:

A
A, B and E only

B
A, C and E only

C
B, D and E only

D
A, B and C only

Solution

Explanation:

A. The charge stored in it increases: This is correct because the capacitance increases and Q = CV with V constant.

B. The energy stored in it decreases: This is incorrect; the energy stored increases.

C. Its capacitance increases: This is correct because capacitance C = ε0 A / d increases as d decreases.

D. The ratio of charge to its potential remains the same: This is incorrect; the ratio Q / V is the same as capacitance, which increases.

E. The product of charge and voltage increases: This is correct as it equals the energy stored, which increases.

∴ The correct option is 2) A, C and E only
Question 2
4 / -1

In the following circuit, the equivalent capacitance between terminal A and terminal B is

A
2 μF

B
1 μF

C
0.5 μF

D
4 μF

Solution

Formulas Used:

For capacitors in parallel:

Cparallel = C1 + C2 + C3 + …

Capacitance (C): Ability of a system to store an electric charge. SI unit: Farad (F), Dimensional Formula: [M-1L-2T4A2]

For capacitors in series:

1/Cseries = 1/C1 + 1/C2 + 1/C3 + …

Calculation

Given data: C1 = C2 = C3 = C4 = 2 μF

Check if the bridge is balanced

⇒ C1/C2 = 2 μF / 2 μF = 1

⇒ C3/C4 = 2 μF / 2 μF = 1

Calculate the equivalent capacitance of each series pair

⇒ 1/Cseries = 1/(2 μF) + 1/(2 μF)

⇒ Cseries = 1 / (2 / 2 μF)

⇒ Cseries = 1 μF

Combine the series pairs in parallel

⇒ Ctotal = Cseries + Cseries

⇒ Ctotal = 1 μF + 1 μF

⇒ Ctotal = 2 μF

∴ The equivalent capacitance between terminals A and B is 2 μF.
Question 3
4 / -1

The metre bridge shown is in balance position with If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, what will be balance condition ?

A

B

C

D
no, no null point
Solution
Concept:
- Meter bridge is an instrument that is used to find the unknown resistance of a coil.
- It works on the principle of Wheatstone bridge which works on null deflection, i.e. the ratio of their resistances are equal.
- The bridge is said to be balanced when no current flows through the galvanometer.
- When there is no current flowing through the bridge the potential difference between the points D and B is zero.
Calculation:

We know that Resistance is directly proportional to the length of the wire.

Let the resistances of the wire of length l₁ be R and length l2 be S then the figure in the question can be simplified as shown below like a Wheatstone network where galvanometer is connected between points B & D while the battery is connected between points A & C.

The bridge is under balanced condition

If we now change the position of Galvanometer and the battery the above circuit appears as shown in the below figure where galvanometer is connected between points A & C while the battery is connected between points B & D.

From the above diagram also we can say that even after interchanging the positions of battery and Galvanometer there is no change in the balancing condition and the bridge is balanced i.e.

Hence, the correct option is (2)
Question 4
4 / -1

The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is: (taking h = 6 × 10⁻³⁴Js)

A
100

B
3 × 10²⁰

C
3 × 10¹⁸

D
1000

Solution
Question 5
4 / -1

Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given mass of hydrogen m_h = 1.67 × 10^–27 kg]

A
10^–47 C

B
10^–20 C

C
10^–23 C

D
10^–37 C

Solution

Concept:

Force between electric charges at rest is called an electrostatic force. It is an attractive and repulsive force between particles which depends upon their electric charges. It is also called as Coulomb’s force.

Examples of electrostatic force are attraction of paper to a charged scale by rubbing to hair, lighting etc.

Coulomb's Law:

According to coulomb's law, electrostatic force between two charges depends upon following factors:

Force is directly proportional to magnitude of the charges and inversely proportional to square of distance between them.

Let q1 and q2 are two charges separated by a distance r, then

Nature of force depends upon the charges. Like charges repel and unlike charges attract each other it mean force of repulsion acts between two positive or two negative charges whereas if one charge is positive and other is negative then force is attractive.

Gravitational Force:

Gravitational Force is the force with which each body in this universe attracts other bodies towards itself.

According to Newton's law of Gravitation,

Force is directly proportional to product of their masses and inversely proportional to square of distance between them.

Let m1 and m2 are masses of two bodies separated by a distance r, then gravitational force between them is FG

where G is constant of proportionality and known as gravitational constant. Its value is 6.67 × 10-11 N m2 kg-2 and remains constant in the whole universe.
Question 6
4 / -1

Two different coils of self inductance L₁ and L₂ are placed close to each other so that the effective flux in one coil is completely linked with other. If M is the mutual inductance between them, then

A

B

C

D

Solution
Question 7
4 / -1

An alternating current I in an inductance coil varies with time t according to the graph as shown: Which one of the following graphs gives the variation of voltage with time?

A

B

C

D

Solution

Explanation:

Rate of change of current is constant for one period at a positive value and is constant at a negative value for the second time period.

Therefore emf is a constant positive value for the first half and a constant negative value for the second half.

∴ Option C is correct.
Question 8
4 / -1

A 800 turn coil of effective area 0.05 m² is kept perpendicular to a magnetic field 5 × 10⁻⁵ T. When the plane of the coil is rotated by 90° around any of its coplanar axis in 0.1 s, the emf induced in the coil will be :

A
2 V

B
0.2 V

C
2 × 10⁻³ V

D
0.02 V

Solution

CONCEPT:

Magnetic flux: The magnetic flux is defined as the number of magnetic fields which passes through the closed surface and it is written as;

Induced EMF - An emf is induced in the coil when the magnetic field is pushed in and out of the coil and it is written as;
Question 9
4 / -1

A body floats in a liquid contained in a beaker. The whole system as shown falls freely under gravity. The upthrust on the body due to the liquid is

A
Zero

B
Equal to the weight of the liquid displaced

C
Equal to the weight of the body in air

D
Equal to the weight of the immersed position of the body

Solution

Calculation:

The Upthrust force is equal to V_ρliquid(g − a)

where, a = downward acceleration,

V = volume of liquid displaced

But for free fall a = g

∴ Upthrust = 0
Question 10
4 / -1

Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to r^5/2, then the square of the time period will be proportional to.

A
r³

B
r²

C
r^2.5

D
r^3.5

Solution