Physics Test 207

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Physics Test 207

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Question 1
4 / -1

In an LCR circuit as shown below, both switches are open initially. Later, switch S₁ is closed and S₂ is kept open. (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statements is correct?

A
At t = τ/2, q = CV(1 - e^-1)

B
Work done by the battery is half of the energy dissipated in the resistor

C
At t = τ, q = CV/2

D
At t = 2τ, q = CV(1 - e^-2)

Solution

Switch S₁ is closed and switch S₂ is kept open. Now, capacitor is charging through a resistor R.

Charge on a capacitor at any time t,

q = q₀(1 - e^-t/τ)

q = CV(1 - e^-t/τ) [As q₀ = CV]

At t = τ/2, q = CV(1 - e^-τ/2τ) = CV(1 - e^-1/2)

At t = τ, q = CV(1 - e^-τ/τ) = CV(1 - e^-1)

At t = 2τ, q = CV(1 - e^-2τ/τ) = CV(1 - e^-2)
Question 2
4 / -1

In an ac generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is

A
NABω

B
NABRω

C
NAB

D
NABR

Solution

In an ac generator, emf induced, e = NABω sinωt

Hence, amplitude of induced emf or maximum emf = NABω
Question 3
4 / -1

A rectangular loop has a sliding connector PQ of length ℓ and resistance RΩ and is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I₁, I₂ and I are

A

B

C

D

Solution

Emf induced across PQ is ε = Bℓv. The equivalent circuit diagram is as shown in the figure.

Applying Kirchhoff's first law at junction Q, we get I = I₁ + I₂ ... (i)

Applying Kirchhoff's second law for the closed loop PLMQP, we get

-I₁R - IR + ε = 0

I₁R + IR = Bv ... (ii)

Again, applying Kirchhoff's second law for the closed loop PONQP, we get

-I₂R - IR + ε = 0

I₂R + IR = Bℓv ... (iii)

Adding equations (ii) and (iii), we get

2IR + I₁R + I₂R = 2Bℓv

2IR + R(I₁ + I₂) = 2Bℓv

2IR + IR = 2Bℓv (Using (i))

3IR = 2Bℓv

Substituting this value of I in equation (ii), we get I₁ = Bℓv/3R

Substituting the value of I in equation (iii), we get I₂ = Bℓv/3R
Question 4
4 / -1

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if Q is equal to

A

B

C
-2q

D
+q

Solution

Since the hypotenuse of the triangle is √2a, the net electrostatic energy is
Question 5
4 / -1

A block is kept on a frictionless inclined surface with angle of inclination α. The incline is given an acceleration 'a' to keep the block stationary. Then, 'a' is equal to

A
g tan α

B
g

C
g/tan α

D
g cosec α

Solution

The incline is given an acceleration a.

Acceleration of the block is to the right.

Pseudo acceleration 'a' acts on block to the left.

Equate resolved parts of 'a' and 'g' along the incline.

ma cos α = mg sin α

or, a = g tan α
Question 6
4 / -1

The diameter of a drop of liquid fuel changes with time due to combustion according to the following relationship.

While burning, the drop falls at its terminal velocity under Stokes flow regime. Find the distance that it will travel before complete combustion.

A

B

C

D

Solution

Given, diameter of the liquid drop,

At complete combustion

D = 0

∴ t = t_b

And at t = 0, D = D₀

Thus, the initial diameter of the drop is D₀ and changes according to given equation and the time for compelete combustion is t_b.

From Stokes law, the velocity of drop,
Question 7
4 / -1

The half-life period of a radioactive element X is the same as the mean life time of another radioactive element Y. Initially, they have the same number of atoms. Then,

A
X and Y will always decay at the same rate

B
Y will decay faster than X

C
X will decay faster than Y

D
X and Y will have the same decay rate initially

Solution
Question 8
4 / -1

An object is 1 metre in front of the curved surface of a plano-convex lens whose flat surface is silvered. A real image is formed 120 cm in front of the lens. What is the focal length of the lens?

A
108.1 cm

B
109.1 cm

C
110.1 cm

D
111.1 cm

Solution
Question 9
4 / -1

Consider a two particle system with particles having masses m₁ and m₂. The first particle is pushed towards the centre of mass through a distance d. By what distance should the second particle be moved, so as to keep the centre of mass at the same position?

A
d

B

C

D

Solution

Let m₂ be moved by x, so as to keep the centre of mass at the same position.

∴ m₁d + m₂ (- x) = 0

or, m₁d = m₂x
Question 10
4 / -1

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure P₁ and temperature T₁. The other chamber has volume V₂ and contains ideal gas at pressure P₂ and temperature T₂. If the partition is removed without doing any work on the gas, then the final equilibrium temperature of the gas in the container will be

A

B

C

D

Solution

As this is a simple mixing of gas, PV = nRT for adiabatic as well as isothermal changes.

The total number of molecules is conserved.