BITSAT Physics Test - 9 - JEE MCQ

Heat generated in the resistance:
H = I²RT = mSΔT
If d = density of the wire and
ρ = specific resistance of wire
Thus,

This is our required solution.

The torque on the loop carrying current , placed into a magnetic field B is given by:

where  is area and  is magnetic field
Therefore,
τ ∝ A
For the same perimeter, the area of the circle is greater.
Therefore, the torque acting on the circle will be highest.
Therefore, it is the correct option.

Since the transformer is an AC device, it will not work with the DC circuit.
Therefore, the induced EMF will be zero.
Therefore, it is the correct option.

Kinetic energy of emitted electron is given as:

When the energy of the photon is doubled, then

This is our required solution.

Reduce the network in its equivalent form as follows:

 as wheat stone  is balance here as 4/12 =8/24 = 1/3

Therefore, the equivalent resistance of the circuit will be = 9 + 6 = 15 Ω
V = 5 V
Therefore,

i = 0.33 A
This is our required solution.

We know that:

The voltage across inductor is given by

This is our required solution.

Mirror formula is given by:

Differentiating w.r.t. time we get:

Therefore,

This is our required solution.

For no dispersion:

The negative shows that the nature of the dispersive power should be opposite in nature.
Therefore, the ratio of the dispersive powers of the material of the lenses will be 3 : 5.
Therefore, it is the correct option.

The jumping of an electron from a higher state to a lower state is incorporated by the emission of photons.
By the conservation of momentum,
Momentum of electron = Momentum of the photon emitted

Therefore,

This is our required solution.

Since, when A and B both are logic 1 then no current will flow through the both 50 Ω.
resistors, because both the diodes will not conduct, therefore, whole current will flow through R = 10 K
and output is potential difference across the R = 10 K which will be 6 V in this case means high
Now, When either of the input is low current through terminal which is low will flow through it and most of the current will flow through the 50 Ω resistance since it is very very less than
R = 10 K therefore, Potential across R = 10 K will be less than 1 V. So, when either of input is low output is also low.
When both the inputs are low same case as when either of input is low will be there hence output will also be low.
Therefore, above circuit acts as a AND Logic gate.

As we know, energy of photo electrons,

Capacity of the spherical capacitor having inner radius 'a' and outer radius 'b'.

Now from dimension

The work done does not characterize a thermodynamic state of matter.
Work done is elaborated in such a way that it includes both force exerted on the body and the total displacement of the body. The purpose of this force is to move the body a certain distance in a straight path in the direction of the force.

If w is the width of the road and r is the radius of curvature of the road, maximum velocity of the vehicle on the curved road is:

Here h is the elevation of the outer edge of the road with respect to the inner edge.

The number of magnetic lines of force passing through a given area is known as magnetic flux.
It provides the measurement of the total magnetic field that passes through a given surface area. Here, the area under consideration can be of any size and under any orientation with respect to the direction of the magnetic field.

Magnetic flux symbol: ϕ or ϕ_B  
Magnetic flux formula is given by:
ϕV = B.A = B A cos θ
Where,

• ϕ_B is the magnetic flux 
• B is the magnetic field
• A is the area
• θ the angle at which the field lines pass through the given surface area

Let the two lenses have focal lengths f1 and f2, in meters.

For a separation of  d = 0.25 m

The powers of the lenses in diopters are 2 and 8.

Given:

Input Power = Output Power

As we know that the stopping potential of the photoelectron is equal to the maximum kinetic energy of the photoelectron,
KE_max = 10.4V
Now, in photoelectric effect,
Energy of incident radiation (E_in) = work function + K.E_max 

Now, for o hydrogen atom,
Energy of first energy level, E₁ = -13.6eV
Energy of second energy level , E₂ = -3.6eV
Energy of third energy level, E₃ = -1.5eV
So, a transition from third to first energy level will result in emission of radiation of energy = E₃ - E₁ = 12.1eV which is same as the energy of incident radiation of above photoelectric effect. 
Thus, correct answer is n = 3 to 1.

Before hitting the ground, the velocity v is given by v² − 2gd (quadratic equation and hence parabolic path)
Downwards direction means negative velocity. After collision, the direction become positive and velocity decreases.

As the direction is reversed and speed is decreased, graph (a) represents these conditions correctly.

Given:

For v to be minimum or maximum:

 

(Since, current is increased at same rate)

Given:

now we have to find the error in

The angular momentum is

From the formula for the speed of sound in air

Squaring both sides:

As the frequency of the incident radiation increases, the potential required to stop the ejected electron (Stopping Potential) increases. 
So, v₃ > v₂ > v₁.
The stopping potential varies linearly with the frequency of the incident radiation.

Let  be the spring constant of spring and it gets compressed by length x in equilibrium position. Let m be the mass of the block and F be the upward thrust of water on block. When the block is at rest,

When the vessel moves downwards with acceleration a(<g) the effective downward acceleration = g - a. Now upthrust is reduced say it becomes F"

In figure, then

So, the spring length will increase.

Given:
AB is an isothermal
BC is an isochoric 
AC is an adiabatic
Therefore the P-V diagram will be 
for A to B- PV = constant
for B to C- V = constant
for A to C- PV^r = constant
As slope of adiabatic AC is more than the slope of isothermal AB, and BC is isochoric (i.e., at constant volume)

For E = constant,