BITSAT Physics Test - 6 - JEE MCQ

We require a voltmeter, an ammeter and resistance wires to verify Ohm's law. A voltmeter can be formed by connecting a high resistance R_L in series with the galvanometer. An ammeter can be formed by connecting a small resistance R_S in parallel with a galvanometer. The configuration of ammeter and voltmeter should be connected with the test resistance in series and parallel connections, respectively. The battery should be connected in series with the test resistance. Therefore, it is the correct option.

Hence θ = 30°
Hence, this is the required solution.

For a moving coil galvanometer, 
C = Restoring force
B = Magnetic field strength
N = Number of turns in coil
A = Area of the coilFor increasing the sensitivity of the galvanometer, θ should be increased. Therefore,N can be increased.A can be increased.B can be increased and can be achieved by using a strong magnet or a magnetised/iron core. Therefore, it is the correct option.

Let the n^th minima of 300 nm wavelength coincide with the m^th minima of 540 nm wavelength light.
Therefore,

Therefore, 5th minima of 540 nm wavelength coincides with 3rd minima of 300 nm light.
Location of the minima for 300 nm light,

Next dark fringe of 300 nm wavelength will coincide at n = 8.

Distance between two minima = 15 mm
Therefore,

This is the required solution.

Wavelength of the emitted radiation is given by:

From 3rd to 2nd

From 4th to 2nd

Hence, this is the required solution.

In the circuit given, symmetry exists in the upper and lower arms. The simplified circuit can be drawn as:

Therefore,


This is the required solution.

For a loop carrying current, the strength of the magnetic field is given by:


In the figure, the current I is divided into two arms. Further,

Here, R = Resistance of arc L = Length of arc θ = Angle substandard by arc at the centre Therefore, Iθ = constant Thus, the magnetic field at the centre of the loop will be opposite and equal to each other due to the arcs PAQ and PBQ.Hence, magnetic field strength at the centre of the loop will be zero.Thus, it is the correct option.

Magnetic field at the centre of the loop of radius 50 cm will be given by:
 where R is the radius of the bigger loop.
Thus, B = 9.65 × 10^-7 T
Also, magnetic flux linked with the smaller loop, φ = BS
φ = 9.65 × 10^-7 × 3.14(0.5)² = 7.57 × 10^-7 Wb
Or φ = 75.7 μWb
Hence, this is the required solution.

For mirror

For convex lens

Magnification M,

Current through the circuit = Power/voltage

Potential difference across R

From conservation of energy
undefined m_nc² = Kn+Ke
The required neutrino energy will be minimum. When the neutron and position are both emitted with zero kinetic energy.
(E_v)_min + 938.2 MeV = 939.5 MeV + 0.5 MeV 
(E_v)_min = 1.8 MeV

In a nuclear fission reaction, high energy neutrons are emitted. These high energy neutrons are slowed down by moderators. As a result of this, neutrons are in thermal equilibrium with surrounding molecules of moderator. These neutrons are called thermal neutrons.

The energies of the orbital in any species with only one electron can be calculated by Bohr's equation.

where,
R = Rydberg constant
h = Plank constant
c = Speed of light
Z = Nuclear charge
n = Bohr orbit
In general, both energy and radius decrease as the nuclear charge increases as
E∝ Z²  [From energy equation]
So, The energy of an orbital in an atom does not remain the same with an increase in the positive charge in its nucleus.

(i) When ₉₂U²³⁵ undergoes fission, 0.1 % of its original mass is changed into energy.
(ii) Most of energy released appears in the form of kinetic energy of fission fragments.
(iii) The energy released in U²³⁵ fission in about 200 MeV.
(iv) By fission of ₉₂U²³⁵, on the average 2.5 neutrons are liberated.

Given threshold frequency 
Work function of given metal surface 

Energy of photon E(= hv) = 8 eV
From equation E = KE_max + ϕ₀  
⇒ KE_max = E− ϕ₀ = 8 − 6 = 2 eV

As λ is increased, there will be a value of λ above which photo-electron will be cease to come out, so photo-current will becomes zero.

∵ 2 molecules of deuterium are fused, then released energy = Q
Hence, energy released per molecule = Q/2
Now, we know that number of molecules in one mole
= 6.02 × 10²³
Hence number of molecules in two moles = 2 × 6.02 × 10²³
Hence, energy released when two mole of deuterium are fused

In α−decay,
here, Element changes.

here, Element changes.

here, Element changes.
In γ−decay,
This decay is a radioactive process in which the excited nucleus comes down to its ground energy level by emitting photons. The element does not change in this process.

The magnetic field at a centre of H-atom due to moving electron is

∴ 


and 
On putting the values in Eqn .(i), we get

Given, mass of the ball B B (m) = 2.5 kg
mass of rod(m_rod) = 1 kg
spring coefficient(k) = 18 kN m⁻¹ 
Let x₀ be the extention in the spring in equilibrium condition.

Balancing torque in this condition about point A‐

After slightly displacing the spring by x distance, again writing torque equation about point A‐

Moment of inertia of the system about point A-

Putting value of I_A and mgl in equation (ii)‐

Comparing it with standard equation of angular SHM α = −ω²θ

The time period of the pendulum


Initially the centre of mass of the sphere is at the centre of the sphere. As the water slowly flows out of the hole at the bottom, the CM of the liquid (hollow sphere) first goes downward and then again upwards. Hence, the effective length of the pendulum first increases and then decreases.

Given, U = U₀(1 – cos αx)   ...(i)
Potential field is defined for a conservative field.
In a conservative potential field, foece 
From Eq. (i), we get

= U₀[+ α sin αx]
= U₀α sin αx
∴  
For small value of α x,
sin αx ≈ α x
∴  F = – U₀ α (αx)
= –U₀α²x                ...(ii)
Clearly F∝ – x
As, U₀, α being constant
∴ Motion is SHM
F = – mω²x                   ..(iii)
Comparing Eqs. (ii) and (iii), we get
mω² = U₀α² 

The periodic time of a simple pendulum is given by, 

When taken to height 2R

k₁ and k₂ are parallel and k₃ and k₄ are parallel. The two combinations are in series with each other

Potential energy is given by,

Time period of oscillation,

Therefore,  Loss in time = 4.5 x 10^-6 x 0.5
= 2.25 x 10^-6 s

For springs having spring constants k₁ and k₂ connected in parallel, their equivalent spring constant K_parallel is given by K_parallel = k₁ + k₂ 
And, for two springs having spring constants k₁ and k₂ connected in series, their equivalent spring constant K_series is given as

Now in the given problem, two springs each of spring constant k₁ are in parallel and therfore their equivalent spring constant is K_parallel = k₁ + k₁ = 2k₁  and this is in series with spring of constant k₂, so equivalent spring constant,

The time period of a pendulum is given by the formula 
Where 'T' is the time period, 'l' is the length of the pendulum, 'g' is the acceleration due to gravity.
Since the lift is moving with a constant velocity downwards, there will be no change in the acceleration of the pendulum. Hence time period will remain same. 

Here, m₁ is removed leaving only m₂.  
Therefore, angular frequency is 
Let x₁ be the extension when only m₂ is left. Then, kx₁ = m₂g 

Similarly, let x₂ be the extension in equilibrium when both m₁ and m₂ are suspended. Then,

From equation (i) and (ii) amplitude of oscillation