BITSAT Practice Test - 8 - JEE MCQ

The mass of the considered element is, 

The electrons in the outermost shell of an atom determines the electrical and chemical properties of element. They take part in the formation of chemical bonds, form free charge carriers in semiconductors, show photoelectric effect etc. The electrons in outermost shell are called valence electrons.

By dividing both the equations, we get,

9m + 9 = 25

16m = 9

m = 9/16.

The minimum value of magnetic field

Here r = 90°.

From Snell's law,

From (i) and (ii),

From (i) and (iii),

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



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

P₁ = P₂ = P
m₁v₁ = m₂v₂

Area of rectangle = 6.01 × 12 = 72.12 m²

​Result should have only two significant numbers (same as in 12 m). So, correct area will be 72 m².

Points B and F are in the same phase as the distance between them is equal to the wavelength. The phase of a periodic function of some real variable, is the angle illustrating the number of periods spanned by that variable. It is denoted and expressed in such a scale that it varies by one full turn as the variable goes through each period.

Speed in horizontal direction remains constant during whole journey because there is no acceleration in this direction.
Loss of gravitation potential energy = gain in kinetic energy


Hence, the speed with which he touches the cliff B is

From the concept of image formation by plane mirror, virtual image will be formed at 10 cm10 cm behind the mirror. 
As shown in figure.

​​​​​​Here we can see image acts as object for eye and thus object distance is, u = 40 cm
So eye must have to focus on image of the object, i.e at 40 cm

Case is shown in figure, 

Now we can see, 
= QE = mg

Where density of material is ρ and accleration due to gravity is g
Similarly for other case, 

As light moves from air to water, its speed decreases due to refractive index of water.

Fringe width, as λ decreases on entering in water, β also decreases.

λ = wavelength in medium,

D = separation between slit and screen,

d = separation between slits.

We know that, p = 
% error in pressure = (% error in F) + 2(error in L)
= (4%) + 2(3%)
= 10%

The horizontally acting tension, T = Mg
The downward vertical force = (M + m)g
So, resultant force, which is choice (d).

As the collision is completely inelastic, so change in momentum is ρAv².
Resolving it into components as shown, we have

Velocity of a particle will be maximum when its kinetic energy is maximum and corresponding to this, potential energy will be minimum.
 J

For minimum value,


Hence, minimum potential energy of the particle is
 J


Hence, maximum velocity of the particle
 m/s

Let ω be the angular velocity acuired by the system (rod + bullet) immediately after the collision.
Since no external torque acts, the angular momentum of the system is conserved. Thus
mvL = Iω ....(1)
where I is the moment of inertia of the system about an axis passing through O and perpendicular to the rod. Thus
I = M.I. of rod about O + M.I. of bullet stuck at its lower end about O
= 1/3 ML² + mL² = 1/3 (M + 3m)L² ...(2)
Using Eq. (1) in Eq. (2), we have
mvL = 1/3 (M + 3m)L²ω
or ω = 
Hence the correct choice is (c).

Let A and B be the two extreme positions of the particle with O as the mean position. Displacements to the right of O are taken as positive while those to the left of O are taken as negative

Let the displacement of the particle is SHM be given by

Let us suppose that at time t = 0, the particle is at extreme position B. Setting x = A at t = 0 in Eq. (i)
we have

Now let us say that the particle reaches point C at t = t₁ and point D at t = t₂. At C, the displacement x(t₁) = + 12.5 cm and at D, it is x(t₂) = -12.5 cm (see Fig.).
So from (ii) we have

Notice that cos ωt₂ = -0.5 even for t₂ = 4π/3 .
This value of t₂ does not correspond to the minimum time because this is the time at which the particle, moving to left, reaches A and then returns to D.

When the temperature changes from 30^oC to 10^oC, then change in length of the wire is 
= 
Now, Y =  or F = , where a is the area of cross-section of the wire
 F or T = 
= 4.76  10⁷ aN
Now, speed of transverse wave, v = 
But, 
 v = 
= 72 ms^-1

Longest diagonal of a regular hexagon is 2a.
Distance of the point of intersection of diagonals from each of its vertices = Side of the hexagon = a.
The potential at this point due to each charge = 
Therefore, total potential = , which is choice (c).

Let I₁ and I₂ be the currents in branches BAD and DCB, respectively as shown in the figure. Let I₃ be the current along DB through the 2 Ω resistor.

Applying Kirchhoff's first law at function D, we have
I₁ = I₂ + I₃ or I₃ = I₁ - I₂
Applying Kirchhoff's second law to loops BADB and DCBD, we have
2I₁ + I₁ + 2(I₁ - I₂) = 2 - 1 = 1
or, 5I₁ - 2I₂ = 1 (i)
and 3I₂ + I₂ - 2(I₁ - I₂) = 3 - 1 = 2
or, 6I₂ - 2I₁ = 2 (ii)
From Eqs. (i) and (ii), we get 
 
The negative sign shows that the current through the 2 Ω resistor flows along BD and not along DB as assumed. So, the correct choice is (d).

Let θ be the angular deflection (in radians) when a current i is passed through the coil.
Then, restoring torque = Cθ
When the coil is in equilibrium, Deflecting torque = Restoring torque, i.e. iNAB = Cθ
 Current sensitivity is 
Hence, the correct choice is (a).

Let the distance from R to MN be x. Then the area of the loop between MN and R is xd and the magnetic flux linked with the loop is Bxd.
As the rod moves, the emf induced in the loop is given by
|e| = d/dt (Bxd) = Bd dx/dt = Bvd
where v = velocity of MN.

Given u = -2 km = -2000 m and f₀ = 1.2 m.
The lens formula is


Since f₀ << u, f₀ + u ≈ u. Hence
v = f₀ = 1.2 m
Therefore, magnification produced by the objective is

∴ Height of image = I × m₀ = 10 m × - 
= -0.006 m = -6 mm
The negative sign indicates the image is inverted.
Hence the correct choice is (c).

According to Bohr's postulate, 
Putting 
Moment of inertia, I = 
_{KE rotational = 
= 
⁼}