BITSAT Practice Test - 10 - JEE MCQ

Given angular momentum about O and P are I_O and I_P, respectively.
Angular momentum about P: 



Also, angular moment about O,

Thus, the direction of  is constant.
But  will continuously change its direction.
Thus,  ​​will change its direction at every instant and thus varies.
But  = constant.

Given,
velocity of the particle is 
time instance is 1 sec.
The acceleration of the particle is the rate of change of velocity with respect to time thus, acceleration is given by,

Acceleration of the particle remains constant at all times.

Let us suppose that original length simple pendulum is L and time period is now if the acceleration due to gravity is g.
Now the frequency should be 
Given the ratio of frequencies,it means,

Efficiency of the carnot engine,

where,

η = efficiency

T_f = temperature of sink

T_i = temperature of source

Temperature should be in Kelvin.

For 50% efficiency,

For 60% efficiency,

If a particle of the mass M, revolving in a circular path of the radius R, with the speed v, then the centripetal force F is,  acting towards the centre along the radius. In the following diagram, according to the question,

 Centripetal forces are the same, so,  
By equation (1) and (2), 

As the block is released from rest, it means the initial velocity, (u) = 0

and let block cover a distance d at a time (t₁),

a = gsinθ.

We know,

Substituting values, we get,

When the block slides from a rough surface, then,

The coefficient of kinetic friction is applicable here as the block moves over the inclined plane.

The coil will orient in a position of torque equilibrium. 
As the torque on the loop is given by  so torque will be zero when area vector  is parallel to magnetic field i.e. when the plane of the coil is perpendicular to the magnetic field. This is so because area vector is perpendicular to the plane of the loop.

Since, from Rutherford - Soddy law, the number of atoms left after half-lives is given by

Where, N₀ is the original number of atoms.
The number of half-lives, 

Relation between effective disintegration constant (λ) and half life (T)


Effective half life,

Here, the potential gradient for the potentiometer will be, 
Now, the potential difference across the given resistance will be, V = IR = 0.2 A × 3.5 Ω = 0.7
If the balancing length for the given resistance is l, then, 
V = xl
⇒ 0.7 = 0.2 × l

Particle taken t₁ time to reach from 0 to 3h and also t₁ to reach from 3h to 4h and yet again t₁ time from 4h to 3h.
∴ v_x3t₁ − 10 × 3t₁ = v_xt₁
⇒ v_x = 15 m s⁻¹

Let T = temperature of the junction J
L = length and A = area of each rod
Heat conducted in first rod = heat conducted by second rod + heat conducted by third rod

At the time of collision net force on the system along horizontal is zero, so, from the conservation of linear momentum, we have 
mv + 0 = (m + m)V, let V be the common velocity of both blocks.

At the amplitude means maximum displacement. Total kinetic energy at mean position is converted to total potential energy at the extreme position.

A full-wave rectifier is defined as a type of rectifier that converts both halves of each cycle of an alternating wave (AC signal) into a pulsating DC signal.

Therefore, the fundamental frequency of output will be doubled the fundamental frequency of input.
f_o = 2f_in ⇒ f_o = 2 × 60 = 120 Hz

Equivalent resistance of the circuit is given as R_eq = 10 + 20 = 30 Ω. 
Current flowing in the circuit is given as 
Voltage drop across the potentiometer wire is given as V = ε − IR, where ε is the emf of battery.

Potential gradient of potentiometer wire is given as 

As the two capacitors of equal capacitance 10 μF in the upper branch are in series, so their equivalent capacitance is 5 μF. So, given circuit can be redrawn as

Now as the 5 μF and 10 μF capacitors are in parallel, so their equivalent capacitance is 15 μF.
The capacitors 5 μF and 15 μF are in series, so potential across 5 μF capacitor is given by

Let velocity at the end of 50 m be u m/s.
u² = 2gh = 2 x 9.8 x 50 = 980 units
Let further displacement be s.


 Total height = h + s = 50 + 243 = 293m

When the masses are accelerating, there is a tension in the string. When a mass m is added to m₁ such that the acceleration is zero, the system of masses (m₁ + m) will slide on the surface with a uniform speed and then, there is no tension in the string. This will happen if the downward force m₂g equals the force of friction μ(m₁ + m)g on blocks m₁ and m.
 11 kg
Hence, the correct choice is (c).

Scale reading is due to the force acting on its pan. This force is due to the rate of change of momentum given to it because of colliding particles.
Velocity with which the particle collides with the pan, v =  =  = _^2.8√5 ms^-1
= 12.52 x 10^-3 kg ms^-1
 Momentum imparted to pan per second = 12.52 x 10^-3 x 100
= 1.252 kg = 1252 g (weight)

dv/dx = x³ - x = 0
⇒ x(x² - 1) = 0
⇒ x = 0, ±1
 = (3x² - 1) = Positive for x = ±1
i.e. V is minimum at x = ±1
⇒ V_max = - J
⇒ K_max = 2 -  =  J
 × 1 × v² = 
⇒ v = 3/√2 m/sec

Let the rocket be at a distance x from the moon when the gravitational force on it is zero.
Its distance from earth = r - x. Gravitational force on the rocket due to earth is

where m is the mass of the rocket, Gravitational force on the rocket due to moon is

Since the two forces are in opposite directions, the net force on the rocket will be zero if F_e = F_m.
Equating the two we get

which gives x = r/10.
Hence the correct choice is (c).

If the length of capillary tube is l, the pressure difference is given by

where η is the coefficient of viscosity of water. If r becomes 2r and Q becomes Q/2, we have

Hence the correct choice is (a).

The speed of the wave is


v = λf
Frequency of the wave does not change, so

Electric field E = -
Force on charge (-q) = -qE = +8qx
At x = 0.5 m, force = 8 x 2 x 10^-6 x 0.5 = 8 x 10^-6 N
Hence, the correct option is (d).

Charge will flow from A to B until their potentials become equal. If charge q flows from A to B, then 
or, Q - q = a/b, which gives q = 
Hence, charge left on A = Q - q = Q - 
Hence, the correct choice is (d).

The magnetic field at P(a, 0, a) due to the loop is equal to the vector sum of the magnetic fields produced by loops ABCDA and AFEBA as shown in the figure.

Magnetic field due to loop ABCDA will be along  and due to loop AFEBA, along .
Magnitude of magnetic field due to both the loops will be equal.
Therefore, direction of resultant magnetic field at P will be 

Let P be the position of the mark. Q is the position of its image.
Since the incident ray PA lies in a medium of refractive index μ₂, and is refracted into a medium of refractive index μ₁, our formula becomes


where u = -2 R = -10 cm,
R = -5 cm, μ₂ = 1.5 and μ₁ = 1
Putting these values in the above formula, we have

which gives v = -20 cm
∴ Distance of image Q from O = 20 - 5 = 15 cm

so a= 1 , b=1 so a+b =2