SRMJEEE Physics Mock Test - 2 - JEE MCQ

Given, Centripetal force = mv^2​/r = -k/r²​
Kinetic Energy =​ mv²/2 = k/2r
Potential Energy = −∫^r_∞ ​Fdr
the lower limit has been taken as ∞ because potential energy is zero at infinty

⇒ −∫^r_∞ (k/r²)​dr

⇒ −k ∫^r_∞ ​r⁻²dr

⇒ −k |r^-1/-1|^r_∞

⇒ −k/r​

Total energy = k/2r​ − k/r ​= −k/2r​

According to principle of continuity, for streamline flow of fluid through a tube of non-uniform cross-section the rate of flow of fluid (Q) is same at every point in the tube.
i.e, Av = constant 
⇒  A_1​v₁ ​= A₂​v₂​
Therefore, the rate of flow of fluid is same at M and N. 

The focal length of a combination of lenses, is given by, 
f = R​/2(μ−1)
Substituting, the values R = −10 cm and μ = 1.5 
We get, f = −10 cm
Here, -ve sign shows that it will behave like a concave mirror.

Comparing with a standard Equation of SHM ,
y = a sin(wt + φ)

ω = 4π, and θ = φ = initial phase.

So frequency is as follows :
freq. = 4π/2π
⇒ f = 2 Hz

So final answer is 2 Hz frequency.

We know,
Potential energy stored in the spring, 
Therefore,

This is the required solution.

When there is no damping , the damping constant b = 0 and at resonace, the amplitude of forced oscillations is infinite.

Torque = r x F

r = î - j

so torque = (i - j) x -Fk

= -F(-j - i)

= F( i + j)

Paramagnetic materials possess a permanent magnetic dipole moment due to incomplete cancellation of electron spin and orbital magnetic moment.

Energy of the body at point P
= mgh = 2 × 10 × 40 = 800 J
Energy of the particle at point Q

Work done against friction = Change in the mechanical energy of the body
= 800 - 500
= 300 J
This is the required solution.

5g − T₂ = 5a        ....(i)
T₂ − T₁ − 3g = 3a      ...(ii)
T₁ − g = a    ...(iii)
Adding Eqs (i) and (iii),
−T₂ + T₁ + 4g = 6a
Adding this to Eq. (ii), we get
g = 9a or a = g/9

Mass of first particle is m₁​ = m and its velocity before collision  u_1​ = v

Mass of second particle is m₂​ = 2m and its velocity before collision  u₂ ​= 0 

Let the velocity of combined system just after collision  be V.

Using conservation of linear momentum :  

m₁​u_1​ + m₂​u₂ ​= (m_1​ + m₂​)V

∴  (m)(v) + (2m)(0) = (m + 2m)V

⟹  V = v/3​

We know that the moment of inertia of a body is given as:
l = MR²
So, the unit of moment of interia will be kg m²

As 2m > m
The acceleration of the two block system will be as shown in figure.

The equation for block of mass 2m is
2mg − T = 2ma  ...(i)
Similarly, for block of mass m
T − mg sin30° = ma ...(ii)
From Eqs. (i) and (ii), we have

⟹ a = g/2

From the graph,


For same stress,

Therefore, Young's modulus of P > Young's modulus of Q
Also, the breaking stress in P is more than breaking stress in Q. Therefore, tensile stress of P is greater than that of Q. Therefore, it is the correct option.

Let the velocity of the ball before the collision be u.
Velocity of the bodies after the collision is given by:
Using conservation of momentum
mu = mv₁ + mv₂
u = v₁ + v₂
Coefficient of restitution,

Hence,

Therefore, the ratio of velocities is given by:

Therefore, it is the correct option.

Initial velocity v₁ = v
Final velocity v₂ = -v
Initial momentum p₁ = mv
Final momentum p₂ = m(-v) = -mv
Change in momentum
Δp = p₁ - p₂
⇒ mv - (-mv)
⇒ 2mv

The ratio of shearing stress to the corresponding shearing strain is called shear modulus or modulus of rigidity.

ma = mg sin θ − f
or f = mg sinθ − ma

Along OE​
Velocity, VOE​=3 m/s

Since, applied force is perpendicular to OE so the velocity VOE​ will be constant.

So, displacement along OE in 4 sec SOE​=3×4=12 m

Along OF​

Force applied, F = 4 N

Mass of the body, m = 2 kg

So, acceleration a = mF ​= 2 m/s²

Displacement along OF in time t = 4 sec

S_OF​ = ut + ​at²/2

S_OF ​= 0 + 2×4²/2

S_OF​ = 16 m

After t = 4 sec, the distance of the body from O to OP

OP² = 12² + 16²

OP² = 144 + 256

OP² = 400

OP = 20 m

Given, u = 7m/s

a = 4m/s²

n = 5

The distance covered by the body in n^th sec is given by

S_n​ = u + a/2​(2n−1)

S₅​ = 7 + 24​(2×5−1)

S_5​ = 25 m

The viscous drag on a spherical body is given as F = 6πηRV.

It is directly proportional to V. Here η is the coefficient of viscosity, R is the radius of the sphere and V is its velocity.

It is given that C_p/C_v = γ = 5/3​

For an adiabatic process,

P₁​V₁^γ​ = P_2​V₂^γ​

⟹ ​P₂/P₁ ​​= (​V₁/V₂​​)^5/3

= (8​)^5/3 = 32

Equation of projectile is given as:


This is the required solution.

Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction
F = nmv = 40X0.05X6 = mg
M = (40x0.05x6)/16 = 1.2 kg

We know that P = V² / R
Thus as P = 100W and V = 200V we get R = 40000/100 = 400Ω
Now as we know V = IR
We get I = V / R
= 100 / 400
= ¼ A

The Huygens-Fresnel principle states that every point on a wavefront is a source of wavelets. These wavelets spread out in the forward direction, at the same speed as the source wave. The new wavefront is a line tangent to all of the wavelets.

Let, XX' be the axis of symmetry and YY' be the axis perpendicular to XX'. 

Let us consider a circular disc S of width dx at a distance x from YY' axis. 

Mass per unit length of the cylinder is M​/l. 

Thus, the mass of disc is M​dx/l
Moment of inertia of this disc about the diameter of the rod = (M​dx/l)R²/4​.
Moment of inertia of disc about YY' axis given by parallel axes theorem is = (M​dx/l)R^2​/4 + (M​dx/l)x²
Moment of inertia of cylinder,

Time period of a simple pendulum,
T = (2π√l)/√g​​
for freely falling system effective, g = 0
So, T = ∞
∴ pendulum does not oscillate at all.

V = Vj​
Hence, B and E should be perpendicular to each other and also perpendicular to the velocity. Hence, B and E should be in z-axis and x-axis 
If B is towards + z axis , Force = q(V \times B ) is in + x direction . To balance this , E has to be in -x direction.