JEE SRME Physics Mock Test - 7 Free Online Test 2026

According to Newton's law of cooling,

For the first condition
..... (i)
and for the seconds condition
...... (ii)
By soving Eqs. (i) and (ii) we get t = T minutes.

As ω₂ = ω₁ + αt  ∴40π = 20π + α × 10 
or α = 2π rads⁻¹

Number of rotations completed

In pure rolling mechanical energy remains conserved,
therefore speed will be same in both cases.
Time of descent on an inclined plane for a rolling body is,

The higher the inclination, the smaller the time of descent.

To have 1000 V tolerance, you need to stack up four 8μF capacitor in series. The equivalent capacitance of 4 such capacitors is:
⟹ Ceq​ = (8μF)/4​ = 2μF
To make a total of 16 μF that tolerates 1000V, you need 8 set of them arranged in parallel such that
⟹Ceq′​ = 8 × 2μF = 16μF
∴ Total number of capacitors = 8×4 = 32

Explanation:

• Statement A: This statement is correct because according to Hooke's law, the elongation in a spring is directly proportional to the mass connected with the spring balance. Therefore, more mass will result in more elongation in the spring balance.
• Statement B: This statement is correct as well. When more mass is added to one pan of a beam balance, more mass is required on the other pan to keep the beam horizontal and balanced. This is based on the principle of torque and the concept of moments in physics.
• Statement D: This statement is also true. According to Newton's second law of motion, the acceleration of an object is inversely proportional to its mass for a given force. Therefore, the greater the mass of a body, the lesser its acceleration will be when the same force is applied.

Therefore, all of the statements are absolutely correct about mass.

Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction
F = n m v

It is evident from Kepler’s first law of planetary motion that “The orbit of a planet is an ellipse with the Sun at one of the two foci.”

At resonance, the impedance of the circuit is equal to the resistance in the circuit, i.e. Z = R
∴ power factor of the circuit, cos φ = R/Z = R/R = 1

Both the cathode rays and the visible light rays (because of having particle nature) affect the photographic plates when incident on it. So, cathode rays and visible light rays similar in that sense.

Work done during cyclic process ABCDA
= area enclosed by ABCDA



= 150 J

Given, Initial temperature (T₁) : 27°C = 300 K
Final temperature (T₂) = 621°C = 900 K

Initial volume (V₁) = 4 m³.

Rotational kinetic energy,

ie, moment of inertia about an axis of a body is twice the rotational kinetic energy.

If the angular acceleration of the wheel is α, acceleration of the body a = Rα.
Using torque equation
TR = Iα
And for the block,

By solving the above equations,

Therefore, the angular acceleration is less than 

Here mass, m = 60kg

and coefficient of friction, μ = 0.5 

and g =10m/s²

Force applied by fireman, F = 600N

Friction Force, f = μmg = 0.5×60×10 = 300N

Net movement force,

F − f = ma
⟹ 600 − 300 = ma
⟹ a = 300/60​
⟹ a = 5m/s²

Hence the acceleration with which the fireman slide down is 5m/s²

Moment of inertia of whole disc about an axis through centre of disc and perpendicular to its plane is 
As one quarter of disc is removed, new mass,

Since, in this case, instantaneous axis of rotation is always below the centre of mass. This is possible only when point of contact moves with a velocity equal to centre of mass.

A = N + Z, where A is mass number and Z is atomic number and N is number of neutrons
So A > Z except for H-atom in which A = Z as N = 0

Rom the body of mass , the distance of the point where gravitational field is zero

net torque about bottommost point

then for accelaration of com ,
Friction force will act towards right.

The angular displacement of a particle can be calculated using the formula for motion with constant acceleration:

• Initial angular velocity (ω₀) is 0 rad/s since the particle is at rest.
• Final angular velocity (ω) is 36 rad/s.
• Time (t) is 6 seconds.

Using the formula for angular displacement (θ):

• θ = ω₀t + 0.5 × α × t²

Since ω = ω₀ + αt, solve for angular acceleration (α):

• 36 = 0 + α × 6
• Solve for α: α = 6 rad/s²

Substitute α back into the displacement formula:

• θ = 0 × 6 + 0.5 × 6 × 6²
• θ = 0.5 × 6 × 36
• θ = 108 rad

Thus, the total angle turned is 108 rad.

Moment of inertia about axis AA' is


Torque of Mg will maximum when ‘OC’ is horizontal

• Because specific resistance depends upon value of α the temperature coefficient of the resistance. the value α for conductors is positive and for semiconductors the value of α is negative.
• Hence By increasing the temperature, the specific resistance of a conductor increases and that of a semiconductor decreases.

The energy of a photon is directly proportional to its frequency with proportionality constant being the Plank's constant h.

Hence E = hν = hc/λ​  (Since c=λν)

⟹ E = hc/λ

For resonance:
X_L ​= X_C​
ωL = 1/ωC
ω = 1/√2C​​
2πf = 1/√2C​
For same resonant frequency:
LC should be constant.
So, LC = 4CL_1​
⇒ L1 = LC/4C​ = L/4​

The radius of the first bohr orbit = a_0​ According to 3rd Postulate of bohr atomic model

L = mvr = rh/2π​; where n=1,2,3

⇒ For nth orbit
r_n ​= a₀​n²
Radius of second orbit = a₀(2)² = 4a₀

Young’s Modulus = 2 × Rigidity Modulus × (1+ poisson’s Ratio) 
Given, Young’s Modulus = 2.6 × Rigidity Modulus
2.6 = 2×(1+σ)
1.3 = 1 + σ
σ = 0.3​