JEE Main Physics Test- 3 Free Online Test 2026

In the given equation = dimensionless. Therefore, unit of t must be same as that of p.

Option D is correct.

When adding measured quantities, the absolute uncertainties add.

Let l₁ = 11.05 cm and l₂ = 11.05 cm. Then l = l₁ + l₂ = 22.10 cm.

Each uncertainty is Δ1 = Δ2 = 0.02 cm. So Δ = Δ1 + Δ2 = 0.02 + 0.02 = 0.04 cm.

Therefore the combined length is (22.10 ± 0.04) cm, which corresponds to Option D.

As we know,

This implies that   are parallel but in opposite directions.
So, the angle between vectors  will be π.

Now retain only the positive sign.

By using equation of trajectory y = gx²/2u² for given condition

Concept:

Elastic potential energy: It is the potential energy created when an object undergoes an elastic deformation due to a force applied to it.

• This energy is stored in the object as long as the force is in action and the object goes back to its original shape when the force is removed.
• A spring undergoes a similar situation when a force acting on it causes a displacement due to elastic deformation.

The potential energy (U) required to stretch a string by x distance is given by

where k is the spring constant.

Calculation:

Given:

Let the potential energies of the spring be U₁ and U₂ when they are stretched be x₁ and x₂ respectively.

Given that: U₁ = U

When x₁ = 2 cm:

On dividing equations (1) and (2), we get

⇒ U₂ = 25U₁ = 25U 

Due to inertia the stone continues in tangential direction to the circular motion.

Conservation of momentum (gives)
mu + 0 = mv₁​ + mv_2​
⇒ v₁​ + v₂ ​= u ...(i)
Newton's experimental formula,
v₁​ − v₂​ = −e[u₁ ​− u₂​] gives
v₁ ​− v₂ ​= −eu ...(ii)
From Eqs. (i) and (ii), we get

As boy walk from left to right on the block, the block will recoil towards left. The friction acting on the block will act towards right. Hence the centre of mass of the boy and block will shift towards right.

For a rod, the Moment of Inertia (MOI) about an axis passing through its center of mass and perpendicular to its plane is:

Let the radius of gyration about the same axis be k,
I = mk² (ii)
From both equations:

If r is decreased, K will increase but U and E will decrease.

Energy of a satellite is given by 
For a satellite very close to earth, r = R

Increase in kinetic energy = decrease in potential energy

P∀ = nRT
n = P/RT
⇒ no of molecule = n ⋅ N_{A ​}

It means line of smaller slope represent greater volume of gas. For the given problem figure
Point 1 and 2 are on the same line so they will represent same volume i.e. V₁ = V₂
Similarly point 3 and 4 are on the same line so they will represent same volume i.e. V₃ =V₄
But V₁ > V₃ (= V₄ ) or V₂ > V₃ (= V₄ ) as slope of line 1-2 is less than 3-4.

Heat is transferred to the working material (Q₁) and heat is rejected during (Q₂).
The thermal efficiency is ηth = W/Q₁.
Applying first law, we have, 
W = Q₁ − Q₂.  
ηth = 1 − Q₂/Q₁.
Hence, correct answer is (b).

R₁ = Initial radius = R 

R₂ = final radius = 2R 

The two conductors are thermally in parallel. Therefore, equivalent thermal resistance R is given by:- 

7th harmonic means 7 loops. 
In one loop, there are 2 nodes and 1 antinode. In 7 loops, there will be 8 nodes and 7 antinodes.

Frictional Force Between the Blocks: The frictional force between the two blocks can be calculated using the formula:
f_friction​ = μ ⋅ N
Where:

• μ = 0.6 is the coefficient of friction.
• N is the normal force, which is equal to the weight of block A.

For block A (1 kg):
N = mA​ ⋅ g = 1⋅9.8 = 9.8N
Now, calculate the maximum frictional force:
f_friction = 0.6⋅9.8 = 5.88 N
Maximum Acceleration Without Slipping: The maximum force that can be applied to block A without it slipping over block B is the frictional force we calculated. This force will cause an acceleration in both blocks (since they are stacked together).
Using Newton’s second law:

This is the maximum acceleration the system can have without the blocks slipping over each other.
Force on the System Due to Spring: The spring exerts a restoring force given by Hooke’s law:
F_spring = k ⋅ x
Where:

• k = 900 N/m is the spring constant.
• x is the elongation in the spring.

This spring force will accelerate both blocks together, so the total mass being accelerated is:

Applying Newton’s second law for the whole system:

The maximum elongation of the spring such that the blocks do not slip over each other is approximately 1 cm (rounded to the nearest whole number).

Understanding the dynamics of the system:

• When the sphere is projected up the incline with both translational velocity (v₀​) and rotational velocity (ω₀​), the sphere experiences friction that affects both its motion and rotation.
• Since v₀ > ω_0​, the sphere is slipping, and friction will work to reduce both linear velocity and angular velocity.

Forces acting on the sphere: The main forces are:

• The gravitational component along the incline, mgsin⁡θ, which acts to decelerate the sphere.
• The frictional force f=μN, where N = mgcos⁡θ is the normal force.
• The friction provides torque, which affects the rotational motion of the sphere.

Equations of motion:

• The linear deceleration due to the frictional force and gravitational force is:
  
  Where
• The angular deceleration is due to the torque provided by friction:
  
  Where is the moment of inertia of the sphere, and α\alphaα is the angular deceleration.

Condition for stopping:

• The sphere stops when both its linear velocity and rotational velocity become zero.
• The time ttt it takes for the velocities to reduce to zero can be derived by analyzing the combined effect of linear and angular decelerations.

Solving for k:

• From the total time of rise formula given in the image, the expression includes terms involving the initial velocities and accelerations due to the forces acting on the sphere.
• The final expression given is:
  
• To find k, we compare this with standard results for the motion of rolling bodies on inclined planes.

Result: After solving and matching terms, we find that the value of k = 17.