SRMJEEE Subject Wise & Full Length Physics Mock Test - 5 Free Online Test

When the same tension is applied to wires of the same material, the elongation depends on the ratio of their length (L) to the square of their diameter (d²). The wire with the smallest value of L/d² will experience the minimum elongation.

Let's analyze the given options:

a) Length = 300 cm, Diameter = 3 mm

L/d² = 300 / (3²) = 33.33

b) Length = 100 cm, Diameter = 1 mm

L/d² = 100 / (1²) = 100

c) Length = 200 cm, Diameter = 2 mm

L/d² = 200 / (2²) = 50

d) Length = 50 cm, Diameter = 0.5 mm

L/d² = 50 / (0.5²) = 200

Therefore, the wire with Length = 300 cm, Diameter = 3 mm (option a) will experience the minimum elongation when the same tension is applied.

Time period of second's pendulum = 2 s
Change in time period 
Therefore,

Therefore, 
New Time Period T' = ΔT ⇒ T' = 2.000167 s
Therefore, time lost in a week,

Hence, this is the required solution.

Using maximum power transfer theorem
Power delivered to the load will be maximum when the internal resistance of the cell is equal to the load resistance.
r = 10Ώ

The electric charge will always reside on the surface of the sphere. As the diameters of the both spheres are same, both can be equally charged.
Also, the charge density and electric field intensity will also be same for both the spheres.
Therefore, it is the correct option.

 When the lift accelerates upwards with an acceleration of g/3, the effective acceleration is 
Therefore, the new time period is 


Hence the correct choice is (2).

As we know,

Let the intial temperature be T, pressure be P and volume be V for isotheral process

For adiabatic process

This is our required solution.

Let each plate move a distance 'x' from its initial position.

Now, lets consider that charge 'q' flows in the loop.
Using Kirchoff's Voltage Law, we get

Given, mass of the child (m) = 30 kg
Time taken (t)  = 20 min
Fall in temperature = (101 - 98)^oF

Specific heat of human body (s) = 4.2 x 103 J/kg - oC
Latenl heat of evaporation (L) = 580 cal/g = 580 x 10³ cal/kg
= (580 x 10³ x 4.2) J/kg
Heat given by body during fall in temperature
Q₁ = ms ΔT
Let m' be the mass of sweat evaporates from the human body.
Heat taken in evaporation
Q₂ = m' L
But  Q₁ = Q₂
∴ msΔT = m′L
or 


∴ Rate of evaporation of sweat 

= 0.00431 kg/min
= 4.31 g/min.

Here, r = 10 cm = 0.1 m
T = 1000 K
σ = 5.67 × 10⁻⁸ W m⁻¹ °C⁻⁴.
If A is the surface area of the sphere then, heat energy incident per second on the sphere,
E = A × (σT⁴) = 4πr² × σT⁴
= 4π × (0.1)² × 5.67 × 10⁻⁸ × (1,000)⁴ = 7,125 J s⁻¹  

= 1/2 = 0.5

From Wein's displacement law,


From Boltzmann's law

Work done in a process will be zero if the volume remains constant throughout the process(Isochoric process). If the volume changes in a process, the work done in the process will not be zero even if the final volume and the initial volume are equal(Example: Vertical semi-circle in a PV graph).
Therefore, the Assertion is false but Reason is true.

Here, T₁ = 27℃ = (27 + 273)K = 300 K
T₂ = 327℃ (327 + 273)K = 600 K
According to Stefan's law,
E = eAδT⁴

⇒ E₂ = 16E₁

Given, triple point of water on scale A = 200 A
Triple point of water on scale B = 350 B
We know that triple point of water on absolute scale = 273.16 K
∴ 200 A = 350 B = 273.16 K

If  T_A and T_B are the triple point of water on two scales A and B, then

According to Wien's law,  As the temperature of body increases, frequency corresponding to maximum energy in radiation (v_m) increases. This is shown in graph (C).

Work done by a gas is equal to the area under the curve on volume axis.
Since cycle is anticlockwise so negative work done is more than the positive work done. S o overall work done will be negative.
Hence option (B) is the correct answer of the given diagram.

Till the particle reaches the centre of the planet, force on both the bodies are in direction of their respective velocities, hence kinetic energies of both keep on increasing. After the particle crosses the centre of the planet, forces on both are retarding in nature. Hence, as the particle passes on both are retarding in nature. Hence, as the particle passes through the centre of the planet. Sum of kinetic energies of both the bodies is maximum. Therefore Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I.

For Earth to revolve around the Sun, there is need of centripetal force to complete its revolution. The gravitational force between the Sun and the Earth acts as the necessary centripetal force to make the revolution of Earth around the Sun possible and the direction of this centripetal force is always towards the centre of motion. So, as the sun is located in the centre, the necessary centripetal force for revolutionary motion and gravitational force is directed towards the Sun.

Mass of spaceship m = 1000 kg
Mass of sun M_s = 2 × 10³⁰ kg
Mass of mars M_m = 6.4 × 10²³ kg
Radius of mars R_m = 3395 km = 3.395 × 10⁶ m
Radius of the orbit of mars (r) = 2.28 × 10⁸ km = 2.28 × 10¹¹ m and 
G = 6.67 × 10⁻¹¹ N m² kg⁻²
The spaceship is present in the gravitational field of the sun as well as in the gravitational field of the mars.

Therefore, the potential energy of the spaceship due to the gravitational field of the sun

The potential energy of the spaceship due to the gravitational field of the mars

Therefore, the total potential energy of the spaceship

The potential energy of the spaceship outside the solar system = 0.
Therefore, the energy imparted to the spaceship required just rockets out of the solar system

Gravitational potential energy is given as,

Therefore, the gravitational potential energy increases.

Below the surface of the earth g ∝ r and above the surface of earth. And above the earth surface it inversly propotional to r². Therefore, the graph (a) is correct

According to Question
Acceleration due to gravity at 60^o latitude = g₆₀ = 0 ms⁻²
Acceleration due to gravity at pole = g = 10 m s⁻²
Now Relation between acceleration due to gravity (g) and latitude (λ)
g' = g − R_eω² cos²λ
Here g'= Acceleration due to gravity at λ latitude
g = Acceleration due to gravity at pole
R_e = 6400 km = Radius of earth
R_e = 6400 km = 6.4 × 10⁶ m
λ = Latitude of position
ω = Angular velocity of earth
Apply the above g and λ relation
g₆₀ = g − R_eω² cos² 60^o
Put the all known values in the above relation

Total energy of satallite is 
where M is the mass of planet , m is the mass of satallite and r is the distance between centre of mass of planet and satallite.
Total energy of the satallite is negative, which means that the satallite will never be able to escape from the gravitational pull of the planet.

To determine the angular velocity at which the Earth's rotation makes the gravitational force at the equator zero, follow these steps:

• The centrifugal force needs to balance the gravitational force at the equator.
• The formula for centrifugal force is: F_c = mω²r.
• Gravitational force is: F_g = mg.
• Set F_c = F_g to find ω.

Since mg = mω²r, solve for ω:

• Cancel m from both sides: g = ω²r.
• Rearrange to find ω: ω = √(g/r).

Substitute the given values:

• g = 9.8 m/s²
• r = 6400 × 10³ m
• Calculate ω: ω = √(9.8 / (6400 × 10³))

After calculation, the angular velocity ω is approximately 1.24 × 10^-3 rad/s.

Given, the power of objective lens, 
P_o​ = 0.5D 
The power of eye-piece lens, 
P_e ​= 20D 
The magnifying power of an astronomical telescope 
M = f_o​​/f_e
or = ​P_e/P_o​​    [∴ P = 1/f​]
= 20/0.5
= 40

Δϕ = 90^∘ = a​/2

Δx = 0.8m

we know that Δϕ = (2π/α).Δx

So λ = (2π.Δx)/Δϕ ​= (2π×0.8)/​(π/2)

⇒ λ = 3.2m​

v = 120 Hz & v = vλ

velocity v = 120×3.2 = 384m/s​