BITSAT Physics Test - 10 - JEE MCQ

MSD = 0.1 cm

Least count = MSD - VSD = 0.1 - 0.09 = 0.01 cm
Total reading = MSD + VSD × LC

This is our required solution.

In the figure vc represents the velocity of the car and v_p that of the parcel. M is the position of the man.
From parallelogram law, the direction of the resultant velocity v_r must be along the direction along which the man is standing.
It follows from Fig. that angle θ is given by

Hence, required angle = 90 + 45 = 135°

From the FBD of the block,

Putting the value of F, mg and θ in the above equation, we get

Relative velocity of object approaching the wall = v - 0 = v m/s
Relative velocity of separation of object = 0 - u
Coefficient of restitution = 0.8
Therefore,

Therefore, the return velocity of the ball = -0.8v m/s

Given that v = 20 m/s and time t = 6 sec
We know that,
s = vt
Then,
s = 20 x 6
s = 120m
Now, heat is generated due to the sliding of body at rough horizontal plane.
The heat is equal to the work done against friction.
Q = Work done against friction

Let the first astronaut jump in the direction of motion of the spaceship and the second astronaut jump in the direction opposite to the motion of the spaceship.
Let the two astronauts meet at time t sec.
Therefore, distance travelled by astronaut moving in the direction of the spaceship,

Distance travelled by spaceship in time t sec,

Distance travelled by second astronaut in t sec = t²
Distance travelled by the first astronaut w.r.t. spaceship = t² - 5t
Distance travelled by the second astronaut w.r.t. spaceship = t² + 5t
Since the length of the chamber = 18 m
Therefore,
t² + 5t + t² - 5t = 18
2t² = 18
t = 3 sec

From the FBD of the mass m,
mg = T
From the FBD of mass 2m,
2mg = T + kx
Putting the value of T from above equation,
2mg = mg + kx
mg = kx
Or,
x = mg/k
This is our required solution.

Initial momentum of the cart = 15 x 180 = 2700 kgm/s
Final momentum of the cart = 15 x (180 - 35) = 2175 kgm / s
Change in momentum = 2700 - 2175 = 525
Time taken = 10 seconds
By Newton's law of motion, rate of change of momentum = Force
Therefore,

Therefore, it is the correct option.

Before the bounce, height of the ball = 16 m
As the ball is dropped, the initial velocity of the ball is zero. Let the velocity of the ball just before striking the surface be u.
Thus, u² = 32g = 320
After the bounce, the velocity of the ball at a height of 5 m above the ground is v.
Therefore,

Therefore, the KE of the ball at the height of 5 m will be:

This energy of the ball will be transferred into the spring in the form of elastic potential energy.
Therefore,

The compression in the spring will be 6.63 m.
This is the required solution.

Let M be the mass of the Sun and m be the mass of the planet.
We know that orbital speed of the planet is given by,

Therefore, time period of the revolution is given by

By Kepler's law of planetary motion,

Therefore,

Thus, it is the correct option.

Ideal gas equation is PV = N RT

we know that density 

The graph will be

If l and r be the length and radius of the tube and p the pressure difference, then from Poiseuille's formula, volume of liquid flowing per second is given by

∴ Fluid resistance 
When capillaries are connected in series then equivalent fluid resistance is

In rolling without slipping through the distance L down the incline, the height of the rolling object changes by "h".
So, the gravitational potential energy changes by mgh.


As we know that moment of inertia of solid sphere is

∴ The velocity of the solid sphere is

The moment of inertia of the hollow sphere is

∴ The velocity of the hollow sphere is

On dividing equation 1 and 2 , we get

Given:
I = 1.2t + 3
Charge (q) is given by q = ∫ Idt
Integrating the expression using

The surface area of cylindrical surface is 2πrl, where r is radius of base and l is length of cylinder. 
We know that with increasing distance from source, the total energy or power transmitted remains same, but intensity decreases. For any source of power P, intensity I at distance r from it will be I = P/S, where S is surface area.
(i) For spherical wave front s = 4πr² (surface) 

(ii) For cylindrical wave front S = 2πrl
so

Let V₀ and V_t be volumes of mercury at 0°C and t°C , respectively;
A₀ and A_t are areas of cross section of tube at 0°C and t°C , respectively.
Apparent increase in volume:

Mass of the ring (M) = 5 kg , diameter of the ring (d) = 20 cm , radius of the ring (r) = 10 cm = 10⁻¹ m 
v = 4200 rpm = 4200/60 rps = 70 rps
The moment of inertia of a uniform circular ring is

The relation between the angular momentum and moment of inertia is given by

Let θ be the temperature of B

Substituting values  θ = 880°C
Rate of heat flow from B to D 

Speed in m/sec

Let initial Kinetic energy = KE₁ = E
Given: Final kinetic energy 

 The relation between the momentum and the kinetic energy is given by:

Final momentum (P′) will be:

Increase in momentum (ΔP) = P' - P

Coulombic force on electron, F_e = a_em_e
Similarly, for proton F_p = a_pm_p
Here F_e = F_p 

Wavelength associated with a moving particle λ = m^pv^qh^r 

After solving we get,
p = -1, q = -1, r = 1

We know that,

∴ the speed of the smaller object is higher than that of bigger.
The momentum of the smaller object:


The lighter body will have smaller momentum.

Using conservation of charge 
q_in = q_out
Current (rate of flow of charge) remains constant in a wire.

Therefore, for non-uniform cross-section (different values of A ) drift speed will be different Sections. Only current (or rate of flow of charge) will be same.

We shall use ω = ω₀ + αt
ω initial anugular speed in rad / s = 2 π × angular speed in

Similarly ω = final angular speed in rad/s

∴ Angular acceleration

The angular acceleration of the engine 4πrad/s²

The phase difference =1.57 − 0.5 = 1.07
[ or using sin  We get the same result.]

So gravitational potential is non-uniform and negative.

If a collision does occur, the velocities of the two spheres will interchange. If a collision does not occur, they retain their original velocities. As the spheres are identical, an observer cannot tell from the velocities of the two spheres whether a collision occurred or not.

Given:
A uniform force,
Mass, m = 2 kg
Initial position of particle, 
Final position of particle, 
Net Displacement of particle,

Work done by the force on the particle,