BITSAT Physics Test - 1 - JEE MCQ

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​

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

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​

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

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.

• 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.

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

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

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

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.”

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.

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

Initial volume (V₁) = 4 m³.

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.

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

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 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.

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

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

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

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​

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, 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

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.

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

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,

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.

Torque = r x F

r = î - j

so torque = (i - j) x -Fk

= -F(-j - i)

= F( i + j)