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In LCR circuit, the capacitance is changed from C to 4C. For the same resonant frequency the inductance should be changed from L to
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 Hatom in which A = Z as N = 0
The radius of the first orbit of the hydrogen atom is a₀. The radius of the second orbit will be
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 }= a_{0}n^{2}
Radius of second orbit = a_{0}(2)^{2} = 4a_{0}
If rigidity modulus is 2.6 times of youngs modulus then the value of poission's ratio is
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
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.
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
By increasing the temperature, the specific resistance of a conductor and a semiconductor
The power factor of a series LCR circuit when at resonance is
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
Minimum no. of 8 μF and 250 V capacitors, used to make a combination of 16 μF and 1000 V, is
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
A fireman of mass 60 kg slides down a pole. He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5, with what acceleration will the fireman slide down (g=10m/s^{2})
Here mass, m = 60kg
and coefficient of friction, μ = 0.5
and g =10m/s^{2}
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/s^{2}
Hence the acceleration with which the fireman slide down is 5m/s^{2}
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.”
A body takes T minutes to cool from 62 degree C to 61 degree C when the surrounding temperature is 30 degree C. The time taken by the body to cool from 46 degree C to 45.5 degree C is
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.
An ideal gas is heated from 27°C to 627°C at constant pressure. If initial volume was 4 m^{3}, then the final volume of the gas will be
Given, Initial temperature (T_{1}) : 27°C = 300 K
Final temperature (T_{2}) = 621°C = 900 K
Initial volume (V_{1}) = 4 m^{3}.
The monoenergetic beams of electrons moving along + y direction enters a region of uniform electric and magnetic fields. If the beam goes straight through, then these simultaneously fields B̅ and E̅ are directed respectively
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 zaxis and xaxis
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.
If a substance is to be paramagnetic, its atoms must have a
Paramagnetic materials possess a permanent magnetic dipole moment due to incomplete cancellation of electron spin and orbital magnetic moment.
Horizontal tube of nonuniform crosssection has radii of 0.1m and 0.05m respectively at M and N. For a streamline flow of liquid the rate of liquid flow is
According to principle of continuity, for streamline flow of fluid through a tube of nonuniform crosssection the rate of flow of fluid (Q) is same at every point in the tube.
i.e, Av = constant
⇒ A_{1}v_{1 }= A_{2}v_{2}
Therefore, the rate of flow of fluid is same at M and N.
The viscous drag on a spherical body moving with a speed ν in a viscous medium is directly proportional to
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 tangential stress to the shearing strain is called
The ratio of shearing stress to the corresponding shearing strain is called shear modulus or modulus of rigidity.
A particle of charge q and mass m moves in a circle of radius r around an infinitely long line charge of linear charge density + λ . Then time period will be:(Where )
The initial velocity of a body moving along a straight line is 7 m/s. It has a uniform acceleration of 4 m/s^{2}. The distance covered by the body in the 5th second of its motion is
Given, u = 7m/s
a = 4m/s^{2}
n = 5
The distance covered by the body in n^{th }sec is given by
S_{n} = u + a/2(2n−1)
S_{5} = 7 + 24(2×5−1)
S_{5 }= 25 m
A particle is moving in a circle with uniform speed υ. In moving from a point to another diametrically opposite point
Initial velocity v_{1} = v
Final velocity v_{2} = v
Initial momentum p_{1 }= mv
Final momentum p_{2 }= m(v) = mv
Change in momentum
Δp = p_{1}  p_{2}
⇒ mv  (mv)
⇒ 2mv
If a particle of mass m is moving in a horizontal circle of radius r with a centripetal force (k/r^{2}), the total energy is
Given, Centripetal force = mv^{2}/r = k/r^{2}
Kinetic Energy = mv^{2}/2 = k/2r
Potential Energy = −∫^{r}_{∞ }Fdr
the lower limit has been taken as ∞ because potential energy is zero at infinty
⇒ −∫^{r}_{∞ }(k/r^{2})dr
⇒ −k ∫^{r}_{∞ }r^{−2}dr
⇒ −k r^{1}/1^{r}_{∞}
⇒ −k/r
Total energy = k/2r − k/r = −k/2r
A body of mass 2 kg has an initial velocity of 3 metres per second along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of the body from O after 4 seconds will be
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/s^{2}
Displacement along OF in time t = 4 sec
S_{OF} = ut + at^{2}/2
S_{OF }= 0 + 2×4^{2}/2
S_{OF} = 16 m
After t = 4 sec, the distance of the body from O to OP
OP^{2 }= 12^{2 }+ 16^{2}
OP^{2} = 144 + 256
OP^{2} = 400
OP = 20 m
An astronomical telescope has objective and eyepiece lens of powers 0.5 D and 20 D respectively. Its magnifying power will be
Given, the power of objective lens,
P_{o }= 0.5D
The power of eyepiece 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
A simple pendulum executing simple harmonic motion is falling freely along with the support. Then
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.
In the absence of damping, the amplitude of forced oscillation at resonance is
When there is no damping , the damping constant b = 0 and at resonace, the amplitude of forced oscillations is infinite.
The moment of inertia of a cylinder of radius R, length l and mass M about an axis passing through its centre of mass and normal to its length, is
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 Mdx/l
Moment of inertia of this disc about the diameter of the rod = (Mdx/l)R^{2}/4.
Moment of inertia of disc about YY' axis given by parallel axes theorem is = (Mdx/l)R^{2}/4 + (Mdx/l)x^{2}
Moment of inertia of cylinder,
Two bodies of different masses of 2 kg and 4 kg are moving with velocities 2 m/s and 10 m/s towards each other due to mutual gravitational attraction. What is velocity of their centre of mass ?
The radius of curvature of a thin planoconvex lens is 10 cm (of curved surface) and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be
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.
A force of acts on O, the origin of the coordinate system. The torque about the point (1, 1) is
Torque = r x F
r = î  j
so torque = (i  j) x Fk
= F(j  i)
= F( i + j)
A bulb rated at (100W200V) is used on a 100V line. The current in the bulb is
We know that P = V^{2} / R
Thus as P = 100W and V = 200V we get R = 40000/100 = 400Ω
Now as we know V = IR
We get I = V / R
= 100 / 400
= ¼ A
A monoatomic gas is suddenly compressed to 1/8 of its volume adiabatically. The ratio of pressure of the gas now to that of its original pressure is
(Given, the ratio of the specific heats of the given gas to be 5/3)
It is given that C_{p}/C_{v} = γ = 5/3
For an adiabatic process,
P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ}
⟹ P_{2}/P_{1} = (V_{1}/V_{2})^{5/3}
= (8)^{5/3 }= 32
5 mole of hydrogen gas is heated from 30^{o}C to 60^{o}C at constant pressure. Heat given to the gas is (given R = 2 cal/mole degree)
Hydrogen is diatomic.
Hence, molar heat capacity, Cp = 3.5R
Also change in temperature = ΔT = 30K, no.of moles, n = 5
Hence, heat absorbed = ΔQ = nC_{p}ΔT = 5×3.5×2×30 = 1050cal
We know that the moment of inertia of a body is given as:
l = MR^{2}
So, the unit of moment of interia will be kg m^{2}
Wave optics is based on wave theory of light put forward by Huygen and modified later by
The HuygensFresnel principle states that every point on a wavefront is a source of wavelets. These wavelets spread out in the forward direction, at the same speed as the source wave. The new wavefront is a line tangent to all of the wavelets.
Huygen's wave theory explain the propagation of the wavefront.
If the equation of SHM is y = a sin (4 π t + φ) how much is its frequency?
Comparing with a standard Equation of SHM ,
y = a sin(wt + φ)
ω = 4π, and θ = φ = initial phase.
So frequency is as follows :
freq. = 4π/2π
⇒ f = 2 Hz
So final answer is 2 Hz frequency.
The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is 90^{o}. Then the velocity of wave will be
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
A solid disc of mass M is just held in air horizontally by throwing 40 stones per sec vertically upwards to strike the disc each with a velocity of 6 ms⁻^{1}. If the mass of each stone os 0.05 kg what is the mass of the disc. (g = 10 ms⁻^{2})
Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction
F = nmv = 40X0.05X6 = mg
M = (40x0.05x6)/16 = 1.2 kg
A particle of mass m moving with velocity ν collides with a stationary particle of mass 2 m and sticks to it. The speed of the system, after collision, will be
Mass of first particle is m_{1} = m and its velocity before collision u_{1 }= v
Mass of second particle is m_{2} = 2m and its velocity before collision u_{2} = 0
Let the velocity of combined system just after collision be V.
Using conservation of linear momentum :
m_{1}u_{1 }+ m_{2}u_{2} = (m_{1 }+ m_{2})V
∴ (m)(v) + (2m)(0) = (m + 2m)V
⟹ V = v/3
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